VIRESCIT VULNERE VERITAS  
THE ELEMENTS OF GEOMETRY of the most ancient Philosopher EUCLID of Megara.  
Faithfully (now first) translated into the English tongue, by H. Billingsley, Citizen of London. Whereunto are annexed certain Scholics, Annotations, and Inventions, of the best Mathematicians, both of time past, and i● this our age.  
With a very fruitful preface made by M. I Dee, specifying the chief Mathematical Sciences, what they are, and whereunto commodious● where, also, are disclosed certain new Secrets. Mathematical and Mechanical, until these our days greatly miss:  
Imprinted at London by john day.   

The Translator to the Reader.  
THere is (gentle Reader) nothing (the word of God only set apart) which so much beautifieth and adorneth the soul and mind of man, as doth the knowledge of good arts and sciences: as the knowledge of natural and moral Philosophy. The one setteth before our eyes, the creatures of God, both in the heavens above, and in the earth beneath: in which as in a glass, we behold the exceeding majesty and wisdom of God, in adorning and beautifying them as we see: in giving unto them such wonderful and manifold proprieties, and natural workings, and that so diversly and in such variety: farther in maintaining and conserving them continually, whereby to praise and adore him, as by S. Paul we are taught. The other teacheth us rules and precepts of virtue, how, in common life amongst men, we aught to walk uprightly: what duties pertain to ourselves, what pertain to the government or good order both of an household, and also of a city or common wealth. The reading likewise of histories, conduceth not a little, to the adorning of the soul & mind of man, a study of all men commended: by it are seen and known the arts and doings of infinite wise men gone before us. In histories are contained infinite examples of heroical virtues to be of us followed, and horrible examples of vices to be of us eschewed. Many other arts also there are which beautify the mind of man: but of all other none do more garnish & beautify it, than those arts which are called Mathematical. Unto the knowledge of which no man can attain, without the perfect knowledge and instruction of the principles, grounds, and Elements of Geometry. But perfectly to be instructed in them, requireth diligent study and reading of old ancient authors. Amongst which, none for a beginner is to be preferred before the most ancient Philosopher Euclid of Megara. For of all others he hath in a true method and just order, gathered together whatsoever any before him had of these Elements written: inventing also and adding many things of his own: whereby he hath in due form accomplished the art: first giving definitions, principles, & grounds, whereof he deduceth his Propositions or conclusions, in such wonderful wise, that that which goeth before, is of necessity required to the proof of that which followeth. So that without the diligent study of Euclides Elements, it is impossible to attain unto the perfect knowledge of Geometry, and consequently of any of the other Mathematical sciences. Wherefore considering the want & lack of such good authors hitherto in our English tongue, lamenting also the negligence, and lack of zeal to their country in those of our nation, to whom God hath given both knowledge, & also ability to translate into our tongue, and to publish abroad such good authors, and books (the chief instruments of all learnings): seeing moreover that many good wits both of gentlemen and of others of all degrees, much desirous and studious of these arts, and seeking for them as much as they can, sparing no pains, and yet frustrate of their intent, by no means attaining to that which they seek: I have for their sakes, with some charge & great travail, faithfully translated into our vulgar tongue, & set abroad in Print, this book of Euclid. Whereunto I have added easy and plain declarations and examples by figures, of the definitions. In which book also ye shall in due place find manifold additions, Scholies, Annotations, and Inventions: which I have gathered out of many of the most famous & chief Mathematicies, both of old time, and in our age: as by diligent reading it in course, ye shall well perceive. The fruit and gain which I require for these my pains and travail, shall be nothing else, but only that thou gentle reader, will gratefully accept the same: and that thou mayest thereby receive some profit: and moreover to excite and stir up others learned, to do the like, & to take pains in that behalf. By means whereof, our English tongue shall no less be enriched with good Authors, then are other strange tongues: as the Dutch, French, Italian, and Spanish: in which are read all good authors in a manner, found amongst the Greeks or Latins. Which is the chiefest cause, that amongst them do flourish so many cunning and skilful men, in the inventions of strange and wonderful things, as in these our days we see there do. Which fruit and gain if I attain unto, it shall encourage me hereafter, in such like sort to translate, and set abroad some other good authors, both pertaining to religion (as partly I have already done) and also pertaining to the Mathematical Artes. Thus gentle reader farewell.   

TO THE UNFEIGNED LOVERS of truth, and constant Students of Noble Sciences, JOHN DEEof London, heartily wisheth grace from heaven, and most prosperous success in all their honest attempts and exercises.  
Divine Plato, the great Master of many worthy Philosophers, and the constant avoucher, and pithy persuader of unum, Bonum, and Ens: in his School and Academy, sundry times (besides his ordinary Scholars) was visited of a certain kind of men, alured by the noble fame of Plato, and the great commendation of his profound and profitable doctrine. But when such Hearers, after long hearkening to him, perceived, that the drift of his discourses issued out, to conclude, this unum, Bonum, and Ens, to be Spiritual, Infinite, aeternal, Omnipotent, etc. Nothing being alleged or expressed, How, worldly goods: how, worldly dignity: how, health, Strength or justiness of body: nor yet the means, how a merue●lous sensible and bodily bliss and felicity hereafter, might be attained: Straightway, the fantasies of those hearers, were dampt: their opinion of Plato, was clean changed: yea his doctrine was by them despised: and his school, no more of ●hem visited. Which thing, his Scholar, Aristotle, narrowly considering, found the cause thereof, to be, For that they had no forwarning and information, in general, whereto his doctrine tended. For, so, might they have had occasion, either to have forborn his school haunting: (if they, then, had misliked his Scope and purpose) or constantly to have continued therein: to their full satisfaction: if such final scope & intent, had been to their desire. Wherefore, Aristotle, ever, after that, used in brief, to forewarn his own Scholars and hearers, both of what matter, and also to what end, he took in hand to speak, or teach. While I consider the diverse trades of these two excellent Philosophers (and am most sure, both, that Plato right well, otherwise could teach: and that Aristotle might boldly, with his hearers, have dealt in like sort as Plato did) I am in no little pang of perplexity: Because, that, which I mislike, is most easy for me to perform (and to have Plato for my example.) And that, which I know to be most commendable: and (in this first bringing, into common handling, the Arts Mathematical) to be most necessary: is full of great difficulty and sundry dangers. Yet● neither do I think it me●e, for so strange matter (as now is meant to be published) and to so strange an audience, to be bluntly, at first, put forth, without a peculiar Preface: Nor (Imitating Aristotle) well can I hope, that according to the ampleness and dignity of the State Mathematical, I am able, either plainly to prescribe the material bounds: or precisely to express the chief purposes, and most wonderful applications thereof. And though I am sure, that such as did shrink from Plato his s●hole, after they had perceived his final conclusion, would in these things have been his most diligent hearers) so infinitely might their desires, in fine and at length, by our Arts Mathematical be satisfied) yet, by this my preface & forewarning, Aswell all such, may (to their great behoof) the sooner, hither be alured: as also the Pythagorical, and Platonical perfect scholar, and the constant profound Philosopher, with more ease and speed, may (like the Bee,) gather, hereby, both wax and honey.  
Wherefore, seeing I find great occasion (for the causes alleged, and farther, in respect of my Art Mathematic general) to use a certain forewarning and preface, whose content shallbe, that mighty, most pleasant, and fruitful Mathematical Tree, with his chief arms and second (grifted) branches: 
The intent of this Preface.  Both, what every one is, and also, what commodity, in general, is to be looked for, aswell of griff as stock: And forasmuch as this enterprise is so great, that, to this our time, it never was (to my knowledge) by any achieved: And also it is most hard, in these our dreary days, to such rare and strange Arts, to win due and common credit: Nevertheless, if, for my sincere endeavour to satisfy your honest expectation, you will but lend me your thankful mind a while: and, to such matter as, for this time, my pen (with speed) is able to deliver, apply your eye or ear attentifely: perchance, at once, and for the first saluting, this Preface you will find a lesson long enough. And either you will, for a second (by this) be made much the apt: or shortly become, well able yourselves, of the lions claw, to conjecture his royal symmetry, and farther property. Now then, gentle, my friends, and country men, Turn your eyes, and bend your minds to that doctrine, which for our present purpose, my simple talon is able to yield you.  
All things which are, & have being, are found under a triple diversity general.  
For, either, they are deemed Supernatural, Natural, or, of a third being. Things Supernatural, are immaterial, simple, indivisible, incorruptible, & unchangeable. Things Natural, are material, compounded, divisible, corruptible, and changeable. Things Supernatural, are, of the mind only, comprehended: Things Natural, of the sense exterior, are able to be perceived. In things Natural, probability and conjecture hath place: But in things Supernatural, chief demonstration, & most sure Science is to be had. By which properties & comparasons of these two, more easily may be described, the state, condition, nature and property of those things, which, we before termed of a third being: which, by a peculiar name also, are called Things Mathematical. For, these, being (in a manner) middle, between things supernatural and natural: are not so absolute and excellent, as things supernatural: Nor yet so base and gross, as things natural: But are things immaterial: and nevertheless, by material things able somewhat to be signified. And though their particular Images, by Art, are agg●egable and divisible: yet the general Forms, notwithstanding, are constant, unchangeable, untransformable, and incorruptible. Neither of the sense, can they, at any time, be perceived or judged. Nor yet for all that, in the royal mind of man, first conceived. But, surmounting the imperfection of conjecture, weening and opinion: and coming short of high intellectual conception, are the Mercurial fruit of dianoetical discourse, in perfect imagination subsisting. A marvelous neutrality have these things Mathematical. and also a strange participation between things supernatural, immortal, intellectual, simple and indivisible: and things natural, mortal, sensible, compounded and divisible. Probability and sensible proof, may well serve in things natural: and is commendable: In Mathematical reasonings, a probable Argument, is nothing regarded: nor yet the testimony of sense, any whit credited: But only a perfect demonstration, of truths certain, necessary, and invincible: universally and necessarily concluded: is allowed as sufficient for an Argument exactly and purely Mathematical."  
Of Mathematical things, are two principal kinde●● namely, Number, 
Number.  and Magnitude. Number, we define, to be, a certain Mathematical Sum, of Vnits. 
Note the word, unit, to express the Greek Mona●, & not Unity: as we hau● all, commonly, till now, used.  And, an unit, is that thing Mathematical, Indivisible, by participation of some likeness of whose property, any thing, which is in de●de, or is counted One, may reasonably be called One. We accounted an unit, a thing Mathematical, though it be no Number, and also indiuisible● because, of it, materially, Number doth consist● which, principally, is a thing Mathematical. Magnitude is a thing Mathematical, 
Magnitude.  by participation of some likeness of whose nature, any thing is judged long, broad, or thick. A thick Magnitude we call a Solid, or a Body. What Magnitude so ever, is Solid or" Thick, is also broad, & long. A broad magnitude, we call a Superficies or a Plain. Every plain magnitude, hath also length. A long magnitude, we term a Line. A Line is neither thick nor broad, but only long: Every certain Line, hath two ends: The ends of a line, are Points called. 
A point.  A Point, is a thing Mathematical, indivisible, which may have a certain determined situation. If a Point move from a" determined situation, the way wherein it moved, is also a Line: mathematically produced. whereupon, of the ancient mathematicians, a Line is called the race or course of a Point. 
A Line.  A Point we define, by the name of a thing Mathematical: though it be no Magnitude, and indivisible: because it is the proper end, and bound of a Line: which is a true Magnitude. 
Magnitude.  And Magnitude we may define to be that thing Mathematical, which is divisible for ever, in parts divisible, long, broad or thick. Therefore though a Point be no Magnitude, yet Terminatively we reckon it a thing Mathematical (as I said) by reason it is properly the end, and bound of a line.  
Neither Number, nor Magnitude, have any materiality. First, we will consider of Number, and of the Science Mathematical, to it appropriate, called Arithmetic: and afterward of Magnitude, and his Science, called Geometry. But that name contenteth me not: whereof a word or two hereafter shall be said. How Immaterial and free from all matter, Number is, who doth not perceive? ye●, who doth not wonderfully wonder at it: For, neither pure Element, nor Aristoteles, Quinta Essentia, is able to serve for Number, as his proper matter. Nor yet the purity and simpleness of Substance Spiritual or Angelical, will be found proper enough thereto. And therefore the great & godly Philosopher Anitius Boetius, said: Omnia quaecunque a primava rerum na●●ra constructa sunt, Numerorum videntur ratione formata. Hoc enim fuit principale in animo Conditoris Exemplar. That is: All things (which from the very first original being of things, have been framed and made) do appear to be Form by the reason of Numbers. For this was the principal example or pattern in the mind of the Creator. O comfortable allurement, O ravishing persuasion, to deal with a Science, whose Subject, i● so Ancient, so pure, so excellent, so surmounting all creatures, so used of the Almighty and incomprehensible wisdom of the Creator, in the distinct creation of all creatures● in all their distinct parts, properties, natures, and virtues, by order, and most absolute number, brought, from Nothing, to the Formality of their being and state. By Numbers property therefore, of us, by all possible means, (to the perfection of the Science) learned, we may both wind and draw ourselves into the inward and deep search and vow, of all creatures distinct virtues, natures, properties, and Former: And also, farther, arise, clime, ascend, and mount up (with Speculative wings) in spirit, to behold in the Glass of Creation, the Form of Forms, the Exemplar Number of all things Numerable: both visible and inuisible● mortal and immortal, Corporal and Spiritual. Part of this profound and divine Science, had joachim the Prophesier attained unto: by Numbers Formal, Natural, and Rational, forcing, concluding, and forshewyng great particular events, long before their coming. His books yet remaining, hereof, are good proof: And the noble Earl of Mirandula, (besides that,) a sufficient witness: that joachim, in his prophecies, proceeded by no other way, then by Numbers Formal. And this Earl himself, in Rome, 
Ano. 1488.  set up 900. Conclusions, in all kind of Sciences, openly to be disputed of: and among the rest, in his Conclusions Mathematical, (in the eleventh Conclusion) hath in Latin, this English sentence. By Numbers, a way is had, to the searching out, and understanding of every thing, able to be known. For the verifying of which Conclusion, I promise' to answer to the 74. questions, under written, by the way of Numbers. Which Conclusions, I omit here to rehearse: aswell avoiding superfluous prolixity: as, because joannes Picus, works, are commonly had. But, in any case, I would wish that those Conclusions were read diligently, and perceived of such, as are earnest Observers and Considerers of the constant law of numbers: which is planted in things Natural and Supernatural: and is prescribed to all Creatures, inviolably to be kept. For, so, besides many other things, in those Conclusions to be marked, it would appear, how sincerely, & within my bounds, I disclose the wonderful mysteries, by numbers, to be attained unto.  
Of my former words, easy it is to be gathered, that Number hath a triple state: One, in the Creator: an other in every Creature (in respect of his complete constitution:) and the third, in Spiritual and Angelical Minds, and in the Soul of man. In the first and third state, Number, is termed Number Numbering. But in all Creatures, otherwise, Number, is termed Number Numbered. And in our Soul, Number beareth such a sway, and hath such an affinity therewith: that some of the old Philosophers taught, Man's Soul, to be a Number moving itself. And in deed, in us, though it be a very Accident: yet such an Accident it is, that before all Creatures it had perfect being, in the Creator, Sempiternally. Number Numbering therefore, is the discretion discerning, and distincting of things. But in God the Creator, This discretion, in the beginning, produced orderly and distinctly all things. For his Numbering, then, was his Creating of all things. And his Continual numbering, of all things, is the Conservation of them in being: And, where and when he will lack an unit: there and then, that particular thing shallbe Discreated. Here I stay. But our Severalling, distincting, and Numbering, createth nothing but of Multitude considered, maketh certain and distinct determination. And albeit these things be weighty and truths of great importance, yet (by the infinite goodness of the Almighty ternary,) Artificial Methods and easy ways are made, by which the zealous Philosopher, may win near this Riverish Ida ● this Mountain of Contemplation: and more than Contemplation. And also, though Number, be a thing so Immaterial, so divine, and aeternal: yet by degrees, by little and little, stretching forth, and applying some likeness of it, as first, to things Spiritual: and then, bringing it lower, to things sensibly perceived: as of a momentany sound iterated: then to the lest things that may be seen, numerable: And at length, (most grossly,) to a multitude of any corporal things seen, or felt: and so, of these gross and sensible things, we are trained to learn a certain Image or likeness of numbers: and to use Art in them to our pleasure and profit. So gross is our conversation, and dull is our apprehension: while mortal Sense, in us, ruleth the common wealth of our little world. Hereby we say, Three Lions, are three: or a ternary. Three Eagles, are three, or a ternary. Which 
☞  Ternaries, are each, the Union, knot, and Uniformity, of three discrete and distinct Vnits. That is, we may in each ternary, thrice, severally point, and show a part, One, One, and One. Where, in Numbering, we say One, two, Three. But how far, these visible One's, do differre from our Indivisible Vnits (in pure Arithmetic, principally considered) no man is ignorant. Yet from these gross and material things, may we be led upward, by degrees, so, informing our rude Imagination, toward the coceiving of Numbers, absolutely (Not supposing, nor admixting any thing created, Corporal or Spiritual, to support, contain, or represent those Numbers imagined:) that at length, we may be able, to find the number of our own name, gloriously exemplified and registered in the book of the Trinity most blessed and aeternal.  
But farther understand, that vulgar Practisers, have Numbers, otherwise, in sundry Considerations: and extend their name farther, then to Numbers, whose lest part is an Vnit. For the common Logist, Reckenmaster, or Arithmeticien, in his using of Numbers: of an unit, imagineth less partest and calleth them Fractions. As of an unit, he maketh an half, and thus noteth it, ½ and so of other, (infinitely diverse) parts of an unit, Yea and farther, hath, Fractions of Fractions. etc. And, forasmuch, as, Addition, Substraction, Multiplication, Division and Extraction of roots, are the chief, and sufficient parts of Arithmetic: 
Arithmetic.  which is, the Science that demonstrateth the properties, of Numbers, and all operations, in numbers to be performed. How often, therefore, 
Note.  these five sundry sorts of Operations, do, for the most part, of their execution, differre from the five operations of like general property and name, in our Whole numbers practisable, So often, (for a more distinct doctrine) we, vulgarly accounted and name it, an other kind of Arithmetic. And by this reason: the Consideration, doctrine, and working, in whole numbers only: where, of an unit, is no less part to be allowed: is named (as it were) an Arithmetic by itself. And so of the Arithmetic of Fractions. In like sort, the necessary, wonderful and Secret doctrine of Proportion, and proportionalytie hath purchased unto itself a peculiar manner of handling and working: and so may seem an other form of Arithmetic. Moreover, the Astronomers, for speed and more commodious calculation, have devised a peculiar manner of ordering numbers, about their circular motions, by Sexagenes, and Sexagesmes. By Signs, Degrees and Minutes etc. which commonly is called the Arithmetic of Astronomical or Physical Fractions. That, have I briefly noted, by the name of Arithmetic Circular. By cause it is also used in circles, not Astronomical. etc. Practice hath led Numbers farther, and hath framed them, to take upon them, the show of Magnitudes property: Which is Incommensurabilitie and Irrationalitie. (For in pu●e Arithmetic, an unit, is the common Measure of all Numbers.) And, here, Numbers are become, as lines, Plains and Solides: some times Rational, some times irrational ● And have proper and peculiar characters, (as √ ●. √ ●. and so of other. Which is to signify Ro●e Square, Rote Cubik: and so forth:) & proper and peculiar fashions in the five principal parts: Wherefore the practiser, esteemeth this, a diverse Arithmetic from the other. Practice bringeth in, here, diverse compounding of Numbers: as some time, two, three, four (or more) Radical numbers, diversly knit, by signs, o● Moore & Less: as thus √ ● 12 + √ ● 15. Or ●hus √ ●● 19 + √ ● 12- √ ● 2● etc. And some time with whole numbers, or fractions of whole Number, among them as 20 + √ ●●4● √ ● + 33- √ ● 10, √ ●● 44 + 12- + √ ●9. And so, infinitely, may hap the variety. After this: Both the one and the other hath fractions incident: and so is this Arithmetic greatly enlarged, by diverse exhibiting and use of Compositions and mixtynges. Consider how● I (being desirous to deliver the student from error and Cavillation) do give to this Practice, the name of the Arithmetic of Radical numbers: Not of Irrational or Surd Numbers● which other while, are Rational: though they have the Sign of a Rote before them, which, Arithmetic of whole Numbers most usual, would say they had no such Root: and so accounted them Surd Numbers: which generally spoken, is untrue: as Euclides tenth book may teach you. Therefore to call them, generally, Radical Numbers, (by reason of the sign √. prefixed,) is a sure way: and a sufficient general distinction from all other ordering and using of Numbers: And yet (beside all this) Consider: the infinite desire of knowledge, and incredible power of man's Search and Capacity: how, they, jointly have waded farther (by mixting of speculation and practice) and have found out, and attained to the very chief perfection (almost) of Numbers Practical use. Which thing, is well to be perceived in that great Arithmetical Art of AEquation: commonly called the Rule of Coss. or Algebra. The Latins termed it, Regulam Rei & Census, that is, the Rule of the thing and his value. With an apt name: comprehending the first and last points of the work. And the vulgar names, both in Italian, French and Spanish, depend (in naming it,) upon the signification of the Latin word, Res: A thing: unleast they use the name of Algebra. And therein (commonly) is a double error. The one, of them, which think it to be of Geber his inventing: the other of such as call it Algebra. For, first, though Geber for his great skill in Numbers, Geometry, Astronomy, and other marvelous Arts, might have seemed able to have first devised the said Rule: and also the name carrieth with it a very near likeness of Geber his name: yet true it is, that a Greek Philosopher and Mathematicien, named Diophantus, before Geber his time, wrote 13. books thereof (of which, six are yet extant: and I had them to 
* Anno. 1550.  use, of the famous Mathematicien, and my great friend, Petrus Mon●aureus:) And secondly, the very name, is Algiebar, and not Algebra: as by the Arabien Avicen, may be proved: who hath these precise words in Latin, by Andrea's Alpagus (most perfect in the Arabik tongue) so translated. Scientia faciendi Algiebar & Almachabel. i. Scientia inveniendi numerum ignotum, per additionem Numeri, & divisionem & aequationem. Which is to say: The Science of working Algiebar and Almachabel, that is, the Science of finding an unknown number, by Adding of a Number, & Division & aequation. Here have you the name: and also the principal parts of the Rule, touched. To name it, The rule, or Art of AEquation, doth signify the middle part and the State of the Rule. This Rule, hath his peculiar Characters: and the principal parts of Arithmetic, to it appertaining, do differe from the other Arithmetical operations. This Arithmetic, hath Numbers Simple, Compound, Mixed: and Fractions, accordingly. This Rule, and Arithmetic of Algiebar, is so profound, so general and so (in manner) containeth the whole power of Number's Application practical: that man's wit, can deal with nothing more profitable about numbers: nor match, with a thing, more meet for the divine force of the Soul, (in human Studies, affairs, or exercises) to be tried in. Perchance you looked for, (long ere now,) to have had some particular proof, or evident testimony of the use, profit and Commodity of Arithmetic vulgar, in the Common life and trade of men. Thereto, then, I will now frame myself: But herein great care I have, lest length of sundry profess, might make you dame, that either I did misdoute your zealous mind to virtues school: or else mistrust your able wits, by some, to guess much more. A proof then, four, five, or six, such, will I bring, as any reasonable man, therewith may be persuaded, to love & honour, yea learn and exercise the excellent Science of Arithmetic.  
And first: who, nearer at hand, can be a better witness of the fruit received by Arithmetic, than all kind of Merchants? Though not all, alike, either need it, or use it. How could they forbear the use and help of the Rule, called the Golden Rule? Simple and Compounde● both forward and backward? How might they miss Arithmetical help in the Rules of Felowshyp● either without time, or with timer and between the Merchant & his ●actor● The Rul●● of Ba●tering in wares only or part in wares, and part in money, would they gladly want? Our Merchant ventures, and travailers over Sea, how could they order their doings justly and without loss, unleast certa●ne and general Rules for Exchange of money, and Rechaunge, were, for their use, devised? The Rule of Alligation, in how sundry cases, doth it conclude for them, such precise verities, as neither by natural wit, nor other experience, they, were able, else, to know? And (with the Merchant then to make an end) how ample & wonderful is the Rule of False positions? especially as it is now, by two excellent mathematicians (of my familer acquaintance in their life time) enlarged? I mean Gemma Frisius, and Simon jacob. Who can either in brief conclude, the general and Capital Rules? or who can Imagine the myriads of sundry Cases, and particular examples, in Act and earnest, continually wrought, tried and concluded by the forenamed Rules, only? How sundry other Arithmetical practices, are commonly in merchants hands, and knowledge: They themselves, can, at large, testify.  
The Mintmaster, and Goldsmith, in their Mixture of Metals, either of diverse kinds, or diverse values: how are they, or may they, exactly be directed, and marvelously pleasured, if Arithmetic be their guide? And the honourable Physicians, will gladly confess themselves, much beholding to the Science of Arithmetic, and that sundry ways: But chief in their Art of Graduation, and compound Medicines. And though Galenus, Auer●ois, Arnoldus, Lullus, and other have published their positions, aswell in the quantities of the Degrees above Temperament, as in the Rules, concluding the new Form resulting: yet a more precise, commodious, and easy Method, is extant: 
R. B.  by a Countryman of ours (above 200 years ago) invented. And forasmuch as I am uncertain, who hath the same: or when that little Latin treatise, (as the Author writ it,) shall come to be Printed: (Both to declare the desire I have to pleasure my Country, wherein I may: and also, for very good proof of Numbers use, in this most subtle and fruitful, Philosophical Conclusion,) I intend in the mean while, most briefly, and with my farther help, to communicate the pith thereof unto you.  
First describe a circle: whose diameter let be an inch. Divide the Circumference into four equal parts. Fron the Centre, by those 4. sections, extend 4. right lines: each of 4. inches and a half long: or of as many as you list, above 4. without the circumference of the circle: So that they shall be of 4. inches long (at the lea●t) without the Circle● Make good evident marks, at every inches end. If you list, you may subdivide the inches again into 10. or 12. smaller parts, equal. At the ends of the lines, writ the names of the 4. principal elemental Qualities. hot and Cold●, one against the other. And likewise Moist and Dry, one against the other. And in the Circle writ Temperate. Which Temperature hath a good Latitude: as appeareth by the Complexion of man. And therefore we have allowed unto it, the foresaid Circle: and not a point Mathematical or Physical.  
And though I, here, speak only of two things Miscible: and most commonly, more than three, four, five or six, (etc.) are to be Mixed: (and in one Compound to be reduced● & the Form resulting of the same, to serve the turn) yet these Rules are sufficient: duly repeated and iterated. 
Note.  In proceeding first, with any two: and then, with the Fonne Resulting, and an other: & so forth: For, the last work, concludeth the Form resulting of them all: I need nothing to speak, of the Mixture (here supposed) what it is. Common Philosophy hath defined it, saying, Mixtio est miscibilium, alteratorum, per minima coniunctorum, Vnio. Every word in the definition, is of great importance. I need not also spend any time, to show, how, the other manner of distributing of degrees, doth agreed to these Rules. Neither need I of the farther use belonging to the Cross of Graduation (before described) in this place declare, unto such as are capable of that, which I have all ready said. Neither yet with examples specify the Manifold variety's, by the foresaid two general Rules, to be ordered. The witty and Studious, here, have sufficient: And they which are not able to attain to this, without lively teaching, and more in particular: would have larger discoursing, then is meet in this place to be dealt withal: And other (perchance) with a proud snuff will disdain this litl●● and would be unthankful for much more. I, therefore conclude: and wish such as have modest and earnest Philosophical minds, to laud God highly for this: and to Marvel, that the profoundest and subtlest point, concerning Mixture of Forms and Qualities Natural, is so Match● and married with the most simple, easy, and short way of the noble Rule of Algiebar. Who can remain, therefore unpersuaded, to love, allow, and honour the excellent Science of Arithmetic? For, here, you may perceive that the little finger of Arithmetic, is of more might and contriving then a hundred thousand men's wits, of the middle sort, are able to perform, or truly to conclude, with out help thereof.  
Now will we f●rder, by the wise and valiant captain, be certified, what help he hath, by the Rules of Arithmetic: in one of the Arts to him appertaining: And of the Greeks named 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉. 
〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉.  That is, the Skill of Ordering Soldiers in Battle ray after the best manner to all purposes. This Art so much dependeth upon Number's use, and the Mathematicals, that AElianus (the best writer thereof,) in his work, tö the Emperor Hadrianus, by his perfection, in the Mathematicals, (being greater, than other before him had,) thinketh his book to pass all other the excellent works, written of that Art, unto his days. For, of it, had written AEneas: Gyneas of Thessaly: Pyrrhus Epirota: and Alexander his son: Clearchus: Pausanias: Euangelus: Polybi●s, familiar friend to Scipio: Eupolemus: Iphicrates, Possidonius: and very many other worthy Captains, Philosophers and Princes of Immortal fame and memory: Whose fairest flower of their garland (in this feat) was Arithmetic: and a little perceiverance, in Geometrical Figures. But in many other cases doth Arithmetic stand the captain in great stead. As in proportioning of victuals, for the Army, either remaining at a stay: or suddenly to be increased with a certain number of Soldiers: and for a certain time. Or by good Art to diminish his company, to make the victuals, longer to serve the remanent, & for a certain determined time: if need so require. And so in sundry his other accounts, Reckeninges, Measurynges, and proportionynges, the wise, expert, and Circumspect ●apitaine will affirm the Science of Arithmetic, to be one of his chief Counsellors, directors and aiders. Which thing (by good means) was evident to the Noble, the Courageous, 
☞  the loyal, and Courteous john, late Earl of Warwick. Who was a young Gentleman, thoroughly known to very few. Albeit his lusty valiantness, force, and Skill in Chivalrous feats and exercises: his humbleness, and frendelynes to all men, were things, openly, of the world perceived. But what roots (otherwise,) virtue had fastened in his breast, what Rules of godly and honourable life he had framed to himself: what vices, (in some then living) notable, he took great care to eschew: what manly virtues, in other noble men, (flourishing before his eyes,) he Sythingly aspired after: what prowesses he purposed and meant to achieve: with what feats and Arts, he began to furnish and fraught himself, for the better service of his King and Country, both in peace & war. These (I say) his Heroical Meditations, forecas●inges and determinations, not twain, (I think) beside myself, can so perfectly, and truly report. And therefore, in Conscience, I count it my part, for the honour, preferment, & procuring of virtue (thus, briefly) to have put his Name, in the Register of Fa●● Immortal.  
To our purpose. This john, by one of his acts (besides many other: both in England and France, by me, in him noted.) did dislose his hearty love to virtuous Sciences: and his noble intent, to excel in Martial prowess: When he, with humble request, and instant Soliciting: got the best Rules (either in time passed by Greek or Roman, or in our time used: and new Stratagems therein de●ised) for ordering of all Companies, sums and Number● of men, (Many or few) with one kind of weapon, or more, appointed: with Artillery, or without: on horseback, or on foot: to give, or take onset: to seem many, being few● to s●em few, being many. To march in battle or jornay: with many such feats, to Fought f●●ld, Skarmoush, or Ambush appartaining: And of all these, lively desi●nementes (most curiously) to be in velame parchment described: with Notes & peculiar marks, as the Art requireth: and all these Rules● and descriptions Arithmetical, 
This noble Earl, died Anno. 1554. scarce of 24. years of age● having no issue by his wife: Daughter to the Duke of Somerset.  enclosed in a rich Case of Gold, he used to wear about his necke● as his jewel most precious, and counselor most trusty. Thus, Arithmetic, of him was shrined in gold: Of Number's fruit, he had good hope. Now, Numbers therefore innumerable, in Numbers praise, his sh●●ne shall find.  
What need I, (for farther proof to you) of the Schoolmasters of justice, to require testimony: how needful, how fruitful, how skilful a thing Arithmetic is? I meane● the Lawyers of all sorts. Undoubtedly, the civilians, can marvelously declare: how, neither the Ancient Roman laws, without good knowledge of Number's art, can be perceived: Nor (justice in infinite Cases) without due proportion, (narrowly considered,) is able to be executed● How justly, & with great knowledge of Art, did Pap●●●●anus institute a law of partition, and allowance, between man and wife a●●● a ●●●orce: But how Acc●rsius, Bald●s, Bartolus, jason, Alex●nder, and finally Alciat●● (being otherwise, notably well learned) do jumble, guess, and err, from the ●quity●●rt ●nd I●●ent of the lawmaker: Arithmetic can detect, and convince: and clearly, make the truth to shine. Good Bartolus, tired in the examining & proportioning of the matter: ●nd with Accursius Gloss, much cumbered: burst out, and said: Nulla est i● 〈◊〉 libr●, 〈◊〉 glossa difficili●r: Evils c●mputationem nec Scholas●ici ●ec Doct●res intelligent. etc. That is● In the whole book, there is no ●losse harder than thi●● Whose account or reckoning, neither the Scholars, nor the Doctors understand. etc. What can they say of I●lianus law, Si ita S●ript●m. etc. Of the Testators will justly performing, between the wife, Son and daughter? How can they perceive the ●●●●●tie of Aphricanus, Arithmetical Reckoning, where he treateth of Lex Falcid●a ● How ●●n they deliver him, from his Reprovers: and their maintainers: as I●●●●es, Accursius Hippolytus and Alciatus? How ●ustly and artificially, was African●s reckoning made? Proportionating to the Sums bequeathed, the Contributions of each part? Namely, for the hundred presently received, 17 1/7. And for the hundred, received after ten months, 12 6/7: which make the 30: which were to be contributed by the legatari●s to the heir. For, what proportion, 100 hath to 75: the same hath 17 1/7 to 12 6/7: Which is Sesquitertia: that is, as 4, to 3. which make 7. Wonderful many places, in the Civil law, require an expert Arithmeticien, to understand the deep judgement, & Just determination of the Ancient Roman Lawmakers. But much more expert aught he to be, who should be able, to decide with equity, the infinite variety of Cases, which do, or may happen, under every one of those laws and ordinances Civil. Hereby, easily, ye may now conjecture: that in the Canon law: and in the laws of the Realm (which with us, bear the chief Authority), justice and equity might be greatly preferred, and skilfully executed, through due skill of Arithmetic, and proportions appertaining. The worthy Philosophers, and prudent lawmakers (who have written many books De Republica: How the best state of Common wealths might be procured and maintained,) have very well determined of justice: (which, not only, is the Base and foundation of Common weals: but also the total perfection of all our works, words, and thoughts:) defining it, to be that virtue, 
justice.  by which, to every one, is rendered, that to him appertaineth. God challengeth this at our hands, to be honoured as God: to beloved, as a father: to be feared as a Lord & master. Our neighbour's proportion, is also prescribed of the Almighty lawmaker: which is, to do to other, even as we would be done unto. These proportions, are in justice necessary in duty, commendable: and of Common wealths, the life, strength, stay and flourishing. Aristotle in his ethics (to fatch the seed of justice, and light of direction, to use and execute the sam●) was fain to fly to the perfection, and power of Numbers: for proportions Arithmetical and Geometrical. Plato in his book called Epinomis (which book, is the Threasury of all his doctrine) where, his purpose is, to seek a Science, which, when a man had it, perfectly: he might seem, and so be, in deed, Wise. He, briefly, of other Sciences discoursing, findeth them, not able to bring it to pass: But of the Science of Numbers, he saith. Illa, qua numerum mortalium generi d●●n, id profecto efficiet● Deum antem aliquem, magis quam fortunam, ●d sa●●tem nostram, hoc m●nus nobis arbitror contulisse. etc. Nam ipsum ●onorum omnium Authorem, cur non maximi boni, Prudentiae dico, causam arbitramur? That Science, verily, which hath taught mankind number, shall be able to bring it to pass. And, I think, a certain God, rather than fortune, to have given us this gift, for our bliss. For, why should we not judge him, who is the Author of all good things, to be also the cause of th● greatest good thing, namely, Wisdom? There, at length, he proveth Wisdom to be attained, by good Skill of Numbers. With which great Testimony, and the manifold profess, and reasons, before expressed● you may be sufficiently and fully persuaded: of the perfect Science of Arithmetic, to make this account: That of all Sciences, 
☞  next to theology, it is most divine, most pure, most ample and general, most profound, most subtle, most commodious and most necessary. Whose next Sister, is the Absolute Science of Magnitudes: of which (by the Direction and aid of him, whose Magnitude is Infinite, and of us Incomprehensible) I now intend, so to writ, that both with the Multitude, and also with the Magnitude of marvelous and fruitful verities, you (my friends and Countrymen) may be stirred up, and awaked, to behold what certain Arts and Sciences, (to our unspeakable behoof) our heavenly father, hath for us prepared, and revealed, by sundry Philosophers and mathematicians.  
BOth, Number and Magnitude, have a certain Original seed, (as it were) of an incredible property: and of man, never able, Fully, to be declared. Of Number, an unit, and of Magnitude, a Point, do seem to be much like Original pretence of just content, and me●sure● 〈◊〉 lan●es and groundes●●re●● 〈◊〉, disquietness, murder, and war did (full often) ensue: ●ill●● by God's 〈◊〉 and 〈◊〉 industry, The perfect Science of Lines, Plains, and Solides (lik●● divine 〈◊〉,) gave unto every man, his own. The people then, by this art pleasured, and greatly relieved, in their lands just measuring: & other Philosophers● writing Rules for land measuring● between them both, thus, con●●rmed the name of G●●m●tria, that is, (according to the very etymology of the word) Land measuring. Wherein, the people knew no farther, of Magnitudes use, but in Plai●●●: and the Philosophers, of them, had no feethearers, or Scholers●●a●der to disclose unto, then of ●lat, plain Geometry. And though, these Philosophers, knew of farther use, and best understood the etymology of the word, yet this name G●●●etria, was of them applied generally to all sorts of Magnitudes: unleast, otherwhile, of Plato, and Pythagoras● When they would precisely declare their own doctrine. Then, was 
* Plato. 7. de Rep.  Geometria, with them, Studium quod circa planum versatur. But, well you may perceive by Euclides Elements, that more ample is our Science, then to measure Plains: and nothing less therein is taught (of purpose) than how to measure Land. another name, therefore, must needs be had, for our Mathematical Science of Magnitudes: which regardeth neither clod, no● turf neither hill, nor dale● neither ●●●th nor heaven: but is absolute Megeth●logia: not creeping on ground, and dissellng the eye, with pole perch, rod or line: 
☞  but lifting the heart above the heavens, by 〈◊〉 lines, and immortal beames● meeteth with the reflections, of the light incomprehensible● and so procureth joy, and perfection unspeakable. Of which true use of our Meg●thica, or Megethologia, Divine Plato seemed to have good taste, and iudgement● and (by the name of Geometry) so noted it● and warned his Scholars thereof: as, in his seventh Dialog, of the Common wealth, may evidently be seen. Where (in Latin) thus it is: right well translated: 〈…〉 quid Geometri● 〈…〉 loquuntur, qui in ipsa vers●●t●●. In English, 〈◊〉 o● Verily (saith Plat●) whosoever have, (but even very little) tasted of Geometry, will not deny unto us, this: but that this Science, is of an other condition, quite contrary to that, which they that are exercised in it, do speak of it. And there it followeth● of our Geometry, Quod q●●ritur cognoscendi illius gratia, quod se●per est, 〈…〉 & Interit. Geometria, eius quod est semper, Cognit●● est. 〈…〉 Veritatem, ●nimum: atque ita, ad Philosophandum 〈…〉 qua, nunc, contra quam decet, and inferiors 〈…〉 etc. 〈…〉 praecipiendum est, ut qui praclarissimam hanc 〈…〉 spernant. Name & quoe pr●ter ipsius propositu●, 〈…〉 sunt. etc. It must needs be confessed (saith Plat●) That [〈◊〉] i● learned, for the knowing of that, which is 〈◊〉: and not of that, which, in time, both is bred and is brought to an end. etc. Geometry is the knowledge of that which is everlasting. It will lif● up therefore (O Gentle Sir) our mind to the Verity: and by that means, it will prepare the Thought, to the Philosophical love of wisdom: that we may turn or con●ert, toward heavenly things [〈…〉] which now, otherwise then b●comme●●● us, we cast 〈…〉 or inferior things etc. Chief, therefore, Commandment must be given, that such as do ●nhabit this most honourable City, by no means, despise Geometry. For even those things [done by ●●] which, in manner, seam to be, beside the purpose of Geometry: are of by untrue measuring and surveying of Land or Woods, any way. And, this I am sure: that the Value of the difference, between the truth and such Surveys, would have been able to have found (forever) in each of our two Universities, an excellent Mathematical Reader: to each, allowing (yearly) a hundred Marks of lawful money of this realm: which, in deed, would seem requisite, here, to be had (though by other ways provided for) as well, as the famou●●●●ersit●e of Paris, hath two Mathematical Readers: and each, two hundredth French Crowns yearly, of the French Kings magnificent liberality only. Now, again, to our purpose returning: Moreover, of the former knowledge Geometrical, are grown the Skills of Geography, chorography, hydrography, and S●●●rithmetrie.  
Geographie teacheth ways, by which, in sundry forms, (as Spharike, Plain or other), the Situation of Cities, Towns, Villages, Forts, castles, Mountains, Woods, Havens, Rivers, creeks, & such other things, upon the outface of the earthly Globe (either in the whole, or in some principal member and portion thereof contained) may be described and designed, in commensurations Analogical to Nature and verity: and most aptly to our view, may be represented. Of this Art how great pleasure, and how manifold commodities do come unto us, daily and hourly: of most men, is perceived. While, some, to beautify their Halls, Parlours, Chambers, galleries, Studies, or Libraries with: other some for things past, as battles fought, earthquakes, heavenly fyringes, & such occurentes, in histories mentioned: thereby lively, as it were, to view the place, the region adjoining, the distance from us: and such other circumstances. Some other, presently to view the large dominion of the Turk: the wide Empire of the Moschovite: and the little morsel of ground, where Christendom (by profession) is certainly known. Little, I say, in respect of the rest. etc. Some, either for their own journeys directing into far lands: or to understand of other m●n● trauaile●. To conclude, some, for one purpose: and some, for another, liketh, loveth, ge●●●th, and vse●●, Maps, charts, & Geographical Globes. Of whose use, to speak sufficiently, would require a book peculiar.  
chorography 〈…〉  
hydrography, delivereth to our knowledge, on Globe or in Plain, the perfect Analogical description of the Ocean Sea coasts, through the whole world ● or in the chief and principal parts thereof: with the ●les and chief particular places of dangers, contained within the bounds, and Sea coasts described: as, of Quicksands, Banks, Pits, Rocks, Races, Countertides, Whorlepooles. etc. This, dealeth with the Element of the water chief: as Geographie did principally take the Element of the earths description (with his appurtenances) to task. And besides this, hydrography, requireth a particular Register of certain Landmarks (where marks may be had) from the sea, well able to be skried, in what point of the Seacumpase they appear, and what apparent form, Situation, and bigness they have, in respect of any dangerous place in the sea, or near unto it, assigned: And in all Coasts, what Moon, maketh full Sea: and what way, the Tides and Ebbs, come and go, the Hydrographer aught to record. The Soundinge likewise: and the Channels ways: their number, and depths ordinarily, at ebb and flood, aught the Hydrographer, by observation and diligence of Measuring, to have certainly known. And many other points, are belonging to perfect hydrography, and for to make a Rutter, by: of which, I need not here speak: as of the describing, in any place, upon Globe or Plain, the 32. points of the Compass, truly: (whereof, scarcely four, in England, have right knowledge: because, the lines thereof, are no strait lines, nor Circles.) Of making due projection of a Sphere in plain. Of the Variation of the Compass, from true north: And such like matters (of great importance, all) I leave to speak of, in this place: because, I may seam (all ready) to have enlarged the bounds, and duty of an Hydographer, much more, than any man (to this day) hath noted, or prescribed. Yet am I well able to prove, all these things, to appertain, and also to be proper to the Hydrographer. The chief use and end of this Art, is the Art of Navigation; but it hath other diverse uses: even by them to be enjoyed, that never lack sight of land.  
Stratarithmetrie, is the Skill, (appertaining to the war,) by which a man can set in figure, analogical to any Geometrical figure appointed, any certain number or sum of men: of such a figure capable: (by reason of the usual spaces between Soldiers allowed: and for that, of men, can be made no Fractions. Yet, nevertheless, he can order the given sum of men, for the greatest such figure, that of them, can be ordered) and certify, of the overplus: (if any be) and of the next certain sum, which, with the overplus, will admit a figure exactly proportional to the figure assigned. By which Skill, also, of any army or company of men: (the figure & sides of whose orderly standing, or array, is known) he is able to express the just number of men, within that figure contained: or (orderly) able to be contained. 
* Note.  And this figure, and sides thereof, he is able to know: either being by, and at hand: or a far of. Thus far, stretcheth the description and property of Stratarithmetrie: sufficient for this time and place. It differreth from the Feat tactical, De acic●us instruendis ● because, 
The difference between Strataruhmetrie and Tacticie.  there, is necessary the wisdom and foresight, to what purpose he so ordereth the men: and Skilful ability, also, for any occasion, or purpose, to devise and use the aptest and most necessary order, array and figure of his Company and Sum of men. By figure, I mean: as, either of a Perfect Square, Triangle, Circle, Ouale, long square, (of the Greeks it is called Eteromekes) Rhombe, Rhomboïd, Lunular, Ring, Serpentine, and such other Geometrical figures: Which, in wars, have been, and are to be used: for commodiousness, necessity, and advantage etc. And no small skill aught he to have, that should make true report, or near the truth, of the numbers and Sums, of footmen or horsemen, in the Enemies ordering. A far of, to make an estimate, between near terms of Moore and Less, is not a thing very rife, among those that gladly would do it. Great policy may be used of the Captains, 
I.D. F●end●● you will find it hard, to performs my description of ●his F●ate. But by Ch●r●graphie● you may help yourself some ●hat: wher● th● Figure's known (in Sid●●●nd Angles) are not Regular: And where● Resolution into Triangles can s●●u●. etc. And yet you will find it strange to deal thus generally with Arithmetical figures: and, that for Battle ●ay. Their co●tent●●● differ so much from like geometrical Figur●s.  (at times feet, and in places convenient) as to use Figures, which make greatest show, of so many as he hath: and using the advantage of the three kinds of usual spaces: (between footmen or horsemen) to take the largest: or when he would seem to have few, (being many:) contrary wise, in Figure, and space. The Herald, Pursuivant, Sergeant Royal, captain, or who soever is careful to come near the truth herein●, besides the judgement of his expert eye, his skill of Ordering tactical, the help of his Geometrical instrument: Ring, or Staff Astronomical: (commodiously framed for carriage and use.) He may wonderfully help himself, by perspective Glasses. In which, (I trust) our posterity will prove more skilful and expert, and to greater purposes, then in these days, can (almost) be credited to be possible.  
Thus have I lightly passed over the Artificial Feats, chief depending upon vulgar Geometry: & commonly and generally reckoned under the name of Geometry. But there are other (very many) Methodical Arts, which, declining from the purity, simplicity, and Immateriality, of our Principal Science of Magnitudes: do yet nevertheless use the great aid, direction, and Method of the said principal Science, and have proper names, and distinct: both from the Science of Geometry, (from which they are derived) and one from the other. As Perspective, Astronomy, Music, cosmography, Astrology, Statike, Anthropographie, trochilic, Helicosophie, Pneumatithmie, Menadrie, Hypogeiodie, Hydragogie, Horometrie, Zographie, Architecture, Navigation, Thaumaturgike and Archemastrie. I think it necessary, orderly, of these to give some peculiar descriptions: and withal, to touch some of their commodious uses, and so to make this Preface, to be a little sweet, pleasant nosegay for you: to comfort your Spirits, being almost out of courage, and in despair, (through brutish brute) Weening that Geometry, had but served for building of an house, or a curious bridge, or the roof of Westminster hall, or some witty pretty devise, or engine, appropriate to a Carpenter, or a joiner etc. That the thing is far otherwise, than the world, (commonly) to this day, hath deemed, by word and work, good proof willbe made.  
Among these Arts, by good reason, Perspective aught to be had, ere of Astronomical Apparences, perfect knowledge can be attained. And because of the prerogative of Light, being the first of God's Creatures: and the eye, the light of our body, and his Sense most mighty, and his organ most Artificial and Geometrical: At Perspective, we will begin therefore. Perspective, is an Art Mathematical, which demonstrateth the manner, and properties, of all Radiations Direct, Broken, and Reflected. This Description, or Notation, is briefs but it reacheth so far, as the world is wide. It concerneth all Creatures, all Actions, and passions, by Emanation of beams performed. Beams, or natural lines, (here) I mean, not of light only, or of colour (though they, to eye, give show, witness, and proof, whereby to ground the Art upon) but also of other Forms, both Substantial, and Accidental, the certain and determined active Radiall em●nations. By this Art (omitting to speak of the highest points) we may use our eyes, and the light, with greater pleasure: and perfecter judgement: both of things, in l●ght seen; & of other: which by like order of Lights Radiations, work and produce their effects. We may be ashamed to be ignorant of the cause, why so sundry ways our eye is deceived, and abused: as, while the eye weeneth a round Globe or Sphere (being far of) to be a flat and plain Circle, and so likewise judgeth a plain Square, to be round: supposeth walls parallels, to approach, a far of: roof and flower parallels, the one to bend downward, the other to rise upward, at a little distance from you. Again, of things being in like swiftness of moving, to think the nearer, to move faster: and the farther, much slower. Nay, of two things, whereof the one (incomparably) doth move swifter than the other, to dame the slower to move very swift, & the other to stand: what an error is this, of our eye? Of the rainbow, both of his Colours, of the order of the colours, of the bigness of it, the place and heith of it, (etc.) to know the causes demonstrative, is it not pleasant, is it not necessary? of two or three Sons appearing: of Blazing Stars: and such like things: by natural causes, brought to pass, (and yet nevertheless, of farther matter, Significative) is it not commodious for man to know the very true cause, & occasion Natural? Yea, rather, is it not, greatly, against the Sovereignty of Man's nature, to be so overshot and abused, with things (at hand) before his eyes? as with a Peacocks tail, and a doves neck: or a whole ore, in water, holden, to seem broken. Things, far of, to seem near: and near, to seem far of. Small things, to seem great: and great, to seem small. One man, to seem an Army. Or a man to be curstly afraid of his own shadow. Yea, so much, to fear, that, if you, being (alone) near a certain glass, and proffer, with dagger or sword, to foin at the glass, you shall suddenly be moved to give back (in manner) by reason of an Image, 
A marvelous Glass. ☞  appearing in the air, between you & the glass, with like hand, sword or dagger, & with like quickness, foining at your very eye, likewise as you do at the Glass. Strange, this is, to hear of: but more marvelous to behold, than these my words can signify. And nevertheless by demonstration optical, the order and cause thereof, is certified: even so, as the effect is consequent. Yea, thus much more, dare I take upon me, toward the satisfying of the noble courage, that longeth ardently for the wisdom of Causes Natural: as to let him understand, that, in London, he may wish his own eyes, have proof of that, which I have said herein. A Gentleman, (which, for his good service, 
S.W.P.  done to his Country, is famous and honourable: and for skill in the Mathematical Sciences, and Languages, is the Odd man of this kind. etc.) even he, is able: and (I am sure) will, very willingly, let the Glass, and proof be seen: and so I (here) request him: for the increase of wisdom, in the honourable: and for the stopping of the mouths malicious: and repressing the arrogancy of the ignorant. You may easily guess, what I mean. This Art of Perspective, is of that excellency, and may be led, to the certifying, and executing of such things, as no man would easily believe: without Actual proof perceived. I speak nothing of Natural Philosophy, which, without Perspective, can not be fully understanded, nor perfectly attained unto. Nor, of Astronomy: which, without Perspective, can not well be grounded: Nor Astrology, naturally Verified, and avouched. That part hereof, which dealeth with Glasses (which name, Glass, is a general name, in this Art, for any thing, from which, a Beam reboundeth) is called Catoptrike and hath so many uses, both marvelous, and profitable: that, both, it would hold me to long, to no●● therein the principal conclusions, all ready known: And also (perchance) some things, might lack due credit with you: And I, thereby, to lose my labor● and you, to slip into light judgement 
☞  ● Before you have learned sufficiently the power of Nature and Arte.  
NOw, to proceed: Astronomy, is an Art Mathematical, which demonstrateth the distance, magnitudes, and all natural motions, apparences, and passions proper to the Planets and fixed Starts: for any time past, present and to come: in respect of a certain Horizon, or without respect of any Horizon. By this Art we are certified of the distance of the Starry Sky, and of each Planete from the Centre of the Earth: and of the greatness of any Fixed star seen, or Planete, in respect of the earths greatness. As, we are sure (by this Art) that the Solidity, Massines and Body of the Son, containeth the quantity of the whole Earth and Sea, a hundred three score and two times, less by 1/● one eight part of the earth. But the Body of the whole earthly globe and Sea, is bigger than the body of the Moon, three and forty times less by 1/● of the Moon. Wherefore the Son is bigger than the Moon, 7000 times, less, by 59 ●●/164 that is, precisely 6940 ●5/●● bigger than the Moon. And yet the unskilful man, would judge them a like big. Wherefore, of Necessity, the one is much farther from us, than the other. The Son; when he is farthest from the earth (which, now, in our age, is, when he is in the 8. degree, of Cancer) is, 1179 Semidiameters of the Earth, distant. And the Moon when she is farthest from the earth, is 68 Semidiameters of the earth and 1/● The nearest, that the Moon cometh to the earth, is Semidiameters 52 ¼ The distance of the Starry Sky is, from us, in Semidiameters of the earth 20081 1/● Twenty thousand fourscore, one, and almost a half. Subtract from this, the Moans nearest distance, from the Earth: and thereof remaineth Semidiameters of the earth 20029 1/● Twenty thousand nine and twenty and a quarter. 
Note.  So thick is the heavenly Palace, that the Planets have all their exercise in, and most marvelously perform the Commandment and Charge to them given by the omnipotent Majesty of the king of kings. This is that, which in Genesis is called Ha' Rakia. Consider it well. The Semidiameter of the earth; coteineth of our common miles 3436 ●/● three thousand, four hundred thirty six and four eleventh parts of one mile: Such as the whole earth and Sea, round about, is 21600. One and twenty thousand six hundred of our miles. Allowing for every degree of the greatest circle, three score miles. Now if you way well with yourself but this little parcel of f●ute Astronomical, as concerning the bigness, Distances of Son, Moon, Sterry Sky, and the huge massines of Haraldus Rakia, will you not find your Consciences moved, with the kingly Prophet, to sing the confession of God's Glory, and say, The Heavens declare the glory of God, and the Firmament [Ha-Rakia] showeth forth the works of his hands. And so forth, for those five first staves, of that kingly Psalm. Well, well, It is time for some to lay hold on wisdom, and to judge truly of things: and not so to expound the Holy word, all by Allegories: as to Neglect the wisdom, power and Goodness of God, in, and by his Creatures, and Creation to be seen and learned. By parables and Analogies of whose natures and properties, the course of the Holy Scripture, also, declareth to us very many, Mysteries. The whole Frame of God's Creatures; (which is the whole world,) is to us, a bright glass: from which, by reflection, reboundeth to our knowledge and perceiverance, Beams, and Radiations● representing the Image of his Infinite goodness Omnipotency, and wisdom. And we thereby, are taught and persuaded to Glorify our Creator as God: and be thankful therefore. Can the Heathenistes find these uses, of these most pure, beautiful, and Mighty Corporal Creatures: and shall we after that the true Son of right wiseness is risen above the Horizon, of our temporal Hemispharie, and hath so abundantly streamed into our hearts, the direct beams of his goodness, mercy, and grace: Whose heat All Creatures feel: Spiritual and Corporally Visible and Invisible: Shall we (I say) look upon the Heaven, Stars, and Planets, as an Ox and an Ass doth: not further careful or inquisitive, what they are: why were they Created, How do they execute that they were Created for? seeing, All Creatures, were for our sake created: and both we, and they, Created, chief to glorify the Almighty Creator: and that, by all means, to us possible. Nolite ignorare (saith Plat● in Epinomis) Astronomiam, Sapientissimun quiddam esse. Be ye not ignorant, Astronomy to be a thing of excellent wisdom. Astronomy, was to us, from the beginning commended, and in manner commanded by God himself. In as much as he made the Son, Moon, and Stars, to be to us, for Signs, and knowledge of Seasons, and for Distinctions of Days, and years. Many words need not. But I wish, every man should way this word, Signs. And besides that, confer it also with the tenth Chapter of Hieremie. And though Some think, that there, they have found a rod: Yet Modest Reason, will be indifferent judge, who aught to be beaten therewith, in respect of our purpose. Leaving that: I pray you understand this: that without great diligence of Observation, examination and Calculation, their periods and courses (whereby Distinction of Seasons, years, and New Moans might precisely be known) could not exactly be certified. Which thing to perform, is that Art, which we here have Defined to be Astronomy. Whereby, we may have the distinct Course of Times, days, years, and Ages: aswell for Consideration of Sacred Prophecies, accomplished in due time, foretold: as for high Mystical Solemnities holding: And for all other humane affairs, Conditions, and covenants, upon certain time, between man and man● with many other great uses: Wherein, (verily), would be great incertainty, Confusion, untruth, and brutish barbarousness: without the wonderful diligence and skill of this Art: continually learning, and determining Times, and periods of Time, by the Record of the heavenly book, wherein all times are written: and to be read with an Astronomical staff, in stead of a festue.  
Music, of Motion, hath his Original cause: Therefore, after the motions most swift, and most Slow, which are in the Firmament, of Nature performed: and under the Astronomers Consideration: now I will Speak of an other kind of Motion, producing sound, audible, and of Man numerable. Music I call here that Science, which of the Greeks is called Harmonice. Not meddling with the Controversy between the ancient Harmonistes, and Ca●onistes. Music is a Mathematical Science, which teacheth; by sense and reason, perfectly to judge, and order the diversities of sounds, high and low. Astronomy and Music are Sisters; saith Plato. As, for Astronomy, the eyes: So, for Harmonious Motion, the cares were made. But as Astronomy hath a more divine Contemplation, and commodity, then mortal eye can perceive: So, is Music to be considered, that the 
1.  Mind may be pref●r●ed, before the ear. And from audible sound, we aught to ascend, to the examination: which numbers are Harmonious, and which not. And why, either, the one are: or the other are not. I could at large, in the heavenly 
2.  motions and distances, describe a meruallous Harmony, of Pythagoras. Harp with ●ight strings. Also, somewhat might be said of Mercurius 
3.  two haps, each of four Strings Elemental. And very strange matter, might be alleged of the Harmony, to our 
5.  Spiritual part appropriate. As in Pt●lom●us third book, in the fourth and sixth Chapters may appear. 
6.  And what is the cause of the apt bond, and friendly fellowship, of the Intellectual and Mental part of us, with our gross & corruptible body: but a certain Mean, and Harmonious Spirituality, with both participating, & of both (i● a manner) 〈◊〉? In the 
7.  T●ne of Man's voice, and also 
8.  the sound of Instrument, what might be said, of Harmony: Not common Musicien would lightly believe. But of the sundry Mixture (as I may term it) and concourse, 
I.D. Read in Aristotle his 8. book of politics: the 5, 6, and 7. chapters. Where you shall have some occasion farther to think of Music, than commonly is thought.  diverse collation, and Application of these Harmonies: as of three, four, five, or more: marvelous have the effects been: and yet may be found, and produced the like: with some proportional consideration for our time, and being: in respect of the State, of the things then: in which, and by which, the wondrous effects were wrought. Democritus and Theophrastus affirmed, that, by Music, griefs and diseases of the Mind, and body might be cured, or inferred. And we find in Record, that Terpander, Arion, Ismenias, Orpheus, Amphion, David, Pythagoras, Empedocles, Asclepiades and Timotheus, by Harmonical Consonancy, have done, and brought to pass, things, more than marvelous, to here of. Of them then, making no farther discourse, in this place: Sure I am, that Common Music, commonly used, is found to the Musiciens and Hearers, to be so Commodious and pleasant, That if I would say and dispute, but thus much: That it were to be otherwise used, than it is, I should find more repreevers, than I could find privy, or skilful of my meaning. In things therefore evident, and better known, than I can express: and so allowed and liked of, (as I would wish, some other things, had the like hap) I will spare to enlarge my lines any farther, but consequently follow my purpose.  
Of cosmography, I appointed briefly in this place, to give you some intelligence. cosmography, is the whole and perfect description of the heavenly, and also elemental part of the world, and their homologal application, and mutual collation necessary. This Art, requireth Astronomy, Geographie, hydrography and Music. Therefore, it is no small Art, nor so simple, as in common practice, it is (slightly) considered. This matcheth Heaven, and the Earth, in one frame, and aptly applieth parts Correspondent: So, as, the Heavenly Globe, may (in practice) be duly described upon the Geographical, and Hydrographical Globe. And there, for us to consider an Equinoctial Circle, an Ecliptic line, Colours, Poles, Stars in their true Longitudes, Latitudes, Declinations, and Verticalitie: also Climes, and Parallels: and by an Horizon annexed, and revolution of the earthly Globe (as the Heaven, is● by the Pr●mouan●, carried about in 24● equal Hours) to learn the risings and Settinge of Stars (of Virgil in his Georgikes: of Hes●od: of Hypocrates in his Medicinal Sphere, to Perdic●● King of the Macedonians: of Diocles, to King Antigonus, and of other famous Philosophers prescribed) a thing necessary, for due manuring of the earth, for Navigation, for the Alteration of man's body● being, whole, Sick, wounded, or bruised. By the Revolution, also, or moving of the Globe Cosmographical, the Rising and Setting of the Son: the lengths, of days and nights: the Hours and times (both night and day) are known: with very many other pleasant and necessary uses: Whereof, some are known: but better remain, for such to know and use who of a spark of true fire, can make a wonderful bonfire, by applying of due matter, duly. 
☜   
Of Astrology, here I make an Ar●e, several from astronomy ● not by new devise, but by good reason and authority: for, Astrology, is an Art Mathematical, which reasonably demonstrateth the operations and effects, of the natural beams, of light, and secret influence: of the Stars and Planets: in every element and elemental body: at all times, in any Horizon assigned. This Art is furnished with many other great Arts and experiences: As with perfect Perspective, Astronomy, cosmography, Natural Philosophy of the 4. Elements, the Art of Graduation, and some good understanding in Music: and yet moreover, with an other great Art, hereafter following, though I, here, set this before, for some considerations me moving. Sufficient (you see) is the stuff, to make this rare and secret Art, of: and hard enough to frame to the Conclusion Syllogistical. Yet both the manifold and continual travails of the most ancient and wise Philosophers, for the attaining of this Art: and by examples of effects, to confirm the same: hath left unto us sufficient proof and witness: and we, also, daily may perceive, That man's body, and all other Elemental bodies, are altered, disposed, ordered, pleasured, and displeasured, by the Influential working of the Sun, Moon, and the other Stars and Planets. And therefore, saith Aristotle, in the first of his Meteorological books, in the second Chapter: Est autem necessariò Mundus iste, supernis lationibus ferè continuus. ut, inde, vis eius universa regatur. Ea siquidem Causa prima putanda omnibus est, unde motus principium existit. That is: This [Elemental] World is of necessity, almost, next adjoining, to the heavenly motions: That, from thence, all his virtue or force may be governed. For, that is to be thought the first Cause unto all: from which, the beginning of motion, is. And again, in the tenth Chapter. Op●rtet igitur & horum principia sumamus, & causas omnium similiter. Principium igitur vy movens, praecipuumque & omnium primum, Circulus ille est, in quo manifest Solis latio, etc. And so forth. His Meteorological books, are full of arguments, and effectual demonstrations, of the virtue, operation, and power of the heavenly bodies, in and upon the four Elements, and other bodies, of them (either perfectly, or unperfectly) composed. And in his second book, De Generatione & Corruption, in the tenth Chapter. Quocirca & prima lati●, Or●us & Interitus causa non est: Sed obliqui Circuli latio: ea namque & continua est, & duobus motibus fit: In English, thus. Wherefore the uppermost motion, is not the cause of Generation and Corruption, but the motion of the Zodiac: for, that, both, is continual, and is caused of two movings. And in his second book, and second Chapter of his Physikes. Homo namque generat hominem, atque Sol. For Man (saith he) and the Son, are cause of man's generation. Authorities may be brought, very many: both of 1000 2000 yea and 3000. years Antiquity: of great Philosophers, Expert, Wise, and godly men, for that Conclusion: which, daily and hourly, we men, may discern and perceive by sense and reason: All beasts do feel, and simply show, by their actions and passions, outward and inward; All Plants, Herbs, Trees, Flowers, and Fruits. And finally, the Elements, and all things of the Elements composed, do give Testimony (as Aristotle said) that their Whole Dispositions, virtues, and natural motions, depend of the Activity of the heavenly motions and Influences. Whereby, beside the specifical order and form, due to every seed: and beside the Nature, proper to the Individual Matrix, of the thing produced: What shall be the heavenly Impression, the perfect and circumspect Astrologien hath to Conclude. Not only (by Apotelesmes) 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 but by Natural and Mathematical demonstration 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉. Whereunto, what Sciences are requisite (without exception) I partly have here warned: And in my Brop●de●●●● besides other matter there disclosed) I have Mathematically furnished up the whole Method: To this our age, not so carefully handled by any, that ever I saw, or heard of. I was, (for 
* Anno. 1548 and 1549. in Lovayn.  ●1. years ago) by certain earnest disputations, of the Learned G●●ardus M●rc●t●●, and 〈◊〉 Goga●a, (and other,) thereto so provoked: and (by my constant and invincible zeal to the verity) in observations of Heavenly Influences (to the Min●te of time,) than, so diligent: And chief by the Supernatural influence, from the Star of jacob, so directed; That any Modest and Sober Student, carefully and diligently sel●ing for the Truth, will both find & confess, therein, to be the Verity, of these my words: And also become a Reasonable Reformer, of three Sorts of people: about these Influential Operations, greatly erring from the truth. Whereof, the one, is Light Believers, the other, 
Note.  Light Despisers, and the third Light Practisers. The first, & most common Sort, think the Heaven and Stars, to be answerable to any their doubts or desires: which is not so: and, in deed, they, to much, over reach. The Second sort think no Influential virtue (from the heavenly bodies) to bear any Sway in Generation and Corruption, in this Elemental world. And to the Sun, Moon and Stars (being so many, so pure, so bright, so wonderful big, so far in distance, so manifold in their motions, so constant in their periods. etc.) they assign a sleight, simple office or two, and so allow unto them (according to their capacities) as much virtue, and power Influential, as to the Sign of the Sun, Moon, and seven Stars, hanged up (for Signs) in London, for distinction of houses, & such gross helps, in our worldly affairs: And they understand not (or will not understand) of the other workings, and virtues of the Heavenly Sun, Moon, and Stars: not so much, as the Mariner, or Husband man● no, not so much as the Elephant doth, as the Cynocephalus, as the Porpentine do●h: nor will allow these perfect, and incorruptible mighty bodies, so much virtual Radiation, & Force, as they see in a little piece of a Magnes stone: which, at great distance, showeth his operation. And perchance they think, the Sea & Rivers (as the Thames) to be some quick thing, and s● to ebb, a●d slow, run in and out, of themselves, at ●hei● own fantasies. God help God help. Surely, these men come to short: and either are to dull: or wilfully blind: or, perhaps, to malicious. The third man, is the common and vulgar Astrologien, or Practiser: who, being not duly, artificially, and perfectly furnished: yet, either for vain glory, or gain: or like a simple dolt, & blind Bayard● both in matter and manner, erreth: to the discredit of the Wary, and modest Astrologien: and to the robbing of those most noble corporal Creatures, of their Natural Virtue: being most mighty: most beneficial to all elemental Generation, Corr●p●ion and the appa●●●nances● and most Harmonious in thei● Monarchy: For which things, being known, and modestly used: we might highly ●nd continually glorify God, with the princely Prophet saying. The Heavens declare the Glory of God: who made the Heavens in his wisdom: who made the Son, for to have dominion of the day: the Moon and Stars to have dominion of the night: whereby, Day to day ●●●●reth tal●●● and night, to night declareth knowledge. Praise him, all ye St●rr●s, and Light. Amen.  
IN order, now followeth, of Statike, somewhat to say, what we mean by ●hat name● and what 〈…〉 doth, on 〈◊〉 Art, depend. Statike, is an Art Mathematical, which demonstra●●th the causes of heaviness, and lightness of all things: and of motions and properties, to heaviness and lightness belonging● And for as much as by the Bilanx, or Balance (as the chief sensible Instrument,) Experience of these demonstrations may be had: we call this Art, Statike: that is, the Experiments of the Balance. O; that men witted, what profit, (all manner of ways) by this Art might grow, to the able examiner, and diligent practiser. Thou only, knowest all things precisely (O God) who hast made weight and Balance, thy judgement: who hast created all things in Number, weight, and Measure: and hast weighed the mountains and hills in a Balance: who hast poised in thy hand, both Heaven and earth. We therefore warned by the Sacred word, to Consider thy Creatures: and by that consideration, to win a glyms (as it were,) or shadow of perceiverance, that thy wisdom, might, and goodness is infinite, and unspeakable, in thy Creatures declared: And being farther advertised, by thy merciful goodness, that, three principal ways, were, of the, used in Creation of all thy Creatures, namely, Number, weight and Measure, And for as much as, of Number and Measure, the two Arts (ancient, famous, and to humane uses most necessary,) are, all ready, sufficiently known and extant: This third key, we beseech thee (through thy accustomed goodness,) that it may come to the needful and sufficient knowledge, of such thy Servants, as in thy workmanship, would gladly find, thy true occasions (purposely of the used) whereby we should glorify thy name, and show forth (to the weaklings in faith) thy wondrous wisdom and Goodness. Amen.  
Marvel nothing at this pang (godly friend, you Gentle and zealous Student.) another day, perchance, you will perceive, what occasion moved me. Here, as now, I will give you some ground, and withal some show, of certain commodities, by this Art arising. And because this Art is rare, my words and practices might be to dark: unleast you had some light, holden before the matter: and that, best will be, in giving you, out of Archimedes demonstrations, a few principal Conclusions, as followeth.  
1. 
The Superficies of every Liquor, by itself consisting, and in quiet, is Spherical: the centre whereof, is the same, which is the centre of the Earth.  2. 
If Solid Magnitudes, being of the same bigness, or quantity, that any Liquor is, and having also the same Weight: be let down into the same Liquor, they will settle downward, so, that no part of them, shall be above the Superficies of the Liquor: and yet nevertheless, they will not sink utterly down, or drown.  3. 
If any Solid Magnitude being Lighter than a Liquor, be let down into the same Liquor, it will settle down, so far into the same Liquor, that so great a quantity of that Liquor, as is the part of the Solid Magnitude, settled dow●e into the same Liquor ● is in Weight, equal, to the weight of the whole Solid Magnitude.  4. 
Any Solid Magnitude, Lighter than a Liquor, forced down into the same Liquor, will move upward, with so great a power, by how much, the Liquor having equal quantity to the whole Magnitude, is heavier than the same Magnitude.  5. 
Any Solid Magnitude, heavier than a Liquor, being let down into the same Liquor, will sink down utterly: And willbe in that Liquor, Lighter by so much, as is the weight or heaviness of the Liquor, having bygnes or quantity, equal to the Solid Magnitude.  6. 
If any Solid Magnitude, Lighter than a Liquor, 
I.D. The Cutting of a Sphere according to any proportion assigned may by this proposition be done Mechanically by tempering Liquour to a certain weight in respect of the weight of the Sphere 〈◊〉 Swy●●●ng.  be let down into the same Liquor, the weight of the same Magnitude, will be, to the Weight of the Liquor. (Which is equal in quantity to the whole Magnitude,) in that proportion, that the part, of the Magnitude settled down, is to the whole Magnitude.   
BY these verities, great Errors may be reformed, in Opinion of the Natural Motion of things, Light and Heavy. Which errors, are in Natural Philosophy (almost) of all men allowed: to much trusting to Authority: and false Suppositions. As, Of any two bodies, the heavier, to move downward faster than the lighter. This error, is not first by me, Noted: 
A common errors noted.  but by one john Baptist de Benedictis. The chief of his propositions, is this: which seemeth a Paradox.  
If there be two bodies of one form, and of one kind, equal in quantity or unequal, they will move by equal space, 
A paradox.  in equal time: So that both their movings be in air, or both in water: or in any one Middle.  
Hereupon, in the feat of Gunning, certain good discourses (otherwise) may receive great amendment, and furtherance. 
N. T.  In the intended purpose, also, allowing somewhat to the imperfection of Nature: 
The wonderful use of these Propositions.  not answerable to the preciseness of demonstration. Moreover, by the foresaid propositions (wisely used.) The Air, the water, the Earth, the Fire, may be nearly, known, how light or heavy they are (Naturally) in their ●●●gned parts: or in the whole. And then, to things Elemental, turning your practice: you may deal for the proportion of the Elements, in the things Compounded. Then, to the proportions of the Humours in Man: their weights: and the weight of his bones, and flesh. etc. Than, by weight, to have consideration of the Force of man, any manner of way: in whole or in part. Then, may you, of Ships water drawing, diversly, in the Sea and in fresh water, have pleasant consideration: and of weighing up of any thing, sunken in Sea or in fresh water etc. And (to lift up your head a fit:) by weight, you may, as precisely, as by any instrument else, measure the Diameters of Son and Moon. etc. Friend, I pray you, way these things, with the just Balance of Reason. And you will find Marvels upon Marvels: And esteem one Drop of Truth (yea in Natural Philosophy) more worth, then whole Libraries of Opinions, undemonstrated: or not answering to Nature's Law, and your experience. Leaving these things, thus: I will give you two or three, light practices, to great purpose● and so finish my Annotation statical. In Mathematical matters, by the Mechaniciens aid, we will behold, here, the Commodity of weight. Make a Cube, of any one Uniform: 
The practice statical, to know the proportion, between the Cube, and the Sphere.  and through like heavy stuff: of the same Stuff, make a Sphere or Globe, precisely, of a Diameter equal to the Radical side of the Cube. Your stuff, may be wood, Copper, Tin, Led, Silver. etc. (being, as I said, of like nature, condition, and like weight throughout.) And you may, by Say Balance, have prepared a great number of the smallest weights: which, by those Balance can be discerned or tried: and so, have proceeded to make you a perfect Pyle, company & Number of weights: to the weight of six, eight, or twelve pound weight: most diligently tried, all. And of every one, the Content known, in your least weight, that is wayable. [They that can not have these weights of preciseness: may, by Sand, Uniform, and well dusted, make them a number of weights, somewhat near preciseness: by halfing ever the Sand: they shall, at length, come to a lest common weight. Therein, I leave the farther matter, to their discretion, whom need shall pinch.] The Venetians consideration of weight, may seem precise enough: 
I D.  by eight descents progressionall, 
* For, so, have you. 256. parts of a Grain.  halfing, from a grain. Your Cube, Sphere, apt Balance, and convenient weights, being ready: fall to work. ●. First, way your Cube. Note the Number of the weight. Way, after that, your Sphere. Note likewise, the Number of the weight. If you now found the weight of your Cube, to be to the weight of the Sphere, as 21. is to 11: Then you see, how the Mechanicien and Experimenter, without Geometry and Demonstration, are (as nearly in effect) taught the proportion of the Cube to the Sphere: as I have demonstrated it, in the end of the twelfth book of Euclid. Often, try with the same Cube and Sphere. Then, change, your Sphere and Cube, to an other matter: or to an other bigness: till you have made a perfect universal Experience of it. Possible it is, that you shall win to nearer terms, in the proportion.  
When you have found this one certain Drop of Natural verity, proceed on, to Infer, and duly to make assay, of matter depending. As, because it is well demonstrated, that a Cylinder, whose heith, and Diameter of his base, is equal to the Diameter of the Sphere, is Sesquialter to the same Sphere (that is, as 3. to 2:) To the number of the weight of the Sphere, add half so much, as it is: and so have you the number of the weight of that Cylinder. Which is also Comprehended of our former Cube: So, that the base of that Cylinder, is a Circle described in the Square, which is the base of our Cube. But the Cube and the Cylinder, being both of one heith, have their Bases in the same proportion, in the which, they are, one to an other, in their Massines or Solidity. But, before, we have two numbers, expressing their Massines, Solidities, and Quantities, by weight: wherefore, we have 
* The proportion of the Square to the Circle inscribed.  the proportion of the Square, to the Circle, inscribed in the same Square. And so are we fallen into the knowledge sensible, and Experimental of Archimedes great Secret: of him, by great travail of mind, sought and found. Wherefore, to any Circle given, you can give a Square equal: 
* The Squaring of the Circled. Mechanically.  as I have taught, in my Annotation, upon the first proposition of the twelfth book, And likewise, to any Square given, you may give a Circle equal: 
* To any Squir● g●uen● to 〈…〉.  If you describe a Circle, which shall be in that proportion, to your Circle inscribed, as the Square is to the same Circle: This, you may do, by my Annotations, upon the second proposition of the twelfth book of Euclid, in my third Problem there. Your diligence may come to a proportion, of the Square to the Circle inscribed, nearer the truth, then is the proportion of 14. to 11. And consider, that you may begin at the Circle and Square, and so come to conclude of the Sphere, & the Cube, what their proportion is: as now, you came from the Sphere to the Circle. For, of Silver, or Gold, or Latton Lamyns or plates (through one hole drawn, as the manner is) if you make a Square figure● & way it: and then, describing thereon, the Circle inscribed: & cut of, & file away, precisely (to the Circle) the overplus of the Square: you shall then, weighing your Circle, see, whether the weight of the Square, be to your Circle, as 14. to 11. As I have Noted, in the beginning of Euclides twelfth book. etc. after this resort to my last proposition, upon the last of the twelfth. And there, help yourself, to the end. And, here, Note this, 
Note Squaring of the Circle without knowledge of the proportion between Circumference and Diameter.  by the way. That we may Square the Circle, without having knowledge of the proportion, of the Circumference to the Diameter: as you have here perceived. And otherways also, I can demonstrate it. So that, many have cumberd themselves superfluously, by travailing in that point first, which was not of necessity, first: and also very intricate. And easily, you may, (and that diversly) come to the knowledge of the Circumference: the Circles Quantity, being first known. Which thing, I leave to your consideration: making haste to dispatch an other Magistral Problem: and to bring it, nearer to your knowledge, and readier dealing with, than the world (before this day,) had it for you, that I can tell of. And that is, A Mechanical Doubling of the Cube: etc. 
To Double the Cube readily: by Art Mechanical: depending upon Demonstration Mathematical.  Which may, thus, be done: Make of Copper plates, or Tyn plates, a foursquare upright Pyramid, or a Cone: perfectly fashioned in the hollow, within. Wherein, let great diligence be used, to approach (as near as may be) to the Mathematical perfection of those figures. At their bases, let them be all open: every where, else, most close, and just to. From the vertex, to the Circumference of the base of the Cone: & to the sides of the base of the Pyramid: Let 4. strait lines be drawn, in the inside of the Cone and Pyramid: 
I D. The 4. sides of this Pyrami● must be 4. Isosceles Triangles ● like and squall.  making at their fall, on the perimeters of the bases, equal angles on both sides themselves, with the said perimeters. These 4. lines (in the Pyramid: and as many, in the Cone) divide: one, in 12. equal parts: and an other, in 24. an other, in 60, and an other, in 100 (reckoning up from the vertex.) Or use other numbers of division, 
I D.  as experience shall reach you● Then, 
* In all workinger with this Pyramid or Cone. Let their Situations be in all Pointe● and Conditions, a like, o● all one: while you are about ●ne work. Else you will 〈◊〉.  set your Cone or Pyramid, with the vertex downward, perpendicularly, in respect of the Base. (Though it be otherways, it hindereth nothing.) So let them most steadyly be stayed. Now, if there be a Cube, which you would have Doubled. Make you a pretty Cube of Copper, Silver, Led, Tin, Wood, Stone, or Bone. Or else make a hollow Cube, or Cubi● coffin, of Copper, Silver, Tin, or Wood etc. These, you may so proportion in respect of your Pyramid or Cone, that the Pyramid or Cone, will be able to contain the weight of them, in wa●●, 3. or 4. times: at the lest: what stuff so ever they be made of● Let not your Solid angle, at the vertex, be to sharp: but that the water may come with ease, to the very vertex, of your hollow Cone or Pyramid. Put one of your Solid Cubes in a Balance apt: 
I D.  take the weight thereof exactly in water. power that water, (without loss) into the hollow Pyramid or Cone, quietly. Mark in your lines, what numbers the water Cutteth: Take the weight of the same Cube against in the same kind of water, which you had before: put that 
* Consider well when you must put your wate●● together: and when, you must empty you● first water● out of your Pyrami● or Cone. El● you will 〈◊〉.  also, into the Pyramid or Cone, where you did put the first. Mark now again, in what number or place of the lines, the water Cutteth them. Two ways you may conclude your purpose: it is to weet, either by numbers or lines. By numbers: as, if you divide the side of your Fundamental Cube into so many equal parts, as it is capable of, conveniently, with your ease, and preciseness of the division. For, as the number of your first and less line (in your hollow Pyramid or Cone,) is to the second or greater (both being counted from the vertex) so shall the number of the side of your Fundamental Cube, be to the number belonging to the Radical side, of the Cube, double to your Fundamental Cube: Which being multiplied Cubik wise, will soon show itself, whether it be double or not, to the Cubik number of your Fundamental Cube. By lines, thus: As your less and first line, (in your hollow Pyramid or Cone,) is to the second or greater, so let the Radical side of your Fundamental Cube, be to a fourth proportional line, by the 12. proposition, of the sixth book of Euclid. Which fourth line, shall be the Rote Cubik, or Radical side of the Cube, double to your Fundamental Cube: which is the thing we desired. For this, may I (with joy) say, EYPHKA, EYPHKA, EYPHKA: thanking the holy and glorious Trinity: having greater cause thereto, then 
* 
vitrvuius. Lib. 9 Cap. 3.  
☞ God b● thanked ●or this Inventions & the fruit ensuing.   Archimedes had (for finding the fraud used in the kings Crown, of Gold): as all men may easily judge: by the diversity of the fruit following of the one, and the other. Where I spoke before, of a hollow Cubik Coffin: the like use, is of it: and without weight. Thus. Fill it with water, precisely full, and pour that water into your Pyramid or Cone. And here note the lines cutting in your Pyramid or Cone. Again, fill your coffin, like as you did before. Put that Water, also, to the first● Mark the second cutting of your lines. Now, as you proceeded before, so must you here proceed. 
* Note.  And if the Cube, which you should Double, be never so great: you have, thus, the proportion (in small) between your two little Cubes: And then, the side, of that great Cube (to be doubled) being the third, will have the fourth, found, to it proportional: by the 12. of the sixth of Eu●lide.  

Note, as concerning the Spherical Superficies of the water.  Note, that all this while, I forget not my first Proposition statical, here rehearsed: that, the Supersicies of the water, is Spherical. Wherein, use your discretion: to the first line, adding a small hear breadth, more: and to the second, half a hear breadth more, to his length. For, you will easily perceive, that the difference can be no greater, in any Pyramid or Cone, of you to be handled. Which you shall thus try. For ●inding the swelling of the water above level. Square the Semidiameter, 
☞  from the Centre of the earth, to your first Water's Superficies. Square then, half the Subtendent of that watery Superficies (which Subtendent must have the equal parts of his measure, all one, with those of the Semidiameter of the earth to your watery Superficies): Subtracte this square, from the first: Of the residue, take the Rote Square. That Ro●e, Subtracte from your first Semidiameter of the earth to your watery Superficies: that, which remaineth, is the heith of the water, in the middle, above the level. Which, you will find, to be a thing insensible. And though it were greatly sensible, 
* Note.  yet, by help of my sixth Theorem upon the last Proposition of Euclides twelfth book, noted: you may reduce all, to a true Level. But, farther diligence, of you is to be used, against accidental causes of the waters swelling: as by having (somewhat) with a moyet Sponge, before, made moist your hollow Pyramid or Cone, will prevent an accidental cause of Swelling, etc. Experience will teach you abundantly: with great ease, pleasure, and commodity.  

Note this Abridgement of Doubling 〈◊〉 Cube. ●●.  Thus, may you Double the Cube Mechanically, Triple it, and so forth, in any proportion. Now will I Abridge your pain, cost, and Care herein. Without all preparing of your Fundamental Cubes: you may (alike) work this Conclusion. For, that, was rather a kind of Experimental demonstration, than the shortest way: and all, upon one Mathematical Demonstration depending. Take water (as much as conveniently will serve your turn: as I warned before of your Fundamental Cubes bigness) Way it precisely. Put that water, into your Pyramid or Cone. Of the same kind of water, then take again, the same weight you had before: put that likewise into the Pyramid or Cone. For, in each time, your marking of the lines, how the Water doth cut them, shall give you the proportion between the Radical sides, of any two Cubes, whereof the one is Double to the other: working as before I have taught you: 
* Note. * ☜  saving that for you Fundamental Cube his Radical side: here, you may take a right line, at pleasure.  
Yet farther proceeding with our drop of Natural truth: you may (now) give Cubes, one to the other, in any proportion given: 
To give Cubes one to the other in any proportion, Rational or Irrational.  Rational or Irrational: on this manner. Make a hollow Parallelipipedon of Copper or Tin: with one Base wanting, or open: as in our Cubike Coffin. Fron the bottom of that Parallelipipedon, raise up, many perpendiculars, in every of his four sides. Now if any proportion be assigned you, in right lines: Cut one of your perpendiculars (or a line equal to it, or less than it) likewise: by the 10. of the sixth of Euclid. And those two parts, set in two sundry lines of those perpendiculars (or you may set them both, in one line) making their beginnings, to be, at the base: and so their lengths to extend upward. Now, set your hollow Parallelipipedon, upright, perpendicularly, steady. Pour in water, handsomely, to the heith of your shorter line. Pour that water, into the hollow Pyramid or Cone. Mark the place of the rising. Settle your hollow Parallelipipedon again. Pour water into it: unto the heith of the second line, exactly. Pour that water 
* Emptying the first.  duly into the hollow Pyramid or Cone: Mark now again, where the water cutteth the same line which you marked before. For, there, as the first marked line, is to the second: So shall the two Radical sides be, one to the other, of any two Cubes: which, in their Solidity, shall have the same proportion, which was at the first assigned: were it Rational or Irrational.  
Thus, in sundry ways you may furnish yourself with such strange and profitable matter: which, long hath been wished for. And though it be Naturally done and Mechanically: yet hath it a good Demonstration Mathematical. 
The demonstrations of this Doubling of the Cube, and o● the rest.  Which is this● Always, you have two Like Pyramids: or two Like Cones, in the proportions assigned: and like Pyramids or Cones, are in proportion, one to the other, in the proportion of their homologal sides (or lines) tripled. Wherefore, if to the first, and second lines, found in your hollow Pyramid or Cone, you join a third and a fourth, in continual proportion: that fourth line, shall be to the first, as the greater Pyramid or Cone, is to the less: by the 33● of the eleventh of Euclid. If Pyramid to Pyramid, 
I.D.  or Cone to Cone, be double, then shall 
* Here 〈…〉 of the water.  Line to Line, be also double, etc. But, as our first line, is to the second, so is the Radical side of our Fundamental Cube, to the Radical side of the Cube to be made, or to be doubled: and therefore, to those twain also, a third and a fourth line, in continual proportion, joined: will give the fourth line in that proportion to the first, as our fourth Pyramidal, or Conike line, was to his first: but that was double, or triple, etc. as the Pyramids or Cones were, one to an other (as we have proved) therefore, this fourth, shallbe also double or triple to the first, as the Pyramids or Cones were one to an other: But our made Cube, is described of the second in proportion, of the four proportional lines: therefore 
* By the 33. of the eleventh books of Euclid.  as the fourth line, is to the first, so is that Cube, to the first Cube: and we have proved the fourth line, to be to the first, as the Pyramid or Cone, is to the Pyramid o● Cone: Wherefore the Cube is to the Cube, 
I.D.  as Pyramid is to Pyramid, or Cone is to Cone. But we 
* And your diligence in practice, can ●o (in weight of wate●● perform it: Therefore. now, you ar● able to ●eue good reason of your whole doing.  Suppose Pyramid to Pyramid, or Cone to Cone, to be double or triple. etc. Therefore Cube, is to Cube, double, or triple, etc. Which was to be demonstrated. And of the Parallelipipedom, it is evident, that the water Solid Parallelipipedons, are one to the other, as their heithes are, seeing they have one base. Wherefore the Pyramids or Cones, made of those water Parallelipipedons, are one to the other, as the lines are (one to the other) between which, our proportion was assigned. But the Cubes made of lines, after the proportion of the Pyramidal or Conik homologal lines, are one to the other, as the Pyramids or Cones are, one to the other (as we before did prove) therefore, the Cubes made, shallbe one to the other, as the lines assigned, are one to the other: Which was to be demonstrated. Note. 
* Note this Corollary.  This, my Demonstration is more general, then only in Square Pyramid or Cone: Consider well. Thus, have I, both Mathematically and Mechanically, been very long in words: yet (I trust) nothing tedious to them, who, to these things, are well a●fected. And verily I am forced (avoiding prolixity) to omit sundry such things, easy to be practised: which to the Mathematicien, would be a great Treasure: and to the Mechanicien, no small gain. 
* The great Commodities following of these new Inventions.  Now may you, Between two lines given, find two middle proportionals, in Continual proportion: by the hollow Parallelipipedon, and the hollow Pyramid, or Cone. Now, any Parallelipipedon rectangle being given: three right lines may be found, proportional in any proportion assigned, of which, shall be produced a Parallelipipedon, equal to the Parallelipipedon given. Hereof, I noted somewhat, upon the 36. proposition, of the 11. book of Euclid. Now, all those things, which vitrvuius in his Architecture, specified able to be done, by doubling of the Cube● Or, by finding of two middle proportional lines, between two lines given, may easily be performed. Now, that Problem, which I noted unto you, in the end of my Addition, upon the 34. of the 11. book of Euclid, is proved possible. Now may any regular body, be Transformed into an other, etc. Now, any regular body: any Sphere, yea any Mixed Solid: and (that more is) Irregular Solides, may be made (in any proportion assigned) like unto the body, first given. Thus, of a Manneken, (as the Dutch Painters term it) in the same Symmetry, may a Giant be made: and that, with any gesture, by the Manneken used: and contrariwise. Now, may you, of any Mould, or Model of a Ship, make one, of the same Mould (in any assigned proportion) bigger or lesser. Now, may you, of any 
* ☞  Gun, or little piece of ordinance, make an other, with the same Symmetric (in all points) as great, and as little, as you william. Mark that: and think on it. Infinitely, may you apply this, so long sought for, and now so easily concluded: and withal, so willingly and frankly communicated to such, as faithfully deal with virtuous studies. Thus, can the Mathematical mind, 
Such is the Fruit of the Mathematical Sciences and Artes.  deal Speculatively in his own Art: and by good means, Mount above the clouds and sterres● And thirdly, he can, by order, Descend, to frame Natural things, to wonderful uses: and when he list, retire home into his own Centre: and there, prepare more Means, to Ascend or Descend by: and, all, to the glory of God, and our honest delectation in earth.  
Although, the Printer, hath looked for this preface, a day or two, yet could I not bring my pen from the paper, before I had given you comfortable warning, and brief instructions, of some of the Commodities, by Statike, able to be reaped: In the rest, I will therefore, be as brief, as it is possible: and with all, describing them, somewhat accordingly. And that, you shall perceive, by this, which in order cometh next. For, whereas, it ●s so ample and wondered, that, an whole year long, one might find fruitful matter therein, to speak of: and also in practice, is a Treasure endless: yet will I glanse over it, with words very few.  
THis do I call Anthropographie. Which is an Art restored, and of my preferment to your Service. I pray you, think of it, as of one of the chief points, of Human knowledge. Although it be, but now, first Confirmed, with this new name: yet the matter, hath from the beginning, been in consideration of all perfect Philosophers. Anthropographie, is the description of the Number, Measure, weight, figure, Situation, and colour of every diverse thing, contained in the perfect body of MAN: with certain knowledge of the Symmetry, figure, weight, Characterization, and due local motion, of any parcel of the said body, a●signed● and of Numbers, to the said parcel appertaining. This, is the one part of the Desinition, meet for this place: Sufficient to notify, the particularity, and excellency of the Art: and why it is, here, ascribed to the Mathematicals. If the description of the heavenly part of the world, had ●●eculier Art, called Astronomy: If the description of the earthly Globe, hath h●●●●culier art, called Geographie. If the Matching of both, hath his peculiar Art, called cosmography: Which is the Description of the whole, and universal frame of the world: Why should not the description of him, who is the Less world: and, from the beginning, called Microcosmus (that is. The Less World. 
MAN is the Less World. ) And for whose sake, and service, all bodily creatures else, were created: Who, also, participateth with Spirits, and Angels: and is made to the Image and similitude of God: have his peculiar Art? and be called the Art of Arts: rather, then, either to want a name, or to have to base and impropre a name? You must of sundry professions, borrow or challenge home, peculiar parts hereof: and farther proceed: as, God, Nature, Reason and Experience shall inform you. The Anatomists will restore to you, some part: The Physiognomistes, some: The Chyromantistes some. The Metaposcopistes, some: The excellent, Albert Durer, a good part: the Art of Perspective, will somewhat, for the Eye, help forward: Pythagoras, Hipocrates, Plato, Galenus, Meletius, & many other (in certain things) will be Contributaries. And farther, the Heaven, the Earth, and all other Creatures, will each show, and offer their Harmonious service, to fill up, that, which wanteth hereof: and with your own Experience, concluding: you may Methodically register the whole, for the posterity: Whereby, good proof will be had, of our Harmonious, and microcosmical constitution. 
* ☜ Microcosmus.  The outward Image, and view hereof: to the Art of Zographie and Painting, to Sculpture, and Architecture: (for Church, House, Fort, or Ship) is most necessary and profitable: for that, it is the chief base and foundation of them. Look in 
* Lib. 3. Cap. 1.  vitrvuius, whether I deal sincerely for your behoove, or no. Look in Albertus Durerus, De Symmetria humani Corporis. Look in the 27. and 28. Chapters, of the second book, De occulta Philosophia. Consider the Ark of Noe. And by that, wade farther. Remember the Delphicall Oracle NOSCE TEIPSUM (Know thyself) so long ago pronounced: of so many a Philosopher repeated: and of the Wisest attempted: And then, you will perceive, how long ago, you have been called to the School, where this Art might be learned. Well. I am nothing afraid, of the disdain of some such, as think Sciences and Arts, to be but Seven. Perhaps, those Such may, with ignorance, and shame enough, come short of them Seven also: and yet nevertheless they can not prescribe a certain number of Arts: and in each, certain unpassable bounds, to God, Nature, and man's Industry. New Arts, daily rise up: and there was no such order taken, 
☞  that, All Arts, should in one age, or in one land, or of one man, be made known to the world. Let us embrace the gifts of God, and ways to wisdom, in this time of grace, from above, continually bestowed on them, who thankfully will receive them: Et bonis Omnia Cooperabuntur in bonum.  
trochilic, is that Art Mathematical, which demonstrateth the properties of all Circular motions, Simple and Compound. And because the fruit hereof, vulgarly received, is in Wheels, it hath the name of trochilic: as a man would say, Wheel Art. By this art, a Wheel may be given which shall move once about, in any time assigned. Two Wheels may be given, whose turnings about in one and the same time, (or equal times), shall have, one to the other, any proportion appointed. By Wheels, may a strait line be described: Likewise, a spiral line in plain, conical Section lines, and other Irregular lines, at pleasure, may be drawn. These, and such like, are principal Conclusions of this Art: and help forward many pleasant and profitable Mechanical works: As Milles, 
Saw Milles.  to See great and very long Deal boards, no man being by. Such have I seen in Germany: and in the City of prague: in the kingdom of Bohemia: Coining Milles, Hand Milles for Corn grinding: And all manner of Milles, and Wheel work: By Wind, Smoke, Water, weight, Spring, Man or Beast, moved. Take in your hand, Agricola ●ere Metallica: and then shall you (in all Mines) perceive, how great need is, of Wheel work. By Wheels, strange works and incredible, are done as will, in other Arts hereafter, appear. A wonderful example of farther possibility, and present commodity, was seen in my time, in a certain Instrument: which by the Inventer and Artificer (before) was sold for xx. Talentes of Gold: and then had (by misfortune) received some injury and hurt: And one janellus of Cremona did mend the same, and presented it unto the Emperor Charles the fifth. Hieronymus Cardanus, can be my witness, that therein, was one Wheel, which moved, and that, in such rate, that, in 7000. years only, his own period should be finished. A thing almost incredible: But how far, I keep me within my bounds: very many men (yet alive) can tell.  
Helicosophie, is near Sister to trochilic: and is, An Art Mathematical, which demonstrateth the designing of all spiral lines in Plain, on Cylinder, Cone, Sphere, Conoid, and Sphaeroid, and their properties appertaining. The use hereof, in Architecture, and diverse Instruments and Engines, is most necessary. For, in many things, the Screw worketh the feat, which, else, could not be performed. By help hereof, it is 
* Atheneus Lib. 5. cap. 8.  recorded, that, where all the power of the City of Syracuse, was not able to move a certain Ship (being on ground) mighty Archimedes, setting to, his Skruish Engine, caused Hiero the king, by himself, at ease, to remove her, as he would. Whereat, the King wondering: 
Proclus. Pag. 18.  〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉. From this day, forward (said the King) Credit aught to be given to Archimedes, what soever he saith.  
Pneumatithmie demonstrateth by close hollow Geometrical Figures, (regular and irregular) the strange properties (in motion or stay) of the Water, Air, Smoke, and Fire, in their continuity, and as they are joined to the Elements next them. This Art, to the Natural Philosopher, is very profitable: to prove, that Vacuum, or Emptiness is not in the world. And that, all Nature, abhorreth it so much: that, contrary to ordinary law, the Elements will move or stand. As, Water to ascend: rather then between him and Air, Spac● or place should be left, more than (naturally) that quantity of Air requireth, or can fill. Again, Water to hung, and not descend: rather then by descending, to leave Emptiness at his back. The like, is of Fire and Air: they will descend: when, either, their continuity should be dissolved: or their next Element forced from them. And as they will not be extended, to discontinuitie: So, will they not, nor yet of man's force, can be priest or penned, in space, not sufficient and answerable to their bodily substance. Great force and violence will they use, to enjoy their natural right and liberty. 
To go to the bottom of the Sea without danger.  Hereupon, two or three men together, by keeping Air under a great Cauldron, and forcing the same down, orderly, may without harm descend to the Sea bottom: and continued there a time etc. Where, Note, how the thicker Element (as the Water) giveth place to the th'inner (as, is the air:) and receiveth violence of the thinner, in manner. etc. Pumps and all manner of bellows, have their ground of this Art: and many other strange devices. As Hydraulica, Organs going by water. etc. Of this Feat, (called commonly Pneumatica,) goodly works are extant, both in Greek, and Latin. With old and learned School men, it is called Scientia de pleno & vacuo.  
Menadrie, is an Art Mathematical, which demonstrateth, how, above Nature's virtue and power simple: Virtue and force may be multiplied: and so, to direct, to lift, to pull to, and to put o● cast fro, any multiplied or simple, determined Virtue, Weight or Force: naturally, not, so, directible or movable. Very much is this Art furdred by other Arts as, in some points, by Perspective: in some, by Statike: in some, by trochilic: and in other, by Helicosophie: and Pneumatithmie. By this Art, all Grains, Gybbettes, & Engines to lift up, or to force any thing, any manner way, are ordered: and the certain cause of their force● is known: As, the force which one man hath with the Duche waghen Rack: therewith, to set up again, a mighty waghen laden, being overthrown. The force of the Crossbow Rack, is certainly, here, demonstrated. The reason, why one man, doth with a leaver, lift that, which Six men, with their hands only, could no●, so easily do. By this Art, in out common Cranes in London, where power is to Crane up, the weight of 2000 pound: by two Wheels more (by good order added) Art concludeth, that there may be Craned up 200000. pound weight etc. So well knew Archimedes this Art: that he alone, with his devices and engines, (twice or thrice) spoiled and discomfited the wholey Army and Host of the romans, besieging Syracuse, Marcus Marcellas the Consul, 

Plut●●●bus in Marco M●rcello.  
Sy●asius in Epistolis.  
Polybius.  
Plinius.  
Quint●lianus.  
T. Livius.   being their General captain. Such huge Stones, so many, with such force, and so far, did he with his ●ngynes hail among them, out of the City. And by Sea likewise: though their Ships might come to the walls of Syracuse, yet he utterly confounded the Roman Navy. What with his mighty Stones hurling: what with Pikes of 
* Athena●s.  18 foot long, made like sliaftes: which he forced almost a quarter of a mile. What, with his catching hold of their Ships, and hoisting them up above the Water, and suddenly letting them fall into the Sea again: what with his 
* Gale●us. Anthemius.  Burning Glasses● by which he fired their other Ships a far-off: what, with his other policies, devices, and engines, he so manfully acquit himself: that all the Force, courage, and policy of the romans (for a great season) could nothing prevail, for the winning of Syracuse. Whereupon, the Romans named Archimedes, Briareus, and Centimanus. Zonara's maketh mention of one Proclus, who so well had perceived Archimedes Art of Menadrie ● and had so well invented of his own, that with his Burning Glasses, 
Burning Glasses.  being placed upon the walls of Bysance, he multiplied so the heat of the Sun, and directed the beams of the same against his enemy's Navy with such force, and so suddenly (like lightning) that he burned and destroyed both man and ship. And Dion specifieth of Priscus, a Geometricien in Bysance, who invented and used sundry Engines, of Force multiplied: Which was cause, that the Emperor Senerus pardoned him, his life, after he had won Bysance: Because he honoured the Art, wit, and rare industry of Priscus. But nothing inferior to the invention of these engines of Force, was the invention of Guns. 
Guns.  Which, from an English man, had the occasion and order of first inventing: though in an other land, and by other men, it was first executed. And they that should see the record, where the occasion and order general, of Gunning, is first discoursed of, would think: that, small things, flight, and common: coming to wise men's consideration, and industrious men's handling, may grow to be of force incredible.  
Hypogeiodie, is an Art Mathematical, demonstrating, how, under the Spherical Superficies of the earth, at any depth, to any perpendicular line assigned (whose distance from the perpendicular of the entrance: and the Azimuth, likewise, in respect of the said entrance, is known) certain way may be prescribed and gone: And how, any way above the Superficies of the earth designed, may under earth, at any depth limited, be kept: going always, perpendicularly, under the way, on earth designed: And, contrariwise, Any way, (strait or crooked,) under the earth, being given: upon the utface, or Superficies of the earth, to Line out the same: So, as, from the Centre of the earth, perpendiculars drawn to the Spherical Superficies of the earth, shall precisely fall in the Correspondent points of those two ways. This, with all other Cases and circumstances herein, and appurtenances, this Art demonstrateth. This Art, is very ample in variety of Conclusions● and very profitable sundry ways to the Common Wealth. The occasion of my Inventing this Art, was at the request of two Gentlemen, who had a certain work (of gain) under ground: and their grounds did join over the work: and by reason of the crookedness, divers depths, and heithes of the way under ground, they were in doubt, and at controversy, under whose ground, as then, the work was: The name only (before this) was of me published, De Itin●r● Subterranco: The rest, be at Gods william. For pioneers, Miners, Diggers for Metals, Stone, Cole, and for secret passage● under ground, between place and place (as this land hath diverse) and for other purposes, any man may easily perceive, both the great fruit of this Art, and also in this Art, the great aid of Geometry.  
Hydragogie, demonstrateth the possible leading of Water, by Nature's law, and by artificial help, from any head (being a Spring, standing, or running Water) to any other place assigned. Long, hath this Art been in use: and much thereof written: and very marvelous works therein, performed● as may ye● appear, in Italy: by the Ruins remaining of the Aqueductes. In other places, of Rivers leading through the Main land, Navigable many a Mile. And in other places, of the marvelous forcinges of Water to Ascend which all, declare the great skill, to be required of him, who should in this Art be perfect, for all occasions of waters possible leading. To speak of the allowance of the Fall, for every hundred foot: or of the Ventills (if the waters labour be far, and great) I need not: seeing, at hand (about us) many expert men can sufficiently testify, in effect, the order: though the Demonstration of the Necessity thereof, they know not: Not yet, if they should be led, up and down, and about Mountains, from the head of the Spring: and then, a place being assigned: and of them, to be demanded, how low or high, that last place is, in respect of the head, from which (so crookedly, and up and down) they be come: Perhaps, they would not, or could not, very readily, or nearly assoil that question. Geometry therefore, is necessary to Hydragogie. Of the sundry ways to force water to ascend, either by Tympane, Kettell mills, Screw, Ctesibike, or such like: in vitrvuius, Agricola, (and other,) fully, the manner may appear. And so, thereby, also be most evident, how the Arts, of Pneumatithmie, Helicosophie, Statik●, trochilic, and Menadrie, come to the furniture of this, in Speculation, and to the Commodity of the Common Wealth, in practice.  
Horometrie, is an Art Mathematical, which demonstrateth, how, at all times appointed, the precise usual denomination of time, may be known, for any place assigned. These words, are smooth and plain eas●e English, but the reach of their meaning, is farther, than you would lightly imagine. Some part of this Art, was called in old time, Gnomonice: and of late, Ho●ologiographia:) and in English, may be termed, Dialling. Ancient is the use, and more ancient, is the Invention. The use, doth well appear to have been (at the lest) above two thousand and three hundred year ago: in 
4. Reg. 20.  King Acha● Dial, then, by the Sun, showing the distinction of time. By Sun, Moon, and Stars, this Dialling may be performed, and the precise Time of day or night known. But the demonstrative delineation of these dials, of all sorts, requireth good skill, both of Astronomy, and Geometry Elemental, Spherical, Phaenomenall, and Conikall. Then, to use the grounds of the Art, for any regular Superficies, in any place offered: and (in any possible apt position thereof) th●ron, to describe (all manner of ways) how, usual hours, may be (by the suns shadow) truly determined: will be found no sleight Painters work. So to Paint, and prescribe the suns Motion, to the breadth of a hear. In this Feat (in my youth) I Invented a way, How in any horizontal, Mural, or Equinoctial Dial, etc. At all hours (the Sun shining) the Sign and Degree ascendent, may be known. Which is a thing very necessary for the Rising of those fixed Stars: whose Operation in the Air, is of great might, evidently. I speak no further, of the use hereof. But forasmuch as, Man's affairs require knowledge of Times & moments, when, neither Sun, Moon, or Star, can be seen: Therefore, by Industry Mechanical, was invented, first, how, by Water, running orderly, the Time and hours might be known: whereof, the famous Ctesibius, was Inventor: a man, of vitrvuius, to the Sky (justly) extolled. Then, after that, by Sand running, were hours measured: Then, by trochilic with weight: And of late time, by trochilic with Spring: without weight. All these, by Sun or Stars direction (in certain time) require oversight and reformation, according to the heavenly Equinoctial Motion: besides the inequality of their own Operation. There remaineth (without parabolical meaning herein) among the Philosophers, 
A perpetual Motion.  a more excellent, more commodious, and more marvelous way, than all these: of having the motion of the Primovant (or first equinoctial motion,) by Nature and Arte● Imitated: which you shall (by further search in weightier studies) hereafter, understand more of. And so, it is time to finish this Annotation, of times distinction, used in our common, and private affairs: The commodity whereof, no man would want, that can tell, how to bestow his tyme.  
Zographie, is an Art Mathematical, which teacheth and demonstrateth, how, the Intersection of all visual Pyramids, made by any plain assigned, (the Centre, distance, and lights, being determined) may be, by lines, and due proper colours, represented. A notable Art, is this and would require a whole Volume, to declare the property thereof: and the Commodities ensuing. Great skill of Geometry, Arithmetic, Perspective, and Anthropographie, with many other particular Art●s, hath the Zographer, need of, for his perfection. For, the most excellent Painter, (who is but the propre Mechanicien, & Imitator sensible, of the Zographer) hath attained to such perfection, that Sense of Man and beast, have judged things painted, to be things natural, and not artificial: alive, and not dead. This Mechanical Zographer (commonly called the Painter) is marvelous in his skill: and seemeth to have a certain divine power: As, of friends absent, to make a friendly, present comfort: yea, and of friends dead, to give a continual, silent presence: not only with us, but with our posterity, for many Ages. And so proceeding, Consider, How, in Winter, he can show you, the lively view of summers joy, and riches: and in Summer, exhibit the countenance of Winter's doleful State, and nakedness. Cities, Towns, Forts, Woods, Armies, yea whole Kingdoms (be they never so far, or great) can he, with ease, bring with him, home (to any man's judgement) as Patterns lively, of the things rehearsed. In one little house, can he, enclose (with great pleasure of the beholders,) the portraiture lively, of all visible Creatures, either on earth, or in the earth, living: or in the waters lying, Creeping, sliding, or swimming: or of any ●oule, or fly, in the air flying. Nay, in respect of the Stars, the Sky, the Clouds: yea, in the show of the very light itself (that Divine Creature) can he match our eyes judgement, most nearly. What a thing is this? things not yet being, he can represent so, as, at their being, the Picture shall seam (in manner) to have Created them. To what Artificer, is not Picture, a great pleasure and Commodity's Which of them all, will refuse the Direction and aid of Picture? The Architect, the Goldsmith, and the Arras Weaver: of Picture, make great account. Our lively Herbals, our portraitures of birds, beasts, and fishes: and our curious Anatomies, which way, are they most perfectly made, or with most pleasure, of us beholden? Is it not, by Picture only? And if Picture, by the Industry of the Painter, be thus commodious and marvelous: what shall be thought of Zographie, the Schoolmaster of Picture, and chief governor? Though I mention not Sculpture, in my Table of Arts Mathematical: yet may all men perceive, How, that Picture and Sculpture, are Sisters germane: and both, right profitable, in a Common wealth. and of Sculpture, aswell as of Picture, excellent Artificers have written great books in commendation. Witness I take, of Georgio Vasari, Pittore Aretino: of Pomponius Gauricus ● and other. To these two Arts, (with other,) is a certain odd Art, called Althalmasat, much beholding: more, than the common Sculptor, Entayler, carver, Cut●er, Graver, Founder, or Painter (etc.) know their Art, to be commodious.  
Architecture, to many may seem not worthy, or not meet, 
An objection.  to be reckoned among the Arts mathematical ● To whom, I think good, to give some account of my so doing. Not worthy, (will they say,) because it is but for building, of a house, Palace, Church, Forte, or such like, gross works. And you, also, defined the Arts Mathematical, to be such, as dealt with no Material or corruptible thing: and also did demonstratively proceed in their faculty, by Number or Magnitude. First, you see, that I count, here, Architecture, among those Arts Mathematical, 
The Answer.  which are Derived from the Principals: and you know, that such, may deal with Natural things and sensib●●●a●●er. Of which, some draw nearer, to the Simple and absolute Mathematical Speculation, than other do. And though, the Architect 
☜  procureth, informeth, & directeth, the Mechanicien, to handworke, & the building actual, of house, Castle, or Palace, and is chief judge of the same: yet, with himself (as chief Master and Architect,) remaineth the Demonstrative reason and cause, of the Mechaniciens work in Line, plain, and Solid: by Geometrical, Arithmetical, optical, Musical, Astronomical, Cosmographical (& to be brief) by all the former Derived Arts Mathematical, and other Natural Arts, able to be confirmed and established. If this be so●then, may you think, that Architecture, hath good and due allowance, in this honest Company of Arts Mathematical Derivative, I will, herein, crave judgement of two most perfect Architect●s: the one, being vitrvuius, the Roman: who did writ ten books thereof, to the Emperor Augustus (in whose days our Heavenly Archemaster, was borne): and the other, Leo Baptista Albertus, a Florentine: who also published ten books thereof. Architectura (saith vitrvuius) est Scientia pluribus disciplinis & varijs eruditionibus ornata● cuius judicio probantur omni●, qua ab cateris Artificibus perficiuntur opera. That is. Architecture, is a Science garnished with many doctrines & diverse instructions: by whose judgement, all works, by other workmen finished, are judged. It followeth. Ea nascitur ex Fabrics. & Ratiocinatione. etc. Ratiocinatio aute● est, quae, res fabricatas, Solertia ac ratione proportionis, demonstrare atque explicare potest. Architecture, groweth of Framing, and Reasoning. etc. Reasoning, is that, which of things framed, with forecast, and proportion: can make demonstration, and manifest declaration. Again. Cùm, in omnibus enim rebus, tùm maxim etiam in Architectura, haec duo insunt: qoud significatur, & qoud significa●. Significatur proposita res, de qua dicitur: hanc autem Significat Demonstratio, rationibus doctrinarum explicata. Forasmuch as, in all things: therefore chief in Architecture, these two things are: the thing signified: and that which signifieth. The thing propounded, whereof we speak, is the thing Signified. But Demonstration, expressed with the reasons of diverse doctrines, doth signify the same thing. After thate ut literatus sit, peritus Graphidos, eruditus Geometriae, & Optices non ignarus: instructus Arithmetica: historias complures noverit, Philosophos diligenter audiverit: Musicam sciverit: Medicinae non sit ignarus, responsa jurisperitorun noverit: Astrologiam, Caelique rationes cognitas habeat. An Architect (saith he) aught to understand Languages, to be skilful of Painting, well instructed in Geometri●, not ignorant of Perspective, furnished with Arithmetic, have knowledge of many histories, and diligently have heard Philosophers, have skill of Music, not ignorant of Physic, know the answers of Lawyers, and have Astronomy, and the courses Celestial, in good knowledge. He giveth reason, orderly, wherefore all these Arts, Doctrines, and Instructions, are requisite in an excellent Architect. And (for brevity) omitting the Latin text, thus he hath. Secondly, it is behofefull for an Architect to have the knowledge of Paintings that he may the more easily fashion out, in patterns painted, the form of what work he liketh. And Geometry, giveth to Architecture many helps: and first teacheth the Use of the Rule, and the compass: whereby (chief and easily) the descriptions of Buildings, are despatched in Groundplats: and the directions of Squires, Levels, and Lines. Likewise, by Perspective, the Lights of the heaven, are well led, in the buildings: from certain quarters of the world. By Arithmetic, the charges of Buildings are summed together: the measures are expressed, and the hard questions of Symmetries, are by Geometrical Means and Methods discoursed on. etc. Besides this, of the Nature of things (which in Greek is called 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉) Philosophy doth make declaration. Which, it is necessary, for an Architect, with diligence to have learned: because it hath many and divers natural questions: as specially, in Aqueductes. For in their courses, leadinges about, in the level ground, and in the mountinge, the natural Spirits or breaths are engendered divers ways: The hindrances, which they cause, no man can help, but he, which out of Philosophy, hath learned the original causes of things. Likewise, who soever shall read C●esibius, or Archimedes books, (and of others, who have written such Rules) can not think, as they do: unless he shall have received of Philosophers, instructions in these things. And Music he must needs know: that he may have understanding, both of Regular and Mathematical Music: that he may temper well his Balistes, Catapultes, and Scorpions. etc. Moreover, the Brazen Vessels, which in Theatres, are placed by Mathematical order, in ambries, under the steps: and the diversities of the sounds (which the Grecians call 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉) are ordered according to Musical Symphonies & Harmonies: being distributed in the Circuits, by Diatessaron, Diapente, and Diapason. That the convenient voice, of the players sound, when it came to these preparations, made in order, there being increased: with that increasing, might come more clear & pleasant, to the ears of the lookers on. etc. And of Astronomy, is known the East, West, South, and North. The fashion of the heaven, the equinox, the Solstice, and the course of the stars. Which things, unleast one know: he can not perceive, any thing at all, the reason of Horologies. Seeing therefore this ample Science, is garnished, beautified and stored, with so many and sundry skills and knowledges: I think, that none can justly accounted themselves Architectes, of the sudden. But they only, who from their child's years, ascending by these degrees of knowledges, being fostered up with the attaining of many Languages and Arts, have won to the high Tabernacle of Archicture. etc. And to whom Nature hath given such quick Circumspection, sharpness of wit, and Memory, that they may be very absolutely skilful in Geometry, Astronomy, Music, and the rest of the Arts Mathematical: Such surmount and p●sse the calling, and state, of Architectes: and are become mathematicians. 
A Mathematicien.  etc. And they are found, seldom. As, in times past, was Aristarchus Samius: Philolaus, and Archytas, Tarentynes: Apollonius Pergeus: Eratosthenes Cyreneus: Archimedes, and Scopas, Syracusians. Who also, left to their posterity, many Engines and gnomonical works: by numbers and natural means, invented and declared.  
Thus much, and the same words (in sense) in one only Chapter of this Incomparable Architect Vitrunius, 
Vitrunius.  shall you find. And if you should, but take his book in your hand, and slightly look through it, you would say strait way: This is Geometry, Arithmetic, Astronomy, Music, Anthropographie, Hydragogie, Horometrie etc. and (to conclude) the Storehouse of all workmanship. Now, let us listen to our other judge, our Florentine, Leo Baptista: and narrowly consider, how he doth determine of Architecture. Sed anteque ultra progrediar. etc. But before I proceed any further (saith he) I think, that I aught to express, what man I would have to be allowed an Architect. For, I will not bring in place a Carpenter: as though you might Compare him to the Chief Master● of other Artes. For the hand of the Carpenter, is the Architectes Instrument. 
Who is an Architect.  But I will appoint the Architect to be that man, who hath the skill, (by a certain and marvelous means and way,) both in mind and Imagination to determine: and also in work to finish: what works so ever, by motion of weight, and cuppling and framing together of bodies, may most aptly be Commodious for the worthiest. Uses of Man. And that he may be able to perform these things, he hath need of attaining and knowledge of the best, and most worthy things. etc. The whole Feat of Architecture in building, consisteth in lineaments, and in Framing. And the whole power and skill of lineaments, tendeth to this: that the right and absolute way may he had, of Coapting and joining Lines and angles: by which, the face of the building or frame, may be comprehended and concluded. And it is the property of lineaments, to prescribe unto buildings, and every part of them, an apt place, & certain number: a worthy manner, and a seemly order: that, so, the whole ●orme and figure of the building, may rest in the very lineaments. etc. And we may prescribe in mind and imagination the whole forms, 
* The immateriality of perfect Architecture.  all material stuff being secluded. Which point we shall attain, by Noting and forepointing the angles, and lines, by a ●ure and certain direction and connexion. Seeing then, these things, are thus: Lineament, 
What, Lineament is.  shallbe the ●ertaine and constant prescribing, conceived in mind: made in lines and angles: and finished with a learned mind and wit. We thank you Master Baptist, that you have so aptly brought your Art, and phrase thereof, to have some Mathematical perfection: 
Note.  by certain order, number, form, figure, and Symmetric mental: all natural & sensible stuff set apart. Now, then, it is evident, (Gentle reader) how aptly and worthily, I have preferred Architecture, to be bred and fostered up in the Dominion of the perales' Princess, Mathematica: and to be a natural Subject of hers. And the name of Architecture, is of the principality, which this Science hath, above all other Artes. And Plato affirmeth, the Architect to be Master over all, that make any work. Whereupon, he is neither Smith, not Builder: nor, separately, any Artificer: but the Head, the Provost, the Director, and judge of all Artificial works, and all Artificers. For, the true Architect, is able to teach, Demonstrate, distribute, describe, and judge all works wrought. And he, only, searcheth out the causes and reasons of all Artificial things. Thus excellent, is Architecture: though few (in our days) attain: hereto: yet may not the Art, be otherwise thought on, then in very deed it is worthy. Nor we may not, of ancient Arts, make new and imperfect Definitions in our days: for scarcity of Artificers: Not more, than we may pinch in, the Definitions of Wisdom, or Honesty, or of petition or of Justice. Not more will I consent, to Diminish any whit, of the perfection and dignity, (by just cause) allowed to absolute Architecture. Under the Direction of this Art, are three principal, necessary Mechanical Artes. Namely, housing, Fortification, and Naupegie. housing, I understand, both for Divine Service, and Man's common usage: public, and private. Of Fortification and Naupegie, strange matter might be told you: But perchance, some will be tired, with this Bederoll, all ready rehearsed: and other some, w●●● nicely nip my gross and homely discoursing with you: made in post hast: for fear you should want this true and friendly warning, and taste giving, of the Power Mathematical. Life is short, and uncertain: times are perilous: etc. And still the Printer awaiting, for my pen staying: All these things, with farther matter of Ingratefulnes, give me occasion to pass away, to the other Arts remaining, with all speed possible.  
THe Art of Navigation, demonstrateth how, by the shortest good way, by the aptest Direction, & in the shortest time, a sufficient Ship, between any two places (in passage Navigable,) assigned: may be conducted: and in all storms, & natural disturbances chancing, how, to use the best possible means, whereby to recover the place first assigned. What need, the Master Pilot, hath of other Arts, here before recited, it is easy to know: as, of hydrography, Astronomy, Astrology, and Horometrie. Presupposing continually, the common Base, and foundation of all: namely Arithmetic and Geometry. So that, he be able to understand, and judge his own necessary Instruments, and furniture Necessary: Whether they be perfectly made or no: and also can, (if need be) make them, himself. As Quadrantes, The Astronomers Ring, The Astronomers staff, The Astrolabe universal. An Hydrographical Globe. Charts Hydrographical, true, (not with parallel Meridian's). The Common Sea Compas: The Compass of variation: The Proportional, and paradoxal Compasses (of me Invented, 
Anno. 1559.  for our two Moscovy Master Pilots, at the request of the Company) Clocks with spring: hour, half hour, and three hour Sandglasses: & sundry other Instruments: And also, be able, on Globe, or Plain to describe the paradoxal Compass: and duly to use the same, to all manner of purposes, whereto it was invented. And also, be able to Calculate the Planets places for all times.  
Moreover, with Son Moon or Star (or without) be able to define the Longitude & Latitude of the place, which he is in: So that, the Longitude & Latitude of the place, from which he sailed, be given: or by him, be known. whereto, appertaineth expert means, to be certified ever, of the Ships way. etc. And by foreseeing the Rising, Setting, Nonesteding, or Midnighting of certain tempestuous fixed Stars: or their Conjunctions, and Anglynges with the Planets, etc. he aught to have expert conjecture of Storms, Tempests, and Spouts: and such like Meteorological effects, dangerous on Sea. For (as Plato saith,) Mu●ationes ● Some by weight. whereof Ti●●●●● speaketh. Some, by Strings strained, or Springs, therewith Imitating lively Motions. Some, by other means, as the Images of Mercury: and the brazen head, made by Albertus Magn●●, which did seem to speak. B●●thius was excellent in these feats. To whom, Cassiod●●●● writing, saith. Your purpose is to know profound things: and to show marvels. By the disposition of your Art, Metals do low: Diomedes of brass, doth blow a Trumpet loud: a brazen Serpent hisseth: birds made, sing sweetly. Small things we rehearse of you, who can Imitate the heaven. etc. Of the strange Selfmoving, which, at Saint Denys, by Paris, 
* Anno. 1551  I saw, once or twice (Orontius being then with me, in Company) it were to strange to tell. But some have written it. And yet, (I hope) it is there, of other to be seen. And by Perspective also strange things, are done. As partly (before) I gave you to understand in Perspective. As, to see in the Air, a fit, the lively Image of an other man, either walking to and fro: or standing still. Likewise, to come into an house, and there to see the lively show of Gold, Silver or precious stones: and coming to take them in your hand, to find naught but Air. Hereby, have some men (in all other matters counted wise) foully overshot themselves: misdeaming of the means. 
De his quae Mundo mirabiliter eveniunt. cap. 8.  Therefore said Claudius Caelestinus. hody magna liter●●●rae vi●●s 〈◊〉 magna 〈…〉, opera qu●da●● quasi miranda, supra Natura 〈◊〉 de qu●●●s in 〈…〉 sa●iliter reddidisse●. That is. Now a days, 〈◊〉 see 〈…〉, y●a of great learning and reputation, to judge certain works 〈◊〉 marvellous, above the power of Nature: Of which works, one that were skilful in Perspective might easily have given the Cause. Of Archimedes Sphere, Cicero witnesseth. 
Tusc. ●.  Which is very strange to think on. For when Archimedes (saith he) did fasten in a Sphere, the mo●ynges of the Son, Moon, and of the fi●e other Planets, he did, as the God, which (in Timaeus of Plato) did make the world. That, on● 〈◊〉, should rule motions most unlike in slowness, and swiftness. But a greater cause of marveling we have by Claudianus report hereof. Who affirm in this Archimedes work, to have ●en of Glass. And discourseth of it more at large: which I omit. The Dove of wood, which the Mathematicien Arc●y●a● did make to fly, is by Agellius spoken of. Of Daedalus strange Images, Plat● reporteth. H●mere of Vul●ans Selfmovers, (by secret wheels) leaveth in writing. Aristotle, in his politics, of both, maketh mention. Marvellous was the workmanship, of la●e days, performed by good skill of Tr●chi●ike. etc. For in Noremberge, A sly of jern, being l●t out of the Artifi●ers ●●nd, did (as it were) fly about by the jests, at the table, and at length, as though ●● were weary, return to his master's ha●● again. Moreover, an Artificial E●le, was ordered, to fly out of the same Town, a mighty way, and that a lo●t in the Air, toward the Emperor coming thither: and followed him, being come to the gate of the town●. 
* ☞  Thus, you see, what● Art Mathematical an parforme, when Skill, will, Industry, and ability, are duly applied to pro●e.  

A Digression. Apologetical.  ANd for these, and such like marvelous Acts and Feats, Naturally, Mathematically, and Mechanically, wrought and contrived: aught any honest Student, and Modest Christian Philosopher, be counted, & called a conjuror? Shall the folly of Idiot's, and the Malice of the Scornful, so much prevail, that He, who seeketh no worldly gain or glory at their handes● But only, of God, the treasure of heavenly wisdom, & knowledge of p●re verity: Shall he (I say) in the mean space, be rob and spoiled of his honest name and fame? He that seeketh (by S. Paul's advertisement) in the Creatures Properties, and wonderful virtues, to find just cause, to glorify the aeternal; and Almighty Creator by: Shall that man, be (in) condemned, as a Companion of the Hellhounds, and a Caller, and conjuror of wicked and damned Spirits? He that bewaileth his great want of time, sufficient (to his contentation) for learning of Godly wisdom, and Godly Verities in: and only therein setteth all his delight: Will that man lose and abuse his time, in dealing with the Chief enemy of Christ our Redeemer: the deadly foe of all mankind: the subtle and impudent perverter of Godly Verity: the Hypocritical Crocodile: the Envious Basilisk, continually desirous, in the twink of an eye, to destroy all Mankind, both in Body and Soul, eternally? Surely (for my part, somewhat to say herein) I have not learned to make so brutish, and so wicked a Bargain. Should I, for my xx. or xxv. years Study: for two or three thousand Marks spending: seven or eight thousand Miles going and travailing, only for good learnings sake: And that, in all manner of wethers: in all manner of ways and passages: both early and late: in danger of violence by man: in danger of destruction by wild beasts: in hunger: in thirst: in perilous heats by day, with toil on foot: in dangerous damps of cold, by night, almost bereaving life: (as God knoweth): with lodgings, often times, to small ease: and sometime to less security. And for much more (than all this) done & suffered, for Learning and attaining of Wisdom: Should I (I pray you) for all this, not otherwise, nor more warily: or (by God's mercifulness) no more luckily, have fished, with so large, and costly, a Net, so long time in drawing (and that with the help and advise of Lady Philosophy, & Queen theology): but at length, to have catched, and drawn up, 
* A proverb. Fair fished, and caught a Frog.  a Frog? Nay, a Devil? For, so, doth the Common peevish Prattler Imagine and jangle: And, so, doth the Malicious scorner, secretly wish, & bravely and boldly face down, behind my back. Ah, what a miserable thing, is this kind of Men? How great is the blindness & boldness, of the Multitude, in things above their Capacity? What a Land: what a People: what Manners: what Times are these? Are they become Devils, themselves: and, by false witness bearing against their Neighbour, would they also, become Murderers? Doth God, so long give them respite, to reclaim themselves in, from this horrible slandering of the guiltless: contrary to their own Consciences: and yet will they not cease? Doth the Innocent, forbear the calling of them, juridically to answer him, according to the rigour of the Laws: and will they despise his Charitable patience? As they, against him, by name, do forge, fable, rage, and raise slander, by word & Print: Will they provoke him, by word and Print, likewise, to Note their Names to the World: with their particular devices, fables, beastly Imaginations, and unchristenlike slanders? Well: Well. O (you such) my unkind Country men. O unnatural Country men. O unthankful Country men. O Brainsick, Rash, Spiteful, and Disdainful Country men. Why oppress you me, thus violently, with your slandering of me: Contrary to Verity: and contrary to your own Consciences? And I, to this hour, neither by word, deed, or thought, have been, any way, hurtful, damageable, or injurious to you, or yours? Have I, so long, so dearly, so far, so carefully, so painfully, so dangerously sought & travailed for the learning of Wisdom, & attaining of Virtue: And in the end (in your judgement) am I become, worse, than when I began? Worse, them a Mad man? A dangerous Member in the Common Wealth: and no Member of the Church of Christ? Call you this, to be Learned? Call you this, to be a Philosopher? and a lover of Wisdom? To forsake the strait heavenly way: and to wallow in the broad way of damnation? To forsake the light of heavenly Wisdom: and to lurk in the dungeon of the Prince of darkness? To forsake the Verity of God, & his Creatures: and to fawn upon the Impudent, Crafty, Obstinate Liar, and continual disgracer of God's Verity, to the uttermost of his power? To forsake the Life & Bliss aeternal: and to cleave unto the Author of Death everlasting? that Murderous Tyrant, most greedily awaiting the Pray of Man's Soul? Well: I thank God and our Lord jesus Christ, for the Comfort which I have by the Examples of other men, before my time: To whom, neither in godliness of life, nor in perfection of learning, I am worthy to be compared: and yet, they sustained the very like Injuries, that I do: or rather, greater. Patient Socrates, his Apology will testify: Apuleius his Apologies, will declare the Brutishness of the Multitude. joannes Picus, Earl of Mirandula, his Apology will teach you, of the Raging slander of the Malicious Ignorant against him. joannes Trithemius, his Apology will specify, how he had occasion to make public Protestation: as well by reason of the Rude Simple: as also, in respect of such, as were counted to be of the wisest sort of men. Many could I recite: But I defer the precise and determined handling of this matter: being loath to detect the Folly & Malice of my Native Country men. 
☞  Who, so hardly, can digest or like any extraordinary course of Philosophical Studies: not falling within the compass of their Capacity: or where they are not made privy of the true and secret cause, of such wonderful Philosophical Feats. These men, are of four sorts, chief. The first, I may name, Vain prattling busy bodies: The second, Fond Friends: The third, Imperfectly zealous: and the fourth, Malicious Ignorant. To each of these (briefly, and in charity) I will say a word or two, and so return to my preface. Vain prattling busy bodies, use your idle assemblies, and conferences, otherwise, then in talk of matter, either above your Capacities, for hardness: or contrary to your Consciences, in Verity. Fond Friends, leave of, so to commend your unacquainted friend, upon blind affection: As, because he knoweth more, than the common Student: that, therefore, he must needs be skilful, and a doer, in such matter and manner, as you term Conjuring. Weening, thereby, you advance his fame: and that you make other men, great marueilers of your hap, to have such a learned friend. Cease to ascribe Impiety, where you pretend Amity. For, if your tongues were true, then were that your friend, Untrue, both to God, and his Sovereign. Such Friends and Fondlinges, I shake of, and renounce you: Shake you of, your Folly. Imperfectly zealous, to you, do I say: that (perhaps) well, do you Mean: But far you miss the Mark: If a Lamb you will kill, to feed the flock with his blood. Sheep, with lambs blood, have no natural sustenance: Not more, is Christ's flock, with horrible slanders, duly edified. Nor your fair pretence, by such rash ragged Rhetoric, any whit, well graced. But such, as so use me, will find a fowl Crack in their Credit. Speak that you know: And know, as you aught: Know not, by Hear say, when life lieth in danger. Search to the quick, & let Charity be your guide. Malicious Ignorant, what shall I say to thee? Prohibe linguam tuam a malo. A detractione parcite linguae. 'Cause thy tongue to refrain from evil. Refrain your tongue from slander. Though your tongues be sharpened, Serpent like, & Adder's poison lie in your lips: 
Psal. 140.  yet take heed, and think, betimes, with yourself, Vir linguosus non stabilietur in terra. Virum violentum venabitur malum, donec praecipitetur. For, sure I am, Quia faciet Dominus judicium afflicti: & vindictam pauperum.  
Thus, I require you, my assured friends, and Country men (you mathematicians, Mechaniciens, and Philosophers, Charitable and discrete) to deal in my behalf, with the light & untrue tongued, my envious Adversaries, or Fond friends. And farther, I would wish, that at leysor, you would consider, how Basilius Magnus, layeth Moses and Daniel, before the eyes of those, which count all such Studies Philosophical (as mine hath been) to be ungodly, or unprofitable. Way well S. Stephen his witness of Moses. 
Act. 7. C.  Eruditus est Moses omni Sapientia Aegyptiorun: & erat potens in verbis & operibus suis. Moses was instructed in all manner of wisdom of the Egyptians: and he was of power both in his words, and works. You see this Philosophical Power & Wisdom, which Moses had, to be nothing misliked of the Holy Ghost. Yet Plinius hath recorded, Moses to be a wicked Magicien. And that (of force) must be, either for this Philosophical wisdom, learned, before his calling to the leading of the Children of Israel: or for those his wonders, wrought before King Pharaoh, after he had the conducting of the Israelites. As concerning the first, you perceive, how S. Stephen, at his Martyrdom (being full of the Holy Ghost) in his Recapitulation of the old Testament, hath made mention of Moses Philosophy: with good liking of it: And Basilius Magnus also, avoucheth it, to have been to Moses profitable (and therefore, I say, to the Church of God, necessary). But as concerning Moses wonders, done before King Pharaoh: God, himself, said: Vide ut omnia ostenta, quae posui in manutua, facias coram Pharaone. See that thou do all those wonders before Pharaoh, which I have put in thy hand. Thus, you evidently perceive, how rashly, Plinius hath slandered Moses, 
Lib. 30. Cap. 1.  of vain fraudulent Magic, saying: Est & alia Magices Factio, a Mose, jamne, & jotape, judaeis pendens: sed multis millibus annorum post Zoroastrem. etc. Let all such, therefore, who, in judgement and Skill of Philosophy, are far Inferior to Pliny, take good heed, lest they overshoot themselves rashly, 
☜  in judging of Philosophers strange Act●s ● and the Means, how they are done. But, much more, aught they to beware of forging, devising, and imagining monstrous feats, and wonderful works, when and where, no such were done: not, not any spark or likelihood, of such, as they, without all shame, do report. And (to conclude) most of all, let them be ashamed of Man, and afraid of the dreadful and just judge: both Foolishly or Maliciously to devise: and then, devilishly to father their new fond Monsters on me: Innocent, in hand and heart: for trespacing either against the law of God, or Man, in any my Studies or Exercises, Philosophical, or Mathematical: As in due time, I hope, will be more manifest.  
NOw end I, with Archemastrie. Which name, is not so new, as this Art is rare. For an other Art, under this, a degree (for skill and power) hath been endued with this English name before. And yet, this, may serve for our purpose, sufficiently, at this present. This Art, teacheth to bring to actual experience sensible, all worthy conclusions by all the Arts Mathematical purposed, & by true Natural Philosophy concluded: & both addeth to them a farther scope, in the terms of the same Arts, & also by his proper Method, and in peculiar terms, proceedeth, with help of the foresaid Arts, to the performance of complete Experiences, which of no particular Art, are able (Formally) to be challenged. If you remember, how we considered Architecture, in respect of all common handworkes: some light may you have, thereby, to understand the Sovereignty and property of this Science. Science I may call it, rather, than an Art: for the excellency and Mastership it hath, over so many, and so mighty Arts and Sciences. And because it proceedeth by Experiences, and searcheth forth the causes of Conclusions, by Experiences: and also putteth the Conclusions themselves, in Experience, it is named of some, Scientia Experimentalis. The Experimental Science. Nicolaus Cusanus termeth it so, in his Experiments Statikall, And an other Philosopher, 
R. B.  of this land Native (the flower of whose worthy fame, can never die nor whither) did writ thereof largely, at the request of Clement the sixth. The Art carrieth with it, a wonderful Credit: By reason, it certefieth, sensibly, fully, and completely to the utmost power of Nature, and Arte. This Art, certifieth by Experience complete and absolute: and other Arts, with their Arguments, and Demonstrations, persuade: and in words, prove very well their Conclusions. 
☞  But words, and Arguments, are no sensible certifying: nor the full and final fruit of Sciences practisable. And though some Arts, have in them, Experiences, yet they are not complete, and brought to the uttermost, they may be stretched unto, and applied sensibly. As for example: the Natural Philosopher disputeth and maketh goodly show of reason: And the Astronomer, and the Optical Mechanicien, put some things in Experience: but neither, all, that they may: nor yet sufficiently, and to the utmost, those, which they do, There, then, the Archemaster steppeth in, and leadeth forth on, the Experiences, by order of his doctrine Experimental, to the chief and final power of Natural and Mathematical Artes. Of two or three men, in whom, this Description of Archemastry was Experimentally, verified, I have read and hard: and good record, is of their such perfection. So that, this Art, is no fantastical Imagination: as some Sophister, might, Cum suis Insolubilibus, make a slorish: and dassell your Imagination: and dash your honest desire and Courage, from believing these things, so unheard of, so marvelous, & of such Importance. Well: as you william. I have forewarned you. I have done the part of a friend: I have discharged my Duty toward God: for my small Talon, at his most merciful hands received. To this Science, doth the Science Alnirangiat, great Service. Muse nothing of this name. I change not the name, so used, and in Print published by other: being a name, proper to the Science. Under this, cometh Arsenio Sintrillia, by Artephius, briefly written. But the chief Science, of the Archemaster, (in this world) as yet known, is an other (as it were) OPTICAL Science: whereof, the name shall be told (God willing) when I shall have some, (more just) occasion, thereof, to Discourse.  
Here, I must end, thus abruptly (Gentle friend, and unfeigned lover of honest and necessary verities.) For, they, who have (for your sake, and virtues cause) requested me, (an old forworne Mathematicien) to take pen in hand: (through the confidence they reposed in my long experience: and tried sincerity) for the declaring and reporting somewhat, of the fruit and commodity, by the Arts Mathematical, to be attained unto: even they, Sore against their wills, are forced, for sundry causes, to satisfy the workman's request, in ending forthwith: He, so feareth this, so new an attempt, & so costly: And in matter so slenderly (hitherto) among the common sort of Students, considered or esteemed.  
And where I was willed, somewhat to allege, why, in our vulgar Speech, this part of the Principal Science of Geometry, called Euclides Geometrical Elements, is published, to your handling: being unlatined people, and not University Scholars: Verily, I think it needless.  
For, the Honour, and Estimation of the Universities, and Graduates, is, hereby, nothing diminished. seeing, from, and by their Nurse Children, you receive all this Benefit: how great soever it be.  
 
And surely, the Common and Vulgar Scholar (much more, the Grammarian) before his coming to the University, shall (or may) be, now (according to Plato his Counsel) sufficiently instructed in Arithmetic, and Geometry, for the better and easier learning of all manner of philosophy, Academical, or Peripatetical. And by that means, go more cheerfully, more skilfully, and speedily forward, in his Studies, there to be learned. And, so, in less time, profit more, than (otherwise) he should● or could do.  
 
And great Comfort, with good hope, may the Universities have, by reason of this English Geometry and Mathematical preface, that they (hereafter) shall be the more regarded, esteemed, and resorted unto. For, when it shall be known and reported, that of the Mathematical Sciences only, such great Commodities are enfu●ng (as I have specified): and that in deed, some of you unlatined Students, can be good witness, of such rare fruit by you enjoyed (thereby): as either, before this, was not heard of 〈…〉 fully credited: Well, may all men conjecture, that far greater 〈…〉, to win to the Perfection of all Philosophy, 
Universities.  may in 〈…〉 the Storehouses & treasury of all Sciences, and 〈…〉, 
☜  and most noble State of Common Wealths.  
Besides this, how ma 〈…〉 here, in these Realms of England and Ireland, tha● 〈…〉 & compass: Who, with their own Skill and expe 〈…〉ble (by these good helps and informations) to found 〈…〉 s, strange Engines, and Instruments: for sundry purp 〈…〉 Wealth? or for private pleasure? and for the better maintaining 〈…〉 own estate? I will not (therefore) fight against mine own shadow. For, no man (I am s●re) will open his mouth against this Enterprise. No man (I say) who either hath Charity toward his brother (and would be glad of his furtherance in virtuous knowledge): o● that hath any care & zeal for the bettering of the Common state of this Realm. Neither any, that make account, what the wiser sort of men (Sage and Stayed) do think of them. To ●one (therefore) will I make any Apology, for a virtuous act doing: and for commending, or setting forth Profitable Arts to English men, in the English tongue. But, unto God our Creator, let us all be thankful: for that, As he, of his Goodness, by his power, and in his wisdom, hath Created all things, in Number, weight, 
☞  and Measure: So, to us, of his great Mercy, he hath revealed Means, whereby, to attain the sufficient and necessary knowledge of the foresaid hy● three principal Instruments: Which Means, I have abundantly proved unto you, to be the Sciences and Arts Mathematical.  
And though I have been pinched with straightness of time: that no way, I could so pen down the m●tter (in my Mind), as I determined● hoping of convenient leisure: Yet, if virtuous zeal, and honest Intent provoke and bring you to the reading and examining of this Compendious treatise, I do not doubt, but, as the verity thereof (according to our purpose) will be evident unto you: So the pith and force thereof, will persuade you: and the wonderful fruit thereof, highly pleasure you. And that you may the easier perceive, and better remember, the principal points, whereof my Preface treateth, I will give you the Ground platt of my whole discourse, 
The Ground platt of this preface in a Table.  in a Table annexed from the first to the last, somewhat Methodically contrived.  
If Hast, hath caused my poor pen, any where, to stumble: You will, (I am sure) in part of recompence● (for my earnest and sincere good will to pleasure you). Consider the rockish huge mountains, and the perilous unbeaten ways which (both night and day, for the while) it hath ●oyled and laboured through, to bring you this good News and Comfortable proof of virtues fruit.  
So, I Commit you unto God's Merciful direction, for the rest: heartily beseeching him, to prospero your Studies, and honest Intentes: to his Glory & the Commodity of our Country. Amen.  
Written at my poor H●●se At Mortlake. Anno. 1570● February ●.   

Here have you (according to my promiss) the Groundplat of my MATHEMATICAL preface: annexed to Euclid (now first) published in our English tongue. An. 1570. Febr. 3  
sciences, ●nd Arts Mathematical, are, either 

Principal, which are two, only, 

Arithmetic. 


Simple, Which dealeth with Numbers only: and demonstrateth all their properties and appurtenances: where, an unit, is Indivisible.  
The use whereof, is either, 

In things Supernatural, eternal, & Divine: By Application, Ascending. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.  
In things Mathematical: without farther Application. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.  
In things Natural: both Substantial, & Accidental, Visible, & Invisible. etc. By Application: Descending. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.     

Mixed, Which with aid of Geometry principal, demonstrateth some Arithmetical Conclusion, or Purpose.  
The use whereof, is either, 

In things Supernatural, eternal, & Divine: By Application, Ascending. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.  
In things Mathematical: without farther Application. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.  
In things Natural: both Substantial, & Accidental, Visible, & Invisible. etc. By Application: Descending. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.       
Geometry. 


Simple, Which dealeth with Magnitudes, only: and demonstrat●th all their properties, passions, and appurtenances: whose Point, is Indivisible.  
The use whereof, is either, 

In things Supernatural, eternal, & Divine: By Application, Ascending. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.  
In things Mathematical: without farther Application. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.  
In things Natural: both Substantial, & Accidental, Visible, & Invisible. etc. By Application: Descending. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.     

Mixed, Which with aid of Arithmetic principal, demonstrateth some Geometrical purpose: as EUCLIDES ELEMENTS.  
The use whereof, is either, 

In things Supernatural, eternal, & Divine: By Application, Ascending. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.  
In things Mathematical: without farther Application. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.  
In things Natural: both Substantial, & Accidental, Visible, & Invisible. etc. By Application: Descending. The like Uses and Applications are, (though in a degree lower) in the Arts Mathematical Derivative.         
Derivative from the Principals: o● which, some have 

The names of the Principals: as, 

Arithmetic, vulgar: which considereth 

Arithmetic of most usual whole Numbers: And of Fractions to them appertaining.  
Arithmetic of Proportions.  
Arithmetic Circular.  
Arithmetic of Radical Numbers: Simple, Compound, Mixed: And of their Fractions.  
Arithmetic of Cossike Numbers: with their Fractions: And the great Art of Algiebar.    
Geometry, vulgar: which teacheth Measuring 

At hand 
All lengths. 
Mecometrie.  All Plains: As, Land, Borde, Glass, etc. 
Embadometrie.  All Solids: As, Timber, Stone, Vessels, etc. 
stereometry.    
With distance from the thing Measured, as, 


How far, from the Measurer, any thing is: of him seen, on Land or Water: called Apomecometrie.  
Of which are grown the Feats & Arts of 

Geodesie: more cunningly to Measure and Survey Lands, Woods, Waters. etc.  
Geographie.  
chorography.  
hydrography.  
Stratarithmetrie.     

How high or deep, from the level of the Measurers standing, any thing is: Seen of him, on Land or Water: called Hypsometrie  
Of which are grown the Feats & Arts of 

Geodesie: more cunningly to Measure and Survey Lands, Woods, Waters. etc.  
Geographie.  
chorography.  
hydrography.  
Stratarithmetrie.     

How broad, a thing is, which is in the Measurers' view: so it be situated on Land or Water: called Platometrie.  
Of which are grown the Feats & Arts of 

Geodesie: more cunningly to Measure and Survey Lands, Woods, Waters. etc.  
Geographie.  
chorography.  
hydrography.  
Stratarithmetrie.           
Proper names as, 
Perspective, 
Which demonstrateth the manners and properties of all Radiations: direct, Broken, and Reflected.  Astronomy, 
Which demonstrateth the Distances, Magnitudes, and all Natural motions, Apparences, and Passions, proper to the Planets and fixed Stars: f●r any time, past, present, and to come: in respect of a certain Horizon, or without respect of any Horizon.  Music, 
Which demonstrateth by reason, and teacheth by sense, perfectly to judge and order the diversity of Sounds, hi● or l●w.  cosmography, 
Which, wholly and perfectly maketh description of the Heavenly, and also Elemental part of the World: and of these parts, maketh h●m●l●gall application, and mutual collation necessary.  Astrology, 
Which reasonably demonstrateth the operations and effects of the natural bea●es of light, and 〈◊〉 Influence of the Planets, and fixed Stars, 〈◊〉 every Element and Elemental body: at all times, in any Horizon assigned.  Statike, 
Which demonstrateth the causes of heaviness and lightness of all things: and of the motions and properties to heaviness and lightness belonging.  Anthropographie, 
Which describeth the Number, Measure, weight, Figure, Situation, and colour of every divers thing contained in the perfect● body of ●● AN: and giveth certain knowledge of the Figure, Symmetric, weight, Characterization, & due Local motion of any p●rcell of the said body assigned: and of numbers to the said p●rcell appertaining.  trochilic, 
Which demonstrateth the properties of all Circular motions: Simple and Compound.  Helicosophie, 
Which demonstrateth the designing of all spiral lines: in Plain, on Cylinder, Co●●, Sphere, C●n●id, and Spharo●d: and their properties.  Pneumatithmie, 
Which demonstrateth by close hollow Geometrical figures (Regular and Irregular) the strange properties (in motion or stay) of the Water, Air, Smoke, and Fire, in their Continuiti●, and as they are joined to the Elements next them.  Menadrie, 
Which demonstrateth, how, about Nature's Virtue, and power simple: Virtue and force, may be multiplied: and so to direct, to lif●, to pull to, a●d to put or cast fro, any multiplied, or simple determined Virtue, weight, or Force: naturally, not, so, directible, or movable.  Hypogeiodie, 
Which demonstrateth, how, under the spherical Superficies of the Earth, at ●ny depth, to any perpendicular line assigned (whose distance from the perpendicular of the entrance: and the Azimuth likewise, 〈◊〉 resperse of the said entrance, is known) certain way, may be prescribed and g●ne, etc.  Hydragogie, 
Which demonstr●teth the possible leading of water by Nature's l●●, and by artificial help, fr●● any head (being Spring, standing, or running water) to any other place assigned.  Horometrie, 
Which demonstrateth, how, at all times appointed, the precise, usual denomination of time, ●●y ●e know●n, for any place assigned.  Zographie, 
Which demonstrateth and teacheth, how, the Intersection of all usual 〈…〉 assigned (the Centre, distanc●, and lights b●ing determined) may be, by lines, and proper colour's repre●●●●.  Architecture, 
Which is a Sci●●●● garnished with many doctrines, and 〈…〉, are judged.  Navigation, 
 Thaumaturgike, 
 Archemastrie, 
        

¶ The first book of Euclides Elements.  
IN THIS FIRST BOOK is entreated of the most simple, 
The argum●●● of the first Book.  easy, and first matters and grounds of Geometry, as, namely, of lines, Angles, Triangles, Parallels, Squares, and parallelograms, First of their definitions, showing what they are, After that it teach●th how to draw Parallel lines, and how to form diversly figures of three sides, & four sides, according to the variety of their sides, and Angles & compareth them all with Triangles, & also together the one with the other. In it also is taught how a figure of any form may be changed into a Figure of an other form. And for that it entreateth of these most common and general things, this book is more universal than is the second, third, or any other, and therefore justly occupieth the first place in order: as that without which, the other books of E●clide which follow, and also the works of others which have written in Geometry, cannot be perceived nor understanded. And forasmuch ●s all the demonstrations and proofs of all the propositions in this whole book, depend of these grounds and principles following, which by reason of their plainness need no great declaration, yet to remove all (be it never so little) obscurity, there are here set certain short and manifest expositions of them.  

Definitions.  

1. A sign or point is that, which hath no part. 
Definition of a point.   
The better to understand what man●r of thing a sign or point is, ye must note that the nature and property of quantity (whereof Geometry entreateth) is to be divided, so that whatsoever may be divided into sund●y parts, is called quantity. But a point, although it pertain to quantity, and hath his being in quantity, yet is it no quantity, for that it cannot be divided. Because (as the definition saith) it hath no parts into which it should be divided. So that a point is the lest thing that by mind and understanding can be imagined and conceived: than which, there can be nothing less, as the point A in the margin.  
A sign or point is of Pythagoras Scholars after this manner defined. 
Definition of a point after Pythagoras.  A point is an unity which hath position. Numbers are conceived in mind without any form & figure, and therefore without matter whereon to 〈◊〉 figure, & consequently without place and position. Wherefore unity being a part of number, hath no position, or determinate place. Whereby it is manifest, that ●umb●● i● more simple and pure then is magnitude, and also immaterial: and so unity which i● the beginning of number, is less material than a ●igne or poy●●, which is the beginning of magnitude. For a point is material, and requireth position and place, and ●●●rby differeth from unity.   

●. A line is length without breadth. 
Definition of a li●●.   
There pertain to quanti●●e three dimensions, length, breadth, & thickness, or depth: and by these three are all quantitie● measured & made known. There are also, according to these three dimensions, three kinds of continual quantities: a line, a superficies, or plain, and a body. The first kind, namely, a line is here defined in these words, A line is length without breadth. A point, for that it is no quantity nor hath any parts into which it may be divided, but remaineth indivisible, hath not, nor can have any of these three dimensions. It neither hath length, breadth, nor thickness. But to a line, which is the first kind of quantity, is attributed the first dimension, namely, length, and only that, for it hath neither breadth nor thickness, but is conceived to be drawn in, length only, and by it, it may be divided into parts as many as ye list, equal, or unequal. But as touching breadth it remaineth indivisible. As the line AB, which is only drawn in length, may be divided in the point C equally, or in the point D unequally, and so into as many parts as ye list. There are also of divers other given other definitions of a line; as these which follow.  


another definition of a line.  A line is the moving of a point, as the motion or draft of a pin or a pen to your sense maketh a line.  

An other.  Again, A line is a magnitude having one only space or dimension, namely, length wanting breadth and thickness.    


The ends of a line.  3 The ends or limits of a line, are points.  
For a line hath his beginning from a point, and likewise endeth in a point: so that by this also it is manifest, that points, for their simplicity and lack of composition, are neither quantity, nor parts of quantity, but only the terms and ends of quantity. As the points A, B, are only the ends of the line AB, and no parts thereof. And herein differeth a point in quantity, 
Difference of a point fr●●nity.  from unity in number● for that although unity be the beginning of numbers, and no number (as a point is the beginning of quantity, and no quantity) yet is unity a part of number. 
Unity is a part of number.  For number is nothing else but a collection of unities, and therefore may be divided into them, as into his parts. But a point is no part of quantity, 
A point is no part of quantity.  or of a lyne● neither is a line composed of points, as number is of unities. For things indivisible being never so many added together, can never make a thing divisible, as an instant in time, is neither time, nor part of time, but only the beginning and end of time, and coupleth & joineth parts of time together.   


Definition of a right line.  4 A right line is that which lieth equally between his points.  
As the whole line AB lieth strait and equally between the points AB without any going up or coming down on either side.  
A right line is the shortest extension or draft, that is or may be from one point to an other. Archimedes defineth it thus.  

Definition thereof after Plato.  Plato defineth a right line after this manner: A right line is that whose middle part shadoweth the ex●reme●. As if you put any thing in the middle of a right line, you shall not see from the one end to the other, which thing happeneth not in a crooked line. The Eclipse of the Sun (say Astronomers) then happeneth, when the Sun, the Moon, & our eye are in one right line. For the Moon then being in the midst between us and the Sun, causeth it to be darkened. divers other define a right line diversly, 
another definition.  as followeth.  
A right line is that which standeth firm between his extremes.  

An other.  Again, A right line is that which with an other line of like form cannot make a figure.  
Again, 
An other.  A right line is that which hath not one part in a plain superficies, and an other erected on high.  
Again, 
An other.  A right line is that, all whose parts agreed together with all his other parts.  
Again, 
An other.  A right line is that, whose extremes abiding, cannot be altered.  
Euclid doth not here define a crooked line, 
Why Euclid here defineth not a crooked line.  for it needed not. It may easily be understand by the definition of a right line, for every contrary is well manifested & set forth by his contrary. One crooked line may be more crooked than an other, and from one point to an other may be drawn infinite crooked lines: but one right line cannot be righter than an other, and therefore from one point to an other, there may be drawn but one tied line. As by figure above set, you may see.   

5 A superficies is that, which hath only length and breadth. 
Definition of a superficies.   
A superficies is the second kind of quantity, and to it are attributed two dimensions, namely length, and breadth. 
A superficies may be divided two ways.  As in the superficies ABCD. whose length is taken by the line AB, or CD, and breadth by the line AC. or BD: and by reason of those two dimensions a superficies may be divided two ways, namely by his length, and by his breadth, but not by thickness, for it hath none. For, that is attributed only to a body, which is the third kind of quantity, and hath all three dimensions, length, breadth, and thickness, and may be divided according to any of them.  
Others define a superficies thus: 
another definition of a superficies.  A superficies is the terms or end of a body. As a line is the end and term of a superficies.   

6 Extremes of a superficies, are lines. 
The extremes of a superficies.   
As the ends, limits, or borders of a line, are points, enclosing the line: so are lines the limits, borders, and ends enclosing a superficies. As in the figure aforesaid you may see the superficies enclosed with four lines. The extremes or limits of a body, are superficiesles. And therefore a superficies is of some thus defined: 
Another definition of a superficies.  A superficies is that, which endeth or encloseth a body: as is to be seen in the sides of a die, or of any other body.   

7 A plain superficies is that, which lieth equally between his lines. 
Definition of a plain superficies.   
A plain superficies is the shortest extension or dr●ught from one line to an other: 
Another definition of a plain superficies.  like as a right line is the shortest extension or draft from one point to an other.  
Euclid also leaveth out here to speak of a crooked and hollow superficies, because it may easily be understand by the definition of a plain superficies, being his contrary. And even as from one point to an other may be drawn infinite crooked lines, & but one right line, which is the shortest: so from one line to an other may be drawn infinite crooked superficiesses, & but one plain superficies, which is the shortest. Here must you consider when there is in Geometry mention made of points, 
NOTE.  lines, circles, triangles, or of any other figures, ye may not conceive of them as they be in matter, as in wood, in metal, in paper, or in any such like, for so is there no line, but hath some breadth, and may be deuided● nor points, but that shall have some parts, and may also be divided, and so of others. But you must conceive them in mind, plucking them by imagination from all matter so shall ye understand them truly and perfectly, in their own nature as they are defined. As a line to be long, and not broad: and a point to be so little, that it shall have no part at all.  

Another definition of a plain superficies.  Others otherwise define a plain superficies: A plain superficies is that, which is firmly set between his extremes, as before was said of a right line.  

another definition.  Again, A plain superficies is that, unto all whose parts a right line may well be applied.  

another definition.  Again, A plain superficies is that, which is the shortest of all superficies, which have one & the self extremes: As a rig●t line was the shortest line that can be drawn between two points.  

another definition.  Again, A plain superficies is that, whose middle darkeneth the extremes, as was also said of a right line.   


Definition of a plain angle.  8 A plain angle is an inclination or bowing of two lines the one to the other and the one touching the other, and not being directly joined together.  
As the two lines AB, & BC, incline the one to the other, and touch the one the other in the point B, in which point by reason of the inclination of the said lines, is made the angle ABC. But if the two lines which touch the one the other, be without all inclination of the one to the other, and be drawn directly the one to the other, then make they not any angle at all, as the lines CD, and DE, touch the one the other in the point D, and yet as ye see they make no angle.   


Definition of a rectilined angle.  9 And if the lines which contain the angle be right lines, then is it called a rightlyned angle.  
As the angle ABC, in the former figures, is a rightlined angle, because it is contained of right lines: where note, that an angle is for the most part described by three letters, of which the second or middle letter representeth the very angle, and therefore is set at the angle.  
By the contrary, a crooked lined angle, is that which is contained of crooked lines, which may be diversly figured. Also a mixed angle is that which is caused of them both, namely, of a right line and a crooked, which may also be diversly figured, as in the figures before set ye may see. 
〈◊〉 of angles.  There are of angles three kinds, a right angle, an acute angle, and an obtuse angle, the definitions of which now follow.   


What a right angle, & What also a perpendicular line i●.  10 When a right line standing upon a right line maketh the angles on either side equally then either of those angles is a right angle. And the right line which standeth erected, is called a perpendicular line to that upon which it standeth.  
As upon the right line CD, suppose there do stand an other line AB, in such sort, that it maketh the angles on either side thereof equal: namely, the angle ABC on the one side equal to the angle ABD on the other side: then is each of the two angles ABC, and ABD a right angle, and the line AB, which standeth erected upon the line CD, without inclination to either part is a perpendicular line, commonly called among artificers a plumb line.   


What an obtuse angle ●●.  11 An obtuse angle is that which is greater than a right angle.  
As the angle CBE in the example is an obtuse angle, for it is greater than the angle ABC, which is a right angle, because it containeth it, and containeth moreover the angle ABE.   

12 An acute angle is that, which is less than a right angle. 
What an acute angle is.   
As the angle EBD in the figure before put is an acute angle, for that it is less than the angle ABD, which is a right angle, for the right angle containeth it, and moreover the angle ABE.   

13 A limit or term, is the end of every thing. 
The limit of any thing.   
For as much as of things infinite (as Plato saith) there is no science, 
No science of things infinite  therefore must magnitude or quantity (whereof Geometry entreateth) be finite, and have borders and limits to enclose it, which are here defined to be the ends thereof. As a point is the limit or term of a line, because it is th'end thereof: A line likewise is the limit & term of a superficies: and likewise a superficies is the limit and term of a body, as is before declared.   

14 A figure is that, which is contained under one limit or term, or many. 
Definition of a figure.   
As the figure A is contained under one limit, which is the round line, also the figure B is contained under three right lines. And the figure C under four, and so of others, which are their limits or terms.   

15 A circle is a plain figure, contained under one line, which is called a circumference, 
Definition of a circle.  unto which all lines drawn from one point within the figure and falling upon the circumference thereof are equal the one to the other.  
As the figure here set is a plain figure, that is a figure without grossness or thickness, and is also contained under one line, namely, the crooked line B CD, which is the circumference thereof, it hath moreover in the middle thereof a point, namely, the point A, from which, all the lines drawn to the superficies, are equals as the lines AB, AC, AD, and other how many soever.  
Of all figures a circle is the most perfect, 
A circle the most perfect of all figures.  and therefore is it here first defined.   

16 And that point is called the centre of the circle, as is the point A, which is set in the mids of the former circle. 
The centre of a circle.   
For the more easy declaration, that all the lines drawn from the centre of the circle to the circumference, are equal, ye must note, that although a line be not made of points: yet a point, by his motion or draft, describeth a line. Likewise a line drawn, or moved, describeth a superficies● also a superficies being moved maketh a solid or body. Now then imagine the line AB, (the point A being fixed) to be moved about in a plain superficies, drawing the point B continually about the point A, till it return to the place where it began first to move: so shall the point B, by this motion, describe the circumference of the circle, and the point A being fixed, is the centre of the circle. Which in all the time of the motion of the line, had like distance: from the circumference, namely, the length of the line AB. And for that all the lines drawn from the centre to the circumference are described of that line, they are also equal unto it, & between themselves.   


Definition of a diameter.  17 A diameter of a circle, is a right line, which drawn by the centre thereof, and ending at the circumference on either side, divideth the circle into two equal parts.  
As the line BAC in this circle present is the diameter, because it passeth from the point B, of the one side of the circumference● to the point C, on the other side of the circumference, & passeth also by the point A being the centre of the circle. And moreover it divideth the circle into two equal parts: the one, namely BDC, being on the one side of the line, & the other namely, BEC, on the other side, which thing did Thales Miletius (which brought Geometry out of Egypt into Grece) first observe and prove. For if a line drawn by the centre, do not divide a circle into two equal parts; all the lines drawn from the centre to the circumference should not be equal.   


Definition of a semicircle.  18 A semicircle, is a figure which is contained under the diameter, and under that part of the circumference which is cut of by the diameter.  
As in the circle ABCD the figure BAC is a semicircle, because it is contained of the right line BGC, which is the diameter, and of the crooked line BAC, being that part of the circumference, which is cut of by the diameter BGC. So likewise the other part of the circle, namely BDC ● is a semicircle as the other was.   


Definition of a section of a circle.  19 A section or portion of a circle, is a figure which is contained under a right line, and a part of the circumference, greater or less than the semicircle.  
As the figure ABC, in the example, is a section of a circle, & is greater than half a circle, and the figure ADC, is also a section of a circle, and is less than a semicircle. A section, portion, or part of a circle is all one, and signifieth such a part which is either more or less, than a semicircle: so that a semicircle is not here called a section or portion of a circle. A right line drawn from one side of the circumference of a circle to the other, not passing by the centre, divideth the circle into two unequal parts, which are two sections, of which that which containeth the centre is called th● greater section, and the other is called the less section. As in the example, the part of the circle ABC, which containeth in it the centre E, is the greater section, being greater than the half circle: the other part, namely ADC, which hath not the centre in it, is the less section of the circle, being less than a semicircle.   


Definition of rectilined figures.  20 Rightlined figures are such which are contained under right lines.  
As are such as followeth, of which some are contained under three right lines, some under ●oure, some under five, and some under more.   


Definition of three sided figures.  21 Three sided figures, or figures of three sides, are such which are contained under three right lines.  
As the figure in the example ABC, is a figure of three sides because it is contained under three right lines, namely, under the lines AB, BC, CA  
A figure of three sides, or a triangle, is the first figure in order of all right lined figures, and therefore of all others it is first defined. For under less than three lines, can no figure be comprehended.   

22. Four sided figures or figures of four sides are such, 
Definition of four sided figures.  which are contained under four right lines.  
As the figure here set, is a figure of four sides, for that it is comprehended under four right lines, namely, AB, BD, DC, CA  
Triangles, and four sided figures serve commonly to many uses in demonstrations of Geometry. Wherefore the nature and properties of them, are much to be observed, the use of other figures is more obscure.   

23. Many sided figures are such which have more sides than four. 
Definition of many sided figures.   
Right lined figures having more sides than four, by continual adding of sides may be infinite. Wherefore to define them all severally, according to the number of their ●ides, should be very tedious, or rather impossible. Therefore hath Euclid, comprehended them under one name, and under one definition: calling them many sided figures, as many as have more sides than four: as if they have five sides, six, seven, or mo● Here note ye, that every rightlined figure hath as many angles, as it hath sides, & taketh his denomination aswell of the number of his angles, as of the number of his sides. As a figure contained under three right lines, of the number of his three sides, is called a three sided figure ●euen so of the number of his three angles, it is called a triangle. Likewise a figure contained under four right lines, by reason of the number of his sides, is called a four sided figure: and by reason of the number of his angles, it is called a quadrangled figure, and so of others.   

24. Of three sided figures or triangles, 
Definition of an equilater triangle.  an equilatre triangle is that, which hath three equal sides.  
Triangles have their differences partly of their sides, and partly of their angles. As touching the differences of their sides, there are three kinds. For either all three sides of the triangle are equal, or two only are equal, & the third unequal: 〈…〉 three are unequal the one to the other. The first kind of triangles, namely, that which hath three equal sides, is most simple, and easiest to be known: and is here first defined, and is called an equilater triangle, as the triangle A in the example, all whose sides are of one length.   


Definition of an Isosceles.  25. Isosceles, is a triangle, which hath only two sides equal.  
The second kind of triangles hath two sides of one length, but the third side is either longer or shorter than the other two, as are the triangles here figured, B, C, D  
In the triangle B, the two sides AE and EF are equal the one to the other, and the side AE, is longer than any of them both: Likewise in the triangle C the two sides GH and HK, are squall, and the side GK is greater. Also in the triangle D, the sides LM and MN, are squall, and the side LN is shorter.   


Definition of a Scalenum.  26. Scalenum is a triangle, whose three sides are all unequal.  
As are the triangles E, F, in which there is no one side equal to any of the other. For in the triangle E, the side AC is greater than the side BC, and the side BC is greater than the side AB. Likewise in the triangle F, the side DH, is greater than the side DG, and the side DG, is greater than the side GH.   


An Orthigonium triangle.  27. Again of triangles, an Orthigonium or a rightangled triangle, is a triangle which hath a right angle.  
As there are three kinds of triangles, by reason of the diversity of the sides, so are there also three kinds of triangles, by reason of the variety of the angles. For every triangle either containeth one right angle, & two acute angles: or one obtuse angle, & two acute: or three acute angles: for it is impossible that one triangle should contain● two obtuse angles, or two right angles, or one obtuse angle, and the other a right angle. All which kinds are here defined. First a rightangled triangle which hath in it a right angle. As the triangle BCD, of which the angle BCD, is a right angle.   


An Ambligonium triangle.  28. An ambligonium or an obtuse angled triangle, is a triangle which hath an obtuse angle.  
As is the triangle B, whose angle AC D, is an obtuse angle, and is also a Scalenon, having his three sides unequal: the triangle E, is likewise an Ambligonion, whose angle FGH, is an obtuse angle, & is an Isosceles, having two of his sides equal, namely FG and GH.   


An Oxigonium triangle.  29. An oxigonium or an acute angled triangle, is a triangle which hath all his three angles acute.  
As the triangles A, B, C, in the example, all whose angles are acute: of which A is an equilater triangle, B, an Isosceles, and C a Scalenon. An equilater triangle is most simple, and hath one uniform construction, and therefore all the angles of it are equal, and never hath in it either a right angle, or an obtuse: but the angles of an Isosceles or a Scalenon, may diversly vary. It is also to be noted that in comparison of any two sides of a triangle, the third is called a base. As of the triangle ABC in respect of the two lines AB and AC, the line BC, is the base: and in respect of the two sides AC and CB, the line AB, is the base, and likewise in respect of the two sides CB & BA, the line AC, is the base.   

30 Of four syded figures, 
Definition of a square.  a quadrate or square is that, whose sides are equal, and his angles right.  
As triangles have their difference and variety by reason of their sides and angles: so likewise do figures of four sides, take their variety and difference partly by reason of their sides, & partly by reason of their angles, as appeareth by their definitions. The four sided figure ABCD is a square or a quadrate, because it is a right angled figure, all his angles are right angles, and also all his four sides are equal.   

31 A figure on the one side longer, 
Definition of a long square.  or squarelike, or as some call it, a long square, is that which hath right angles, but hath not equal sides.  
This figure agreeth with a square touching his angles, in that either of them hath right angles, and differeth from it only by reason of his sides, in that the sides thereof be not equal, as are the sides of a square. As in the example, the angles of the figure ABCD, are right angles, but the two sides thereof AB, and CD, are longer than the other two sides AC, & BD.   

32 Rhombus (or a diamond) is a figure having four equal sides, 
Definition of a Diamond figure  but it is not rightangled.  
This figure agreeth with a square, as touching the equality of lines, but differeth from it in that it hath not right angles, as hath the square. As of this figure, the four lines AB, BC, CD, DA, be equal, but the angles thereof are not right angles. For the two angles ABC and ADC, are obtuse angles, greater than right angles, & the other two angles BAD and BCD, are two acute angles less than two right angles. And these four angles are yet equal to four right angles: for, as much as the acute angle waits of a right angle, so much the obtuse angle exceedeth a right angle.   


Definition of a diamondlike figure.  33 Rhomboides (or a diamond like) is a figure, whose opposite sides are equal, and whose opposite angles are also equal, but it hath neither equal sides, nor right angles.  
As in the figure ABCD, all the four sides are not equal, but the two sides AB and CD, being opposite the one to the other, also the other two sides AC and BD, being also opposite, are equal the one to the other. Likewise the angles are not right angles, but the angles CAB, and CDB, are obtuse angles, and opposite and equal the one to the other. Likewise the angles ABD, and ACD, are acute angles, and opposite, and also equal the one to the other.   


Trapezia or tables.  34 All other figures of four sides besides these, are called trapezia, or tables.  
Such are all figures, in which is observed no equality of sides nor angles: as the figures A and B, in the margin, which have neither equal sides, nor equal angles, but are described at all adventure without observation of order, and therefore they are called irregular figures.   


Definition of Parallel lines.  35 Parallel or equidistant right lines are such, which being in one and the self same superficies, and produced infinitely on both sides, do never in any part concur.  
As are the lines AB, and CD, in the example.  
   

Petitions or requests.  

1 From any point to any point, to draw a right line.  
After the definitions, which are the first kind of principles, now follow petitions, which are the second kind of principles: 
What Petitions are.  which are certain general sentences, so plain, & so perspicuous, that they are perceived to be true as soon as they are uttered, & no man that hath but common sense, can, nor will deny them. Of which, the first is that, which is here set. As from the point A, to the point B, who will deny, but easily grant that a right line may be drawn? For two points howsoever they be set, are imagined to be in one and the self same plain superficies, wherefore from the one to the other there is some shortest draft, which is a right line. Likewise any two right lines howsoever they be set, are imagined to be in one superficies, and therefore from any one line to any one line, may be drawn a superficies.   

2 To produce a right line finite, strait forth continually.  
As to draw in length continually the right line AB, who will not grant? For there is no magnitude so great, but that there may be a greater, nor any so little, but that there may be a less. And a line is a draft from one point to an other, therefore from the point B, which is the end of the line AB, may be drawn a line to some other point, as to the point C, and from that to an other● and so infinitely.   

3 Upon any centre and at any distance, ●o describe a circle.  
A plain superficies may in compass be extended infinitely: as from any point to any point may be drawn a right line, by reason whereof it cometh to pass that a circle may be described upon any centre and at any space or distance. As upon the centre A, and upon the space AB, ye may describe the circle BC, & upon the same centre, upon the distance AD, ye may describe the circle DE, or upon the same centre A, according to the distance AF, ye may describe the circle FG, and so infinitely extending your space.   

4 All right angles are equal the one to the other.  
This petition is most plain, and offereth itself even to the sense. For as much as a right angle is 〈…〉 ri●ht line falling perpendicularly upon an other, and no one line can fall more perpendicularly upon a line then an other● therefore no one right angle can be greater 〈◊〉 an other● neither d● the length or shortness of the lines altar the greatness of the angle. For in the example, the right angle ABC, though it be made of much longer lines than the right angle DEF, whose lines are much shorter, yet is that angle no greater than the other. For if ye set the point E ●ust upon the point B ● then shall the line ED, evenly and justly fall upon the line AB ● and the line EF, shall also fall equally upon the line BC, and so shall the angle DEF, be equal to the angle ABC, for that the lines which 'cause them, are of like inclination.  
It may evidently also be seen at the centre of a circle. For if ye draw in a circle two diameters, the one cutting the other in the centre by right angles, ye shall divide the circle into four equal parts, of which each containeth one right angle, so are all the four right angles about the centre of the circle equal.   

5 When a right line falling upon too right lines, doth make ●n one & the self same side, the two inward angles less than two right angles, then shall these two right lines being produced 〈◊〉 length concur on that part, in which are the two angles less than two right angles.  
As if the right line AB, fall upon two right lines, namely, CD and EF, so that it make the two inward angles on the one side, as the angles DHI & FIE, less than two right angles (as in the example they do) the said two lines CD, and EF, being drawn forth in length on that part, whereon the two angles being less● 〈◊〉 too right angle● consist, shall 〈◊〉 length concur and meet together: as in the point D, as it is easy to see. For the parts of the lines towards DF, are more inclined the one to the other, than the parts of the lines towards CE are. Wherefore the more these parts are produced, the more they shall approach near and near, till at length they shall meet in one point. Contrariwise the same lines drawn in length on the other side, for that the angles on that side, namely, the angle CHB, and the angle EIA, are greater than two right angles, so much as the other two angles are less th●n two right angles, shall never meet, but the further they are drawn, the further they shallbe distant the one from the other.   

6 That two right lines include not a superficies.  
If the lines AB and AC, being right lines, should enclose a superficies, they must of necessity be joined together at both the ends, and the superficies must be between them. join them on the one side together in the point A, and imagine the point B to be drawn to the point C, so shall the line AB, fall on the line AC, and cover it, and so be all one with it, and never enclose a space or superficies.    

Common sentences.  

1 Things equal to one and the self same thing: are equal also the one to the other.  

What common sentences are.  After definitions and petitions, now are set common sentences, which are the third and last kind of principles. Which are certain general propositions, commonly known of all men, of themselves most manifest & clear, & therefore are called also dignities not able to be denied of any. Petitions also are very manifest, but not so fully as are the common sentences, 
Difference between petitions & common sentences.  and therefore are required or desired to be granted. Petitions also are more peculiar to the art whereof they are: as those before put are proper to Geometry: but common sentences are general to all things whereunto they can be applied. Again, petitions consist in actions or doing of somewhat most easy to be done: but common sentences consist in consideration of mind, but yet of such things which are most easy to be understanded, as is that before set. As if the line A be equal to the line B, And if the line C be also equal to the line B, then of necessity the lines A and C, shallbe equal the one to the other. So is it in all super●iciesses, angles, & numbers, & in all other things (of one kind) that may be compared together.   

2 And if ye add equal things to equal things: the whole shallbe equal.  
As if the line AB be equal to the line CD, & to the line AB, be added the line BE, & to the line CD, be added also an other line DF, being equal to the line BE, so that two equal lines, namely, BE, and DF, be added to two equal lines AB, & CD: then shall the whole line AE, be equal to the whole line CF, and so of all quantities generally.   

3 And if from equal things, ye take away equal things: the things remaining shall be equal.  
As if from the two lines AB and CD, being equal, ye take away two equal lines, namely, EB and FD, then may you conclude by this common sentence, that the parts remaining, namely, AE, and CF are equal the one to the other and so of all other quantities.   

4 And if from unequal things ye take away equal things: the things which remain shall be unequal.  
As if the lines AB, and CD, be unequal, the line AB, being longer than the line CD, & if ye take from them two equal lines, as EB, and FD: the parts remaining, which are the lines AE and CF, shall be unequal the one to the other, namely, the line AE, shall be greater than the line CF, which is ever true in all quantities whatsoever.   

5 And if to unequal things ye add equal things: the whole shall be unequal.  
As if ye have two unequal lines, namely, AE the greater, and CF the less, & if ye add unto than two equal lines, EB & FD, then may ye conclude that the whole lines composed are unequal: namely, that the whole line AB, is greater than the whole line CD, and so of all other quantities.   

6 Things which are double to one and the self same thing: are equal the one to the other.  
As if the line AB be double to the line EF, and if also the line CD, be double to the same line EF ● then may you by this common sentence conclude, that the two lines AB, & CD, are equal the one to the other. And this is true in all quantities, and that not only, when they are double, but also if they be triple or quadruple, or in what proportion soever it be of the greater inequality. 
What proportion of the greater inequality i●  Which is when the greater quantity is compared to the less.   

7 Things which are the half of one and the self same things are equal the one to the other.  
As if the line AB, be the half of the line EF, and if the line CD, be the half also of the same line EF: then may ye conclude by this common sentence, that the two lines AB and CD, are equal the one to the other. This is also true in all kinds of quantity, and that not only when it is a half, but also if it be a third, a quarter, or in what proportion soever it be of the less in equality. Which is when the less quantity is compared to the greater. 
What proportion of the less inequality i●.    

8 Things which agreed together are equal the one to the other.  
Such things are said to agreed together, which when they are applied the one to the other: or set the one upon the other, the one ●●●●deth not the other in any thing. As if the two triangles ABC, and DEF, were applied the one to the other, and the triangle ABC, were set upon the triangle DEF, if then the angle A, do justly agreed with the angle D, and the angle B, with the angle E, and also the angle C, with the angle F: and moreover ●f the line AB, do justly fall upon the line DE, and the line AC, upon the line DF, and also the line BC, upon the line EF, so that on every part of these two triangles, there is just agreement, then may ye conclude that the two triangles are equal.   

9 Every whole is greater than his part.  

What a Proposition is.  THe principles thus placed & ended, now follow the propositions, which are sentences set forth to be proved by reasoning and demonstrations, and therefore they are again repeated in the end of the demonstration● For the proposition is ever the conclusion, and that which aught to be proved.  

Propositions of two sorts.  Propositions are of two sorts, the one is called a Problem, the other a Theorem.  

What a Problem is.  A Problem, is a proposition which requireth some action, or doing: as the making of some ●igure, or to divide a figure or line, to apply figure to ●igure, to add figures together, or to subtrah one from an other, to describe, to inscribe, to circumscribe one figure within or without another, and such like. As of the first proposition of the first book is a problem, which is thus Upon a right line given not being infinite, to describe an equilater triangle, or a triangle of three equal sides. For in it, besides the demonstration and contemplation of the mind, is required somewhat to be done: namely to make an equilater triangle upon a line given. And in the end of every problem, after the demonstration, is concluded a●ter this man●er, Which is the thing, which was required to be done.  

What a Theorem is.  A Theorem, is a proposition, which requireth the searching ou● and demonstration of some property or passion of some figure: Wherein is only speculation and contemplation of mind, without doing or working of any things As the fifth proposition of the first book, which is thus, An Isosceles or triangle of two equal sides, hath his angles at th● base, equal the one to the other, etc. is a Theorem. For in it is required only to be proved and made plain by reason and demonstration, that these two angles be equal, without further working or doing. And in the end● of ●u●ry Theorem, after the demonstration is concluded after this manner, Which thing was required to be demonstrated or proved.    

The first Problem. The first Proposition. Upon a right line given not being infinite, to describe an equilater triangle, or a triangle of three equal sides.  
SVppose that the right line given be AB. It is required upon the line AB, to describe an equilater triangle, namely, a triangle of three equal sides. 
Construction.  Now therefore making the centre the point A and the space AB, describe (by the third petition) a circle BCD: and again (by the same) making the centre the point B, and the space B●, describe an other circle ACE. And (by the first petition) from the point C, wherein the circles cut the one the other, draw one right line to the point A, and an other right line to the point B. And forasmuch as the point A is the centre of the circle CBD, 
Demonstration  therefore (by the 15. definition) the line AC is equal to the line AB: Again forasmuch as the point B is the centre of the circle CAE, therefore (by the same definition) the line BC is equal to the line BA. And it is proved, that the line AC is equal to the line AB: wherefore either of these lines CA and CB, is equal to the line AB: but things which are equal to one and the same thing, are also equal the one to the other (by the first common sentence) wherefore the line CA, also is equal to the line CB. Wherefore these three right lines CA, AB, and BC are equal the one to the other. Wherefore the triangle ABC is equilater. Wherefore upon the line AB, is described an equilater triangle ABC. Wherefore upon a line given not being infinite, there is described an equilater triangle, Which is the thing, which was required to be done.  
A triangle or any other rectilined figure is then said to be set or described upon a line, when the line is one of the sides of the figure.  
This first proposition is a Problem, because it requireth act or doing, namely, to describe a triangle. And this is to be noted, that every Proposition, whether it be a Problem, or a Theorem, commonly containeth in it a thing given, and a thing required to be searched out: although it be not always so. And the thing given, 
Thing given.  is ever set before the thing required. 
Thing required  In some propositions there are more things given then one, and more things required then one. In some there is nothing given at all.  
Moreover every Problem & Theorem, being perfect and absolute, aught to have all these parts, namely. First the Proposition, 
Proposition.  to be proved. 
Exposition.  Then the exposition which is the explication of the thing given. After that followeth the determination, 
Determination.  which is the declaration of the thing required. Then is set the construction of such things which are necessary either for the doing of the proposition, 
Construction.  or for the demonstration. Afterwards followeth the demonstration, 
Demonstration.  which is the reason and proof of the proposition. And last of all is put the conclusion, 
Conclusion.  which is inferred & proved by the demonstration, and is ever the proposition. But all those parts are not of necessity required in every Problem and Theorem. But the Proposition, demonstration, and conclusion, are necessary parts, & can never be absent: the other parts may sometimes be away.  

Case.  Further in divers propositions, there happen divers cases: which are nothing else, but variety of delineation and construction, or change of position, as when points, lines super●iciesses, or bodies are changed. Which things happen in divers propositions.  

The thing given in this Problem.  NOw then in this Problem, the thing given, is the line given: the thing required, 
The thing required.  to be searched out is, how upon that line to describe an equilater triangle. The Proposition of this Problem is, 
The pr●position.  Upon a right line given not being infinite, to describe an equilater triangle.  

The exposition.  The exposition is, Suppose that the right line given be AB, and this declareth only the thing given. The determination is, 
The determination.  It is required upon the line AB, to describe an equilater triangle: for thereby as you see, is declared only the thing required. The construction beginneth at these wo●ds, 
The construction  Now therefore making the centre the point A, & the space AB, describe (by the third pe●icion) a circle etc., and continueth until you come to these words, And forasmuch at the point A etc. For thethe●to are described circles and lines, necessary both for the doing of the proposition, and also for the demonstration thereof. Which demonstration beginneth at these words: 
The demonstration.  And forasmuch as the point A is the centre of the circle CBD etc.: And so proceedeth till you come to these words, Wherefore upon the line AB is described an equilater triangle ABC. For until you come thither, is, by grounds before set and constructions had, proved, and made evident, that the triangle made, is equilater And then in these words, 
The particular conclusion.  wherefore upon the line AB, is described an equilater triangle ABC, is put the first conclusion. For there are commonly in every proposition two conclusions: the one particular, the other universal: and from the first you go to the last. And this is the first and particular conclusion, ●or that it concludeth, that upon the line AB is described an equilater triangle, which is according to the exposition. After it, 
The universal c●nclus●●n.  followeth the last and universal conclusion, wherefore upon a right line given not being infinite is described an equilater triangle. For whether the line given be greater or less than this line, the ●ame constructions and demonstrations prove the same conclusion. Last of all is added this clause, 
The note where by it 〈◊〉 known ●o be a Problem.  Which is the thing which was required to be done: whereby as we have before noted, is declared, that this proposition is a Problem and not a Theorem. As for variety of cases in this proposition there is none, 
Not cases in this proposition.  for that the line given, can have no diversity of position.  
As you have in this Problem seen plainly set forth the thing given, and the thing required, moreover the proposition, exposition, determination, construction, demonstration, and conclusion (which are general also to many other both Problems and Theorems) so may you by the example thereof distinct them, and search them out in other problems ● and also Theorems.  

Thr●● kinds of demonstration.  This also is to be noted, that there are three kinds of demonstration, The one is called Demonstratio a priori, or composition. The other is called Demonstrati● a posteriori, or resolution. And the third is a demonstration leading to an impossibility.  
A demonstration a priori, 
Demonstration a priori, or composition.  or composition is, when in reasoning, from the principles and first grounds, we pass descending continually, till after many reasons made, we come at the length to conclude that, which we first chief intend. And this kind of demonstration useth Euclid in his booke● for the most part.  
A demonstration a posteriori, 
Demonstration a posteriori, or resolution.  or resolution is, when contrariwise in reasoning, we pass from the last conclusion made by the premises, and by the premises of the premises, continually ascending, till we come to the first principles and grounds, which are indemonstrable, and for their simplicity can suffer no farther resolution.  
A demonstration leading to an impossibility is that argument, whose conclusion is impossible: 
Demonstration leading to an impossibility.  that is, when it concludeth directly against any principle, or against any proposition before proved by principles, or propositions be●ore proved.  
Premises in an argument, are propositions going before the conclusion by which the conclusion is proved. 
Premises what they are.   
Composition passeth from the cause to the effect, or from things simple to things more compounded. Resolution contrariwise passeth from things compounded to things more simple, or from the effect to ●he cause.  
Composition or the first kind of demonstration, which passeth from the principles, may easily be seen in this first proposition of Euclid. 
An example of composition in the first proposition.  The demonstration whereof beginneth thus. And forasmuch as the point A is the centre of the circle CBD, therefore the line AC, is equal to the line AB. This reason (you see) taketh his beginning of a principle, namely, of the definition of a circle. And this is the first reason. 
First reason.  Again forasmuch as B is the centre of the circle CAE, therefore the line BC is equal to the line BA: which is the second reason. 
Second reason.  And it was before proved that the line AC is equal to the line AB, wherefore either of these lines CA & CB is equal to the line AB. And this is the third reason. 
Third reason.  But things which are equal to one & the self same thing, are also equal the one to the other. Wherefore the line CA is equal to the line CB. And this is the fourth argument. 
Fourth reason.  Wherefore these three lines CA, AB, and BC are equal the one to the other which is the conclusion, 
Conclusion.  and the thing to be proved.  
You may also in the same first Proposition, easily take an example of Resolution: 
Example of resolution in the first proposition.  using a contrary order passing backward from the last conclusion of the former demonstration, till you come to the first principle or ground whereon it began. For the last argument or reason in composition, is the first in Resolution: & the first in composition, is the last in resolution. Thus therefore must ye proceed. The triangle ABC is contained of three equal right lines, namely, AB, AC, and BC, and therefore it is an equilater triangle by the definition of an equilater triangle: and this is the first reason. 
First reason.  That the three lines be equal, is thus proved. The lines AC and CB are equal to the line AB, wherefore they are equal the one to the other: and this is the second reason. 
Second reason.  That the lines AB● and BC, are equal is thus proved: The lines AB and AC, are drawn from the centre of the circle ACE, to the circumference of the same: wherefore they are equal by the definition of a circle; and this is the third reason. 
Third reason.  Likewise that the lines AC and AB, 
forth ●eason which is th● end of the whole resolution.  are equal, is proved by the same reason. For the lines AC and AB are drawn from the centre of the circle BCD: wherefore they are equal by the same definition of a circle: this is the fourth reason or syllogism. And thus is ended the whole resolution: for that you are come to a principle, which is indemonstrable, & can not be resolved.  
Of a ●emonstration leading to an impossibility, or to an absurdity, you may have an example in the forth proposition of this book.  

An addition of Campanus.  
But now if upon the same line given, namely, AB, ye will describe the other two kinds of triangles, namely, an Isosceles or a triangle of two equal sides, & a Scalenon, or a triangle of three unequal sides. First for the describing of an Isosceles triangle produce the line AB on either side, 
How to describe an Isosceles triangle.  until it concur with the circumferences of both the circles in the points D and F, and making the centre the point A, describe a circle HFG according to the quantity of the line AF. Likewise making the centre the point B, describe a circle HDG, according to the quantity of the line BD. Now then these circles shall cut the one the other in two points, which let be H, and G: And l●t the ●nd●s of the line given be joined with one of the said sections by two right lines, which let be AG and BG. And forasmuch as these two lines AB and AD are drawn from the centre of the circle CDE unto the circumference thereof, therefore are they equal. Likewise the lines BA and BF, for that they are drawn from the centre of the circle EACF to the circumference thereof, a●e equal. And forasmuch as either of the lines AD and BF is equal to the line AB, therefore they are equal the one to the other. Wherefore putting the line AB common to them both, the whole line BD shallbe equal to the whole line AF. But BD is equal to BG, for they are both drawn from the centre of the circle HDG to the circumference thereof. And likewise by the same reason the line AF is equal to the line AG. Wherefore by the common sentence the lines AG and BG are equal the one to the other, and either of them is greater than the line AB, for that either of the two lines BD and AF is greater than the line AB. Wherefore upon the line given is described an Isosceles or triangle of two equal sides.  

How to describe a Scalenum.  You may also describe upon the self same line a Scale●on, or triangle of three unequal sides, if by two right lines, ye join both the ends of the line gruen to some one point that is in the circumference of one of the two greater circles, so that that point be not in one of the two sections, and that the line DF do not concur with it, when it is on either side produced continually and directly. For let the point K be taken in the circumference of the circle HDG, and let it not be in any of the sections, neither let the line DF concur with it, when it is produced continually and directly unto the circumference thereof. And draw these lines AK and BK, and the line AK shall cut the circumference of the circle HFG. Let it cut it in the point L: now then by the common sentence the line BK shallbe equal to the line AL, for (by the definition of a circle) the line BK is equal to the line BG, and the line AL is equal to the line AG which is equal to the line BG. Wherefore the line AK is greater than the line BK and by the same reason may it be proved that the line BK is greater than the line AB. Wherefore the triangle ABK consisteth of three unequal sides. And so have ye upon the line given, described all the kinds of triangles.  
This is to be noted, that if a man will mechanically and readily, not regarding demonstration upon a line given describe a triangle of three equal sides, 
How to describe an equilater triangle readily & mechanically.  he needeth not to describe the whole foresaid circle, but only a little part of each: namely, where they cut the one the other, and so from the point of the section to draw the lines to the ends of the line geuen● As in this figure here put.  
And likewise, if upon the said line he will describe a triangle of two equal sides, 
How to describe an Isosceles triangle readily.  let him extend the compass according to the quantity that he will have the side to be, whether longer than the line given or shorter: and so draw only a little part of each circle, where they cut the one the other, & from the point of the section draw the lines to the end of the line given. As in the figures here put. Note that in this the two sides must be such, that being joined together, they be longer than the line given.  
And so also if upon the said right line he will describe a triangle of three unequal sides, 
How to describe a Scalenum triangle readily.  let him extend the compass. First, according to the quantity that he will have one of the unequal sides to be, and so draw a little part of the circle, & then extend it according to the quantity that he will have the other unequal side to be, and draw likewise a little part of the circle, and that done, from the point of the section dra● the lines to the ends of the line given, as in the figure here put. Note that in this the two sides must be such, that the circles described according to their quantity may, cut the one the other.    

The second Problem. The second Proposition. Fron a point given, to draw a right line equal to a rightline given.  
SVppose that the point given be A, & let the right line given be BC. It is required from the point A, to draw a right line equal to the line BC. Draw (by the first petition) from the point A to the point B a right line AB: 
Constru●ti●●.  and upon the line AB describe (by the first proposition) an equilater triangle, and let the same be DAB, and extend, by the second petition, the right lines DA & DB, to the points E and F, & (by the third petition) making the centre B and the space BC describe a circle CGH: & again (by the same) making the centre D and the space DG describe a circle GKL. And forasmuch as the point B is the centre of the circle CGH, 
Demonstration.  therefore (by the 15. definition) the line BC is equal to the line BG: and forasmuch as the point D is the centre of the circle GKL: therefore (by the same) the line DL is equal to the line DG: of which the line DA is equal to a line DB (by the proposition going before) wherefore the residue, namely, the line AL is equal to the residue, namely, to the line BG (by the third common sentence) And it is proved that the line BC is equal to the line BG Wherefore either of these lines AL & BC is equal to the line BG. But things which are equal to one and the same thing are also equal the one to the other (by the first common sentence.) Wherefore the line AL is equal to the line BC. Wherefore from the point given, namely, A, is drawn a right line AL equal to the right line given BC: which was required to be done.  
Of Problems and Theoremes● as we have before noted, some have no cases at all, which are those which have only one position and constructions and other some have many and divers cases: which are such propositions which have divers descriptions & constructions, and change their positions. Of which sort is this second proposition, which is also a Problem. This proposition hath two things given: 

Two things given in this proposition.  
Power cases in this proposition.   Namely, a point, and a line: the thing required is, that from the point given wheresoever it be put, be drawn a line equal to the line given. Now this point given may have divers positions For it may be placed either without the right line given, or in some point in it. If it be without it, either it is on the side of it, so that the right line drawn from it to the end of the right line given maketh an angle● or else it is put directly unto it● so that the right line given being produced shall fall upon the point given which is without. But if it be in the line given, then either it is in one of the ends or extremes thereof: or in some place between the extremes. So are there four divers positions of the point in respect of the line. Whereupon follow divers delineations and constructions, and consequently variety of cases.  

The second case.  To the second case the figure here on the side set belongeth. And as touching the order both o● construction and of demonstration it is ●ll one with the first.  
The fourth case as touching construction herein differeth from the two first, 
The fourth case.  for th●t whereas in them you are willed to draw a right line from the point given, namely, A, to the point B which is one of the ends of the line given, here you shall not need to draw that line, for that it is already drawn. As touching the rest, both in construction and demonstration you may proceed as in the two firste● As it is manifest to see in this figure here on the side put.   

The 3. Problem. The 3. Proposition. Two unequal right lines being given, to cut of from the greater, a right line equal to the less.  
SVppose that the two unequal right lines given be AB & C, of which l●t the line AB be the greater. It is required from the line AB being the greater, to cut of a right line equal to the right line C, which is the less line-draw● on (by the second proposition) from the point A a right line equal to the line C, and let the same be AD: and making the centre A and the space AD describe (by the third petition) a circle DEF. And forasmuch as the point A is the centre of the circle DEF, therefore AE is equal to AD, but the line C is equal to the line AD. Wherefore either of these lines AE and C is equal to AD, wherefore the line AE is equal to the line C, wherefore two unequal right lines being given, namely, AB and C, there is cut of from AB being the greater, a right line AE equal to the less line, namely, to C: which was required to be done.  
This proposition, which is a Problem, hath two things given, 

Two 〈◊〉 given in ●●is proposition.  
divers cases in 〈◊〉.   namely, two unequal right lines: the thing required is, from the greater to cut of a line equal to the less. It hath also divers cases. For the lines given either are distinct th'one from the other: or are joined together at one of their ends: or they cut the one the other, or the one cutteth the other in one of the extremes. Which may be two ways. For either the greater cutteth the less, or the less the greater. If they cut the one the other, either ●ch cutteth th'other into equal parts: or into unequal parts: or the one into equal parts, and the other into unequal parts. Which may hap in two sorts, for the greater may be cut into equal parts, and the less into unequal parts: or contrariwise.  

The first case.  When the unequal lines given are distinct the one from the other, the figure before put serveth.  
But if the one cut the other in one of the extremes. 
The third case.  As for example: Suppose that the unequal right lines given be AB and CD, of which let the line CD be the greater: And let the line CD cut the line AB in his extreme C. Then making the centre A and the space AB, describe a circle BE. And upon the line AC describe an equilater triangle (by the ●irst) which let b● AEC: & produce the lines EA and EC. And again making the centre E and the space EF describe a ci●●●e GF. Likewise making the centre C and the space CG, describe a circle GL. Now forasmuch as the line EF is equal to the line EG (●or E is the centre) of which the line EA is equal to the line E●: therefore the residue AF is equals to the residue CG. ●ut the line AF is equal to the line AB, for A is the centre, where●o●e also the line CG is equal to the line AB. But the line CG is also equal ●o the line CL, for the point C is the centre. Wherefore the line AB is equal to the line CL. Wherefore from the line CD is cut of the line CL which is equal to the line AB.  

The fifth case.  Or it is less than the half: and then making the centre C & the space CD describe a circle, which shall cut of from the line AB a line equal to the line CD.  

The s●xt case.  Or it is greater than the half. And then unto the point A put the line AF equal to the line CD, by the second. And making the centre A & the space AF describe a circle, which shall cut of from the line AB a line equal to the line AF, that is, unto the line CD.  
But if it be greater than the half, 
The ninth case.  then again unto the point A put the line AF equal to the line CD (by the second proposition:) & making the centre A, and the space AF describe a circle which shall cut of from the line AB a line equal to the line AF, that is, to the line CD.  
But if they cut the one the other as the lines CD & AB do. 
The tenth case.  Then making the centre B & the space BA describe a circle AF, & draw a line from the point B to the point C, & produce it to the point F. And forasmuch as the two right lines BF and CD are unequal, and the line CD cutteth the line BF by one of his extremes, therefore it is possible to cut of from CD a line equal to the line BF. For how to do it we have before declared, wherefore it is possible from the line CD to cut of a line equal to the line AB: or AB and BF are equal the one to the other.  
This is to be noted, that in all these cases, 
In all these cases the construction and demonstration of the first case will serve.  a man may both as touching construction and demonstration, proceed as in the first case. For it is possible in any position to put to the end of the greater line a line equal to the less line, and so making the centre the said end, and the space the less line, to describe a circle, which shall cut of from the greater line a line equal to the line put, namely, to the less line given, as it is manifest to see in the figures partly here under set, and partly in the beginning of the other side put.   

The first Theorem. The 4. Proposition. If there be two triangles, of which two sides of th'one be equal to two sides of the other, each side to his correspondent side, and having also on angle of the one equal to one angle of the other, namely, that angle which is contained under the equal right lines: the base also of the one shall be equal to the base of the other, and the one triangle shall be equal to the other triangle, and the other angles remaining shall be equal to the other angles remaining, the one to the other, under which are subtended equal sides.  
SVppose that there be two triangles ABC, & DEF, having two sides of the one, namely AB, and AC, equal to two sides of the other, namely, to DE and DF, the one to the other, that is, AB to DE, and AC to DF: having also the angle BAC, equal to the angle EDF. Then I say that the base also BC is equal to the base EF: & that the triangle ABC, is equal to the triangle DEF: and that the other angles remaining are equal to the other angles remaining, the one to the other, under which are subtended equal sides: that is, that the angle ABC is equal to the angle DEF, and that the angle ACB is equal to to the angle DFE. 
Demonstration lea●ing to an absurdity.  For the triangle ABC exactly agreeing with the triangle DEF, and the point A being put upon the point D, & the right line AB upon the right line DE, the point B also shall exactly agreed with the point E: for that (by supposition) the line AB is equal to the line DE. And the line AB exactly agreeing with the line DE, the right line also AC exactly agreeth with the right line DF, for that (by supposition) the angle BAC is equal to the angle EDF. And forasmuch as the right line AC is also (by supposition) equal to the right line DF, therefore the point C exactly agreeth with the point F. Again forasmuch as the point C exactly agreeth with the point F, and the point B exactly agreeth with the point E: therefore the base BC shall exactly agreed with the base EF. For if the point B do exactly agreed with the point E, and the point C with the point F, and the base BC do not exactly agree with the base EF, than two right lines do include a superficies: which (by the 10. common sentence) is impossible. Wherefore the base BC exactly agreeth with the base EF, and therefore is equal unto it. Wherefore the whole triangle ABC exactly agreeth with the whole triangle DEF, & therefore (by the 8. common sentence) is equal unto it. And (by the same) the other angles remaining exactly agreed with the other angles remaining, and are equal the one to the other: that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. If therefore there be two triangles, of which two sides of the one, be equal to two sides of the other, each to his correspondent side, and having also one angle of the one equal to one angle of the other, namely, that angle which is contained under the equal right lines: the base also of the one shall be equal to the base of the other, and the one triangle shall be equal to the other triangle, and the other angles remaining shall be equal to the other angles remaining, the one to the other, under which are subtended eqaull sides: which thing was required to be demonstrated.  
This Proposition which is a Theorem, 
Two 〈…〉  hath two things given: namely, the equality of two sides of the one triangle, to two sides of the other triangle, and the equality of two angles contained under the equal sides. In it also are three things required. 
Three things required in it.  The equality of base to base: the equality of field to field: and the equality of the other angles of the one triangle to the other angles of the other triangle, under which are subtended equal sides.  

How one side i● equal to an other, & so generally how one right line is equal to an other.  One side of a plain figure is equal to an other, and so generally one right line is equal to an other, when the one being applied to the other, their extremes agreed together. For otherwise every right line applied to any right line, agreeth therewith: but equal right lines only, agreed in the extremes.  

How one rectili●ne● angle is equal to an other  One rectilined angle is equal to an other rectilined angle, when one of the sides which comprehendeth the one angle, being set upon one of the sides which comprehendeth the other angle, the other side of the one agreeth with the other side of the other. And that angle is the greater, whose side falleth without: and that the less, whose side falleth within.  

Why this particle 〈◊〉 to his correspondent side, is put.  Where as in this proposition is put this particle each to his correspondent side, (instead whereof often times afterward is used this phrase the one to the other) it is of necessity so put. For otherwise two sides of one triangle added together, may be equal to two sides of an other triangle added together, and the angles also contained under the equal sides may be equal: and yet the two triangles may notwithstanding be unequal. 
How one triangle is equal to an other  Where note that a triangle is said to be equal to an other triangle, when the field or area of the one is equal to the area of the other. And the area of a triangle, 
What the field or area of a triangle is, and so of any rectilined 〈◊〉.  is that space, which is contained within the sides of a triangle. And the circuit or compass of a triangle is a line composed of all the sides of a triangle. And so may you think of all o●her rectilined figures. 
What 〈…〉 of a triangle is, and ●o also of any 〈◊〉 figure.  And now to prove that there may be two triangles, two sides of one of which being added together, may be equal to two sides o● the other added together, and the angles contained under the equal sides may be equal, and yet notwithstanding the two triangles unequal. Suppose that there be two rectangle triangles: namely, ABC, and DEF, and let their right angles be BAC and EDF. And in the triangle ABC let the side AB b● 3. and the side AC 4. which both added together make 7.  
That angle is said to subtend a side of a triangle, 
How an angle is said to subtend a side: and a side an angle.  which is placed directly opposite, & against that side. That side also is said to subtend an angle, which is opposite to the angle. For every angle of a triangle is contained of two sides of the triangle, and is subtended to the third side.  
This is the first Proposition in which is used a demonstration leading to an absurdity, o● an impossibility. 
This proposition proved by a demonstration leading to an absurdity.  Which is a demonstration that proveth not directly the thing intended, by principles, or by things before proved by these principles: but proveth the contrary thereof to be impossible, & so doth indirectly prove the thing intended.   

The 2. Theorem. The 5. Proposition. An Isosceles, or triangle of two equal sides, hath his angles a● the base equal the one to the other. And those equal sides being produced, the angles which are under the base are also equal the one to the other.  
SVppose that ABC be a triangle of two equal sides called Isosceles, having the side AB equal to the side AC. And (by the second petition) produce the lines AB & AC directly to the points D & E. Then I say, that the angle ABC is equal to the angle ACB: and the the angle CBD is equal to th' angle BCE. Take in the line BD a point at all adventures, 
Construction.  and let the same be F, and (by the third proposition) from the greater line, namely, AE, cut of a line equal to AF being the less line, and let the same be AG: and draw a right line from the point F to the point C, 
Demonstration.  and an other from the point G to the point B. Now then for as much as AF is equal to AG, and AB is equal to AC, therefore these two lines FA and AC are equal to these two lines GA and AB, the one to the other, and they contain a common angle, namely, that which is contained under FAG: wherefore (by the fourth proposition) the base FC is equal to the base GB: and the triangle AFC is equal to the triangle AGB, and the other angles remaining, are equal to the other angles remaining the one to the other, under which are subtended equal sides: that is, the angle ACF is equal to the angle ABG, and the angle AFC is equal to the angle AGB. And forasmuch as the whole line AF is equal to the whole line AG, of which the line AB is equal to the line AC, therefore the residue of the line AF, namely, the line BF, is equal to the residue of the line AG, namely, to the line CG (by the third common sentence) And it is proved that CF is equal to BG. Now therefore these two BF & FC are equal to these two CG and GB the one to the other, and the angle BFC is equal to the angle CGB, and they have one base, namely, BC, common to them both: wherefore (by the 4. proposition) the triangle BFC is equal to the triangle CGB, and the other angles remaining are equal to the other angles remaining each to other, under which are subtended equal sides. Wherefore the angle FBC is equal to the angle GCB, and the angle BCF is equal to the angle CBG. Now forasmuch as the whole angle ABG is equal to the whole angle ACF (as it hath been proved) of which the angle CBG is equal to the angle BCF. therefore the angle remaining: namely, ABC is equal to the angle remaining, namely, to ACB (by the third common sentence) And they are the angles at the base of the triangle ABC. And it is proved that the angle FBC is equal to the angle GCB, and they are angles under the base. Wherefore a triangle of two equal sides hath his angles at the base equal the one to the other. And those equal sides being produced, the angles which are under the base are also equal the one to the other: which was required to be proved.  
That in an Isosceles triangle, the two angles above the base are equal, may otherwise be 〈◊〉 on str●red without drawing lines beneath the base somewhat altering the construction. Namely, drawing the lines within the triangle, which before were without it after this manner.  
To prove this also, there is an other demonstration of Pappus much shorter which needeth no kind of addition of any thing at all: as followeth.  

Thales Milesius the inuent●r of this proposition.  The old Philosopher Thales Milesius was the first inventor of this fifth proposition, as also of many other.   

The third Theorem. The sixth Proposition. If a triangle have two angles equal the one to the other: the sides also of the same, which subtend the equal angles, shallbe equal the one to the other.  
SVppose that ABC be a triangle, having the angle ABC equal to the angle ACB. Then I say that the side AB is equal to the side AC. For if the side AB be not equal to the side AC, than one of them is greater. Let AB be the greater. And by the third proposition, 
Construction.  from AB being the greater cut of a line equal to the less line, which is AC● And let the same be D B. And draw a line from the point D to the point C. 
Demonstration leading to an impossibility.  Now forasmuch as the side DB is equal to the side AC, and the line BC is common to them both: therefore these two sides DB and BC are equal to these two sides AC & CB the one to the other. And the angle DBC is by supposytion equal to the angle ACB. Wherefore (by the 4 proposytion) the base DC is equal to the base AB: & (by the same) the triangle DBC is equal to the triangle ACB: namely, the less triangle unto the greater triangle, which is impossible. Wherefore the side AB is not unequal to the side AC. Wherefore it is equal. If therefore a triangle have two angles equal the one to the other: the sides also of the same, which subtende the equal angles, shall be equal the one to the other● which was required to be demonstrated.  

The chiefest and 〈◊〉 proper kind of conversion.  In Geometry is oftentimes used conversion of propositions. As this proposition is the converse of the proposition next before. The chiefest and most proper kind of conversion is, when that which was the thing supposed in the former proposition, is the conclusion of the converse and second proposition: and contrary wise that which was concluded in the first, is the thing supposed in the second As in the fifth proposition it was supposed the two sides of a triangle to be equal, the thing concluded is, that the two angles at the base are equal & in this proposition, which is the converse thereof is supposed that the angles at the base be equal. Which in the former proposition was the conclusion. And the conclusion is, that the two sides subtending the two angles are equal, which in the former proposition was the supposition. This is the chiefest kind of conversion uniform and certain.  
There is an other kind of conversion, 
another kind of conversion not so perfect as the first.  but not so full a conversion nor so perfect as the first is. Which happeneth in composed propositions, that is, in such, which have more suppositions than one, and pass from these suppositions to one conclusion. In the connuerses of such propositions, you pass from the conclusion of the first proposition, with one or more of the suppositions of the same: & conclude some other supposition of the self first proposition: of this kind there are many in Euclid. Thereof you may have an example in the 8. proposition being the converse of the four●h. This conversion is not so uniform as the other; but more divers and uncertain according to the multitude of the things given, or suppositions in the proposition.  
But because in the fifth proposition there are two conclusions, 
Two conclusions in the fifth proposition.  the first, that the two angles at the ba●e be equal: the second, that the angles under the ba●e are equal: this is to be noted, that this sixth proposition is the converse of the ●ame fifth as touching the first conclusion only. 
The sixth proposition i● the converse as touching the first conclusion only. The converse as touching the second conclusion.  You may in like manner make a converse of the same proposition touching the second conclusion thereof. And that after this manner.  
THe two sides of a triangle being produced, if the angles under the base be equal, the said triangle shall be an Isosceles triangle. In which proposition the supposition is, that the angles under the base are equal: which in the fifth proposition was the conclusion● & the conclusion in this proposition is, that the two sides of the triangle are equal, which in the fift proposition was the supposition. But now for proof of the said proposition:  
For suppose that AC be equal to AD: and produce the line CA to the point E: and put the line AE equal to the line DB (by the third proposition) wherefore the whole line CE is equal to the whole line AB (by the second common sentence) Draw a line from the point E to the point B. And forasmuch as the line AB is equal to the line EC, and the line BC is common to them both, and the angle ACB is supposed to be equal to the angle ABC: Wherefore (by the fourth proposition) the triangle EBC is equal to the triangle ABC, namely, the whole to the part: which is impossible.   

The 4. Theorem. The 7. Proposition. If from the ends of one line, be drawn two right lines to any point: there can not from the self same ends on the same side, be drawn two other lines equal to the two first lines, the one to the other, unto any other point.  
FOr if it be possible: then from the ends of one & the self same right line, namely, AB, from the points (I say) A and B, let there be drawn two right lines AC and CB to the point C: and from the same ends of the line AB, let there be drawn two other right right lines AD and DB equal to the lines AC and CB the one to the other, is, each to his correspondent line, and on one and the same side, and to an other point, namely, to D: so that let CA be equal to DA being both drawn from one end, that is, A: & let CB be equal to DB, being both also drawn from one end, that is, B. And (by the first petition) draw a right line from the point C to the point D. 
Demonstration le●●●n● to an absurdity.  Now forasmuch as AC is equal to AD, the angle ACD also is (by the 5. proposition) equal to the angle ADC: wherefore the angle ACD is less than the angle BDC. Wherefore the angle BCD is much less than the angle BDC. Again forasmuch as BC is equal to BD, and therefore also the angle BCD is equal to the angle BDC. And it is proved that it is much less than it: which is impossible. If therefore from the ends of one line, be drawn too right lines to any point: there can not from the self same ends on the same side, be drawn two other lines equal to the two first lines, the one to the other, unto any other point: Which was required to be demonstrated.  
In this proposition the conclusion is a negation, which very rarely happeneth in the mathematical arts. 
Negative conclu●sions rarely used in the mathematical arts.  For they ever for the most part use to conclude affirmatively, & not negatively. For a proposition universal affirmative is most agreeable to sciences, as saith Aristotle, and is of itself strong, and needeth no negative to his proof. But an universal proposition negative must of necessity have to his proof an affirmative, For of only negative propositions there can be no demonstrations. And therefore sciences using demonstration, conclude affirmatively, and very seldom use negative conclusions.  

another demonstration after Campanus.  
Suppose that there be a line AB, from whose ends A and B, let there be drawn two lines AC and BC on one side, which let concur in the point C. Then I say that on the same side there cannot be drawn two other lines, from the ends of the line AB, which shall concur at any other point, so that that which is drawn from the point A shall be equal to the line AC, and that which is drawn from the point B shallbe equal to the line BC. For if it be possible, let there be drawn two other lines on the self same side, which let concur in the point D, and let the line AD be equal to the line AC, & the line BD equal to the line BC. 
divers cases in this demonstration.  Wherefore the point D shall fall either within the triangle ABC, or without. For it cannot fall in one of the sides, for then a part should be equal to his whole. If therefore it fall without then either one of the lines AD and DB shall cut one of the lines AC and CB, or else neither shall cut neither. 
First case.  first let one cut the other and draw a right line from C to D. Now forasmuch as in the triangle ACD, the two sides AC and AD are equal, therefore the angle ACD is equal to the angle ADC, by the fifth proposition: likewise forasmuch as in the triangle BCD, the two sides BC and BD are equal, therefore by the same, the angles BCD & BDC are also equal. And forasmuch as the angle BDC is greater than the angle ADC, it followeth that the angle BCD is greater than the angle ACD, namely, the part greater than the whole: which is impossible.  
But if the point D fall without the triangle ABC, 
Second case.  so that the lines cut not the one the other, draw a line from D to C. And produce the lines BD & BC beyond the base CD, unto the points E & F. And forasmuch as the lines AC and AD are equal, the angles ACD and ADC shall also be equal, by the fifth propositions likewise for as much as the lines BC and BD are equal, the angles under the base, namely, the angles FDC and ECD are equal, by the second part of the same proposition. And for as much as the angle ECD is less than the angle ACD: It followeth that the angle FDC is less than the angle ADC: which is impossible: for that the angle AD C is a part of the angle FD C. And the same inconvenience will follow if the point D fall within the triangle ABC,    

The fift Theorem. The 8. Proposition. If two triangles have two sides of th'one equal to two sides of the other, each to his correspondent side, & have also the base of the one equal to the base of the other: they shall have also the angle contained under the equal right lines of the one, equal to the angle contained under the equal right lines of the other.  
SVppose that there be two triangles ABC and DEF: & let these two sides of the one AB and AC, be equal to these two sides of the other DE, and DF, each to his correspondent side, that is, AB to D E, and AC, to DF, & let the base of the one, namely, BC be equal to the base of the other, namely, to EF. Then I say, that the angle BAC is equal to the angle EDF. For the triangle ABC exactly agreeing with the triangle DE F, 
Demonstration leading to an impossibility.  and the point B being put upon the point E, and the right line BC upon the right line EF: the point C shall exactly agreed with the point F (for the line BC is equal to the line EF) And BC exactly agreeing with EF the lines also BA and AC shall exactly agreed with the lines ED & DF. For if the base BC do exactly agreed with the base FE, but the sides BA & AC do not exactly agreed with the sides ED & DF, but differ as FG & GF do: them from the ends of one line shallbe drawn two right lines to a point, & from the self same ends on the same side shallbe drawn two other lines, equal to the two first lines, the one to the other, and unto an other point: but that is impossible, (by the seventh proposition) Wherefore the base BC exactly agreeing with the base EF, the sides also BA and AC do exactly agree with the sides ED and DF. Wherefore also the angle BAC shall exactly agree with the angle EDF, and therefore shall also be equal to it. If therefore two triangles have two sides of the one equal to two sides of the other, each to his correspondent side, and have also the base of the one equal to the base of the other: they shall have also the angle contained under the equal right lines of the one, equal to the angle contained under the equal right lines of the other: which was required to be proved.  

This proposition is the 〈◊〉 of the forth, b●t not the chiefest kind of conversion.  This Theorem is the converse of the fourth, but it is not the chiefest and principal kind o● conversion. For it turneth not the whole supposition into the conclusion, and the whole conclusion into the supposition. For the fourth proposition whose converse this is, is a compound theorem, having two things given or supposed, which are these: the one, that two sides of the one triangle be equal to two sides of the other triangle: th'other, that the angle contained of the two sides of th'one is equal to the angle contained of the two sides of th'one: but hath amongst other one thing required, which is, that the base of the one, is equal to the base of the other. Now in this 8. proposition, being the converse therof● that the base of the one is equal to the base of th'other, is the supposition, or the thing given: which in the former proposition was the conclusion. And this, that two sides of the one are equal to two sides of the other, is in this proposition also a supposition, like as it was in the former proposition: so that it is a thing given in either proposition. The conclusion of this proposition is that the angle enclosed of the two equal sides of the one triangle is equal to the angle enclosed of the two equal sides of the other triangle: which in the former proposition was one of the things given.  
Philo and his scholas demonstrate this proposition without the help of the former proposition, in this manner.  
First let it fall directly. 
The first case.  And forasmuch as the line DE is equal to the line E G, and DFG, is one right line: therefore DEG is an Isosceles triangle: and so, by the fifth proposition, the angle at the point D is equal to the angle at the point G: which was required to be proved.   

The 4. Problem. The 9 Proposition. To divide a rectiline angle given, into two equal parts.  
SVppose that the rectiline angle given be BAC. It is required to divide the angle BAC into two equal parts. 
Construction.  In the line AB take a point at all adventures, & let the same be D. And (by the third proposition) from the line AC cut of the line AE equal to AD. And (by the first petition) draw a right line from the point D to the point E. And (by the first proposition) upon the line DE describe an equilater triangle and let the same be DFE, and (by the first petition) draw a right line from the point A to the point F. Then I say that the angle BAC is by the line AF divided into two equal parts. 
Demonstration.  For, forasmuch as AD is equal to AE, and AF is common to them both: therefore these two DA and AF, are equal to these two EA and AF, the one to the other. But (by the first proposition) the base DF is equal to the base EF: wherefore (by the 8. proposition) the angle DAF is equal to the angle FAE. Wherefore the rectiline angle given, namely, B AC is divided into two equal parts by the right line AF● Which was required to be done.  
In this proposition is not taught to divide a right lined angle into more parts than two: albeit to divide an angle, so it be a right angle, into three parts, it is not hard. And it is taught of Vi●ellio in his first book of Perspective, 
It is impossible to divide an acute rectiline angle into three equal parts without the help of lines which are of a mixed nature.  the 28. Proposition. ●or to divide an acute angle into three equal parts, is (as saith Proclus) impossible: unless it be by the help of other lines which are of a mixed nature. Which thing Nicomedes did by such lines which are called Concoide● linea, who first searched out the invention, nature, & properties of such lines. And others did it by other means as by the help of quadrant lines invented by Hippias & Nicomedes. Others by Helices' or Spiral lines invented of Archimedes. But these are things of much difficulty and hardness, and not here to be entreated of.  
Here against this proposition may of the adversary be brought an 
* An instance is an objection or a doubt, whereby is letted or troubled the construction, or demonstration, & containeth an unruth, and an impossibility: and therefore it must of necessity be answered unto, and the falsehood thereof made manifest.  instance. For he may cavil that the head of the equilater triangle shall not fall between the two right lines, but in one of them, or without them both. As for example.  
There may also in this proposition be divers cases● 
divers cases in this proposition.  ●f it so happen that there be no space under the base DE to describe an equilater triangle, but that of necessity you must describe it on the same side that the lines AB and AC are. For then the sides of the equilater triangle either exactly agreed with the lines AD and AE, if the said lines AD and AE be equal with the base DE. Or they fall without them, if the lines AD and AE be less than the base DE. Or they fall within them, if the said lines be greater the● the base DE.   

The 5. Problem. The 10. Proposition. To divide a right line given being finite, into two equal parts.  
SVppose that the right line given be AB. It is required to divide the line AB into two equal parts. 
Construction.  Describe (by the first proposition) upon the line AB an equilater triangle and let the same be ABC. And (by the former proposition) divide the angle ACB into two equal parts by the right line CD. Then I say that the right line AB is divided into two equal parts in the point D. 
Demonstration.  For forasmuch as (by the first proposition) AC is equal to CB, and CD is common to them both: therefore these two lines AC & CD are equal to these two lines BC & CD, the one to the other, and the angle ACD is equal to the angle BCD. Wherefore (by the 4. proposition) the base AD is equal to the base BD. Wherefore the right line given AB, is divided into two equal parts in the point D: which was required to be done.  
Apollonius reaches to divide a right line being finite into two equal parts after this manner. 
another way to divide a right line being finish, inue●ted by Apollonius.    

The 6. Problem. The 11. Proposition. Upon a right line given, to raise up from a point given in the same line a perpendicular line.  
SVppose that the right line given be AB, & let the point in it given be C. It is required from the point C to raise up unto the right line AB a perpendicular line. Take in the line AC a point at all adventures, 
Construction.  & let the same be D, and (by the 3. proposition) put unto DC an equal line CE. And by the first proposition) upon the line DE describe an equilater triangle FDE, & draw a line from F to C. Then I say that unto the right line given AB, and from the point in it given, namely, C is raised up a perpendicular line FC. For forasmuch as DC is equal to CE, 
Demonstration.  & the line CF is common to them both: therefore these two DC and CF, are equal to these two EC & CF, the one to the other: and (by the first proposition) the base DF is equal to the base EF: wherefore (by the 8. proposition) the angle DCF is equal to the angle ECF: and they be side angles. But when a right line standing upon a right line doth make the two side angles equal the one to the other, either of those equal angles is (by the. 10. definition) a right angle: & the line standing upon the right line is called a perpendicular line. Wherefore the angle DCF, & thangle FCE are right angles. Wherefore unto the right line given AB, & from the point in it C, is raised up a perpendicular line CF: which was required to be done.  
Although the point given should be set in one of the ends of the right line given, it is easy so do it as it was before. For producing the line in length from the point by the second petition, you may work as you did before. But if one require to erect a right line perpendicularly from the point at the end of the line, without producing the rightlyne, that also may well be done after this manner.  
Appollonius teacheth to raise up unto a line given, from a point in it given, a perpendicular line, after this manner. 

another way to erect a perpendicular line invented by Appolonius.  
Constructions.     

The 7. Problem. The 12. Proposition. Unto a right line given being infinite, and from a point given not being in the same line, to draw a perpendicular line.  
LEt the right line given being infinite be AB, & let the point given not being in the said line AB, be C. It is required from the point given, namely, C, to draw unto the right line given AB, a perpendicular line. 
Construction.  Take on the other side of the line AB (namely, on that side wherein is not the point C) a point at all adventures, and let the same be D. And making the centre C, and the space CD, describe (by the third petition) a circle, and let the same be EFG, which let cut the line AB in the points E and G. And (by the x. proposition) divide the line EG into two equal parts in the point H. And (by the first petition) draw these right lines, CG, CH, and CE. Then I say, that unto the right line given AB, & from the point given not being in it, namely, C, is drawn a perpendicular line CH. 
Demonstration.  For forasmuch as GH is equal to HE, and HC is common to them both; therefore these two sides GH and HC, are equal to these two sides EH & HC, the one to the other● and (by the 15 definition) the base CG is equal to the base CE: wherefore (by the 8. proposition) the angle CHG is equal to the angle CHE: and they are side angles: but when a right line standing upon a right line maketh the two side angles equal the one to the other, either of those equal angles is (by the 10. definition) a right angle, and the line standing upon the said right line is called a perpendicular line. Wherefore unto the right line given AB, and from the point given C, which is not in the line AB, is drawn a perpendicular line CH: which was required to be done.  
This Problem did Oen●pides first find out, 
Oenopides the first inventor of this problem.  considering the necessary use thereof to the study of Astronomy.  
There ar● two kinds of perpendicular lines: 
Two kinds of perpendicular lines, namely, a plain perpendicular line and a solid.  whereof one is a plain perpendicular line, the other is a solid. A plain perpendicular line is, when the point from whence the perpendi●uler line i● drawn, is in the same plain superficies with the line whereunto it is a perpendicular. A solid perpendicular line is, when the point, from whence the perpendicular is drawn, is on high, and without the plain superficies. So that a plain perpendicular line is drawn to a right line: & a solid perpendicular line is drawn to a superficies. A plain● perpendicular line causeth right angles with one only line, namely, with that upon whom it falleth. But a solid perpendicular line causeth right an●le●, not only with one line, but with as many lines as may be drawn in that superficies, by the touch thereof. 
This proposition teacheth to draw a plain perpendicular line.  This proposition teacheth to draw a plain perpendicular line, for it is drawn to one line, and supposed to be in the self same plain superficies.   

The 6. Theorem. The 13. Proposition. When a right line standing upon a right line maketh any angles: those angles shall be either two right angles, or equal to two right angles.  
SVppose that the right line AB standing upon the right line CD do make these angles CBA and ABD. Then I say, that the angles CBA and ABD are either two right angles, or its squall to two right angles. If the angle CBA be equal to the angle ABD: then are they two right angles (by the tenth definition. 
Construction, ) But if not, raise up (by the 11. proposition) unto the right line CD, and from the point given in it, namely 
Demonstration  B, a perpendicular line BE● Wherefore (by the x. definition) the angle CBE and EBD ar● right angles. Now forasmuch as the angle CBE, is equal to these two angles CBA and ABE, put the angle EBD common to them both● wherefore the angles CBE and EBD, are equal to these three angles CBA, ABE, and EBD. Again forasmuch as the angle DBA is equal unto these two angles DBE and EBA, put the angle ABC common to them both: wherefore the angles DBA and ABC, are equal to these three angles, DBE, EBA, and ABC. And it is proved that the angles CBE and EBD are equal to the self same three angles: but things equal to one & the self same things are also (by the first common sentence) equal the o●e to the oath. Wherefore the angles CBE and EBD are equal to the angles DBA & ABC. But the angles CBE and EBD are two right angles: wherefore also the angles DBA and ABC are equal to two right angles. Wherefore when a right line standing upon a right line maketh any angles: those angles shallbe either two right angles, or equal to two right angles: which was required to be demonstrated.  

another demonstration after Pelitarius.  
Suppose that the right line AB do stand upon the right line CD. Then I say, that the two angles ABC and ABD, 
another demonstration after Peli●●riu●.  are either two right angles, or equal to two right angles. For if AB be perpendiculer unto CD: then is it manifest, that they are right angles (by the conversion of the definition) But if it incline towards the end C, then (by the 11. proposition) from the point B, erect unto the line CD a perpendicular line BE. By which construction the proposition is very manifest. For forasmuch as the angle ABD is gre●ter than the right angle DBE by the angle ABE ● and the other angle ABC is less than the right angle CBE by the self same angle ABE: if from the greater b●e taken away the excess, and the same be added to the less, they shall be made two right angles. That is, if from the obtuse angle ABD be taken away the angl● ABE, there shall remain the right angle DBE. And then if the same angle ABE be added to the ●cute angle CBA, there shall be made the right angle CBE. Wherefore it is manifest, that the two angles, namely, the obtuse angle ABD, & the acute angle ABC, are equal to the two right angles CBE and DBE: which was required to be proved.    

The 7. Theorem. The 14. Proposition. If unto a right line, and to a point in the same line, be drawn two right lines, no● both on one and the same side, making the side angles equal to two right angles: those two right lines shall make directly one right line.  
Unto the right line AB, & to the point in it B● let ther● be drawn two right lines BC, and BD, unto contrary sides, making the side angles, namely, ABC & ABD, equal to two right angles. Then I say, that the right lines BD and BC make both one right line. 
Demonstration leading to an absurdity.  For if CB and BD do not make both one right line, let the right line BE be so drawn to BC, that they both make one right line. Now forasmuch as the right line AB standeth upon the right line CBE, therefore the angles ABC and ABE are equal to two right angles (by the 13. proposition) But (by supposition) the angles ABC and ABD are equal to two right angles: wherefore the angles CBA, and ABE, are equal to the angles CBA, and ABD: take away the angle ABC, which is common to them both. Wherefore the angle remaining ABE, is equal to the angle remaining ABD, namely, the less to the greater, which is impossible. Wherefore the line BE is not so directly drawn to BC, that they both make one right line. In like sort may we prove, that no other line, beside BD, can so be drawnet Wherefore the lines CB and BD, make both one right line. If therefore unto a right line, & to a point in the same line, be drawn two right lines, not both on one and the same side, making the side angles equal to two right angles: those two lines shall make directly one right line: which was required to be proved.  

another demonstration after Pelitarius.  
Suppose that there be a right line AB, unto whose point ●, 
another demonstration after Pelitarius.  let there be drawn two right lines CB and BD, unto contrary sides: and let the two angles CBA, and DBA, be either two right angles, or equal to two right angles. Then I say, that the two lines CB and BD, do make directly one right line, namely, CD. For if they do not, then l●t ●E b●●o drawn unto CB, that they both make directly one right line C●●: which shall pass ●ither above the line BD, or under it. First l●t it pass above it. And for a● much as the two angles CBA and ABE, are (by the former proposition) equal to two right angles, and are a part of the two angles, CBA and ABD: but the angles CBA and A●D a●e by (supposition) equal also to two right angles● therefore the par●● is equal to the whole which is impossible. And the like absurdity will follow if CB E pass under the line BD: namely, that the whole shallbe equal to the part● which is also impossible. Wherefore CD is one right line: which was required to be proved.    

The 8. Theorem. The 15. Proposition. If two right lines cut the one the other: the head angles shall be equal the one to the other.  
SVppose that these two right lines AB and CD, do cut the one the other in the point E. Then I say, that the angle AEC, is equal to the angle DEB. 
Demonstration.  For forasmuch as the right line AE, standeth upon the right line DC, making these angles CEA, and AED: therefore (by the 13. proposition) the angles CEA, and AED, are equal to two right angles. Again forasmuch as the right line DE, standeth upon the right line AB, making these angles AED, and DEB: therefore (by the same proposition) the angles AED, and DEB, are equal to two right angles: and it is proved, that the angles CEA, and AED, are also equal to two right angles. Wherefore the angles CEA, and AED, are equal to the angles AED, and DEB. Take away the angle AED, which is common to them both. Wherefore the angle remaining CEA, is equal to the angle remaining DEB. And in like sort may it be proved, that the angles CEB, and DEA, are equal the one to the other. If therefore two right lines cut the one the other, the head angles shallbe equal the one to the other: which was required to be demonstrated.  

Thales Milesius the ●irst inuent●r of this proposition.  Thales Milesius the Philosopher was the first inventor of this Proposition, as witnesseth Eudemius, but yet it was first demonstrated by Euclid. And in it there is no construction at all. 
No construction in this proposition.  For the exposition of the thing given, is sufficient enough for the demonstration.  
Head Angles, 
What head angles are.  are apposite angles, caused of the intersection of two right lines: and are so called, because the heads of the two angles are joined together in one point.  

The converse of this proposition after Pelitarius.  

The converse of this propositio after Pelitarius.  If four right lines being drawn from one point, do make four angles, of which the two opposite angles are equal: the two opposite lines shallbe drawn directly, and make one right line.  
Suppose that there be four right lines AB, AC, AD, and AE, drawn from the point A, making four angles at the point A: of which let the angle BAC be equal to the angle DAE, and the angle BAD to the angle CAE. Then I say, that BE and CD are only two right lines: that is, the two right lines BA and AE are drawn directly, and do make one right line, and likewise the two right lines CA and AD are drawn directly, and do make one right line. For otherwise if it be possible, 
Demonstration leading to an absurdity.  let EF be one right line, and likewise let CG be one right line. And forasmuch as the right line EA standeth upon the right line CG, therefore the two angles EAC and EAG, are (by the 13 proposition) equal to two right angles. And forasmuch as the right line B A standeth upon the right line EF: therefore (by the self same) the two angles EAG and FAG are also equal to two right angles. Wherefore taking away the angle EAG, which is common to them both, the angle EAC, shall (by the third common sentence) be equal to the angle FAG: but the angle EAC is supposed to be equal to the angle BAD. Wherefore the angle BAD is equal to the angle FAG, namely a part to the whole: which is impossible. And the sel●e same absurdity will follow, on what side soever the lines be drawn. Wherefore BE is one line, and CD also is one line: which was required to be proved.   

The same converse after Proclus.  
If unto a right line, and to a point thereof be drawn two right lines, not on one and the same side, in such sort that they make the angles at the top equal: 
The same converse after Pelitarius, which is demonstrated directly.  those right lines shallbe drawn directly one to the other, and shall make one right line.  

What a corollary is.  Of this fifteenth Proposition followeth a corollary. Where note that a Corollary is a Proposition, whose demonstration dependeth of the demonstration of an other Proposition, and it appeareth suddenly, as it were by chance offering itself unto us: and therefore is reckoned as lucre or gain. The Corollary which followeth of this proposition, is thus.  

A Corollary following of this proposition.  If four right lines cut the one the other: they make four angles equal to four right angles.  
This Corollary gave great occasion to find out that wondered proposition invented of Pythagoras, which is thus.  

A wonderful proposition invented by P●●hagoras.  Only three kinds of figures of many angles, namely, an equilater triangle, a right angled figure of four sides, and a figure of six sides, having equal sides and equal angles, can fill the whole space about a point, their angles touching the same point.    

The 9 Theorem. The 16. Proposition. Whensoever in any triangle, the line of one side is drawn forth in length: the outward angle shall be greater than any one of the two inward and opposite angles.  
SVppose that ABC be a triangle: & let one of the sides thereof, namely, BC be produced unto the point D. Then I say, that the outward angle ACD, is greater than any one of the two inward and opposite angles, that is, than the angle CBA, 
Construction.  or then the angle BAC. Divide the line AC (by the 10. proposition) into two equal parts, in the point E. And draw a line from the point B to the point E. And (by the 2. petition) extend BE to the point F. And (by the 2. proposition) unto the line BE put an equal line EF. And (by the first petition) draw a line from F to C: and (by the 2. petition) extend the line AC to the point G. 
Demonstration.  Now forasmuch as the line AE, is equal to the line EC, and BE is equal to EF. therefore these two sides AE and EB, are equal to these two sides CE and EF, the one to the other: and the angle AEB, is (by the 15. proposition) equal to the angle FEC, for they are head angles: wherefore (by the 4. proposition) the base AB is equal to the base FC: And the triangle ABE is equal to the triangle FEC: and the other angles remaining are equal to the other angles remaining, the one to the other, under which are subtended equal sides. Wherefore the angle BAE is equal to the angle ECF. But the angle ECD is greater than the angle ECF. Wherefore the angle ACD, is greater than the angle BAC. In like sort also if the line BC be divided into two equal parts, may it be pro●ed, that the angle BCG, that is, the angle ACD, is greater than the angle ABC. Whensoever therefore in any triangle, the line of one side is drawn forth in length: the outward angle shallbe greater than any one of the two inward and opposite angles: which was required to be demonstrated.  

another demonstration after Pelitarius.  
Here is to be noted, that when the side of a triangle is drawn forth, the angle of the triangle which is next the outward angle, is called an angle in order unto it: and the other two angles of the triangle are called opposite angles unto it.  

another corollary ●ollowing also of the same.  By this Proposition also may this be demonstrated, that if a right line falling upon two right lines, do make the outward angle equal to the inward and opposite angle, those right lines shall not make a triangle, neither shall they concur. For otherwise one & the self same angle should be both greater, and also equal: which is impossible. As for example.    

The 10. Theorem. The 17. Proposition. In every triangle, two angles, which two soever be taken, are less than two right angles.  
SVppose that ABC be a triangle. Then I say that two angles of the said triangle ABC, which two angles soever be taken, are less than two right angles Extend (by the 2. petition) the line BC, 
Construction.  to the point D. 
Demonstration.  And forasmuch as (by the proposition going before) the outward angle of the triangle ABC, namely, the angle ACD is greater than the inward and opposite angle ABC: put the angle ACB common to them both: wherefore the angles ACD and ACB are greater than the angles ABC and BCA. But (by the 13 proposition) the angles ACD and ACB are equal to two right angles. Wherefore the angles ABC and BCA are less than two right angles. In like sort also may we prove, that the angles BAC and ACB are less than two right angles● and also that the angles CAB & ABC are less than two right angles. Wherefore in every triangle, two angles, which two soever be taken, are less than two right angles: which was required to be proved.  
This may also be demonstrated without the help of the former proposition, by the converse of the fifth petition, and by the 13. proposition as you saw was done in the former after Politarius.  
It may also be demonstrated without producing any of the sides of the triangle, a●ter this manner.  

A corollary following this Proposition.  By this proposition also may be proved this corollary, that from one and the self same point to one and the self same right line, can not be drawn two perpendicular lines.   

The 11. Theorem. The 18. Proposition. In every triangle, to the greater side is subtended the greater angle.  
SVppose that ABC be a triangle, having the side AC greater than the side AB. Then I say that the angle ABC is greater than the angle BCA. 
Construction.  For forasmuch as AC is greater than AB, put (by the 3. proposition) unto AB an equal line AD. And (by the first petition) draw a line from the point B to the point D. 
Demonstration●  And forasmuch as the outward angle of the triangle DBC, namely, the angle ADB is greater than the inward and opposite angle DCB (by the 16. proposition,) but (by the 5. proposition) the angle ADB is equal to the angle ABD, for the side AB is equal to the side AD: therefore the angle ABD is greater than the angle AC●. Wherefore the angle ABC is much greater than the angle ACB. Wherefore in every triangle, to the greater side is subtended the greater angle: which was required to be proved.  
Note that that which is here spoken in this propositi●on, 
That which is spoken in this Proposition is to be understanded in one and the self same triangle.  is to be understanded in one and the self same triangle. For it is possible that one and the self same angle may be subtended of a greater line, and of a less line: and one and the self same right line may subtend a greater angle, and a less angle. As for example.  
Again suppose that ABC be an Isosceles triangle. And let BC be less then either of the lines BA and AC. And upon BC describe (by the first) an equilater triangle BCD. And draw a line from A to D, and produce it to the point E. And forasmuch as in the triangle ABD, the outward angle BDE, is greater ●hen the inward & opposite angle BAD (by the 16. proposition) And by the same in the triangle ACD, the outward angle CDE, is greater than the inward & opposite angle CAD ● therefore the whole angle BDC is greater than the whole angle BAC. And one and the self same right line subtendeth both these angles, namely, the greater angle & the less. And it is also proved, that greater right lines & less subtende one and the self same angle. But in one and the self same triangle one right line subtendeth one angle, and the great right line ever subtendeth the great angle, and the less the less, as it was proved in the proposition.   

The 12. Theorem. The 19 Proposition. In every triangle, under the greater angle is subtended the greater side.  
SVppose that ABC be a triangle, ●auyng the angle ABC greater than the angle BCA. Then I say that the side AC is greater than the side AB. For if no●, 
Demonstration lea●in● to an impossibility.  the the side AC is either equal to the side AB, or else it is less than it● The side AC is not equal to the side AB, for then (by the 5● proposition) the angle ABC should be equal to the angle ACB: but (by supposition) it is not. Wherefore the side AC is not equal to the side AB. And the side AC can not be less than the side AB, for then the angle ABC should be less than the angle ACB (by the proposition next going before). But (by supposition it is not) Wherefore the side AC is not less than the side AB. Wherefore the side AC is greater than the side AB. Wherefore in every triangle, under the greater angle is subtended the greater side: which was required to be demonstrated.  

This proposition 〈◊〉 the converse the former.  This proposition is the converse of the proposition next going before. Wherefore as you see, that which was the conclusion in the former, is in this the supposition, or thing given: and that which was there the thing given, is here the thing required or conclusion. And it is proved by an argument leading to an impossibility, as commonly all converses are.  
P●oclus demonstrateth this proposition after an other way: but first he putteth this 
* An Assumpt is 〈…〉 of necessity in the help of a demonstration, the certainty whereof is not so pl●i●e, and therefore ●e●e●h is self first to be demonstrated. An assumpt put by Pr●i●s●er the demonstration of this Proposition.  Assumpt following.  
If an angle of a triangle be divided into two equal parts, and if the line which divideth it being drawn to the base, do divide the same into two unequal parts: the sides which contain that angle shallbe unequal, and that shallbe the greater side, which falleth on the grater side of the base, and that the less which falleth on the less side of the base.  
This assumpt being put, this Proposition is of Proclus thus demonstrated.   

The 13. Theorem. The 20. Proposition. In every triangle two sides, which two sides soever be taken, are greater than the side remaining.  
SVuppose that ABC be a triangle. Then I say that two sides of the triangle ABC, which two sides soever be taken, are greater● then the side remaining that is, the sides BA and AC are greater than the side BC: and the sides AB and BC than the side AC: and the sides AC and BC than the side BA. 
Construction.  Produce (by the 2. petition) the line BA to the point D. And (by the third proposition) unto the line AC put an equal line AD. and draw● a line from the point D to the point C. 
Demonstration.  And forasmuch as the line DA is equal to the line AC, therefore (by the 5. proposition) the angle ADC, is equal to the angle ACD. But the angle BCD is greater than the angle ACD, therefore the angle BCD is greater than the angle ADC. And forasmuch as DCB is a triangle, having the angle BCD greater than the angle ADC, but (by the 18. proposition) under the greater angle is subtended the greater side: wherefore DB is greater than BC. But the line DB is equal to the lines AB and AC (for the line AD is equal to the line AC) wherefore the sides BA and AC, are greater than the side BC. And in like ●orte may we prove, that the sides AB and BC are greater than the side AC: & that the sides BC and CA are greater than the side AB. Wherefore in every triangle two sides, which two sides soever be taken, are greater than the side remaining which was required to be demonstrated.  
This Proposition may also be demonstrated without producing any of the sides, a●ter this manner.  
The same may yet also be demonstrated an other way.  
This proposition may yet moreover be demonstrated by an argument leading to an absurdity, and that after this manner.  

A brief demonstration by the definition of a right line.  A man may also more briefly demonstrate this proposition by Campanus definition of a right line, which as we have before declared is thus: A right line is the shortest extension or drawght that is or may be from one point to another. Wherefore any one side of a triangle, for that it is a right line drawn from some one point to some other one point, is of necessity shorter than the other two sides drawn from and to the same points.  

Not all things manifest to the seize, are strait way manifest to reason and understanding.  Epicurus and such as followed him derided this proposition, not counting it worthy to be added in the number of propositions of Geometry for the easiness thereof, for that it is manifest even to the sense. But not all things manifest to sense, are strait ways manifest to reason and understanding. It pertaineth to one that is a teacher of sciences, by proof and demonstration to tender a certain and undoubted reason, why it so appeareth to the sense● and in that only consisteth science.   

The 14. Theorem. The 21. Proposition. If from the ends of one of the sides of a triangle, be drawn to any point within the said triangle two right lines. those right lines so drawn, shallbe less then the two other sides of the triangle, but shall contain the greater angle.  
SVppose that ABC be a triangle: and from the ends of the side BC, namely, from the points B and C, let there be drawn within the triangle two right lines BD and CD to the point D. Then I say, that the lines BD and CD are less than the other sides of the triangle, namely, than the sides BA and AC: and that the angle which they contain, namely, BDC, is greater than the angle BAC. Extend (by the second petition) the line BD to the point E. 
Demonstration.  And forasmuch as (by the 20. proposition) in every triangle the two sides are greater than the side remaining, therefore the two sides of the triangle ABE, namely, the sides AB and AE, are greater than the side EB. Put the line EC common to them both. Wherefore the lines BA and AC, are greater than the lines BE and EC● Again forasmuch as (by the same) in the triangle CED, the two sides CE and ED, are greater than the side DC, put the line DB common to them both● wherefore the lines CE and ED, are greater than the lines CD and DB. But it is proved that the lines BA and AC, are greater than the lines BE and EC. Wherefore the lines BA and AC, are much greater than the lines BD and DC. Again forasmuch as (by the 16. proposition) in every triangle, the outward angle is greater than the inward and opposite angle, therefore the outward angle of the triangle CDE, namely, BDC, is greater than the angle CED. Wherefore also (by the same) the outward angle of the triangle ABE, namely, the angle CEB is greater than the angle BAC. But it is proved, that the angle BDC is greater than the angle CEB. Wherefore the angle BDC is much greater than the angle BAC. Wherefore if from the ends of one of the sides of a triangle, be drawn to any point within the said triangle two right lines: those right lines so drawn shallbe less than the two other sides of the triangle, but shall contain the greater angle: which was required to be demonstrated.  
In this proposition is expressed, that the two right lines drawn within the triangle, have their beginning at the extremes of the side of the triangle. For from the one extreme of the side of the triangle, and from some one point of the same side, may be drawn two right lines within the triangle, which shall be longer than the two outward lines: which is wonderful and seemeth strange, that two right lines drawn upon a part of a line, should be greater than two right lines drawn upon the whole line. And again it is possible from the one extreme of the side of a triangle, and from some one point of the same side to draw two right lines within the triangle which shall contain an angle less than the angle contained under the two outward lines.  
As touching the first part.   

The 8. Problem. The 22. Proposition. Of three right lines, which are equal to three right lines given, to make a triangle. But it behoveth two of those lines, which two soever be taken, to be greater than the third. For that in every triangle two sides, which two sides soever be taken, are greater than the side remaining.  
SVppose that the three right lines given be A, B, C: of which let two of them, which two soever be taken, be greater than the third, that is, let the lines A, B, be greater than the line C, and the lines A, C, then the line B, and the lines B, C, then the line A. It is required of three right lines equal to the right lines A, B, C, to make a triangle. 
Construction.  Take a right line having an appointed end on the side D, and being infinite on the side E. And (by the 3. proposition) put unto the line A, an equal line DF, and put unto the line B, an equal line FG, and unto the line C, an equal line GH. And making the centre F, and the space DF, describe (by the 3. petition) a circle DKL. Again making the centre G, and the space G H, describe (by the same) a circle HKL; and let the point of the intersection of the said circles be K, and (by the first petition) draw a right line from the point K to the point F, & an other from the point K to the point G. Then I say, that of three right lines equal to the lines A, B, C, is made a triangle KFG. 
Demonstration.  For forasmuch as the point F is the centre of the circle DKL. therefore (by the 15. definition) the line FD is equal to the line FK. But the line A is equal to the line FD Wherefore (by the first common sentence) the line FK is equal to the line A. Again forasmuch as the point G, is the centre of the circle LKH, therefore (by the same definition) the line GK is equal to the line GH● But the line C is equal to the line GH: wherefore (by the first common sentence) the line KG is equal to the line C. But the line FG is by supposition equal to the line B wherefore these three right lines GF, FK, and KG, are equal to these three right lines A, B, C. Wherefore of three right lines, that is, KF, FG, and GK, which are equal to the three right lines given, that is to A, B, C, is made a triangle KFG: which was required to be done.  

another construction, and demonstration after Flussates.  
Suppose that the three right lines be A, B, C. And unto some one of them, namely, 
another construction and demonstration after Flussates.  to C, put an equal line DE, and (by the second proposition) from the point E, draw the line EG, equal to the line B: and (by the same) unto the point D put the line DH equal to the line A. And making the centre the point E, & the space EG, describe a circle FG: likewise making the centre the point D, and the space DH, describe an other circle HF: which circles let cut the one the other in the point F. And draw these lines DF and EF. Then I say that DFE is a triangle described of 3. right lines equals to the right lines A, B, C. For forasmuch as the line DH is equal to the right line A, the line DF, shall also be equal to the same right line A. (For that the lines DH and DF, are drawn from the centre to the circumference). Likewise forasmuch as EG is equal to EF (by the 15. definition) and the right line B is equal to the same right line EG: therefore the right line EF is equal to the right line B: but the right line DE, was put to be equal to the right line C. Wherefore of three right lines ED, DF, and FE, which are equal to three right lines given, A, B, C, is described a triangle: which was required to be done.  

Instances in this Problem.  In this proposition the adversary peradventure will cavil, that the circles shall not cut the one the other (which thing Euclid putteth them to do) But now if they cut not the one the other, either they touch the one the other, or they are dista●nte the one from the other. First if it be possible let them touch the one the other: as in the figure here put (the construction whereof answereth to the construction of Euclid).  

First instance.  And forasmuch as F is the centre of the circle DK, therefore the line DF is equal to the line FN. And forasmuch as the point G, is the centre of the circle HL, therefore the line HG, is equal to the line GM. Wherefore these two lines DF, and GH, are equal to one line, namely, to FG. But they were put to be greater than it: for the lines DF, FG, and GH, were put to be equal to the lines A, B, C, every two of which are supposed to be greater than the third: wherefore they are both greater, and also equal, which is impossible. 
Second instance  Again if it be possible, let the circles be distant the one from the other, as are the circles DK and HL. And forasmuch as F is the centre of the circle DK, therefore the line DF is equal to the line FN. And forasmuch as G is the centre of the circle LH, therefore the line HG is equal to the line GM: wherefore the whole line FG is greater than the two lines DF, and GH, (for the line FG, exceedeth the lines DF, and GH, by the line NM) but it was supposed that the 〈◊〉 ● DF and HG 〈…〉 the● the 〈◊〉 FG: a● also i● 〈◊〉 supposed that the lines A and C, were 〈…〉 line B (for th● 〈◊〉 D● is put to be ●qu●ll to the line A, and the line F● to the line B ● and the line HG to the line G ●) Wh●●●fo●● they are both greater and also equal: which is impossible. Wherefore the circles neither touch the one the other, nor are distant the one from the other. Wherefore of necessity they cut the one the other: which was required to be proved.    

The 9 Problem. The 23. Proposition. Upon a right line given, and to a point in it given: to make a rectiline angle equal to a rectiline angle given.  
SVppose that the right line given be AB, & let the point in it given be A. And let also the rectiline angle given be DCH. It is required upon the right line given AB, and to the point in it given A, to make a rectiline angle equal to the rectiline angle given DCH. 
Construction.  Take in either of the lines CD and CH a point at all adventures, & let the same be D and E. And (by th● first petition) draw a right line from D to E. And of three right lines, AF, FG and GA, which let be equal to the three right lines given, that is, to CD, DE, and EC, make (by the proposition going before) a triangle, and let the same be AFG● so that let the line CD be equal to the line AF, and the line CE to the line AG, and moreover the line DE to the line FG. 
Demonstration.  And forasmuch as these two lines DC and CE are equal to these two lines FA and AG, the one to the other, and the base DE is equal to the base FG: therefore (by the 8. proposition) the angle DCE is equal to the angle FAG. Wherefore upon the right line given AB, and to the poi●● i● i● given namely A, is made a rectiline angle FAG, equal to the rectiline angle given DCH: which was required to be done.  

another construction and demonstration after Proclus.  
Suppose that the right line given be AB: 
another construction and demonstration after Proclus.  & let the point in it given be A, & let the rectiline angle given be CDE. It is required upon the right line given AB, & to the point in it given A, to make a rectiline angle equal to the rectiline angle given CDE. Draw a line from C to E. And produce the line AB on either side to the points F and G. And unto the line CD, put the line FA equal, & unto the line DE let the line AB be equal, & unto the line EC put the line BG equal. And making the centre the point A, & the space AF, describe a circle KF. And again making the centre the point B and th● space BG describe an other circle ●L: which shall of necessity cut the one the other, as we have be●ore proved. Let them cut the one the other in the points M & N. And draw these right lin●s AN, AM, BN, and BM. And forasmuch as FA is equal to AM● and also to AN (by the definition of a circle) but CD is equal to FA, wherefore the lines AM and AN are ech● equal to the line DC. Again forasmuch as BG, is equal to BM, and to BN, and BG is equal to CE: therefore either of these lines BM and BN is equal to the line CE. But the line BA is equal to the line DE. Wherefore these two lines BA & AM, are equal to these two lines D E and DC, the one to the other, and the base BM is equal to the base CE. Wherefore (by the 8. proposition) the angle MAB, is equal to the angle at the point D. And by the same reason the angle NAB, is equal to the same angle at the point D. Wherefore upon the right line given AB, and to the point in it given A, is described a rectiline angle on either side of the line AB: namely, on one side the rectiline angle NAB, and on the other side the rectiline angle MAB, either of which is equal to the rectiline angle given CDE: which was required to be done.   

another construction also, and demonstration after Pelitar●us.  
And if the perpendicular line chance to fall without the angle given, namely, if the angle given be an acute angle, the self same manner of demonstration will serve: but only that in stead of the second common sentence, must be used the 3. common sentence.  
Appollonius putteth another construction & demonstration of this proposition: which (though the demonstration thereof depend of propositions put in the third book, yet for that the construction is very good for him that will readily, and mechanically, without demonstration, describe upon a line given, and to a point in it given, a rectiline angle equal to a rectiline angle given) I thought not amiss here to place it. And it is thus.  

Oenopides the first inventor of this proposition.  Oenopides was the first inventor of this proposition as witnesseth Eudemius.    

The 15. Theorem The 24. Proposition. If two triangles have two sides of the one equal to two sides of the other, each to his correspondent side, and if the angle contained under the equal sides of the one, be greater than the angle contained under the equal sides of the other: the base also of the same, shallbe greater than the base of the other.  
SVppose that there be two triangles ABC, and DEF, having two sides of the one, that is, AB, and AC, equal to two sides of the other, that is, to DE, and DF, each to his correspondent side: that is, the side AB, to the side DE, and the side AC to the side DF: and suppose that the angle BAC be greater than the angle EDF. Then I say that the base BC, is greater than the base EF. For forasmuch as the angle BAC is greater than the angle EDF, 
Construction.  make (by the 23. proposition) upon the right line DE, and to the point in it given D, an angle edge equal to the angle given BAC. And to one of these lines, that is, either to AC, or DF, put an equal line DG. And (by the first petition) draw a right line from the point G, to the point E, and an other from the point F, 
Demonstration.  to the point G. And forasmuch as the line AB is equal to the line DE, and the line AC to the line DG, the one to the other, and the angle BAC is (by construction) equal to the angle edge, therefore (by the 4. proposition) the base BC, is equal to the base EG. Again for as much as the line DG is equal to the line DF, ther● (by the 5. proposition) the angle DGF, is equal to the angle DFG. Wherefore the angle DFG is greater than the angle EGF. Wherefore the angle EFG is much greater than the angle EGF. And forasmuch as EFG is a triangle, having the angle EFG greater than the angle EGF, and (by the 18. pr●position) under the greater angle is subtended the greater side, therefore the side EG is greater than the side EF. But the side EG is equal to the side BC: wherefore the side BC is greater than the side EF. If therefore two triangles have two sides of the one equal to two sides of the other, each to his correspondent side, and if the angle contained under the equal sides of the one, be greater than the angle contained under the equal sides of the other: the base also of the same shallbe greater than the base of the other: which was required to be proved.  
In this Theorem may be three cases. 
Three cases in this pr●position.  For the angle edge, being put equal to the angle BAC, and the line DG, being put equal to the line AC, and a line being drawn from E to G, the line EG shall either fall above the line GF, or upon it, or under it. Euclides demonstration serveth, 
The first cas●. 2.  when the line GE falleth above the line GF, as we have, before manifestly seen.  
But now let the line EG, 
Third case.  fall under the line E F, as in the figure here put. And forasmuch as these two lines AB, and AC, are equal to these two lines DE and DG, the one to the other, and they contain equal angles, therefore (by the 4. proposition) the base BC, is equal to the base EG. And forasmuch as within the triangle DEG, the two linnes DF and FE, are set upon the side DE: therefore (by the 21. proposition) the lines DF and F● are less than the outward lines DG and GE: but the line DG is equal to the line DF. Wherefore the line GE is greater than the line FE. But GE is equal to BC. Wherefore the line BC is greater the the line EF. Which was required to be proved.  
It may peradventure seme● that Euclid should here in this proposition have proved, that not only the bases of the triangles are unequal, but also that the areas of the same are unequal: for so in the fourth proposition, after he had proved the base to be equal, he proved also the areas to be equal. But hereto may be answered, 
Why Euclid etc.  that in equal angles and bases, and unequal angles and bases, the consideration is not like. For the angles and bases being equal, the triangles also shall of necessity be equal, but the angles and bases being unequal, the areas shall not of necessity be equal. For the triangles may both be equal and unequal: and that may be the greater, which hath the greater angle, and the greater base, and it may also be the less. And for that cause Euclid made no mention of the comparison of the triangles. Whereof this also might be a cause, for that to the demonstration thereof are required certain Propositions concerning parallel lines, which we are not as yet come unto. 
After the ●● etc.  Howbeit after the 37● proposition of his book you shall found the comparison of the areas of triangles, which have their sides equal, and their bases and angles at the top unequal.   

The 16. Theorem. The 25. Proposition. If two triangles have two sides of the one equal to two sides of the other, each to his correspondent side, and if the base of the one be greater than the base of the other: the angle also of the same contained under the equal right lines● shall be greater than the angle of the other.  
SVppose that there be two triangles, A, B, C, and DEF, having two sides of tb'one, that is, AB, and AC, equal to two sides of the other, that is to DE, and DF, each to his correspondent side, namely, the side AB to the side DF, and the side AC to the side DF. But let the base BC be greater than the base EF. Then I say, they the angle BAC is greater than the angle EDF. For if not, 
Demonstration leading to an absurdity.  then is it either equal unto it, or less than it. But the angle BAC is not equal to the angle EDF: for if it were equal, the base also BC should (by the 4. proposition) be equal to the base EF: but by supposition it is not. Wherefore the angle BAC is not equal to the angle EDF. Neither also is the angle BAC less than the angle EDF: for than should the base BC be less than the base EF (by the former proposition) But by supposition it is not. Wherefore the angle BAC is not less than the angle EDF. And it is already proved, that it is not equal unto it: wherefore the a●gle BAC is greater than the angle EDF. If therefore two triangles have two sides of the one equal to two sides of the other, each to his correspondent side, & if the base of the one be greater than the base of the other, the angle also of the same contained under the equal right lines shall be greater than the angle of the other: which was required to be proved.  
This proposition is plain opposite to the eight, & is the converse of the four and twenty which went before. 
All converses are commonly indirectly demonstrated.  And it is proved (as commonly all converses are) by a reason leading to an absurdity. But it may after Menelaus Alexandrinus be demonstrated directly, 
another demonstration after Menelaus Alexandrinus.  after this manner.  
Suppose that there be two triangles ABC & DEF: having the two sides AB and AC equal to the two sides DE and DF, the one to the other: and le● the base BC be greater than the base EF. Then ● say that the angle at the point A, is greater than the angle at the point D. For from the base BC 〈◊〉 of (by the third) a line eqnall to the base EF, and let the same be BG. And upon the line GB and to the point B put (by the 23, proposition) an angle equal to the angle DEF: which let be GBH: and let the line ●H be equal to the line DE. And draw a line from H to G, and produce it beyond the point G: which being produce shall fall either upon the angle A, 
divers cases in this demonstration.  or upon the line AB, or upon the line AC. First let it fall upon the angle A. 
First case.  And forasmuch as these two lines BG and BH are equal to these two lines EF and ED, the one to the other, and they contain equal angles (by construction) namely, the angles GBH and DEF: therefore (by the 4. proposition) the base GH is equal to the DF, and the angle BHG to the angle EDF. Again forasmuch as the line BH is equal to the line BA (for the line AB is supposed to be equal to the line DE, unto which line the line BH is put equal) therefore (by the 5. proposition) the angle BHA is equal to the angle BAH: wherefore also the angle EDF is equal to the angle BAH. But the angle BAC is greater than the angle BAH: wherefore also the angle BAC is greater than the angle EDF. 
Second case.   
Hero Mechanicus also demonstrateth it an other way, and that by a direct demonstration.   

The 17. Theorem. The 26. Proposition. If two triangles have two angles of the one equal to two angles of the other, each to his correspondent angle, and have also one side of the one equal to one side of the other, either that side which lieth between the equal angles, or that which is subtended under one of the equal angles: the other sides also of the one, shallbe equal to the other sides of the other, each to his correspondent side, and the other angle of the one shallbe equal to the other angle of the other.  
SVppose that there be two triangles AB C, and DEF, having two angles of the one, that is, the angles ABC, and BCA, equal to two angles of the other, that is, to the angles DEF, and EFD, each to his correspondent angle, that is, the angle ABC, to the angle DEF, and the angle BCA to the angle EFD, and one side of the one equal to one side of the other, first that side which lieth between the equal angles, that is, the side BC, to the EF. Then I say that the other sides also of the one shallbe equal to the other sides of the other, each to his correspondent side, that is, that side AB, to the side DE, and the side AC, to the side DF, and the other angle of the one, to the other angle of the other, that is, the angle BAC to the angle EDF. For if the side AB be not equal to the side DE, the one of them is greater. Let the side AB be greater: and (by the 3. proposition) unto the line DE, put an equal line GB, and draw a right line from the point G, to the point C. Now forasmuch as the line GB, 
Demonstration leading to an absurditis.  is equal to the line DE, and the line BC to the line EF, therefore these two lines GB and BC, are equal to these two lines DE and EF, the one to the other, and the angle GBC is (by supposition) equal to the angle DEF. Wherefore (by the 4. proposytion) the base GC is equal to the base DF, and the triangle GCB is equal to the triangle DEF, and the angles remaining are equal to the angles remaining under which are subtended equal sides. Wherefore the angle GCB is equal to the angle DFE. But the angle DFE is supposed to be equal to the angle BCA. Wherefore (by the first common sentence) the angle BCG is equal to the angle BCA, the less angle to the greater: which is impossible. Wherefore the line AB is not unequal to the line DE. Wherefore it is equal And the the line BC is equal to the line EF: now therefore there are two sides AB and BC equal to two sides DE and EF, the one to the other, and the angle ABC, is equal to the angle DEF. Wherefore (by the 4. proposition) the base AC is equal to the base DF, and the angle remaining BAC is equal to the angle remaining EDF.  
Again suppose that the sides subtending the equal angles be equal the one to the other, let the side I say AB be equal to the side DE. Then again I say, that the other sides of the one are equal to the other sides of the other, each to his correspondent side, that is the side AC to the side DF, and the side BC to the side EF: and moreover the angle remaining, namely, BAC, is equal to the angle remaining, that is, to the angle EDF. For if the side BC be not equal to the side EF, the one of them is greater: let the side BC, if it be possible, be greater. And (by the third proposytion) unto the line EF, put an equal line BH, and draw a right line from the point A to the point H. And forasmuch as the line BH is equal to the line EF, and the line AB to the line DE, therefore these two sides AB and BH, are equal to these two sides DE and EF, the one to the other, and they contain equal angles. Wherefore (by the 4. proposition) the base AH is equal to the base DF, and the triangle ABH, is equal to the triangle DEF, and the angles remaining are equal to the angle● remaining, under which are subtended equal sides. Wherefore the angle BHA is equal to the angle EFD. But the angle EFD is equal to the angle BCA. Wherefore the angle BHA is equal to the angle BCA. Wherefore the outward angle of the triangle AHC, namely, the angle BHA, is equal to the inward and opposite angle, namely, to the angle HCA, which (by the 16 proposition) is impossible. Wherefore the side EF is not unequal to the side BC, wherefore it is equal. And the side AB is equal to the side DE: wherefore these two sides AB and BC, are equal to these two sides DE and EF, the one to the other, and they contain equal angles: Wherefore (by the 4. proposition) the base AC is equal to the base DF: and the triangle ABC, is equal to the triangle DEF, and the angle remaining, namely, the angle BAC is equal to the angle remaining, that is, to the angle EDF. If therefore two triangles have two angles of the one equal to two angles of the other, each to his correspondent angle, and have also one side of the one equal to o●e side of the other, either that side which lieth between the equal angles, or that which is subtended under one of the equal angles: the other sides also of the one shallbe equal to the other sides of the other, each to his correspondent side, and the other angle of the one shallbe equal to the other angle of the other: which was required to be proved.  
Whereas in this proposition it is said, that triangles are equal, which having two angles of the one equal to two angles of the other, the one to the other, have also one side of the one equal to one side of the other, either that side which lieth between the equal angles, or that side which subtendeth one of the equal angles: this is to be noted that without that caution touching the equal side, the proposition shall not always be true. As for example.  
The reason whereof is, for that the equal side in one triangle, subtendeth one of the equal angles, and in the other lieth between the equal angles. So that you see that it is of necessity that the equal side do in both triangles, either subtend one of the equal angles, or lie between the equal angles.  
Of this proposition was Thales Milesius the inventor, 
Thales Milesius the inuent●r of this proposition.  as witnesseth Eudemus in his book of Geometrical enarrations.   

The 18. Theorem. The 27. Proposition. If a right line falling upon two right lines, do make the alternate angles equal the one to the other: those two right lines are parallels the one to the other.  
SVppose that the right line EF falling upon these two right lines AB and CD, do make the alternate angles, namely, the angles AEF & EFD equal the one to the other. Then I say that AB is a parallel line to CD. For if not, than these lines produced shall meet together, either on the side of B and D, or on the side of A & C. 
Demonstration ●eading to an absurdity.  Let them be produced therefore, and let them meet if it be possible on the side of B and D, in the point G. Wherefore in the triangle GEF, the outward angle AEF is equal to the inward and opposite angle EFG, which (by the 16. proposition) is impossible. Wherefore the lines AB and CD being produced on the side of B and D, shall not meet. In like sort also may it be proved that they shall not meet on the side of A and C. But lines which being produced on no side meet together, are parrallell lines (by the last definition:) wherefore AB is a parallel line to CD. If therefore a right line falling upon two right lines, do make the alternate angles equal the one to the other: those two right lines are parallels the one to the other: which was required to be demonstrated.  

 This word alternate is of Euclid in divers places diversly taken: sometimes for a kind of situation in place, and sometime for an order in proportion, in which signification he useth it in the u book, and in his books of numbers. And in the first signification he useth it here in this place, and generally in all his other books, 
 having to do with lines & figures. And those two angles he calleth alternate, which being both contained within two parallel or equidistant lines are neither angles in order, nor are on the one and self same side, but are separated the one from the other by the line which falleth on the two lines: the one angle being above, and the other beneath.   

The 19 Theorem. The 28. Proposition. If a right line falling upon two right lines, make the outward angle equal to the inward and opposite angle on one and the same side, or the inward angles on one and the same side, equal to two right angles: those two right lines shall be parallels the one to the other.  
SVppose that the right line EF, falling upon these two right lines AB and CD, do make the outward angle EGB equal to the inward and opposite angle GHD, or do make the inward angles on one and the same side, that is, the angles BGH and GHD equal to two right angles. Then I say that the line AB is a parallel line to the line CD For forasmuch as the angle EGB is (by supposition) equal to the angle GHD, 
Demonstration  and the angle EGB is (by the 15. proposition) equal to the angle AGH: therefore the angle AGH is equal to the angle GHD: and they are alternate angles. Wherefore (by the 27. proposition) AB is a parallel line to CD.  
Again forasmuch as the angles BGH and GHD are (by supposition) equal to two right angles, & (by the 13. proposition) the angles AGH and BGH, are also equal to two right angles, wherefore the angles AGH and BGH, are equal to the angles BGH and GHD: take away the angle BGH which is common to them both Wherefore the angle remaining, namely, AGH is equal to the angle remaining, namely, to GHD. And they are alternate angles. Wherefore (by the former proposition) AB is a parallel line to CD. If therefore a right line falling upon two right lines, do make the outward angle equal to the inward and opposite angle on one and the same side, or the inward angles on one and the same side, equal to two right angles, those two right lines shall be parallels the one to the other: which was required to be proved.  
Ptolomeus demonstrateth the second part of this proposition, namely, that the two inward angles on one and the same side being equal, the right lines are parellels, after this manner. 
another demonstration of the second part of this proposition after Ptolemies    

The 20. Theorem. The 29. Proposition. A right line line falling upon two parallel right lines: maketh the alternate angles equal the one to the other: and also the outward angle equal to the inward and opposite angle on one and the same side: and moreover the inward angles on one and the same side equal to two right angles.  
And the angle AGH is (by the 15. proposition) equal to the angle EGB. 
Second part.  Wherefore (by the first common sentence) the angle EGB is equal to the angle GHD.  
Put the angle BGH common to them both: wherefore the angles EGB and BGH, 
Third part.  are equal to the angles BGH and GHD. But the angles EGB and BGH are (by the 13. proposition) equal to two right angles. Wherefore the angles BGH and GHD are also equal to two right angles. If a line therefore do fall upon two parallel right lines: it maketh the alternate angles equal the one to the other: and also the outward angle equal to the inward and opposite angle on one and the same side: and moreover the inward angles on one and the same side equal to two right angles: which was required to be demonstrated.  
This proposition is the converse of the two propositions next going before. 
This proposition is the converse of the two former propositions.  For, that which in either of them is the thing sought, or conclusion, is in this the thing given, or supposition. And contrariwise the things which in them were given or suppositions, are in this proved, and are conclusions.  

Pelitarius after this proposition addeth this witty conclusion.  
If two right lines which cut two parallel lines, 
An addition of Pelitarius.  do between the said parallel lines concur in a point, and make the alternate angles equal, or the outward angle equal to the inward and opposite angle on the same side, or finally the two inward angles on one and the self same side equal to two right angles: those two right lines are drawn directly and make one right line.  
Moreover (by the second part of this 29. proposition) the angle AHD shallbe equal to the angle BLK, 
Second part.  namely, the outward angle to the inward and opposite angle on one and the same side. But the same angle AHD is supposed to be equal to the angle BKF: wherefore the angle BKF is equal to the angle BLK. Which (by the self same 16. proposition) is impossible.  
Lastly forasmuch as the angles BHD and BKF are supposed to be equal to two right angles, 
Third part.  & the angles BHD & BLK are also by the last part of this 29 proposition equal to two right angles, therefore the angle BKF shallbe equal to the angle BLK: which again by the self same 16. proposition is impossible.    

The 21. Theorem The 30. Proposition. Right lines which are parallels to one and the self same right line: are also parallel lines the one to the other.  
SVppose that these right lines AB and CD, be parallel lines to the right line EF. Then I say, that the line AB is a parallel line to CD. 
Demonstration.  Let there fall upon these three lines a right line GHK. And forasmuch as the right line GHK falleth upon these parallel right lines AB and EF, therefore (by the proposition going before) the angle AGH is equal to the angle GHF. Again forasmuch as the right line GK falleth upon these parallel right lines EF and CD, therefore (by the same) the angle GHF is equal to the angle GKD. Now than it proved that the angle AGH is equal to the angle GHF, and that the angle GKD is equal to the angle GHF. Wherefore the angle AGK is equal to the angle GKD. And they are alternate angles: wherefore AB is a parallel line to CD. Right lines therefore which are parallels to one and the self same right line, are also parallel lines the one to the other: which was required to be proved.  
Euclid in the demonstration of this proposition, setteth the two parallel lines which are compared to one, in the extremes, and the parallel to whom they are compared, he placeth in the middle, for the easier demonstration. It may also be proved even by a principle only. For if they should concur on any one side, they should concur also with the middle line, and so should they not be parallels unto it, which yet they are supposed to be.  

another case in 〈◊〉 Problems.  But if you will altar their position and placing, and set that line to which you will compare the other two lines, above, or beneath: you may use the same demonstration which you had before. As for example.  

An objection.  But here if a man will object that the lines EK and KF, are parallels unto the line CD, 
Answer.  and therefore are parallels the one to the other. We will answer that the lines EK and KF are parts of one parallel line, and are not two parallel lines. 
Parallel lines are understanded to be produced infinitely.  For parallel lines are understanded to be produced infinitely But EK being produced falleth upon KF. Wherefor● it is one and the self same with it, and not an other. Wherefore all the parts of a parallel line are parallels, both to the right line unto which the whole parallel line is a parallel, and also to all the parts of the same right line. As the line EK is a parallel unto HD, and the line KE to the line CH. For if they be produced infinitely, they will never concur.  
Howbeit there are some which like not, that two distinct parellel lines, should be taken and counted for one parallel line: for that the continual quantity, namely, the line is cut asunder, and cesseth to be one. Wherefore they say, that there aught to be two distinct parallel lines compared to one. And therefore they add to the proposition a correction, in this manner. Two lines being parallels to one line: are either parallels the one the other, or else the one is set directly against the other, so that if they be produced they should make one right line. As for example.   

The 10. Problem. The 31. Proposition. By a point given, to draw unto a right line given, a parallel line.  
SVppose that the point given be A, and let the right line given be BC. It is required by the point given, namely A, to draw vuto the right line BC, a parallel line. 
Construction.  Take in the line BC a point at all adventures, and let the same be D. and (by the first petition) draw a right line from the point A, to the point D. And (by the 23. proposition) upon the right line given AD, and to the point in it given A, make an angle DAE, equal to the angle given ADC. And (by the 14. proposition) put unto the line AE the line AF directly, in such sort that they both make one right line. And forasmuch as the right line AD falling upon the right lines BC and EF, doth make the alternate angles, namely, EAD, and ADC equal the one to the other, therefore (by the 27. proposition) EF is a parallel line to BC. Wherefore by the point given, namely A, is drawn to the right line given BC a parallel line EAF: which was required to be done.  
This proposition is to be understanded of a point given without the line given, and in such sort also, that the same line given being produced, do not fall upon the point given.   

The 22. Theorem. The 32, Proposition. If one of the sides of any triangle be produced: the outward angle that it maketh, is equal to the two inward and opposite angles. And the three inward angles of a triangle are equal to two right angles.  
SVppose that ABC be a triangle, & produce one of the sides thereof namely, CB to the point D. Then I say, that the outward angle ACD is equal to the two inward and opposite angles CAB & ABC: and that the three inward angles of the triangle, that is, the angles ABC, BCA, and CAB are equal to two right angles. 
Construction.  For (by the proposition going before) raise up from the point C, a parallel to the right line AB, and let the same be CE. 
Demonstration.  And forasmuch as AB is a parallel to CE, and upon them falleth the right line AC: therefore the alternate angles BAC and ACE are equal the one to the other. Again forasmuch as AB is a parallel unto CE, and upon them falleth the right line BD, therefore the outward angle ECD is (by the 29. proposition) equal to the inward and opposite angle ABC. And it is proved that the angle ACE is equal to the angle BAC: wherefore the whole outward angle ACD is equal to the two inward and opposite angles, that is, to the angles BAC and ABC. Put the angle ACB common to them both. Wherefore the angles ACD and ACB, are equal to these three angles ABC, BCA, and BAC. But the angles ACD & ACB are equal to two right angles (by the 13. proposition): wherefore the angles ACB, CBA, and CAB are equal to two right angles. If therefore one of the sides of any triangle be produced, the outward angle that it maketh, is equal to the two inward and opposite angles. And the three inward angles of a triangle are equal to two right angles: which was required to be demonstrated.  
Euclid demonstrateth either part of this composed Theorem, by drawing from one angle of the triangle a parallel line to one of the sides of the same triangle, without the triangle. Either part thereof may also be proved without drawing of a parallel line without the triangle, only changing the order of the things required or conclusions. For Euclid first proveth that the outward angle of a triangle (one of his sides being produced) is equal to the two inward and opposite angles: and by that he proveth the second part: namely, that the 3. inward angles of a triangle are equal to two right angles. But here it is contrariwise. For first is proved that the three inward angles of a triangle are equal to two right angles, and by that is proved the other part of the Theorem, namely, that one side of a triangle being produced, the outward angle is equal to the two inward and opposite angles. And that after this manner.  
But the angles ACF and ACE are also (by the 13. proposition) equal to two right angles. Take away the angle ACF which is common, wherefore the angle remaining, namely, the outward angle ACE is equal to the two angles remaining, namely, to the two inward and opposite angles ABC and CAB: which was required to be proved.  
Eudemus affirmeth, that the latter part of this Theorem, The three angles of a triangle are equal to two right angles, was first found out by Pythagoras, 

The latter part of this Theorem first ●ound out by Pythagoras.  
The demonstration thereof after him.   whose demonstration thereof is thus.  

The converse of this proposition is thus.  
If the outward angle of a triangle be equal to the two inward angles opposite against it: one of the sides of the triangles is produced, 
The converse of this proposition.  and the line without the triangle, is drawn directly to the side of the triangle, & maketh one right line with it. And if the three inward angles of a rectiline figure be equal to two right angles, the same rectiline figure is a triangle.  

A corollary.  By the second part of this 29. proposition, namely, three angles of a triangle are equal to two right angles, may easily be known, to how many right angles, the angles within any figure having right lines and many angles are equal. As are figures of four angles, of five angles, of six angles, and so consequently: and infinitely, And this is to be noted, that every right lined figure is resolved into triangle, For that a triangle is the first of all figures. 

Every right lined figure is resolved in triangles.  
A triangle is the first of all figures.  
Into how many triangles a figure may be resolved.   For two lines accomplish no figures Wherefore how many sides the figure hath, into so many triangles may it be resolved, saving two. As if the figure have four sides, it is resolved into two triangles, if it have five sides, into 3. triangles: if 6 sides into 4. triangles, and so consequently, and infinitely. And it is proved that the three angles of every triangle are equal to two right angles. Wherefore if you multiply the number of the triangles, into which the figure is resolved, by two, you shall have the number of right angles, to which the angles of the figure are equal. So the angles of every quadrangled figure are equal to 4. right angles. For it is composed of two triangles. And the angles of a five angled figure are equal to 6. right angles, for it is composed of three triangles, and so forth in like order.  
This thing may also be thus expressed. In any figure of many sides, 
another way to know the number of right angles unto which the angles of any figure are squall.  the number of the angles of the figure doubled, is the number of the right angles to which the angles of the figure are equal, saving four, As for example.  
By that which hath now been declared, 
Another expression of the former Corollary.  it followeth that all the angles of any figure having many sides, taken together, are equal to twice so many right angles, as the figure is in the reaw or order of figures. A triangle is the first figure in order, 
A triangle the first figure in order.  & ● his angles are equal to two right angles, which are twice one. A quadrangle is the second figure in order. 
A quadrangle the second, and so consequently. How the order of figures is gathered.  Wherefore his angles are equal to four right angles, which are twice two. The order of figures is gathered of the sides. For if you take two from the number of the sides of a figure, the number of the sides remaining, is the number of the order of the figure. As if you will know, how many in order is a figure of six sides: from six (which is the number of this sides) take a way two, and there will remain four. Wherefore a figure of six sides is the fourth figure in the order of figures. Then double four, so shall you have eight. Wherefore the angles thereof are equal to eight right angles. And so of all other figures.  

another corollary.  Hereby also it is manifest, that the outward angles of any figure of many sides taken together, are equal to four right angles. For the inward angles together with the outward angles, are equal to twice so many right angles, as there be angles in the figure (by the 13. proposition) But the inward angles are equal to twice so many right angles, as there be angles in the figure, saving fourer as it was before declared. Wherefore the outward angles are always equal to four right angles. As for example.  

another corollary.  It is also manifest, that every pentagon, which is so described, that each side thereof divideth two of the other sides, hath his five angles equal to two right angles.  

another corollary.  By this proposition also it is manifest, that every angle of an equilate triangle is two third parts of a right angle. And that in a triangle of two equal sides having a right angle at the top, either of the the two angles at the base is the half of a right angle. And in a triangle called Sealenum, such a Scalenun (I say) which is made by the drawght of a perpendicular line from any one of the angles of an equilater triangle to the opposite side thereof, one angle is a right angle, an other is two third parts of a right angle, namely, that angle which was also an angle of the equilater triangle, wherefore of necessity the angle remaining is one third part of a right angle. For the three angles of a triangle must be equal to two right angles.  

another Corrallory.  Moreover by this proposition it is manifest, that if there be two triangles, and if two angles of the one be equal to two angles of the other: the angle remaining shall also be equal to the angle remaining. For forasmuch as three angles of any triangle are equal to three angles of any other triangle (for that in each the three angles are equal to two right angles) ● If from each triangle be taken away the two equal angles, the angle remaining shall (by the 3. common sentence) be equal to the angle remaining.  
And here I think it good to show how to divide a right angle into three equal parts, for that the demonstration thereof dependeth of this proposition.    

The 23. Theorem. The 33. Proposition. Two right lines joining together on one and the same side, two equal parallel lines: are also themselves equal the one to the other, and also parallels.  
SVppose that AB and CD be right lines equal, and parallels: and let these two right lines AC and BD join them together, the one on the one side, and the other on the other side. Then I say that the lines AC & BD are both equal, & also parallels. Draw (by the first petition) a right line from the point B to the point C. 
Demonstration.  And forasmuch as AB is a parallel to CD, and upon them falleth the right line BC, therefore the alternate angles ABC and BCD, are equal the one to the other (by the 29. proposition). And forasmuch as the line AB is equal to the line CD, and the line BC is common to them both, therefore these two lines AB and BC, are equal to these two lines BC and CD, and the angle ABC is equal to the angle BCD. Wherefore (by the 4. proproposition) the base BD is equal to the base AC, and the triangle ABC, is equal to the triangle BCD, and the angles remaining are equal to the angles remaining the one to the other, under which are subtended equal sides, wherefore the angle ACB is equal to the angle CBD, and the angle BAC to the angle BDC. And forasmuch as upon these right lines AC and BD falleth the right line BC making the alternate angles, that is the angles ACB and CBD, equal the one to the other, therefore (by the 27. proposition) the line AC is a parallel to the line BD. And it is proved that it is equal unto it. Wherefore two right lines joining together on one and the same side two equal lines which are parallels, are also themselves equal the one to the other, and also parallels: which was required to be proved.   

The 24. Theorem. The 34 Proposition. In parallelograms, the sides and angles which are opposite the one to the other, are equal the one to the other, and their diameter divideth them into two equal parts.  
SVppose that ABCD be a parallelogram and let the diameter ther● of be BC. Then I say that the opposite sides and angles of the parallelogram ACDB are equal the one to the other, and that the diameter BC divideth it into two equal parts. 
Demonstration  For forasmuch as AB is a parallel line unto CD, and upon them falleth a right line BC: therefore (by the 29. proposition) the alternate angles ABC and BCD are equal the one to the other. Again forasmuch as AC is a parallel line to BD, and upon them falleth the right line BC: therefore (by the same) the alternate angles, that is, the angles ACB and CBD are equal the one to the other. Now therefore there are two triangles ABC and BCD, having two angles of the one, namely, the angles ABC and ACB equal to two angles of the other, that is, to the angles BCD and CBD, the one to the other, and one side of the one equal to one side of the other, namely, that side that lieth between the equal angles, which side is common to them both, namely, the side BC. Wherefore (by the 26. proposition) the other sides remaining are equal to the other sides remaining, the one to the other, and the angle remaining is equal to the angle remaining. Wherefore the side AB is equal to the side CD, and the side AC to the side BD, & the angle BAC is equal to the angle BDC. And for as much as the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB: therefore (by the second common sentence) the whole angle ABD is equal to the whole angle ACD. And it is proved that the angle BAC is equal to the angle CDB. Wherefore in parallelograms, the sides and angles which are opposite are equal the one to the other. I say also that the diameter thereof divideth it into two equal parts. For forasmuch as AB is equal to CD, and BC is common to them both, therefore these two AB and BC are equal to these two BC and CD, and the angle ABC is equal to the angle BCD. Wherefore (by the 4. proposition) the base AC is equal to the base BD, and the triangle ABC is equal to the triangle BCD. Wherefore the diameter BC divideth the parallelogram ABCD into two equal parts: which is all that was required to be proved.  
In this Theorems are demonstrated three passions or properties of parallelograms. 
Three passions of parallolegrames demonstrated in this Thoreme.  Namely, that thei● opposite sides are equal: that their opposite angles are equal: and that the diameter divideth the parallelogram into two equal parts. Which is true in all kinds of parallelograms. There are four kinds of parallelograms, 
Four kinds of parallelograms.  a square, a ●igure of one side longer than the other, a Rhombus, or diamond figure, and a Rhom●oides or diamondlike figure. And here is to be noted, that in those parallelograms, all whose angles are right angles (as is a square, and a figure on the one side longer) the diameters do not only divide the figure into two equal parts, but also they are equal the one to the other. As for example.  
But in those parallelograms whose angles are not right angles, as is a Rhombus, and a ●homboides, the diameters be ever unequal. As for example.  
Again. In parallelograms of equal sides, as are a square, and a Rhombus, the diameters do not only divide the figures into two equal parts, but also they divide the angles into two equal parts.  
But in parallogrammes whose sides are not equal, such as are a figure on the one side longer, and a rhomboids it is not so.  

The converse of th●● proposition after Pr●●lus.  The converse of the first and second part of this proposition after Proclus. Is 〈◊〉 figure whatsoever have his opposite sides and angles equal: then is a parallellograms.  


A Corollary taken out of Flussates.  A corollary taken out of Flussates.  
A right line cutting a parallelogram which way soever into two equal parts, shall also divide the diameter thereof into two equal parts.  
For if it be possible let the right line GC divide the parallelogram AEBD into two equal parts, but let it divide the diameter DE into two unequal parts in the point I And l●t the part IE be greater than the part ID. And unto the line ID put the line IO equal (by th3. proposition). 
Demonstration leading to 〈◊〉 absurdity.  And by the point O, draw unto the lines AD and BE a parallel line OF (by the 31. proposition.) Wherefore in the triangles FOI and CDI, two angles of the one are equal to two angles of the other, namely, the angles JOF and IDC (by the 29. proposition), & the angles FIO & CID (by the 15. proposition), & the side ID is equal to the side IO. Wherefore (by the 26. proposition) the triangles are equal. Wherefore the whole triangle EIG is greater than the triangle DIC. And forasmuch as the trapesium GBDC is supposed to be the half of the parrallelograme, and the half of the same parallelogram is the triangle EBD (by this proposition). From the trapesium GBDC and the triangle EBD which are equal, take away the trapesium GBDI which is common to them both, and the residue namely, the triangle DIC shallbe equal to the residue, namely, to the triangle EIG: but it is also less (as hath before been proved): which is impossible. Wherefore a right line dividing a parallelogram into two equal parts, shall not divide the diameter thereof unequally. Wherefore it shall divide it equally: which was required to be proved.   

An addition of Pelitarius.  
Between two right lines being infinite and making an angle given: 
An addition o● Pelitarius.  to place a line equal to a line given, in such sort, that it shall make with one of those lines an angle equal to an other angle given. Now it behoveth that the two angles given be less than two right angles.  
Though this addition of Pelitarius be not so much pertaining to the proposition: yet because it is witty and seemeth somewhat difficult, I thought it good here to anexe it.    

The 25. Theorem. The 35. Proposition. parallelograms consisting upon one and the same base, and in the self same parallel lines, are equal the one to the other.  
SVppose that these parallelograms ABCD and EB CF do consist upon one and the same base, that is, upon BC, and in the self same parallel lines, that is AF, and BC. Then I say, that the parallelogram ABCD is equal to the parallelogram EBCF. 
Demonstration.  For forasmuch as ABCD is a parallelogram, therefore (by the 34. proposition) the side AD, is equal to the side BC, and by the same reason also the side EF is equal to the side BC, wherefore AD is equal to EF: and DE is common to them both. Wherefore the whole line AE is equal to the whole line DF. And the side AB is equal to the side DC: wherefore these two EA and AB are equal to these two FD and DC, the one to the other: and the angle FDC is equal to the angle EAB, namely, the outward angle to the inward angle (by the ●9. proposition): wherefore (by the 4 proposition) the base EB is equal to the base FC, and the triangle EAB is equal to the triangle FDC. Take away the triangle DGE, which is common to them both. Wherefore the residue, namely, the trapesium ABGD is equal to the residue, that is, to the trapesium EGCF. Put the triangle GBC common to them both. Wherefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. Wherefore parallelograms consisting upon one and the same base, and in the self same parallel lines, are equal the one to the other: which was required to be demonstrated.  
parallelograms are said to be in the self same parallel lines, when their bases, and the opposite sides unto them, are one and the self same lines with the parallels.  


Three cases in this proposition.  
The first case.   In this proposition are three cases. For the line BE may cut the line AF, either beyond the point D, or in the point D, or on this side the point D. When it cutteth the line AF beyond the point D the demonstration before put serveth.   

The 26. Theorem. The 36. Proposition. parallelograms consisting upon equal bases, and in the self same parallel lines, are equal the one to the other.  
SVppose that these parallelograms ABCD and EFGH do consist upon equal bases, that is, upon BC and FG, and in the self same parallel lines, that is, AH and BG. Then I say, that the parallelogram ABCD is equal to the parallelogram EFGH. 
Construction.  Draw a right line from the point B to the point E, and an other from the point C to the point H. 
Demonstration  And forasmuch as BC is equal to FG, but FG is equal to EH, therefore BC also is equal to EH, and they are parallel lines, and the lines BE and CH join them together: but two right lines joining together two equal right lines being parallels, are themselves also (by the 33 proposition) equal the one to the other, and parallels. Wherefore EBCH is a parallelogram, and is equal to the parallelogram ABCD for they have both one and the same base, that is, BC. And are in the self same parallel lines, that is, BC & EH. And by the same reason also the parallelogram EFGH is equal to the parallelogram EBCH. Wherefore the parallelogram ABCD is equal to the parallelogram EFGH. Wherefore parallelograms consisting upon equal bases, and in the self same parallel lines, are equal the one to the other: which was required to be proved.  

Three cases in this proposition.  In this proposition also are three cases. For the equal bases may either be utterly separated a sunder: or they may touch a● one of the ends: or they may have one part common to them both.  

The first case.  Euclides demonstration serveth when the bases be utterly separated a sunder. Which yet may happen seven divers ways. 
Every case may happen seven divers ways.  For the bases being separated a sunder, their opposite sides also may be utterly separated a sunder beyond the point D, as the sides AD and EH in the first figure.  
Or they may touch together in one of the ends, and the whole side may be beyond the point D, as the sides AD and EH do in the second figure.  
Or they may justly agreed the one with the other, that is, the points A and D may fall upon the points E and H: as in the fourth figure.  
Or the side AD being produced on this side the point A, part of the opposite side unto the base FG may be on this side the point A, and an other part may be common with the line AD, as in the fifth figure.  
Or one end of the side EH may light upon the point A, and the whole side on this side of it: As in the sixth figure.  
Or the said side EH may utterly be separated a sunder on this side the point A, as in the seventh figure.  
And the two other cases also may in like manner have seven variety's: 
The like variety in each of the other two cases.  as in the figures here underneath and on the other side of this leaf set it is manifest. And here is to be noted, that in these three cases and in all their varieties also, 
Euclides construction and demostration serveth in all these cases, and in their varities also.  the construction & demonstration put by Euclid (namely, the drawing of lines from the point B to the point E & from the point C to the point H, and so proving it by the former proposition) will serve only in the fourth variety of each case, there needeth no farther construction: for that the conclusion strait way followeth by the former proposition.   

The 27. Theorem. The 37. Proposition. Triangles consisting upon one and the self same base, and in the self same paralles: are equal the one to the other.  
SVppose that these triangles ABC and DBC do consist upon one and the same base, namely, BC, and in the self same parallel lines, that is, AD and BC. Then I say, that the triangle ABC is equal to the triangle BDC. 
Construction.  Produce (by the 2. petition) the line AD on each side to the points E and F. And (by the 31. proposition) by the point B, draw unto the line CA a parallel line BE and (by the same) by the point C, draw unto the line BD a parallel line CF. 
Demonstration.  Wherefore EBCA. and DBCF are parallelograms. And the parallelogram EBCA, is (by the 35. proposition) equal to the parallelogram DBCF. For they consist upon one and the self same base, namely, BC, and are in the self same parallel lines, that is, BC and EF. But the triangle ABC is (by the 34. proposition) the half of the parallelogram EBCA, for the diameter AB divideth it into two equal parts: & (by the same) the triangle DBC is the half of the parallelogram DBCF, for the diameter DC divideth it into two equal parts: but the halves of things equal are also equal the one to the other (by the 7. common sentence), wherefore ehe triangle ABC is equal to the triangle DBC Wherefore triangles consisting upon one and the self same base, and in the self same parallels: are equal the one to the other: which was required to be demonstrated.  
Those triangles are said to be contained within the self same parallel lines, 
How triangles are said to be in the self same parallel lines.  which having their bases in one of the parallel lines, have their ●oppes in the other.  
Here as I promised will I show out of Proclus the comparison of two triangles, which having their sides equal, have the bases and angles at the top unequal. 
Comparison of two triangles whose sides being equal, their bases and angles at the top are unequal.  And first I say that the unequal angles at the top being equal to two right angles, the triangles shallbe equal. As for example.  

When they are less than two right angles.  But now let the angles BAC and EDF be less than two right angles: and again let the angle at the point A be greater than the angle at the point D. Then I say that the triangle ABC is greater than the triangle DEF. Let the same construction be also here that was in the two former. And forasmuch as the angles BAC and EDF, that is, the angles edge & EDF, are less than two right angles, but the angles edge and GDK are equal to two right angles, take away the angle FDG which is common to them both, wherefore the angle remaining, namely, EDF is less than the angle remaining, namely, then GDK: that is, the angle HDK (which by the 1●. proposition is equal to the angle ED F) is less than the angle GDK. Wherefore the whole angle GD H is less than double to the angle GD K. But it is double to the angle DGF (as before it was proved): wherefore the angle GD K is greater than the angle DGF. Put the angle DGL equal to the angle GD K (by the 23. proposition) and produce the line GL till it concur with the line EF also produced● & let the concourse be in the point L. And draw a line from D to L. And for as much as the angle D GL is equal to the angle GD K, and they are alternate angles, therefore the line GL is a parallel to D E (by the 27. proposition). Wherefore (by this proposition) the triangles GD E and LD E are equal: but the triangle IDE is greater than the triangle FD E, and the triangle GD E is equal to the triangle ABC. Wherefore the triangle ABC is greater than the triangle D EF: which was required to be proved.   

The 28. Theorem. The 38. Proposition. Triangles which consist upon equal bases, and in the self same parallel lines, are equal the one to the other.  
SVppose that these triangles ABC and DEF do consist upon equal bases, that is, upon BC and EF, and in the self same parallel lines, that is BF and AD. Then I say that the triangle ABC is equal to the triangle ABC is equal to the triangle DEF. 
Construction.  Produce (by the second petition) the line AD on each side to the points G and H. And (by the 31. proposition) by the point B draw unto CA a parallel line BG, and (by the same) by the point F draw unto DE a parallel line FH. 
Demonstration.  Wherefore GBCA and DEFH are parallelograms. But the parallelogram GBCA is (by the 36 proposition) equal to the parallelogram DEFH, for they consist upon equal bases, that is, BC and EF, and are in the self same parallel lines, that is, BF and GH. But (by the 34. proposition) the triangle ABC is the half of the parallelogram GBCA, for the diameter AB divideth it into two equal parts: and the triangle DEF is (by the same) the half of the parallelogram DEFH, for the diameter FD divideth it into two equal parts. But the halves of things equal are (by the 7. common sentence) equal ●he one to the other. Wherefore the triangle ABC is equal to the triangle DEF. Wherefore triangles which consist upon equal bases, and in the self same parallel lines, are equal the one to the other: which was required to be proved.  
In this proposition are three cases. 
Three cases in this proposition.  For the bases of the triangles either have one part common to them both or the base of the one toucheth the base of the other only in a point: or their bases are utterly severed a sunder. And each of these cases may also be diversly, 
Each of these cases also may be diversly.  as we before have seen in parallelograms consisting on equal bases, and being in the self same parallel lines. So that he which diligently noteth the variety that was there put touching them, may also easily frame the same variety to each case in this proposition. Wherefore I think it needles here to repeat the same again: for how soever the bases be put, or the tops, the manner of construction and demonstration here put by Euclid will serve: namely, to draw parallel lines to the sides.  

An addition of Pelitarius.  
To divide a triangle given into two equal parts.  

Note.  And so by this you may divide any triangle into so many parts as are signified by any number that is evenly even: as into 14.16.32.64. etc.   

another addition of Pelitarius. 
another addition of Pelitarius.   
From any point given in one of the sides of a triangle, to draw a line which shall divide the triangle into two equal parts.  
Let the triangle given be BCD: and let the point given in the side BC be A. It is required from the point A to draw a line which shall divide the triangle BCD into two equal parts. 
Construction.  Divide the side BC into two equal parts in the point E. And draw a right line from the point A to the point D. And (by the 31. proposition) by the point E draw unto the line AD a parallel line EF: which let cut the side DC in the point F. And draw a line from the point A to the point F. Then I say that the line AF divideth the triangle BCD into two equal parts: namely, the trapesium ABDF is equal to the triangle ACF. For draw a line from E to D, 
Demonstration  cutting the line AF in the point G. Now than it is manifest (by the 38. proposition) that the two triangles BED and CED are equal (if we understand a line to be drawn by the point D parallel to the line BC, for the bases BE and EC are equal). The two triangles also D EF and AEF are (by the 37. proposition) equal: for they consist upon one and the self same base EF, and are in the self same parallel lines AD and EF. Wherefore taking away the triangle EFG which is common to them both, the triangle AEG shallbe equal to the triangle D FG: wherefore unto either of them add the trapesium CFGE, and the triangle ACF shallbe equal to the triangle DEC. But the triangle DEC is the half part of the whole triangle BCD wherefore the triangle ACF is the half part of the same triangle BCD. Wherefore the residue, namely, the trapesium ABFD is the other half of the same triangle. Wherefore the line AF divideth the whole triangle BCD into two equal parts: which was required to be done.    

The 29. Theorem. The 39 Proposition. Equal triangles consisting upon one and the same base, and on one and the same side: are also in the self same parallel lines.  
SVppose that these two equal triangles ABC and DBC do consist upon one and the same base, namely, BC and on one and the same side. Then I say that they are in the self same parallel lines. Draw a right line from the point A to the point D. Now I say that AD is a parallel line to BC. For if not, then (by the 31. proposition) by the point A draw unto the right line BC a parallel line AE, and draw a right line from the point E to the point C. Wherefore the triangle EBC is (by the 37. proposition) equal to the triangle ABC, for they consist upon one and the self same base, namely, BC, and are in the self same parallels, that is, AE and BC. But the triangle DBC is (by supposition) equal to the triangle ABC● Wherefore the triangle DBC is equal to the triangle EBC, the greater unto the less: which is impossible. Wherefore the line AE, is not a parallel to the line BC. And in like sort may it be proved that no other line besides AD is a parallel line to BC, wherefore AD is a parallel line to BC. Wherefore equal triangles consisting upon one and the same base, and on one and the same side, are also in the self same parallel lines: which was required to be proved.  
This proposition is the converse of the 37. proposition. 
This Theorem the converse of the 37. proposition.  And here is to be noted that if by the point A, you draw unto the line BC a parallel line, the same shall of necessity either light upon the point D, or under it, or above it. If it light upon it, then is that manifest which is required: but if it light under it, then followeth that absurdity which Euclid putteth, namely, that the greater triangle is equal to the less: which self same absurdity also will follow, if it fall above the point D. As for example.  

An addition of Flussates.  
The self same also followeth in parallelograms. 
An addition of Fl●ssases.  For if upon the base AB be set on one & the same side these equal parallelograms ABCD & ABGE, they shall of necessity be in the self same parallel lines. For if not, but one of them is set either within or without, let the parallelogram B● being equal to the parallelogram ABCD be set within the same parallel lines: wherefore the same parallelogram BF being equal to the parallelogram ABCD (by the 35. proposition) shall also be equal to the other parallelogram ABGE (by the first common sentence) For the parallelogram ABGE is by supposition equal to the parallelogram ABCD: wherefore the parallelogram BF being equal to the parallelogram ABGE, the part shall be equal to the whole, which is absurd. The same inconvenience also will follow, if it fall without. Wherefore it can neither fall within nor without. Wherefore equal parallelograms being upon one and the self same base and on one and the same side, are also in the self same parallel lines.   

An addition of Campanus.  

An addition of Campanus.  If a right line divide two sides of a triangle into two equal parts: it shall be equidistant unto the third side.  
Suppose that there be a triangle ABC: and let there be a right line DE, which let divide the two sides AB and BC into two equal parts in the points D and E Than I say, that the line DE is a parallel to the line A C. Draw these two lines AE and DC. Now then imagining a line to be drawn by the point E parallel to the line AB, the triangle BDE shall (by the 38. proposition) be equal to the triangle DAE (for their two bases AD and DB are put to be equal) And by the same reason the triangle BDE is equal to the triangle CED. Wherefore (by the first common sentence) the triangles EAD and ECD are equal, and they are erected on one and the sel●e same base, namely, DE, and on one and the same side. Wherefore (by the 39 proposition) they are in the self same parallel lines, and the line which joineth together their tops is a parallel to their base. Wherefore the lines DE and AC are paralles: which was required to be proved.    

The 30. Theorem. The 40. Proposition. Equal triangles consisting upon equal bases, and in one and the same side: are also in the self same parallel lines.  
SVppose that these equal triangles ABC and CDE do consist upon equal bases, that is, upon BC and CE, and on one and the same side, namely, on the side of A. Then I say that they are in the self same parallel lines. 
Construction.  Draw by the first petition a right line from the point A to the point D. Now I say that AD is a parallel line to BE. For if not, then (by the 31. proposition) by the point A draw unto the line BE a parallel line AF. And draw a right line from the point F to the point E. 
Demonstration leading to an absurdity.  Wherefore (by the 38. proposition) the triangle BAC is equal to the triangle CFE● for they consist upon equal bases, that is BC and CE, and are in the self same parallel lines, namely, BE and AF. But by supposition the triangle ABC is equal to the triangle CDE. Wherefore the triangle DCE is equal to the triangle FCE, namely, the greater unto the less, which is impossible. Wherefore AF is not a parallel line to BE. And in like sort may we prove that no other line besides AD is a parallel line to BE. Wherefore AD is a parallel line to BE Equal triangles therefore consisting upon equal bases, and in one and the same side: are also in the self same parallel lines: which was required to be proved.  
This proposition is the converse of the 38. proposition. 
This proposition is the converse of the 38. propositions.  And in this as in the former proposition, if the parallel line drawn by the point A, should not pass by the point D, it must pass either beneath it, or above it. Euclid setteth forth only the absurdity which should follow if it pass beneath it: but the self same absurdity also will follow if it should pass above it: as it is not hard to see by the gathering thereof in the former proposition. And therefore here I omit it.   

The 31. Theorem. The 41. Proposition. If a parallelogram & a triangle have one & the self same base, and be in the self same parallel lines: the parallelogram shallbe double to the triangle.  
SVppose that the parallelogram ABCD and the triangle EBC have one & the same base, namely, BC, and let them be in the self same parallel lines, namely, BC & AE. Then I say, that the parallelogram ABCD is double to the triangle BEC. Draw (by the first petition) a right line from the point A to the point C. 
Demonstration.  Wherefore (by the 37. proposition) the triangle ABC is equal to the triangle EBC● for they are upon one and the self same base BC, and in the self same parallel lines BC and EA: but the parallelogram ABCD is double to the triangle ABC (by the 34. proposition) for the diameter thereof AC divideth it into two equal parts: wherefore the parallelogram ABCD is double to the triangle EBC. If therefore a parallelogram and a triangle have one and the self same base, and be in the self same parallels, the parallelogram shall be double to the triangle: which was required to be proved.  

Two cases in this proposition.  This proposition hath two cases. For the base being one, the triangle may have his top without the parallelogram, or within. The first case is demonstrated of the author. The second case is thus.  

A corollary.  By this proposition it is manifest that if the base be doubled, the triangle erected upon it shallbe equal to the parallelogram.  

The self same demonstration will serve if the triangle & the parallelogram be upon equal bases.  And if the bases of the triangle and of the parallelogram be equal, the sel●e same demonstration will serve if you draw the diameter of the parallelogram. For the triangles being equal, the parallelogram which is double to the one, shall also be double to the other. And the triangles must needs be equal (by the 38. proposition) for that their bases are equal, and for that they are in the self same parallel lines.  

The converse of this proposition is thus.  
If a parallelogram and a triangle have one and the self same base, or equal bases the one to the other, and be described on one and the same side of the base: ●f the parallelogram be double to the triangle, they shallbe in the self same parallel lines.  

The converse of this proposition.  For if they be not, the whole shallbe equal to his part. For then the top of the triangle must needs fall either within the parallel lines or without. And whether of both soever it do, one and the self same impossibility will follow, if by the top of the triangle be drawn unto the base a parallel line.   

another converse of the same proposition.  
If a parallelogram be the double of a triangle, being both within the self same parallel lines: 
another converse of the same proposition.  then are they upon one and the self same base, or upon equal bases. For if in that case their bases should be unequal, then might strait way be proved, that the whole is equal to his part: which is impossible.  
A trapesium having two sides only parallel lines, 
Comparison of a triangle and a trapesium being upon one & the self same base, and in the self same parallel lines.  is either more than double, or less than double to a triangle contained within the self same parallel lines, and having one and the self same base with the trapesium, or table. ●ust the double it cannot be, for than it should be a parallelogram. A trapesium having two sides parallels hath of necessity the one of them longer than the other: for if they were equal then should the other two sides enclosing them be also equal (by the 33. proposition.) If the greater side of the trapesium be the base of the triangle, then shall the trapesium be less than the double of the triangle And if the less side of the trapesium be the base of the triangle than shall the trapesium be greater than the triangle.    

The 11. Problem. The 42. proposition. Unto a triangle given, to make a parallelogram equal, whose angle shall be equal to a rectiline angle given.  
SVppose that the triangle given be ABC, and let the rectiline angle given be D. It is required that unto the triangle ABC there be made a parallelogram equal, whose angle shall be equal to the rectiline angle given, namely, 
Construction.  to the angle D. Divide (by the 10. proposition) the line BC into two equal parts in the point E. And (by the first petition) draw a right line from the point A to the point E. And (by the 23. proposition) upon the right line given EC, and to the point in it given E, make the angle CEF equal to the angle D. And (by the 31. proposition) by the point A draw unto the line EC a parallel line AH: and let the line EF cut the line AH in the point F. and (by the same) by the point C, draw unto the line EF a parallel line CG. Wherefore FECG is a parallelogram. 
Demonstration  And forasmuch as BE is equal to EC, therefore (by the 38. proposition) the triangle ABE is equal to the triangle AEC, for they consist upon equal bases that is BE and EC, and are in the self same parallel lines, namely, BC and AH. Wherefore the triangle ABC is double to the triangle AEC. And the parallelogram CEFG is also double to the triangle AEC: for they have one & the self same base, namely, EC: and are in the self same parallel lines, that is, EC and AH. Wherefore the parallelogram FECG is equal to the triangle ABC, and hath the angle CEF equal to the angle given D. Wherefore unto the triangle given ABC is made an equal parallelogram, namely, FECG, whose angle CEF is equal to the angle given D: which was required to be done.  

The converse of this proposition after Pelitarius.  
Unto a parallelogram given, 
 to make a triangle equal, hauy●g an angle equal to a rectiline angle given.  
Suppose that the parallelogram given be ABCD, and let the angle given be E. It is required unto the parallelogram ABCD to make a triangle equal having an angle equal to the angle E. Upon the line CD and to the point in it C, describe (by the 23. propoposition) an angle equal to the angle E: which let be DCF: a●d let the line CF cut the line AB being produced, in the point F: and produce the line CD (which is parallel to the line AF) to the point G. And put the line DG equal to the line CD and draw a line from F to G. Then I say that the triangle CFG is equal to the parallelogram ABCD. For forasmuch as (by the 38. proposition) the whole triangle CFG is double to the triangle CDF. A●d (by the 41. proposition) the parallelogram ABCD is double to the same triangle CDF: therefore the parallelogram ABCD and the triangle CFG are equal the one to the other: which was required to be done.    

The 32. Theorem. The 43. Proposition. In every parallelogram, the supplements of those parallelograms which are about the diameter, are equal the one to the other.  
SVppose that ABCD be a parallelogram, and let the diameter thereof be AC: and about the diameter AC let these parallelograms EH and GF consist: and let the supplements be BK and KD. Then I say that the supplement BK is equal to the supplement KD. For forasmuch as ABCD is a parallelogram and the diameter thereof is AC, therefore (by the 34. proposition) the triangle ABC is equal to the triangle ADC. Again forasmuch as AEKH is a parallelogram, and the diameter thereof is AK, therefore (by the same) the triangle AEK is equal to the triangle AHK. And by the same reason also the triangle KFC is equal to the triangle KGC. And forasmuch as the triangle AEK is equal to the triangle AHK, and the triangle KFC to the triangle NGC, therefore the triangles AEK and KGC are equal to the triangles AHK and KFC: and the whole triangle ABC is equal to the whole triangle AD C: wherefore the residue, namely, the supplement BK is (by the 3. common sentence) equal to the residue, namely, to the supplement KD. Wherefore in every parallelogram, the supplements of those parallelograms which are about the diameter, are equal the one to the other: which was required to be proved.  

Supplements & Complements.  supplements or Complements are those figures which with the two parallelograms accomplish the whole parallelogram. Although Pel●tarius for distinction sake putteth a difference between supplements and Complementer. The parallelograms about the diameter he calleth Complements, the other too figures he calleth supplements.  

Three cases in this Theorem.  This theorem hath three cases only. For the parallelograms which consist about the diameter, either touch the one the other in a point: or by a certain part of the diameter are severed the one from the other: or else they cut the one the other. 
The first case.  For the first case is the example of Euclid before set. The second case is thus.  
This is to be noted that in ●●h of those three cases it may so happen, that the parallelograms about the diameter shall not have one angle common with the whole parallelogram, as they have in the former figures. But yet though they have not, the self same demonstration will serve, as it is plain to see in the figures here underneath put. For always, if from things equal be taken away things equal, the residue shallbe equal.  
This proposition P●litarius calleth Gnomicall, and mystical, 
This proposition called Gnomical and mystical.  for that of it (saith he) spring infinite demonstrations, and uses in geometry. And he putteth the converse thereof after this manner.  
If a parallelogram be divided into two equal supplements, 
The converse of this proposition.  and into two compliments whatsoever: the diameter of the two complements shall be set directly, and make one diameter of the whole parallelogram.  
Here is to be noted as I before admonished that Pelitarius for distinction sake putteth a difference between supplements and complements, which difference, for that I have before declared, I shall not need here to repeat again.   

The 12. Problem. The 44. Proposition. Upon a right line given, to apply a parallelogram equal to a triangle given, and containing an angle equal to a rectiline angle given.  
SVppose that the right line given be AB, and let the triangle given be C, and let the rectiline angle given be D. It is required upon the right line given AB, to apply a parallelogram equal to the triangle given C, 
Construction.  and containing an angle equal to the rectiline angle given D. Describe (by the 44. proposition) the parallelogram BGEF equal to the triangle C, and having the angle BGF equal to the angle D. And unto the line EB join the line AB in such sort that they make both one right line. And extend the line FG beyond the point G to the point H. And (by the 31. proposition) by the point A draw to either of these lines BG and EF a parallel line AH. And (by the first petition) draw a right line from the point H to the point B. And forasmuch as upon the parallel lines AH and EF falleth a certain right line HF, therefore (by the ●9 proposition) the angles AHF and HFE are equal to two right angles: wherefore the angles BHG and GFE are less than two right angles: but if upon two right lines fall a right line making the inward angles on one and the same side less than two right angles, those right lines being infinitely produced shall at the length meet on that side in which are the angles less than two right angles (by the 5. petition). Wherefore the lines HB and FE being infinitely produced will at the length mete. Let them be produced, & let them meet in the point K. And (by the 31 proposition) by the point K draw to either of these lines EA and FH a parallel line KL. And (by the 2. petition) extend the lines HA and GB till they concur with the line KL in the points L and M. 
Demonstration  Wherefore HLKF is a parallelogram, and the diameter thereof is HK: and about the diameter HK are the parallelograms AG and ME, and the supplements are LB and BF: wherefore (by the 43. proposition) the supplement LB is equal to the supplement BF: but by construction the parallelogram BF is equal to the triangle C: wherefore also the parallelogram LB is equal to the triangle C. And forasmuch as the line FH is a parallel to the line KL, and upon them lighteth the line GM, therefore (by the 27. proposition) the angle FGB is equal to the angle BML. But the angle FGB is equal to the angle D, therefore the angle BML is equal to the angle D. Wherefore upon the right line given AB is applied the parallelogram LB, equal to the triangle given C, and containing the angle BML equal to the rectiline angle given D: which was required to be done.  
Applications of spaces or figures to lines with excesses or wants is (saith Eudemus) an ancient invention of Pythagoras. 
Applications of spaces with excesses or wants an ancient invention of Pythagoras.   
When the space or figure is joined to the whole line, 
How a figure is said to be applied to a line.  them is the figure said to be applied to the line. But if the length of the space be longer than the line, than it is said to exceed: and if the length of the figure be shorter than the line, so that part of the line remaineth without the figure described, then is it said to want.  
In this problem are three things given. 
Three things given in this proposition.  A right line to which the application is made, which here must be the one side of the parallelogram applied. A triangle whereunto the parallelogram applied must be equal: and an angle whereunto the angle of the parallelogram applied must be equal. And if the angle given be a right angle, them shall the parallelogram applied be either a square, or a figure on the one side longer. But if the angle given be an obtuse or an acute angle, then shall the parallelogram applied be a Rhombus or diamond figure, or else a rhomboids or diamondlike figure.  

The converse of this proposition after Politarius.  
Upon a right line given, to apply unto a parallelogram given an equal triangle having an angle equal to an angle given. 
The converse of this proposition.   
Suppose that the right line given be AB, and let the parallelogram given be CDEF, and let the angle given be G. It is required upon the line AB to describe a triangle equal to the parallelogram CDEF, having an angle equal to the angle G. Draw the diameter CF & produce CD beyond the point D to the point H. And put the line DH equal to the line CD. And draw a line from F to H. Now then (by the 41. proposition) the triangle CHF is equal to the parallelogram CD EF. And (by this proposition) upon the line AB describe a parallelogram ABKL equal to the triangle CHF, having the angle ABL equal to the angle given G: and produce the line BL beyond the point L to the point M. And put the line LM equal to the line BL, and draw a line from A to M. Then I say that upon the line AB is described the triangle ABM, which is such a triangle as is required. For (by the 41. proposition) the triangle ABM is equal to the parallelogram ABKL (for that they are between two parallel lines BM and AK, & the base of the triangle is double to the base of the parallelogram): but ABKL is by construction equal to the triangle CHF: and the triangle CHF is equal to the parallelogram CDEF. Wherefore (by the first common sentence) the triangle ABM is equal to the parallelogram given CD EF, and hath his angle ABM equal to the angle given G: which was required to be done.    

The 13. Problem. The 45 Proposition. To describe a parallelogram equal to any rectiline figure given, and containing an angle equal to a rectiline angle given.  
SVppose that the rectiline figure given be ABCD, and let the rectiline angle given be E. It is required to describe a parallelogram equal to the rectiline figure given ABCD, and containing an angle equal to the rectiline angle given E. 
Construction.  Draw (by the first petition) a right line from the point D to the point B. And (by the 42. proposition) unto the triangle ABD describe an equal parallelogram FH, having his angle FKH equal to the angle E. And (by the 4●. of the first) upon the right line GH apply the parallelogram GM equal to the triangle DBC, having his angle GHM equal to the angle E. 
Demonstration.  And forasmuch as either of those angles HKF and GHM is equal to the angle E: therefore the angle HKF is equal to the angle GHM: put the angle KHG common to them both, wherefore the angles FKH and KHG are equal to the angles KHG and GHM. but the angles FKH and KHG are (by the 29. proposition) equal to two right angles. Wherefore the angles KHG and GHM are equal to two right angles. Now then unto a right line GH, and to a point in the same H, are drawn two right lines KH and HM not both on one and the same side, making the side angles equal to two right angles. Wherefore (by the 14. proposition) the lines KH and HM make directly one right line. And forasmuch as upon the parallel lines. KM and FG falleth the right line HG, therefore the alternate aagles MHG and HGF are (by the 29 proposition) equal the one to the other: put the angle HGL common to them both. Wherefore the angles MHG and HGL are equal to the angles HGF and HGL. But the angles MHG & HGL are equal to two right angles (by the 29. proposition). Wherefore also the angles HGF and HGL are equal to two right angles. Wherefore (by the 14. proposition) the lines FG and GL make directly one right line. And forasmuch as the line KF is (by the 34. proposition) equal to the line HG, and it is also parallel unto it: and the line HG is (by the same) equal to the line ML, therefore (by the first common sentence) the line FK is equal to the line ML, and also a parallel unto it (by the 30. proposition). But the right lines KM and FL join them together. Wherefore (by the 33. proposition) the lines KM and FL are equal the on to the other and parallel lines. Wherefore KFLM is a parallelogram. And forasmuch as the triangle ABD is eqnal to the parallelogram FH, and the triangle DBC to the parallelogram GM● therefore the whole rectiline figure ABCD is equal to the whole parallelogram KFLM. Wherefore to the rectiline figure given ABCD is made an equal parallegrame KFLM, whose angle FKM is equal to the angle given, namely, to E: which was required to be done.  
The rectiline figure given 〈◊〉 in the example of Euclid is a parallelogram. But if the rectiline figure be of many sides, as of 5.6. or more, them must you resolve the figure into his triangles, as hath been before taught in the 32. proposition. And then apply a parallelogram equal to every triangle upon a line given, as before in the example of the author. And the same kind of reasoning will serve that was before, only by reason of the multitude of triangles, you shall have need of oftener repetition of the 29. and 14. propositions to prove that the bases of all the parallelograms made equal to all the triangles make one right line, and so also of the tops of the said parallelograms. Pelitarius addeth unto this proposition this Problem following.  
Two unequal rectiline superfici●c●s being given, to found out the excess of the greater above the less. 
An addition of Pelitarius.   

An other more ready way.  
Let the parallelogram CDEF remain equal to the rectiline figure A, & produce the line CD beyond the point D to the point G. And upon the line DG describe the parallelogram DGHK equal to the rectiline figure B. And produce the lines EC & HK beyond the points C and K till they concur in the point L. And by the point D draw the diameter LDM, which let cut the line HG being produced beyond the point G in the point M, & by the point M draw unto the line HL a parallel MN cutting the line EL in the point N: and by that means is HLMN a parallelogram. Then I say that NF is the eucesse of the rectiline figure A above the rectiline figure B. For forasmuch as the parallelogram HD is equal to the rectiline figure B, & the supplements HD and DN are (by the 43. proposition) equal: therefore DN also is equal to the rectiline figure B, which rectiline figure DN being taken away from the parallelogram CF (which is supposed to be equal to the rectiline figure A) the residue NF shall be the ezcesse of the rectiline figure A above the rectiline figure B: which was required to be done.    

The 14. Problem. The 46. Proposition. Upon a right line given, to describe a square.  
SVppose that the right line given be AB. It is required upon the right line AB, to describe a square. Upon the right line AB, and from a point in it given, namely, A, raise up (by the 11 proposition) a perpendicular line AC. And (by the 3. proposition) unto AB put an equal line AD. And (by the 31. proposition) by the point D draw unto AB a parallel line DE. And (by the same) by the point B draw unto AD a parallel line BE. Wherefore ADEB is a parallelogram. Wherefore the line AB is equal to the line DE, and the line AD to the line BE: but the line AB is equal to the line AD wherefore these four lines BA, AD, DE, EB, are equal the one to the other. Wherefore the parallelogram ADEB consisteth of equal sides. I say also that it is rectangle. For forasmuch as upon the parallel lines AB and DE falleth a right line AD: therefore (by the 29. proposition) the angle: BAD and ADE are equal to two right angles: but the angle BAD is a right angle. Wherefore the angle ADE also is a right angle. But in parallelograms the sides and angles which are opposite are equal the one to the other (by the 34 proposition). Wherefore the two opposite angles ABE and BED are each of them a right angle. Wherefore the parallelogram ABED is rectangle: & it is also proved that it is equilater. Wherefore it is a square, & it is described upon the right line given AB: which was required to be done.  
This is to be noted that if you will mechanically and readily, 
To describe a square mechanically.  not regarding demonstration describe a square upon a line given, as upon the line AB, after that you have erected the perpendicular line CA upon the line AB, and put the line AE equal to the line AB: then open your compass to the wydth of the line AB or AE, & set one foot thereof in the point E, and describe a piece of the circumference of a circle: and again make the centre the point B, and describe also a piece of the circumference of a circle, namely, in such sort that the piece of the circumference of the one may cut the piece of the circumference of the other, as in the point D: and from the point of the intersection, draw unto the points E & B right lines: & so shallbe described a square. As in this figure here put, wherein I have not drawn the lines ED and DB, that the pieces of the circumference cutting the one the other might the plainlier be seen.  

An addition of Proc●us.  
If the lines upon which the squares be described be equal, the squares also are equal.  
Suppose that these right lines AB and CD be equal, 
An addition of Proc●●●.  & upon the line AB describe a square ABEG: and upon the line CD describe a square CDHF. Then I say that the two squares ABEG & CDHF are equal. For draw these right lines GB and HD. And for as much as the right lines AB and CD are equal, & the lines AG and HC are also equal, and they contain eqaul angles, namely, right angles (by the definition of a square) therefore (by the 4. proposition) the base BG is equal to the base HD. And the triangle ABG is equal to the triangle CDH. Wherefore the doubles of the said triangles are equal. Wherefore the square AE is equal to the square CF: which was required to be proved.   

The converse thereof is thus.  
If the squares be equal: 
The converse thereof.  the lines also upon which they are described are equal.  
Suppose that there be two equal squares AF and CG described upon the lines AB & BC. Then I say, that the lines AB and BC are equal. Put the line AB directly to the line BC, that they both make on right line. And forasmuch as the angles are right angles, therefore also (by the 14. proposition) the right line FB is set directly to the right line BG. Draw these right lines FC, AG, AF, and CG. Now for as much as the square AF is equal to the square CG, the triangle also AFB shallbe equal to the triangle CBG: put the triangle BCF common to them both. Wherefore the whole triangle ACF is equal to the whole triangle CFG. Wherefore the line AG is a parallel unto the line CF (by the 38. proposition): for the triangles consist upon one and the self same base, namely CF. Again forasmuch as either of these angles AFG & CBG is the half of a right angle, therefore (by the 17. proposition) the line AF is a parallel to the line CG. Wherefore the right line AF is equal to the right line CG (for the opposite sides of a parallelogram are equal). And forasmuch as there are two triangles ABF and BCG, whose alternate angles are equal, namely, the angle AFB to the angle BGC, and the angle BAF to the angle BCG, and one side of the one is equal to one side of the other, namely, the side which lieth between the equal angles, that is, the side AF to the side CG, therefore (by the 26. proposition) the side AB is equal to the side BC, and the side BF to the side BG. Wherefore it is proved that the squares of the lines AF and CG being equal, their sides also shallbe equal: which was required to be proved.    

The 33. Theorem. The 47. Proposition. In rectangle triangles, the square which is made of the side that subtendeth the right angle, is equal to the squares which are made of the sides containing the right angle.  
SVppose that ABC be a rectangle triangle, having the angle BAC a right angle. Then I say that the square which is made of the line BC is equal to the squares which are made of the lines AB and AC. 
Construction.  Describe (by the 46. proposition) upon the line BC a square BDCE, and (by the same) upon the lines BA and AC describe the squares ABFG and ACKH. And by the point A draw (by the 31. proposition) to either of these lines BD and CE a parallel line AL. And (by the first petition) draw a right line from the point A to the point D, and an other from the point C to the point E. 
Demonstration.  And forasmuch as the angles BAC and BAG are right angles, therefore unto a right line BA, and to a point in it given A, are drawn two right lines AC and AG, not both on one and the same side, making the two side angles equal to two right angles: wherefore (by the 14. proposition) the lines AC and AG make directly one right line. And by the same reason the lines BA and AH make also directly one right line. And forasmuch as the angle DBC is equal to the angle FBA (for either of them is a right angle) put the angle ABC common to them both: wherefore the whole angle DBA is equal to the whole angle FBC. And forasmuch as these two lines AB and BD are equal to these two lines BF and BC, the one to the other, and the angle DBA is equal to the angle FBC: therefore (by the 4. proposition) the base AD is equal to the base FC, and the triangle ABD is equal to the triangle FBC. But (by the 31. propositions) the parallelogram BL is double to the triangle ABD, for they have both one and the same base, namely, BD, and are in the self same parallel lines, that is, BD and AL and (by the same) the square GB is double to the triangle FBC, for they have both one and the self same base, that is, BF, and are in the self same parallel lines, that is, FB and GC. But the doubles of things equal, are (by the sixth common sentence) equal the one to the other. Wherefore the parallelogram BL is equal to the square GB. And in like sort if (by the first petition) there be drawn a right line from the point A to the point E, and an other from the point B to the point K, we may prove that the parallelogram CL is equal to the square HC. Wherefore the whole square BDEC is equal to the two squares GB and HC. But the square BDEC is described upon the line BC, and the squares GB and HC are described upon the lines BA & AC: wherefore the square of the side BC, is equal to the squares of the sides BA and AC. Wherefore in rectangle triangles, the square which is made of the side that subtendeth the right angle, is equal to the squares which are made of the sides containing the right angle: which was required to be demonstrated.  
This most excellent and notable Theorem was first invented of the great philosopher Pythagoras, 
Pythagoras the first inventor of this proposition.  who for the exceeding joy conceived of the invention thereof, offered in sacrifice an Ox, as record Hi●rone, Proclus, Lycius, & Vitruutus. And it hath been commonly called of barbarous writers of the latter time Dulcarnon.  


An addition of P●l●tari●●.  An addition of Pelitarius.  
To reduce two unequal squares to two equal squares.  
Suppose that the squares of the lines AB and AC be unequal. It is required to reduce them to two equal squares. join the two lines AB and AC at their ends in such sort that they make a right angle BAC. And draw a line from B to C. Then upon the two ends B and C make two angles each of which may be equal to half a right angle (This is done by erecting upon the line BC perpendiculer lines, from the points B and C: and so (by the 9 proposition) dividing ●che of the right angles into two equal parts): and let the angles BCD and CBD be either of them half of a right angle. And let the lines BD and CD concur in the point D. Then I say that the two squares of the sides BD and CD, are equal to the two squares of the sides AB and AC. For (by the 6 proposition) the two sides DB and DC are equal, and the angle at the point D is (by the 32. proposition) a right angle. Wherefore the square of the side BC is equal to the squares of the two sides DB and DC (by the 47. proposition): but it is also equal to the squares of the two sides AB and AC (by the self same proposition) wherefore (by the common sentence) the squares of the two sides BD and DC are equal to the squares of the two sides AB and AC: which was required to be done.   

another addition of Pelitarius.  

another addition of Pelitarius.  If two right angled triangles have equal bases. the squares of the two sides of the one are equal to the squares of the two sides of the other.  
This is manifest by the former construction and demonstration.   

another addition of Pelitarius.  

another addition of Pelitarius.  Two unequal lines being given, to know how much the square of the one is greater than the square of the other.  
Suppose that there be two unequal lines AB and BC: of which let AB be the greater. It is required to search out how much the square of AB exceedeth the square of BC. That is I will find out the square, which with the square of the line BC shallbe equal to the square of the line AB. Put the lines A B and BC directly, that they make both one right line: and making the centre the point B, and the space BA describe a circle ADE. And produce the line AC to the circumference, and let it concur with it in the point E. And upon the line AE and from the point C erect (by the 11. proposition) a perpendicular line CD, which produce till it concur with the circumference in the point D: & draw a line from B to D. Then I say, that the square of the line CD, is the excess of the square of the line AB above the square of the line BC. For forasmuch as in the triangle BCD, the angle at the point C is a right angle, the square of the base BD is equal to the squares of the two sides BC and CD (by this 47. proposition). Wherefore also the square of the line AB is equal to the self same squares of the lines BC and CD. Wherefore the square of the line BC is so much less than the square of the line AB, as is the square of the line CD: which was required to search out.   

another addition of Pelitarius. 
another addition of Pelitarius.  The diameter of a square being given, to give the square thereof.  
This is easy to be done. For if upon the two ends of the line be drawn two half right angles, and so be made perfect the triangle then shallbe described half of the square: the other half whereof also is after the same manner easy to be described.  
Hereby it is manifest, that the square of the diameter is double to that square whose diameter it is. 
A corollary.     

The 34. Theorem. The 48. Proposition. If the square which is made of one of the sides of a triangle, be equal to the squares which are made of the two other sides of the same triangle: the angle comprehended under those two other sides is a right angle.  
SVppose that ABC be a triangle, and let the square which is made of one of the sides there, namely, of the side BC, be equal to the squares which are made of the sides BA and AC. Then I say that the angle BAC is a right angle. Raise up (by the 11. proposition) from the point A unto the right line AC a perpendicular line AD. And (by the third proposition) unto the line AB put an equal line AD. And by the first petition draw a right line from the point D to the poin● C. And forasmuch as the line DA is equal to the line AB, the square which is made of the line DA is equal to the square which is made of the line AB Put the square of the line AC, common to them both. Wherefore the squares of the lines DA and AC are equal to the squares of the lines BA and AC. But (by the proposition going before) the square of the line DC is equal to the squares of the lines. AD and AC. (For the angle DAC is a right angle) and the square of BC is (by supposition) equal to the squares of AB and AC. Wherefore the square of DC is equal to the square of BC: wherefore the side DC is equal to the side BC. And forasmuch as AB is equal to AD ●nd AC is common to them both, therefore these two sides DA and AC are equal to these two sides BA and AC, the one to the other, and the base DC is equal to the base BE wherefore (by the 8. proposition) the angle DAC is equal to the angle BAC. But the angle DAC is a right angle, wherefore also the angle BAC is a right angle. If therefore the square which is made of one of the sides of a triangle, be equal to the squares which are made of the two other sides of the same triangle, the angle comprehended under those two other sides is a right angle. which was required to be proved.  

This proposition is the converse of the former.  This proposition is the converse of the former, and is of Pelitarius demonstrated by an argument leading to an impossibility after this manner.   
The end of the first book of Euclides Elements.   

¶ The second book of Euclides Elements.  
IN this second book Euclid showeth, 
The argument of the second book.  what is a Gnomon, and a right angled parallelogram. Also in this book are set forth the powers of lines, divided evenly and unevenly, and of lines added one to an other. The power of a line, 
What is the power of a line.  is the square of the same line: that is, a square, every side of which is equal to the line. So that here are set forth the qualities and proprieties of the squares and right lined figures, which are made of lines & of their parts. 
Many compendious rules of reckoning gathered one of this book, and also many rules of Algebra.  The Arithmetician also our of this book gathereth many compendious rules of reckoning, and many rules also of Algebra, with the equations therein used. The grounds also of those rules are for the most part by this second book demonstrated. 
Two wonderful propositions in this book.  This book moreover containeth two wonderful propositions, one of an obtuse angled triangle, and the other of an acute: which with the aid of the 47. proposition of the first book of Euclid, which is of a rectangle triangle, of how great force and profit they are in matters of astronomy, they know which have travailed in that art. Wherefore if this book had none other profit be side, only for these 2. propositions sake it were diligently to be embraced and studied.  

The definitions.  

1. Every rectangled parallelogram, 
First definition.  is said to be contained under two right lines comprehending a right angle.  
A parallelogram is a figure of four sides, 
What a parallelogram is.  whose two opposite or contrary sides are equal the one to the other. There are of parallelograms four kinds, 
Four kinds of parallelograms.  a square, a figure of one side longer, a Rombus or diamond, and a Romboides or diamond like figure, as before was said in the 33. definition of the first book. Of these four sorts, the square and the figure of one side longer are only right angled parallelograms: for that all their angles are right angles. And either of them is contained (according to this definition) under two right lines whi●h concur together, and 'cause the right angle, and contain the same. Of which two lines the one is the length of the figure, & the other the breadth. The parallelogram is imagined to be made by the draft or motion of one of the lines into the length of the other. As if two numbers should be multiplied the one into the other. As the figure ABCD is a parallelogram, and is said to be contained under the two right lines AB and AC, which contain the right angle BAC, or under the two right lines AC and CD, for they likewise contain the right angle ACD: of which 2. lines the one, namely, AB is the length, and the other, namely, AC is the breadth. And if we imagine the line AC to be drawn or moved directly according to the length of the line AB, or contrary wise the line AB to be moved directly according to the length of the line AC, you shall produce the whole rectangle parallelogram ABCD which is said to be contained of them: even as one number multiplied by an other produceth a plain and right angled superficial number, as ye see in the figure here set, where the number of six or six unities, is multiplied by the number of five or by five unities: of which multiplication are produced 30. which number being set down and described by his unities representeth a plain and a right angled number. Wherefore even as equal numbers multipled by equal numbers produce numbers equal the one to the other: so rectangle parallelograms which are comprehended under equal lines are equal the one to the other.   


Second definition.  2. In every parallelogram, one of those parallelograms, which soever it be, which are about the diameter, together with the two supplements, is called a Gnomon.  
Those particular parallelograms are said to be about the diameter of the parallelogram, which have the same diameter which the whole parallelogram hath. And supplements are such, which are without the diameter of the whole parallelogram. As of the parallelogram ABCD the partial or particular parallelograms GKCF and EBKH are parallelograms about the diameter, for that each of them hath for his diameter a part of the diameter of the whole parallelogram. As CK and KB the particular diameters, are parts of the line CB, which is the diameter of the whole parallelogram. And the two parallelograms AEGK and KHFD, are supplements, because they are without the diameter of the whole parallelogram, namely, CB. Now any one of those partial parallelograms about the diameter together with the two supplements make a gnomon. As the parallelogram EBKH, with the two supplements AEGK and KHFD make the gnomon FGEH. Likewise the parallelogram GKCF with the same two supplements make the gnomon EHFG. And this definition of a gnomon extendeth itself, and is general to all kinds of parallelograms, whether they be squares or figures of one side longer or Rhombus or Romboides. To be short, if you take away from the whole parallelogram one of the partial parallelograms which are about the diameter whether ye will, the rest of the figure is a gnomon.  
Campane after the last proposition of the first book addeth this proposition. 
A proposition added by Campane after the last proposition of the first book.  Two squares being given, to adjoin to one of them a Gnomon equal to the other square: which, for that as than it was not taught what a Gnomon is, I there omitted, thinking that it might more aptly be placed here. The doing and demonstration whereof, is thus.  
Suppose that there be two squares AB and CD: unto one of which, namely, unto AB, it is required to add a Gnomon equal to the other square, namely, to CD. Produce the side BF of the square AB directly to the point E. and put the line FE equal to the side of the square CD. And draw a line from E to A. Now then forasmuch as EFA is a rectangle triangle, therefore (by the 47. of the first) the square of the line EA is equal to the squares of the lines EF & FA. But the square of the line EF is equal to the square CD, & the square of the side FA is the square AB. Wherefore the square of the line AE is equal to the two squares CD and AB. But the sides EF and FA are (by the 21. of the first) longer than the side AE, and the side FA is equal to the side FB. Wherefore the sides EF and FB are longer than the side AE. Wherefore the whole line BE is longer than the line AE. From the line BE cut of a line equal to the line AE, which let be BC. And (by the 46. proposition) upon the line BC describe a square, which let be BCGH: which shallbe equal to the square of the line AE, but the square of the line AE is equal to the two squares AB and DC. Wherefore the square BCGH is equal to the same squares. Wherefore forasmuch as the square BCGH is composed of the square AB and of the gnomon FGAH, the said gnomon shallbe equal unto the square CD: which was required to be done.  

An other more ready way after Pelitarius.  
Suppose that there be two squares, whose sides let be AB and BC. It is required unto the square of the line AB, to add a gnomon equal to the square of the line BC. Set the lines AB and BC in such sort that they make a right angle ABC. And draw a line from A to C. And upon the line AB describe a square which let be ABDE. And produce the line BA to the point F, and put the line BF equal to the line AC. And upon the line BF describe a square which let be BFGH: which shallbe equal to the square of the line AC, when as the lines BF and AC are equal: and therefore it is equal to the squares of the two lines AB and BC. Now forasmuch as the square BFGH is made complete by the square ABDE and by the gnomon FEGD, the gnomon FEGD shallbe equal to the square of the line BC: which was required to be done.     

The 1. Theorem. The 1. Proposition. If there be two right lines, and if the one of them be divided into parts how many soever: the rectangle figure comprehended under the two right lines, is equal to the rectangle figures which are comprehended under the line undivided, and under every one of the parts of the other line.  
SVppose that there be two right lines A and BC and let one of them, namely, BC be divided at all adventures in the points D and E. Then I say that the rectangle figure comprehended under the lines A, and BC, is equal unto the rectangle figure comprehended under the lines A, and BD, & unto the rectangle figure which is comprehended under the lines A and DE, and also unto the rectangle figure which is comprehended under the lines A and EC. 
Construction.  For from the point Brayse up (by the 11. of the first) unto the right line BC a perpendicular line BF, & unto the line A (by the third of the first) put the line BG equal, and by the point G (by the 31. of the first) draw a parallel line unto the right line BC and let the same be GM, and (by the self same) by the points D, E, and C, draw unto the line BG these parallel lines DK, 
Demonstration  EL and CH. Now then the parallelogram BH is equal to these parallelograms BK, DL, and EH. But the parallelogram BH is equal unto that which is contained under the lines A and BC. (For it is comprehended under the lines GB & BC, and the line GB is equal unto the line A) And the parallelogram BK is equal to that which is contained under the lines A and BD: (for it is comprehended under the line GB and BD, and BG is equal unto A) And the parallelogram DL is equal to that which is contained under the lines A and DE (for the line DK, that is, BG is equal unto A) And moreover likewise the parallelogram EH is equal to that which is contained under the lines A & EC. Wherefore that which is comprehended under the lines A & BC is equal to that which is comprehended under the lines A & BD, & unto that which is comprehended under the lines A and DE, and moreover unto that which is comprehended under the lines A and EC. If therefore there be two right lines, and if the one of them be divided into parts how many soever, the rectangle figure comprehended under the two right lines, is equal to the rectangle figures which are comprehended under the line undivided and under every one of the parts of the other line: which was required to be demonstrated.  
Because that all the Propositions of this second book for the most part are true both in lines and in numbers, and may be declared by both: therefore have I have added to every Proposition convenient numbers for the manifestation of the same. And to the end the studious and diligent reader may the more fully perceive and understand the agreement of this art of Geometry with the science of Arithmetic, and how near & dear sisters they are together, so that the one cannot without great blemish be without the other, I have here also joined a little book of Arithmetic written by one Barlaam, a Greek author a man of great knowledge. In which book are by the author demonstrated many of the self same proprieties and passions in number, which Euclid in this his second book hath demonstrated in magnitude, namely, the first ten propositions as they follow in order. Which is undoubtedly great pleasure to consider, also great increase & furniture of knowledge. Whose Propositions are set orderly after the propositions of Euclid, every one of Barlaam correspondent to the same of Euclid. And doubtless it is wondered to see how these two contrary kinds of quantity, quantity discrete or number, and quantity continual or magnitude (which are the subjects or matter● of Arithmitique and Geometry) should have in them one and the same proprieties common to them both in very many points, and affections, although not in all. For a line may in such sort be divided, that what proportion the whole hath to the greater part the same shall the greater part have to the less. But that can not be in number. For a number can not so be divided, that the whole number to the greater part thereof, shall have that proportion which the greater part hath to the less, as jordane very plainly proveth in his book of Arithmetic, which thing Campane also (as we shall afterward in the 9 book after the 15. proposition see) proveth. And as touching these ten first propositions of the second book of Euclid, demonstrated by Barlaam in numbers, they are also demonstrated of Campane after the 15. proposition of the 9 book, whose demonstrations I mind by God's help to set forth when I shall come to the place. They are also demonstrated of jordane that excellet learned author in the first book of his Arithmetic. In the mean time I thought it not amiss here to set forth the demonstrations of Barlaam, for that they give great light to the second book of Euclid, besides the inestimable pleasure, which they bring to the studious considerer. And now to declare the first Proposition by numbers. I have put this example following.  
Take two numbers the one undivided as 74. the other divided into what parts and how many you list, as 37. divided into 20. 10. 5. and 2● which altogether make the whole 37. Then if you multiply the number undivided, namely, 74, into all the parts of the number divided as into 20. 10. 5. and 2. you shall produce 1480. 740. 370. 148. which added together make 2738: which self number is also produced if you multiply the two numbers first given the one into the other. As you see in the example on the other side set.  
 
So by the aid of this Proposition is gotten a compendious way of multiplication by breaking of one of the numbers into his parts: which oftentimes serveth to great use in workings chi●●ly in the rule of proportions. The demonstration of which proposition followeth in Barlaam. But ●irst are put of the author these principles following.  


Barlaam.  ¶ Principles.  
1. A number is s●yd to multiply an other number: when the number multiplied is so oftentimes added to itself, as there be unities in the number, which multiplieth: whereby is produced a certain number which the number multiplied measureth by the unities which are in the number which multiplieth.  
2. And the number produced of that a multiplication is called a plain or superficial number.  
3. A square number is that which is produced of the multiplicatian of any number into itself.  
4. Every less number compared to a greater is said to be a part of the greater, whether the less measure the greater, or measure it not.  
5. Numbers, whom one and the self same number measureth equally, that is, by one and the self same number are equal the one to the other.  
6. Numbers that are equemultipl●ces to one and the self same number, that is, which contain one and the same number equally and alike, are equal the one to the other.   

The first Proposition.  
Two numbers being given, if the one of them be divided into any numbers how many soever: 
Barlaam.  the plain or superficial number which is produced of the multiplication of the two numbers first given the one into the other, shall be equal to the superficial numbers which are produced of the multiplication of the number not divided into every part of the number divided.  
Suppose that there be two numbers AB and C. And divide the number AB into certain other numbers how many soever, as into AD, DE, and EB. Then I say that the superficial number which is produced of the multiplication of the number C into the number AB is equal to the superficial numbers which are produced of the multiplication of the number C into the number AD, and of C into DE, and of C into EB. For let F be the superficial number produced of the multiplication of the number C into the number AB, and let GH be the superficial number produced of the multiplication of C into AD: And let high be produced of the multiplication of C into DE: a●d finally of the multiplication of C into EB let there be produced the number IK. Now forasmuch as AB multiplying the number C produced the number F: therefore the number C measureth the number F by the unities which are in the number AB. And by the same reason may be proved that the number C doth also measure the number GH, by the unities which are in the number AD, and that it doth measure the number high by the unities which are in the number DF and finally that it measureth the number IK by the unities which are in the number EB. Wherefore the number C measureth the whole number GK by the unities which are in the number AB. But it before measured the number F by the unities which are in the number AB, wherefore either of these numbers F and GK is equemultiplex to the number C. But numbers which are equemultiplices to one & the self same numbers are equal the one to the other (by the 6. definition). Wherefore the number F is equal to the number GK. But the number F is the superficial number produced of the multiplication of the number C into the number AB: and the number GK is composed of the superficial numbers produced of the multiplication of the number C not divided into every one of the numbers AD, DE, and EB. If therefore there be two numbers given and the one of them be divided etc. Which was required to be proved.    

The 2. Theorem. The 2. Proposition. If a right line be divided by chance, the rectangles figures which are comprehended under the whole and every one of the parts, are equal to the square which is made of the whole.  
SVppose that the right line AB be by chance denied in the point C. Then I say that the rectangle figure comprehended under AB and BC together with the rectangle comprehended under AB and AC is equal unto the square made of AB. 
Construction.  Describe (by the 46. of the first) upon AB a square ADEB: and (by the 31 of the first) by the point C draw a line parallel unto either of these lines AD a●d BE, and let the same be C●. 
Demonstration  Now is the parallelogram AE equal to the parallelograms AF and CE, by the first of this book. But AE is the square made of AB. And AF is the rectangle parallelogram comprehended under the lines BA and AC: for it is comprehended under the lines DA and AC: but the line AD is equal unto the line AB. And likewise the parrallelogramme CE is equal to that which is contained under the lines AB and BC: for the line BE is equal unto the line AB. Wherefore that which is contained under BA and AC together with that which is contained under the lines AB and BC, is equal to the square made of the line AB. If therefore a right line be divided by chance, the rectangle figures which are comprehended under the whole, and every one of the parts, are equal to the square which is made of the whole: which was required to be demonstrated.  

another demonstration of Campane.  
Suppose that the line AB be divided into the lines AC, CD, and DB. Then I say that the square of the whole line AB, which let be AEBF, is equal to the rectangle figures which are contained under the whole and every one of the parts: for take the line K, which let be equal to the line A B. Now then by the first proposition the rectangle figure contained under the lines AB and K, is equal to the rectangle figures contained under the line K and all the parts of the line AB. But that which is contained under the lines K and AB is equal to the square of the line AB, and the rectangle figures contained under the line K and all the parts of AB, are equal to the rectangle figures contained under the line AB and all the parts of the line AB: for the lines AB and K are equal: wherefore that is manifest which was required to be proved.   

An example of this Proposition in numbers.  
Take a number, as 11. and divide it into two parts, namely, 7. and 4: and multiply 11. into 7, and then into 4. and there shallbe produced 77. and 44: both which numbers added together make 121. which is equal to the square number produced of the multiplication of the number 11. into himself, as you see in the example.  
 
The demonstration whereof followeth in Barlaam.  

The second Proposition.  

Barlaam.  If a number given be divided into two other numbers: the superficial numbers, which are produced of the multiplication of the whole into either part, added together, are equal to the square number of the whole number given.  
Suppose that the number given be AB: and let it be divided into two other numbers AC and CB. Then I say that the two superficial numbers, which are produced of the multiplication of AB into AC, and of AB into BC, those two superficial numbers (I say) being added together, shallbe equal to the square number produced of the multiplication of the number AB into itself. For let the number AB multiplying itself produce the number D. Let the number AC also multiplying the number AB produce the number EF: again let the number CB multipliing the self same number AB produce the number FG. Now forasmuch as the number AC multiplying the number AB produced the number EF: therefore the number AB measureth the number EF by the unities which are in AC. Again forasmuch as the number CB multiplied the number AB and produced the number FG: therefore the number AB measureth the number FG by the unities which are in the number CB. But the same number AB before measured the number EF by the unities which are in the number AC. Wherefore the number AB measureth the whole number ●G by the unities which are in AB. Farther forasmuch as the number AB multiplying itself produced the number D: therefore the number AB measureth the number D by the unities which are in himself. Wherefore it measureth either of these numbers, namely, the number D, and the number EG, by the unities which are in himself. Wherefore how multiplex the number D is to the number AB, so multiplex is the number EG to the same number AB. But numbers which are equemultiplices to one and the self same number● are equal the one to the other. Wherefore the number D is equal to the number EG. And the number D is the square number made of the number AB, and the number EG is composed of the two superficial numbers produced of AB into BC, and of BA into AC. Wherefore the square number produced of the number AB is equal to the superficial numbers, produced of the number AB into the number BC, and of AB into AC, added together. If therefore a number be divided into two other numbers etc. which was required to be proved.     

The 3. Theorem. The 3. Proposition. If a right line be divided by chance: the rectangle figure comprehended under the whole and one of the parts, is equal to the rectangle figure comprehended under the parts, & unto the square which is made of the foresaid part.  
SVppose that the right line given AB be divided by chance in the point C. Then I say that the rectangle figure comprehended under the lines AB and BC is equal unto the rectangle figure comprehended under the lines AC and CB, and also unto the square which is made of the line BC. 
Construction.  Describe (by the 46. of the first) upon the line BE square CDEB: and (by the second petition) extend ED unto F. And by the point A draw (by the 31. of the first) a line parallel unto either of these lines CD and BE, and let the same be AF. Now the parallelogram AE is equal unto the parallelograms AD and CE. 
Demonstration  And AE is the rectangle figure comprehended under the lines AB and BC. For it is comprehended under the lines AB and BE, which line BE is equal unto the line BC. And the parallelogram AD is equal to that which is contained under the lines AC and CB: for the line DC is equal unto the line CB And DB is the square which is made of the line CB. Wherefore the rectangle figure comprehended under the lines AB and BC is equal to the rectangle figure comprehended under the lines AC and CB & also unto the square which is made of the line BC. If therefore a right line be divided by chance, the rectangle figure comprehended under the whole and one of the parts, is equal to the rectangle figure comprehended under the parts, and unto the square which is made of the foresaid part: which was required to be proved.  

An example of this Proposition in numbers.  
Suppose a number, namely, 14. to be divided into two parts 8. and 6. The whole number 14. multiplied into 8. one of his parts, produceth 112: the parts 8. & 6. multiplied the one into the other produce 48, which added to 64 (which is the square of 8. the former part of the number) amounteth also to 112: which is equal to the former sum. As you see in the example.  
 
The demonstration hereof followeth in Barlaam.  

The third proposition.  

Barlaam.  If a number given be divided into two numbers: the superficial number which is produced of the multiplication of the whole into one of the parts, is equal to the superficial number which is produced of the parts the one into the other, and to the square number produced of the foresaid part.  
Suppose that the number given be AB, which let be divided into two numbers AC and CB. Then I say that the superficial number which is produced of the multiplication of the number AB into the number BC, is equal to the superficial number which is produced of the multiplication of the number AC into the number CB, and to the square number produced of the number CB. For let the number AB multiplying the number CB produce the number D. And let the number AC multiplying the number CB produce the number EF: and finally let the number CB multiplying himself produce the number FG. Now forasmuch as the number AB multiplying the 〈…〉 〈…〉 was required to be proved.     

The 4. Theorem. The 4. Proposition. If a right line be divided by chance, the square which is made of the whole line is equal to the squares which are made of the parts, & unto that rectangle figure which is comprehended under the parts twice.  
SVppose that the right line AB be by chance divided in the point C. Then I say that the square made of the line AB is equal unto the squares which are made of the lines AC and CB, and unto the rectangle figure contained under the line● AC and CB awise. 
Construction.  Describe (by the ●●. of the first) upon the line AB a square ADEB: and draw a line from B to D, and (by the 31. of the first) by the point C draw a line parallel unto either of these lines AD and BE c●tting the diameter BD in the point G, and let the same be CF. And (by the point G (by the self same) draw a line parallel unto either of these lines AB and DE, and let the same be HK. 
Demonstration.  And forasmuch as the line CF is a parallel unto the line AD, and upon them falleth a right line BD: therefore (by the 29. of the first) the outward angle CGB is equal unto the inward and opposite angle ADB. But the angle ADB is (by the 5. of the first) equal unto the angle ABD: for the side BA is equal unto the side AD (by the definition of a square). Wherefore the angle CGB is equal unto the angle GBC: wherefore (by the 6. of the first) the side BC is equal unto the side CG. But CB is equal unto GK● and CG is equal unto KB wherefore GK is equal unto KB. Wherefore the figure CGKB consisteth of four equal sides. I say also that it is a rectangle figure. For forasmuch as CG is a parallel unto BK & upon them falleth a right line CB, therefore (by the 29. of the 1.) the angles KBC, and GCB are equal unto two right angles. But the angle KBC is a right angle, wherefore the angle. BCG is also a right angle. Wherefore (by the 34. of the first) the angles opposite unto them, namely, CGK, and GKB are right angles. Wherefore CGKB is a rectangle figure. And it was before proved that the sides are equal. Wherefore it is a square, and it is described upon the line B ●. And by the same reason also HF is a square, and is described upon the line HG, that is, upon the line AC. Wherefore the squares HF and CK are made of the lines AC and CB. And forasmuch as the parallelogram AG is (by the 43. of the first) equal unto the parallelogram GOE And AG is that which is contained under AC and CB, for CG is equal unto CB, wherefore GE is equal to that which is contained under AC and CB. Wherefore AG and GE are equal unto that which is comprehended under AC and CB twice. And the squares HF and CK are made of the lines AC and CB. Wherefore these four rectangle figures HF, CK, AG, and GE are equal unto the squares which are made of the lines AC and CB, and to the rectangle figure which is comprehended under the lines AC and CB twice. But the rectangle figures HF, CK, AG, and GE are the whole rectangle figure ADEB which is the square made of the line AB. Wherefore the square which is made of the line AB is equal to the squares which are made of the lines AC and CB, and unto the rectangle figure which is comprehended under the lines AC and CB awise. If therefore a right line be divided by chance, the square which is made of the whole line, is equal to the squares which are made of the parts, & unto the rectangle figure which is comprehended under the parts awise: which was required to be proved.  

another demonstration.  
Hereby it is manifest that the parallelograms which consist about the diameter of a square must needs be squares. 
A Corollary.   
This proposition is of infinite use chiefly in surde numbers. By help of it is made in them addition & substraction, also multiplication in Binomials & residuals. And by help hereofalso is demoustrated that kind of equation, which is, when there are three denominations in natural order, or equally distant, and two of the greater denominations are equal to the third being less On this proposition is grounded the extraction of square roots. And many other things are also by it demonstrated.   

An example of this Proposition in numbers.  
Suppose a number namely, 17. to be divided into two parts 9 and 8. The whole number 17. multiplied into himself, produceth 289. The square numbers of 9 and 8. are 81. and 64: the numbers produced of the multiplication of the parts the one into the other twice are 72. and 72: which two numbers added to the square numbers of 9 and 8. namely, to 81. and 64. make also 289. which is equal to the square number of the whole number 17. As you see in the example.  
 
The demonstration whereof followeth in Barlaam.  

The fourth Proposition.  

Barlaam.  If a number given be divided into two numbers: the square number of the whole, is equal to the square numbers of the parts, and to the superficial number which is produced of the multiplication of the parts the one into the other twice.  
Suppose that the number given be AB: which let be divided into two numbers AC and CB. Then I say that the square number of the whole number AB, is equal to the squares of the parts, that is, to the squares of the numbers AC and CB, and to the superficial number produced of the multiplication of the numbers AC and CB the one into the other twice. Let the square number produced of the multiplication of the whole number AB into himself be D. And let CA multiplied into himself produce the number EF: And CB multiplied into itself let it produce GH: and finally of the multiplication of the numbers AC and CB the one into the other twice let there be produced either of these superficial numbers FG and HK. Now forasmuch as the number AC multiplying itself produced the number EF: therefore the number AC measureth the number EF by the unities which are in itself. And forasmuch as the number CB multiplied the number CA and produced the number FG: therefore the number AC measureth the number FG by the unities which are in the number CB. But it before also measured the number EF by the unities which are in itself. Wherefore the number AB multiplying the number AC produceth the number EG. And therefore the number EG is the superficial number produced of the multiplication of the number BA into the number AC. And by the same reason may we prove that the number GK is the superficial number produced of the multiplication of the number AB into the number BC. Farther the number D is the square of the number AB. But if a number be divided into two numbers, the square of the whole number is equal to the two superficial numbers which are produced of the multiplication of the whole into either the parts (by the 2. Theorem.) Wherefore the square number D is equal to the superficial number EK. But the number EK is composed of the squares of the numbers AC and CB, and of the superficial number which is produced of the multiplication of the number AC and CB the one into the other twice: & the number D is the square of the whole number AB. Wherefore the square number produced of the multiplication of the number AB into himself, is equal to the square numbers of the parts, that is, to the square numbers of the numbers AC and CB, and to the superficial number produced of the multiplication of the numbers AC and CB, the one into the other twice. If therefore a number given be divided into two numbers etc. Which was required to be proved.     

The 5. Theorem. The 5. Proposition. If a right line be divided into two equal parts, & into two unequal parts: the rectangle figures comprehended under the unequal part●● of the whole, together with the square of that which is between the sections, is equal to the square which is made of the half.  
SVppose that the right line AB be divided into two equal parts in the point C, and into two unequal parts in the point D. Then I say that the rectangle figure comprehended under AD and DB together with the square which is made of CD, is equal to the square which is made of CB. 
Construction.  Describe (by the 46. of the first) upon CB a square, and let the same be CEFB. And (by the first petition) draw a line from E to B. And by the point D draw (by the 3●. of the first) a line parallel unto each of these lines CE and BF cutting the diameter BE in the point H, and let the same be DG. And again (by the self same) by the point H draw a line parallel unto each of these lines AB and EF, and let the same be KO: and let KO be equal unto AB. And again (by the self same) by the point A draw a line parallel unto either of these lines CL and BO, 
Demonstration  and let the same be AK. And forasmuch as (by the 43. of the first) the supplement CH is equal to the supplement HF● put the figure DO common unto them both. Wherefore the whole figure CO is equal to the whole figure DF. But the figure CO is equal unto the figure AL, for the line AC is equal unto the line CB. Wherefore the figure AL also is equal unto the figure DF. Put the figure CH common unto them both. Wherefore the whole figure AH is equal unto the figures DL and DF. But AH is equal to that which is contained under the lines AD and DB, for DH is equal unto DB. And the figures FD & DL are the Gnomon MNX. Wherefore the Gnomon MNX is equal to that which is contained under AD and DB. Put the figure LG common unto them both, which is equal to the square which is made of CD. Wherefore the Gnomon MNX and the figure LG are equal to the rectangle figure comprehended under AD and DB and unto the square which is made of CD. But the Gnomon MNX, and the figure LG are the whole square CEFB, which is made of BC. Wherefore the rectangle figure comprehended under AD and DB, together with the square which is made of CD, is equal to the square which is made of CB. If therefore a right line be divided into two equal parts, and into two unequal parts, the rectangle figure comprehended under the unequal parts of the whole, together with the square of that which is between the sections, is equal to the square which is made of the half: which was required to be proved.  
This Proposition also is of great use in Algebra. By it is demonstrated that equation wherein the greatest and lest karectes or numbers are equal to the middle.  

An example of this proposition in numbers.  
Take any number as 20: and divide it into two equal parts 10. and 10. and then into two unequal parts as 13. and 7. And take the difference of the half to one of the unequal parts which is 3. And multiply the unequal parts, that is, 13 and 7. the one into the other, which make 91. take also the square of 3. which is 9 and add it to the foresaid number 91: and so shall there be made 100 Then multiply the half of the whole number into himself, that is, take the square of 10. which is 100 which is equal to the number before produced of the multiplication of the unequal parts the one into the other, & of the difference into itself which is also 100 As you see in the example.  
The demonstration whereof followeth in Barlaam.  

The fifth proposition.  
If an even number be divided into two equal parts, and again also into two unequal parts: the superficial number which is produced of the multiplication of the unequal parts the one into the other, together with the square of the number set between the parts, is equal to the square of half the number.  
Suppose that AB be an even number: which let be divided into two equal numbers AC and CB, and into two unequal numbers AD and DB. Then I say, that the square number which is produced of the multiplication of the half number CB into itself, is equal to the superficial number produced of the multiplication of the unequal numbers AD and DB the one into the other, and to the square number produced of the number CD which is set between the said unequal parts. Let the square number produced of the multiplication of the half number CB into itself be E. And let the superficial number produced of the multiplication of the unequal numbers AD and DB the one into the other, be the number FG: and let the square of the number DC which is set between the parts be GH. Now forasmuch as the number BC is divided into the numbers BD and DC, therefore the square of the number BC, that is, the number E, is equal to the squares of the numbers BD and DC, and to the superficial number which is composed of the multiplication of the numbers BD and DC the one into the other twice, (by the 4. proposition of this book) Let the square of the number BD be the number KL: & let NX be the square of the number DC: and finally of the multiplication of the numbers BD and DC the one into the other twi●e, let be produced either of these numbers LM and MN. Wherefore the whole number KX is equal to the number E ● And forasmuch as the number BD multiplying itself produced the number KL, therefore it measureth it by the unities which are in itself. Moore over forasmuch as the number CD multiplying the number BD produced the number LM, therefore also DB measureth LM by the unities which are in the number CD: but it before measured the number KL by the unities which are in itself. Wherefore the number DB measureth the whole number KM by the unities which are in CB. But the number CB is equal to the number CA Wherefore the number DB measureth the number KM by the unities which are in CA Again forasmuch as the number CD multipli●ng the number DB produced the number MN: therefore the number DB measureth the number MN by the unities which are in the number CD: but it before measured the number KM by the unities which are in the number AC. Wherefore the number BD measureth the whole number KN by the unities which are in the number AD. Wherefore the number FG 〈◊〉 equal to the number KN. For numbers which are equemultiplices to one and the self same number, are equal the one to the other. But the number GH is equal to the number NX: for either of them is supposed to be the squ●re of the number CD. Wherefore the whole number KX is equal to the whole number FH. But the number KX is equal to the number E ● Wherefore also the number FH is equal to the number E. And the number FH is the superficial number produced of the multiplication of the numbers AD and DB the one into the other together with the square of the number DC. And the number E is the square of the number CB. Wherefore the superficial number produced of the multiplication of the unequal parts AD and DB the one into the other, together with the square of the number DC which is set between those unequal parts, is equal to the square of the number CB, which is the half of the whole number AB. If therefore an even number be divided into two equal parts, etc. which was required to be proved.     

The 6. Theorem. The 6. Proposition. If a right line be divided into two equal parts, and if unto it be added an other right line directly, the rectangle figure contained under the whole line with that which is added, & the line which is added together with the square which is made of the half, is equal to the square which is made of the half line and of that which is added as of one line.  
SVppose that the right line AB be divided into two equal parts in the point C: & let there added unto it an other right line DB directly, that is to say, which being joined unto AB make both one right line AD. Then I say, that the rectangle figure comprehended under AD and DB, together with the square which is made of BC is equal to the square which, is made of DC. Describe (by the 46. of the 1.) upon CD a square CEFD, and (by the first petition) draw a line from D to E: and (by the 31. of the first) by the point B draw a line parallel unto either of these lines EC & DF, cutting the diameter DE in the point H, and let the same be BG, & (by the self same) by the point H draw to either of these lines AD and EF a parallel line KM: and moreover by the point A draw a line parallel to either of these lines CL and DM: and let the same be AK. And forasmuch as AC is equal unto CB, therefore (by the 36. of the first) the figure AL is equal unto the figure CH. But (by the 43. of the first) CH is equal un-the figure HF, wherefore AL is equal unto HF. Put the figure CM common to them both, wherefore the whole line AM is equal unto the gnomon NXO. But AM is that which is contained under AD and DB: for DM is equal unto DB: wherefore the gnomon NXO, is equal unto the rectangle figure contained under AD and DB. Put the figure LG common to them both, which is equal to the square which is made of CB. Wherefore the rectangle figure which is contained under AD and DB together with the square which is made of CB is equal to the gnomon NXO, and unto LG. But the gnomon NXO and LG are the whole square CEFD which is made of CD. Wherefore the rectangle figure contained under AD and DB together with the square which is made of CB is equal to the square which is made of CD. If therefore a right line be divided into two equal parts, and unto it be added an other right line directly: the rectangle figure contained under the whole line with that which is added, and the line which is added, together with the square which is made of the half, is equal to the square which is made of the half line and of that which is added as of one line: which was required to be demonstrated.  
By this Proposition (besides many other uses) is in Algebra demonstrated that equation wherein the two less numbers be equal to the number of the greatest denomination.  

An example of this proposition in numbers.  
Take any even number as 18. and add unto it any other number as 3. which make in all 21. And multiply 21. into the number added, namely, into 3. which maketh 63. Take also the half of the whole even number, that is, of 18. which is 9 And multiply 9 into itself which maketh 81. which add unto 63. (the number produced of the whole even number, and the number added into the number added) and you shall make 144. Then add 9 the half of the whole even number unto 3. the number added which maketh 12. And multiply 12. into itself, that is, take the square of 12. which is 144. which is equal to the number composed of the multiplication of the whole number and the number added into the number added, and of the square of the number added, which is also 144. As you see in the example.  
 
The demonstration whereof followeth in Barlaam.   

The sixth Proposition.  
If an even number be divided into two equal numbers, and unto it be added some other number● the superficial number which is made of the multiplication of the number composed of the whole number and the number added, into the number added, together with the square of the half number, is equal to the square of the number composed of the half and the number added.  
Suppose that AB be an even number, and let it be divided into two equal numbers AC and CB: and unto it let there be added an other number BD. Then I say that the super●iciall number produced of the multiplication of the number AD into the number DB is equal to the square of the number CD. For let the square number of the number CD ● be the number E, and let the superficial number produced of the multiplication of the number AD into the number DB be the number FG: and finally let the square number of CB be the number GH. And forasmuch as the square of the number CD is (by the 4. proposition) equal to the squares of the numbers DB and BC together with the superficial number which is produced of the multiplication of the numbers DB and BC the one into the other twice. Let the square of the number BD be the number KL: and let the superficial numbers produced of the multiplication o● the numbers DB and BC the one into the other twice be either of these number● LM and MN● and finally let the square of the number BC be the number NX. Wherefore the whole number KX shall be equal to the square of the number CD. But the square of the number CD is ●he ●●mber E. Wherefore the number KX is equal to the number E. And forasmuch as the number BD multiplying it self produced the number KL: therefore the number BD measureth the number KL, by the unities which are in itself, but it also measureth the number LM by the unities which are in the number CB. Wherefore the number DB measureth the whole number KM by the unities which are in the number CD. The number DB also measureth the number MN by the unities which are in the number CB: & the number CB is equal to the number CA by supposition. Wherefore the number DB measureth the whole number KN by the unities which are in the number AD. But the number DB doth also measure the number FG by the unities which are in the number AD: for by supposition the number FG is the superficial number produced of the multiplication of the numbers AD and DB the one into the other. Wherefore the number FG is equal to the number KN. But the number HG is equal to the number NX: for either of them is the square number of the number CB. Wherefore the whole number FH is equal to the number KX: and the number KX is proved to be equal to the number E. Wherefore the number FH shall also be equal to the number E. And the number FH is the superficial number produced of the multiplication of the numbers AD and DB the one into the other, together with the square of the number CB: and the number E is the square of the number CD. Wherefore the superficial number produced of th● multiplication of the numbers AD and DB the one into the other, together with the square of the number CB; is equal to th● square of the number CD. If therefore an even number etc.    

The 7. Theorem. The 7. Proposition. If a right line be divided by chance, the square which is made of the whole together with the square which is made of one of the parts, is equal to the rectangle figure which is contained under the whole and the said part twice, and to the square which is made of the other part.  
SVppose that the right line AB be divided by chance in the point C. Then I say that the square which is made of AB, together with the square which is made of BC, is equal unto the rectangle figure which is contained under the lines AB and BC twice, and unto the square which is made of AC. Describe (by the 46. of the first) upon AB a square ADEB, and make complete the figure. And forasmuch as (by the 43. of the first) the ●igure AG is equal unto the figure GOE Put the figure CF common to them both: wherefore the whole figure AF is equal to the whole figure CE. Wherefore the figures AF and CE are double to the figure AF. But the figures AF and CE are the gnomon KLM, and the square CF: wherefore the gnomon KLM, and the square CF is double to the figure AF. But the double to AF is that which is contained under AB and BC twice, for BF is equal unto BC. Wherefore the gnomon KLM and the square CF is equal unto the rectangle figure contained under AB and BC twice. Put the figure DG common unto them both, which is the square made of AC. Wherefore the gnomon KLM and the squares BG and GD are equal unto the rectangle figure which is contained under AB & BC twice, & unto the square which is made of AC. But the gnomon KLM, & the squares BG, & DG are the whole square BADE, & the part or square CF, which squares are made of the lines AB & of BC, therefore the squares which are made of AB & BC are equal unto the rectangle figure which is contained under AB and BC twice, and also unto the square of AC. If therefore a right line be divided by chance: the square which is made of the whole together with the square which is made of one of the parts, is equal to the rectangle figure which is contained under the whole and the said part twice, and to the square which is made of the other part: which was required to be demonstrated.  

Flussates addeth unto this Proposition this Corollary.  
The squares of two unequal lines do exceed the rectangle figures contained under the said lines by the square of the excess whereby the greater line exceedeth the less.  
For if the line AB be the greater, and the line BC the less, it is manifest that the squares of AB and BC are equal to the rectangle figure contained under the lines AB and BC twice, and moreover to the square of the line AC, whereby the line AB exceedeth the line BC.  
By this proposition most wonderfully was found out the extraction of root squares in irrational numbers, beside many other strange things.   

An example of this proposition in numbers.  
Take any number as 13. and divide it into two parts as into 4. & 9 Take the square of 13. which is 169. take also the square of 4. which is 16. and add these two squares together which make 185. Then multiply the whole number 13. into 4. the foresaid part twice, and you shall produce 52. and 52● take also the square of the other part, that is, of 9 which is 81. And add it to the productes of 13. into 4. twice, that is, unto 52. and 52. and those three numbers added together shall make 185. which is equal to the number composed of the squares of the whole and of one of the parts, which is also 185. As you see in the example.  
 
The demonstration whereof followeth in Barlaam.   

The seventh proposition.  
If a number be divided into two numbers: the square of the whole number together with the square of one of the parts, is equal to the superficial number produced of the multiplication of the whole number into the foresaid part twice, together with the square of the other part.  
   

The 8. Theorem. The 8. Proposition. If a right line be divided by chance, the rectangle figure comprehended under the whole and one of the parts four times, together with the square which is made of the other part, is equal to the square which is made of the whole and the foresaid part as of one line.  
SVppose that there be a certain right line AB, and let it be divided by chance in the point C. Then I say that the rectangle figure comprehended under AB and BC four times, together with the square which is made of AC is equal to the square made of AB and BC as of one line. Extend the line AB (by the second petition). And (by the third of the first) unto CB put an equal line BD. And (by the 46. of the first) describe upon AD a square AEFD. And describe a double figure. And forasmuch as CB is equal unto BD, but CB is equal unto GK (by the 34. of the first) and likewise BD is equal unto KN, wherefore GK also is equal unto KN: and by the same reason also PR, is equal unto RO. And forasmuch as BC is equal unto BD, and GK unto KN, therefore (by the 36. of the first) the figure CK is equal unto the figure KD, and the figure GR is equal unto the figure RN. Butler (by the 43. of the 1) the figure CK is equal unto the figure RN: for they are the supplements of the parallelogram CO. Wherefore the figure KD also is equal unto the figure NR. Wherefore these figures DK, CK, GR, RN, are equal the one to the other. Wherefore those four are quadruple to the figure CK. Again forasmuch as CB is equal unto BD, but BD is equal unto BK, that is, unto CG. And CB is equal unto GK that is unto GP: therefore CG is equal unto GP. And forasmuch as CG is equal unto GP, and PR is equal unto RO, therefore the figure AG is equal unto the figure MP, and the figure PL is equal unto the figure RF. But the figure MP is (by the 43. of the first) equal unto the figure PL, for they are the supplements of the parallelogram ML: wherefore the figure also AG is equal unto the figure RF. Wherefore these four figures AG, MP, PL, and RF are equal the one to the other: wherefore those four are quadruple to the figure AG. And it is proved, that these four figures CK, KDGR, RN, are quadruple to the figure CK. Wherefore the eight figures which contain the gnomon STV, are quadruple to the figure AK. And forasmuch as the figure AK, is that which is contained under the lines AB and BD, for the line BK is equal unto the line BD: therefore that which is contained under the lines AB and BD four times is quadruple unto the figure AK. And it is proved that the gnomon STV is quadruple to AK● Wherefore that which is contained under the lines AB and BD four times is equal unto the gnomon STV. Put the figure XH which is equal to the square made of AC common unto them both. Wherefore the rectangle figure comprehended under the lines AB and BD four times together with the square which is made of the line AC, is equal to the gnomon STV, and unto the figure XH. But the gnomon STV: and the figure XH are the whole square AEFD, which is made of AD: wherefore that which is contained under the lines AB and BD four times together with the square which is made of AC, is equal to the square which is made of AD. But BD is equal unto BC. Wherefore the rectangle figure contained four times under AB and BC together with the square which is made of AC, is equal unto the square which is made of AD, that is, unto that which is made of AB and BC as of ou● line. If therefore a right line be divided by chance, the rectangle figure comprehended under the whole and one of the parts four times, together with the square which is made on the other part, is equal to the square which is made of the whole and the foresaid part, as of one line, which was required to be demonstrated.  

An example of this Proposition in numbers.  
Take any number as 17. and divide it into two parts, as into 6. and 11. And multiply 17. into 6. namely one of the parts four times, and 〈…〉 produce 102. 102. 102. and 102. Take also the square of 11. the other part, which is 121: and add it unto the four numbers produced of the whole 17. into the part 6. four times, & you shall make 529. Then add the whole number 17. to the foresaid part 6. which make 23. & take the square of 23. which is 529. which is equal to the number composed of the whole into the said part four times, and of the square of the other part, which number composed is also 529. As you see in the example.  
 
The demonstration whereof followeth in Barlaam.   

The eight proposition.  
If a number be divided into two numbers, the superficial number produced of the multiplication of the whole into one of the parts four times, together with the square of the other part, is equal to the square of the number composed of the whole number and the foresaid part.  
   

The 9 Theorem. The 9 Proposition. If a right line be divided into two equal parts, and into two unequal parts, the squares which are made of the unequal parts of the whole, are double to the squares, which are made of the half line, and of that line which is between the sections.  
SVppose that a certain right line AB be divided into two equal parts in the point C, and into two unequal parts in the point D. Then I say that the squares which are made of the lines AD and DB, are double to the squares which are made of the lines AC and CD. 
Constr●ction.  For (by the 11. of the first) erect from the point C to the right line AB a perpendicular line CE. And let CE (by the 3. of the first) be put equal unto either of these lines AC & CB: and (by the first petition) draw lines from A to E, and from E to B. And (by the 31. of the first) by the point D draw unto the line EC a parallel line, and let the same be DF: and (by the self same) by the point F draw unto AB a line parallel, and let the same be FG. And (by the first petition) draw a line from A to F. 
Demonstration.  And forasmuch as AC is equal unto CE, therefore (by the 5. of the first) the angle EAC is equal unto the angle CEA. And forasmuch as the angle at the point C is a right angle: therefore the angles remaining EAC, and AEC, are equal unto one right angle, wherefore each of these angles EAC and AEC is the half of a right angle. And by the same reason also each of these angles EBC and CEB is the half of a rig●t angle. Wherefore the whole angle AEB is a right angle. And forasmuch as the angle GEF is the half of a right angle, but EGF is a right angle. For (by the 29 of the first) it is equal unto the inward and opposite angle, that is, unto ECB: wherefore the angle remaining EFG is the half of a right angle. Wherefore (by the 6. common sentence) the angle GEF is equal unto the angle EFG. Wherefore also (by the 6. of the first) the side EG is equal unto the side FG. Again forasmuch as the angle at the point B is the half of a right angle, but the angle FDB is a right angle, for it also (by the 29. of the first) is equal unto the inward and opposite angle ECB. Wherefore the angle remaining BFD is the half of a right angle. Wherefore the angle at the point B is equal unto the angle DFB. Wherefore (by the 6. of the first) the side DF is equal unto the side DB. And forasmuch as AC is equal unto CE, therefore the square which is made of AC is equal unto the square which is made of CE. Wherefore the squares which are made of CA and CE are double to the square which is made of AC. But (by the 47. of the first) the square which is made of EA is equal to the squares which are made of AC and CE (For the angle ACE is a right angle) wherefore the square of AE is double to the square of AC. Again forasmuch as EG, is equal unto GF, the square therefore which is made of EG is equal to the square which is made of GF. Wherefore the squares which are made of GE and GF are double to the square which is made of GF. But (by the 47. of the first) the square which is made of EF is equal to the squares which are made of EG and GF. Wherefore the square which is made of EF is double to the square which is made of GF. But GF is equal unto CD. Wherefore the square which is made of EF is double to the square which is made of CD. And the square which is made of AE is double to the square which is made of AC. Wherefore the squares which are made of AE and EF are double to the squares which are made of AC and CD. But (by the 47. of the first) the square which is made of AF is equal to the squares which are made of AE and EF (For the angle AEF is a right angle). Wherefore the square which is made of AF is double to the squares which are made of AC & CD. But (by the 47. of the first) the squares which are made of AD and DF are equal to the square which is made of AF. For the angle ot the point D is a right angle. Wherefore the squares which are made of AD and DF are double to the squares which are made of AC and CD. But DF is equal unto DB. Wherefore the squares which are made of AD and DB, are double to the squares which are made of AC and CD. If therefore a right line be divided into two equal parts and into two unequal parts, the squares which are made of the unequal parts of the whole, are double to the squares which are made of the half line, and of that line which is between the sections: which was required to be proved.  

¶ An example of this proposition in numbers.  
Take any even number as 12. And divide it first equally as into 6. and 6. & then unequally as into 8. & 4. And take the difference of the half to one of the unequal parts which is 2. And take the square numbers of the unequal parts 8, and 4, which are 6●● and 16: and add them together, which make 80. Then take the squares of the half 6. and of the difference 2: which are 36, and 4: which added together make 40. Unto which number, the number composed of the squares of the unequal parts, which i● 80, i● double. As you see in the example.  
 
The demonstration whereof followeth in Barlaam.   

The ninth Proposition.  
If a number be divided into two equal numbers, and again be divided into two inequal parts: the square numbers of the unequal numbers, are double to the square which is made of the multiplication of the half number into itself, together with the square which is made of the number set between them.  
   

The 10. Theorem. The 10. Proposition. If a right line be divided into two equal parts, & unto it be added an other right line directly: the square which is made of the whole & that which is added as of one line, together with the square which is made of the line which is added, these two squares (I say) are double to these squares, namely, to the square which is made of the half line, & to the square which is made of the other half line and that which is added, as of one line.  
SVppose that a certain right line AB be divided into two equal parts in the point C. And unto it let there be added an other right line directly, namely, BD. Then I say, that the squares which are made of the lines AD and DB are double to the squares which are made of the lines AC and CD. 
Construction.  Raise up (by the 11. of the first) from the point C unto the right line ACD a perpendicular line, and let the same be CE. And let CE (by the 3. of the first) be made equal unto either of these lines AC and CB. And (by the first petition) draw right lines from E to A, and from E to B. And (by the 31. of the first) by the point E, draw a line parallel unto CD, and let the same be EF. And (by the self same) by the point D draw a line parallel unto CE and let the same be DF. And forasmuch as upon these parallel lines CE & DF lighteth a certain right line EF, therefore (by the 29. of the first) the angles CEF and EFD are equal unto two right angles. Wherefore the angles FEB, and EFD are less than two right angles. But lines produced from angles less than two right angles (by the fifth petition) at the length meet together. Wherefore the lines EB and FD being produced on that side that the line BD is, will at the length meet together. Produce them and let them meet together in the point G. And (by the first petition) draw a line from A to G. 
Demonstration.  And forasmuch as the line AC is equal unto the line CE, the angle also AEC is (by the 5. of the first) equal unto the angle EAC. And the angle at the point C is a right angle. Wherefore each of these angles EAC & and AEC is the half of a right angle. And by the same reason each of these angles CEB, and EBC is the half of a right angle: Wherefore the angle AEB is a right angle. And forasmuch as the angle EBC is the half of a right angle, therefore (by the 15. of the first) the angle DBC is the half of a right angle. But the angle BD● is a right angle (for it is equal unto the angle DCE, for they are alternate angles) Wherefore the angle remaining DGB is the half of a right angle. Wherefore (by the 6 common sentence of the first) the angle DGB is equal to the angle DBG. Wherefore (by the 6. of the first) the side BD is equal unto the side GD. Again forasmuch as the angle EGF is the half of a right angle: and the angle at the point F is a right angle: for (by the 34. of the first) it is equal unto the opposite angle ECD. Wherefore the angle remaining FEG is the half of a right angle. Wherefore the angle EGF is equal to the angle FEG. Wherefore (by the 6. of the first the side, FE is equal unto the side FG. And forasmuch as EC is equal unto CA, the square also which is made of EC is equal to the square which is made of CA Wherefore the squares which are made of CE and CA are double to the square which is made of AC. But the square which is made of EA is (by the 47. of the first) equal unto the squares which are made of EC and CA Wherefore the square which is made of EA is double to the square which is made of AC● Again forasmuch as GF is equal unto EF, the square also which is made of GF is equal to the square which is made of FE. Wherefore the squares which are made of GF and EF are double to the square which is made of EF. Butler (by the 47. of the first) the square which is made of EG is equal to the squares which are made of GF and EF. Wherefore the square which is made of EG is double to the square which is made of EF. But EF is equal unto CD, wherefore the square which is made of EG is double to the square which is made of CD. And it is proved the square which is made of EA is double to the square which is made of AC. Wherefore the squares which are made of AE and EG are double to the squares which are made of AC and CD. But (by the 47. of this first) the square which is made of AG is equal to the squares which are made of AE and EG. Wherefore the square which is made of AG is double to the squares which are made of AC and CD. But unto the square which is made of AG are equal the squares which are made of AD and DG. Wherefore the squares which are made of AD and DG are double to the squares which are made of AC and DC. But DG is equal unto DB. Wherefore the squares which are made of AD and DB are double to the squares which are made of AC and DC. If therefore a right line be divided into two equal parts, and unto it be added an other line directly, the square which is made of the whole and that which is added, as of one line, together with the square which is made of the line which is added, these two squares (I say) are double to these squares, namely, to the square which is made of the half line, and to the square which is made of the other half line and that which is added, as of one line: which was required to be proved.  

¶ An other demonstration after Pelitarius.  
Suppose that the line AB be divided into two equal parts in the point C. And unto it let there be added an other right line directly, namely, BD. Then I say that the square of AD together with the square of BD is double to the squares of AC and CD.  
The Supplement EH is equal to the parallelogram HP. And the square AH together with the lesser supplement, NO, is equal to the other supplement HD (by the first common sentence so oftentimes repeated as is need) wherefore the two supplements EH and HD are equal to the square AH and to the Gnomon KHLPQO. If therefore unto either of them be added the square QF: the two supplements EH and HD together with the square of QF shall be equal to the square AH, & to the Gnomon KHLPQO and to the square QF. But these three figures do make the two squares AH and HF. Wherefore the two supplements EH and HD together with the square QF are equal to the two squares AH and HF, which was the second thing to be proved. Wherefore the two squares AH and HF being taken twice are equal to the whole square DE together with the square of QF. Wherefore the square DE together with the square QF is double to the squares AH and HF: which was required to be proved.   

¶ An example of this Proposition in numbers.  
Take any even number as 18: and take the half of it which is 9 and unto 18. the whole, add any other number as 3. which maketh 21. Take the square number of 21. (the whole number and the number added) which maketh 441. Take also the square of 3 (the number added) which is 9 which two squares added together make 450. Then add the half number 9 to the number added 3. which maketh 12. And take the square of 9 the half number and of 12. the half number and the number added which squares are 81. and 144. and which two squares also added together make 225: unto which sum the foresaid number 450. is double. As you see in the example.  
 
The demonstration whereof followeth in Barlaam.   

The tenth Proposition.  
If an even number be divided into two equal numbers, and unto it be added any other number: the square number of the whole number composed of the number and of that which is added, and the square number of the number added: these two square numbers (I say) added together, are double to these square numbers, namely, to the square of the half number, and to the square of the number composed of the half number and of the number added.  
Suppose that the number AB being an even number be divided into two equal numbers AC and CB: and unto it let be added an other number BD. Then I say, that the square numbers of the numbers AD and DB are double to the square numbers of AC and CD. For forasmuch as the number AD is divided into the numbers AB and BD: therefore the square numbers of the numbers AD and DB are equal to the superficial number produced of the multiplication of the numbers AD and DB the on into the other twice, together with the square of the number AB (by the 7 proposition) But the square of the number AB is equal to four squares of either of the numbers AC or CB (for AC is equal to the number CB): wherefore also the squares of the numbers AD and DB are equal to the superficial number produced of the multiplication of the numbers AD and DB the one into the other twice, and to four squares of the number BC or CA And forasmuch as the superficial number produced of the multiplication of the numbers AD and DB the one into the other, together with the square of the number CB, is equal to square of the number CD (by the 6 proposition): therefore the number produced of the multiplication of the numbers AD and DB the one into the other twice together with two squares of the number CB, is equal to two squares of the number CD. Wherefore the squares of the numbers AD and DB are equal to two squares of the number CD, and to two squares of the number AC. Wherefore they are double to the squares of the numbers AC and CD. And the square of the number AD is the square of the whole and of the number added. And the square of DB is the square of the number added: the square also of the number CD is the square of the number composed of the half and of the number added: If therefore an even number be divided. etc. Which was required to be proved.  
   

The 1. Problem. The 11. Proposition. To divide a right line given in such sort, that the rectangle figure comprehended under the whole, and one of the parts, shall be equal unto the square made of the other part.  
SVppose that the right line given be AB. Now it is required to divide the line AB in such sort, that the rectangle figure contained under the whole and one of the parts, shall be equal unto the square which is made of the other part. Describe (by the 46. of the first) upon AB a square ABCD. 
Construction.  And (by the 10. of the first) divide the line AC into two equal parts in the point E, and draw a line from B to E. And (by the second petition) extend CA unto the point F. And (by the 3. of the first) put the line EF equal unto the line BE. And (by the 46. of the first) upon the line AF describe a square FGAH. And (by the 2. petition) extend GH unto the point K. Then I say that the line AB is divided in the point H in such sort, that the rectangle figure which is comprehended under AB and BH is equal to the square which is made of AH. 
Demonstration  For forasmuch as the right line AC is divided into two equal parts in the point E, and unto it is added an other right line AF. Therefore (by the 6. of the second) the rectangle figure contained under CF and FA together with the square which is made of AE is equal to the square which is made of EF. But EF is equal unto EB. Wherefore the rectangle figure contained under CF, and FA together with the square which is made of EA is equal to the square which is made of EB. But (by 47. of the first) unto the square which is made of EB are equal the squares which are made of BA and AE. For the angle at the point A is a right angle. Wherefore that which is contained under CF and FA, together with the square which is made of AE, is equal to the squares which are made of BA and AE. Take away the square which is made of AE which is common, to them both: Wherefore the rectangle figure remaining contained under CF and FA is equal unto the square which is made of AB. And that which is contained under the lines CF and FA is the figure FK. For the line FA is equal unto the line FG. And the square which is made of AB is the figure AD. Wherefore the figure FK is equal unto the figure AD. Take away the figure AK which is common, to them both. Wherefore the residue, namely, the figure FH is equal unto the residue, namely, unto the figure HD. But the figure HD is that, which is contained under the lines AB and BH, for AB is equal unto BD. And the figure FH is the square which is made of AH. Wherefore the rectangle figure comprehended under the lines AB and BH is equal to the square which is made of the line HA. Wherefore the right line given AB is divided in the point H, in such sort that the rectangle figure contained under AB and BH is equal to the square which is made of AH: which was required to be done.  
This proposition hath many singular uses. Upon it dependeth the demonstration of that worthy Problem the 10. Proposition of the 4. book: 
Many and singular uses of this proposition.  which teacheth to describe an Isosceles triangle, in which either of the angles at the base shall be double to the angle at the top. Many and divers uses of a line so divided shall you find in the 13. book of Euclid.  
This is to be noted that this Proposition can not as the former Propositions of this second book be reduced unto numbers. 
This proposition can not be reduced unto numbers.  For the line EB hath unto the line AE no proportion that can be named, and therefore it can not be expressed by numbers. For forasmuch as the square of EB is equal to the two squares of AB and AE (by the 47. of the first) and AE is the half of AB, therefore the line BE is irrational. For even as two equal square numbers joined together can not make a square number: so also two square numbers, of which the one is the square of the half root of the other, can not make a square number. As by an example. Take the square of 8. which is 64. which doubled, that is, 128. maketh not a square number. So take the half of 8. which is 4. And the squares of 8. and 4. which are 64. and 16. added together likewise make not a square number. For they make 80. who hath no root square. Which thing must of necessity be if this Problem should have place in numbers.  
But in Irrational numbers it is true, and may by this example be declared.  
Let 8. be so divided, that that which is produced of the whole into one of his parts shall be equal to the square number produced of the other part. Multiply 8. into himself and there shall be produced 64. that is, the square ABCD. Divide 8. into two equal parts, that is, into 4, and 4. as the line AE or EC. And multiply 4. into himself, and there is produced 16, which add unto 64, and there shall be produced 80: whose root is √ ● 80: which is the line EB or the line EF by the 47. of the first. And forasmuch as the line EF is √ ● 80. & the line EA is 4. therefore the line AF is √ ● 80— 4. And so much shall the line AH be. And the line BH shall be 8— √ ● 80— 4, that is, 12— √ ● 80. Now then 12— √ ● 80 multiplied into 8 shall be as much as √ ● 80— 4. multiplied into itself. For of either of them is produced 96— √ 5120.   

The 11. Theorem. The 12. Proposition. In obtuseangle triangles, the square which is made of the side subtending the obtuse angle, is greater than the squares which are made of the sides which comprehend the obtuse angle, by the rectangle figure, which is comprehended twice under one of those sides which are about the obtuse angle, upon which being produced falleth a perpendicular line, and that which is outwardly taken between the perpendicular line and the obtuse angle.  
SVppose that ABC be an obtuseangle triangle having the angle BAC obtuse, and from the point B (by the 12. of the first) draw a perpendicular line unto CA produced and let the same be BD. Then I say that the square which is made of the side BC, is greater than the squares which are made of the sides BA and AC, by the rectangle figure comprehended under the lines CA and AD twice. 
Demonstration.  For forasmuch as the right line CD is by chance divided in the point A, therefore (by the 4. of the second) the square which is made of CD is equal to the squares which are made of CA and AD, and unto the rectangle figure contained under CA and AD twice. Put the square which is made of DB common unto them both. Wherefore the squares which are made of CD and DB are equal to the squares which are made of the lines CA, AD, and DB, and unto the rectangle figure contained under the lines CA and AD twice. But (by the 47. of the first) the square which is made of CB is equal to the squares which are made of the lines CD and DB. For the angle at the point D is a right angle. And unto the squares which are made of AD and DB (by the self same) is equal the square which is m●de of AB. Wherefore the square which is made of CB, is equal to the squares which are made of CA and AB and unto the rectangle figure contained under the lines CA and AD twice. Wherefore the square which is made of CB, is greater than the squares which are made of CA and AB by the rectangle figure contained under the lines CA and AD twice. In obtuseangle triangles therefore, the square which is made of the side subtending the obtuse angle, is greater than the squares which are made of the sides which comprehend the obtuse angle, by the rectangle figure which is comprehended twice under one of those sides which are about the obtuse angle, upon which being produced falleth a perpendicular line, and that which is outwardly taken between the perpendicular line and the obtuse angle: which was required to be demonstrated.  
Of what force this Propositions and the Proposition following, touching the measuring of the obtuseangle triangle and the acuteangle triangle, with the aid of the 47. Proposition of the first book touching the rightangle triangle, he shall well perceive, which shall at any time need the art of triangles in which by three things known is ever searched out three other things unknown, by help of the table of arks and cords.   

The 12. Theorem. The 13. Proposition. In acuteangle triangles, the square which is made of the side that subtendeth the acute angle, is less than the squares which are made of the sides which comprehend the acute angle, by the rectangle figure which is comprehended twice under one of those sides which are about the acuteangle, upon which falleth a perpendicular line, and that which is inwardly taken between the perpendicular line and the acute angle.  
SVppose that ABC be an acuteangle triangle having the angle at the point B acute, & (by the 12. of the first) from the point A draw unto the line BC a perpendicular line AD. Then I say that the square which is made of the line AC is less than the squares which are made of the line CB and BA, by the rectangle figure contained under the lines CB and BD twice. 
Demonstration  For forasmuch as the right line BC is by chance divided in the point D, therefore (by the 7. of the second) the squares which are made of the lines CB and BD are equal to the rectangle figure contained under the lines CB and DB twice and unto the square which is made of line CD. Put the square which is made of the line DA common unto them both. Wherefore the squares which are made of the lines CB, BD, and DA, are equal unto the rectangle figure contained under the lines CB and BD twice, and unto the squares which are made of AD and DC. But to the squares which are made of the lines BD and DA is equal the square which is made of the line AB: for th'angle at the point D is a right angle. And unto the squares which are made of the lines AD and DC is equal the square which is made of the line AC (by the 47. of the first): wherefore the squares which are made of the lines CB and BA are equal to the square which is made of the line AC, and to that which is contained under the lines CB and BD twice. Wherefore the square which is made of the line AC being taken alone, is less than the squares which are made of the lines CB and BA by the rectangle figure, which is contained under the lines CB and BD twice. In rectangle triangles therefore the square which is made of the side that subtendeth the acute angle, is less than the squares which are made of the sides which comprehend the acute angle, by the rectangle figure which is comprehended twice under one of those sides which are about the acute angle, upon which falleth a perpendicular line, and that which is inwardly taken between the perpendicular line and the acute angle: which was required to be proved.  

¶ A Corollary added by Orontius.  

A Corollary.  Hereby is easily gathered, that such a perpendicular line in rectangle triangles falleth of necessity upon the side of the triangle, that is, neither within the triangle, nor without. But in obtuseangle triangles it falleth without, and in acuteangle triangles within. For the perpendicular line in obtuseangle triangles, and acuteangle triangles can not exactly agreed with the side of the triangle: for then an obtuse & an acuteangle should be equal to a right angle, contrary to the eleventh and twelfth definitions of the first book. Likewise in obtuseangle triangles it can not fall within, nor in acuteangle triangles without: for then the outward angle of a triangle should be less than the inward and opposite angle, which is contrary to the 16. of the first.  

This Proposition true in all kinds of triangles.  And this is to be noted, that although properly an acuteangle triangle, by the definition thereof given in the first book, be that triangle, whose angles be all acute: yet forasmuch as there is no triangle, but that it hath an acute angle, this proposition is to be understanded, & is true generally in all kinds of triangles whatsoever, and may be declared by them, as you may easily prove.    

The 2. Problem. The 14. Proposition. Unto a rectiline figure given, to make a square equal.  
SVppose that the rectiline figure given be A. It is required to make a square equal unto the rectiline figure A. 
Construction.  Make (by the 45. of the first) unto the rectiline figure A an equal rectangle parallelogram BCDE. Now if the line BE be equal unto the line ED, then is the thing done which was required: for unto the rectiline figure A is made an equal square BD. But if not, one of these lines BE & is ED the greater. Let BE be the greater, and let it be produced unto the point F. And (by the 3. of the first) put unto ED an equal line EF. And (by the 10. of the first) divide the line BF into two equal parts in the point G. And making the centre the point G, and the space GB or GF describe a semicircle BHF. And (by the 2. petition) extend the line DE unto the point H. 
Demonstration  And (by the 1. petition) draw a line from G to H. And forasmuch as the right line FB is divided into two equal parts in the point G, and into two unequal parts in the point E, therefore (by the 5. of the second) the rectangle figure comprehended under the lines BE and EF together with the square which is made of the line EG, is equal to the square which is made of the line GF. But the line GF is equal unto the line GH. Wherefore the rectangle figure comprehended under the lines BE and EF together with the square which is made of the line GE is equal to square which is made of the line GH. But unto the square which is made of the line GH are equal the squares which are made of the lines HE and GE (by the 47. of the first.) Wherefore that which is contained under the lines BE and EF together with the square which is made of GE is equal to the squares which are made of HE and GOE Take away the square of the line EG common to them both. Wherefore the rectangle figure contained under the lines BE & EF is equal to the square which is made of the line EH. But that which is contained under the lines BE and EF is the parallelogram BD, for the line EF is equal unto the line ED. Wherefore the parallelogram BD is equal to the square which is made of the line HERALD But the parallelogram BD is equal unto the rectiline figure A. Wherefore the rectiline figure A is equal to the square which is made of the line HERALD Wherefore unto the rectiline figure given A, is made an equal square described of the line EH: which was required to be done.   
¶ The end of the second Book of Euclides Elements.   

¶ The third book of Euclides Elements.  

The argument of this book.  THis third book of Euclid entreateth of the most perfect figure, which is a circle. Wherefore it is much more to be esteemed then the two books going before, in which he did set forth the most simple proprieties of rightlined figures. For sciences take their dignities of the worthiness of the matter that they entreat of. But of all figures the circle is of most absolute perfection, whose proprieties and passions are here set forth, and most certainly demonstrated. Here also is entreated of right lines subtended to arks in circles: also of angles set both at the circumference and at the centre of a circle, and of the variety and differences of them. Wherefore the reading of this book, is very profitable to the attaining to the knowledge of chords and arcs. It teacheth moreover which are circles contingent, and which are cutting the one the other: and also that the angle of contingence is the lest of all acute rightlined angles: and that the diameter in a circle is the longest line that can be drawn in a circle. Farther in it may we learn how, three points being given how soever (so that they be not set in a right line), may be drawn a circle passing by them all three. Again, how in a solid body, as in a Sphere, Cube, or such like, may be found the two opposite points. Which is a thing very necessary and commodious: chief for those that shall make instruments serving to Astronomy, and other arts.  

Definitions. 
The first definition.  Equal circles are such, whose diameters are equal, or whose lines drawn from the centres are equal.  
The circles A and B are equal, if their diameters, namely, EF and CD be equal: or if their semidiameters, which are lines drawn from the centre to the circumference● namely AF and BD be equal.  
By this also is known the definition of unequal circles. 
Definition of unequal circles.   
Circles whose diameters or semidiameters are unequal, are also unequal. And that circled which hath the greater diameter or semidiameter, is the greater circle: and that circled which hath the less diameter or semidiameter, is the less circle.   

A right line is said to touch a circle, 
Second definition.  which touching the circle and being produced cutteth it not.  
As the right line EF drawn from the point E, and passing by a point of the circle, namely, by the point G to the point F only toucheth the circle GH, and cutteth it not, nor entereth within it. For a right line entering within a circle, cutteth and divideth the circle. As the right line KL divideth and cutteth the circle KLM, and entereth within it: and therefore toucheth it in two places. But a right line touching a circle, which is commonly called a contingent line, 
A contigent line.  toucheth the circle only in one point.   

Circles are said to touch the one the other, 
Third definition.  which touching the one the other, cut not the one the other.  
As the two circles AB and BC touch the one the other. For their circumferences touch together in the point B. But neither of them cutteth or divideth, the other. Neither doth any part of the one enter within the other. 
The touch of circles is 〈◊〉 in one po●●● only.  And such a touch of circles is ever in one point only: which point only is common to them both. As the point B is in the conference of the circle AB, and also 〈…〉 the ●●●●●ference of the circle BC.  

Circles may touch together two ma●●● of ways.  Circles may touch together two manner of ways, either outwardly the one wholly without the other: or else the one being contained within the other.  
As the circles DE and DF: of which the one DE containeth the other, namely DF: and touch the one the other in the point D: and that only point is common to them both: neither doth the one enter into the other. If any part of the one enter into any part of the other, than the one cutteth and divideth the other, and toucheth the one the other not in one point only as in the other before, but in two point●s, and have also a superficies common to them both. As the circles GHK and HLK cut the one the other in two points H and K: and the one entereth into the other: Also the superficies HK is common to them both: For it is a part of the circle GHK, and also it is a part of the circle HLK.   


Fourth definition.  Right lines in a circle are said to be equally distant from the centre, when perpendicular lines drawn from the centre unto those lines are equal. And that line is said to be more distant, upon whom falleth the greater perpendicular line.  
As in the circle ABCD whose centre is E, the two lines AB and CD have equal distance from the centre E: because that the line EF drawn from the centre E perpendicularly upon the line AB, and the line EG drawn likewise perpendilarly from the centre E upon the line CD are equal the one to the other. But in the circle HKLM whose centre is N the line HK hath greater distance from the centre N than hath the line LM: for that the line ON drawn from the centre N perpendicularly upon the line HK is greater than the line NP which is drawn from the centre N perpendicularly upon the line LM.  
So likewise in the other figure the lines AB and DC in the circle ABCD are equidistant from the centre G ● because the lines OG and GP perpendicularly drawn from the centre G upon the said lines AB and DC are equal. And the line AB hath greater distance from the centre G than hath the the line EF, because the line OG perpendicularly drowen from the centre G to the line AB is gre●ter than the line HG which is perpendicularly drawn from the c●●tre G to the line EF.   


Fift definition.  A section or segment of a circle, is a figure comprehended under a right line and a portion of the circumference of a circle.  
As the figure ABC is a section of a circle because it is comprehended under the right line AC and the circumference of a circle ABC. Likewise the figure DEF is a section of a circle, for that it is comprehended under the right line DF, and the circumference DEF. And the figure ABC for that it containeth within it the centre of the circle is called the greater section of a circle: and the figure DEF is the less section of a circle, because it is wholly without the centre of the circle as it was noted in the 16. Definition of the first book.   

An angle of a section or segment, is that angle which is contained under a right line and the circumference of the circle. 
sixth definition.   
As the angle ABC in the section ABC is an angle of a section, because it is contained of the circumference BAC and the right line BC. Likewise the angle CBD is an angle of the section BDC because it is contained under the circumference BDC, and the right line BC. And these angles are commonly called mixed angles, 
Mixed angles.  because they are contained under a right line and a crooked. And these portions of circumferences are commonly called arks, 
arks.  and the right lines are called chords, 
Chords.  or right lines subtended. And the greater section hath ever the greater angle, and the less section the less angle.   

An angle is said to be in a section, 
Seventh definition.  when in the circumference is taken any point, and from that point are drawn right lines to the ends of the right line which is the base of the segment, the angle which is contained under the right lines drawn from the point, is (I say) said to be an angle in a section.  
As the angle ABC is an angle is the section ABC, because from the point B being a point in the circumference ABC are drawn two right lines BC and BA to the ends of the line AC which is the base of the section ABC. Likewise the angle ADC is an angle in the section ADC, because from the point D being in the circumference ADC are drawn two right lines, namely, DC & DA to the ends of the right line AC which is also the base to the said section ADC. So you see, it is not all one to say, 
Difference of an angle of a Section, and of an angle in a Section.  an angle of a section, and an angle in a section. An angle of a section consisteth of the touch of a right line and a crooked. And an angle in a section is placed on the circumference, and is contained of two right lines. Also the greater section hath in it the less angle, and the less section hath in it the greater angle.   

But when the right lines which comprehend the angle do receive any circumference of a circle, 
Eight definition.  than that angle is said to be correspondent, and to pertain to that circumference.  
As the right lines BA and BC which contain the angle AB C, and receive the circumference ADC therefore the angle ABC is said to subtend and to pertain to the circumference ADC. And if the right lines which cause the angle, concur in the centre of a circle: then the angle is said to be in the centre of a circle. As the angle EFD is said to be in the centre of a circle, for that it is comprehended of two right lines FE and FD: which concur and touch in the centre F. And this angle likewise subtendeth the circumference EGD: which circumference also, is the measure of the greatness of the angle EFD.   

A Sector of a circle is (an angle being set at the centre of a circle) a figure contained under the right lines which make that angle, 
Ninth definition.  and the part of the circumference received of them.  
As the figure ABC is a sector of a circle, for that it hath an angle at the centre, namely the angle BAC, & is contained of the two right lines AB and AC (which contain that angle and the circumference received by them.   


Tenth definition.  Like segments or sections of a circle are those, which have equal angles, or in whom are equal angles.  

Two definitions.  Here are set two definitions of like sections of a circle. The one pertaineth to the angles which are set in the centre of the circle and receive the circumference of the said sections: 
First.  the other pertaineth to the angle in the section, which as before was said is ever in the circumference. As if the angle BAC, being in the centre A and received of the circumference BLC be equal to the angle FEG being also in the centre E and received of the circumference FKG, then are the two sections BCL and FGK like by the first definition. By the same definition also are the other two sections like, namely BCD, and FGH, for that the angle BAC is equal to the angle FEG.  

Second.  Also by the second definition if B AC being an angle placed in the circumference of the section BCA be e-angle EDF being an angle in the section EFD placed in the circumference, there are the two sections BCA, and EFD like the one to the other. Likewise also if the angle BGC being in the section BCG be equal to the angle EHF being in the section EHF the two sections BCG and EFH are like. And so is it of angles being equal in any point of the circumference.  
Euclid defineth not equal Sections: 
Why Euclid defineth not equal Sections.  for they may infinite ways be described. For there may upon unequal right lines be set equal Sections (but yet in unequal circles) For from any circle being the greater, may be cut of a portion equal to a portion of an other circle being the less. But when the Sections are equal, and are set upon equal right lines, their circumferences also shallbe equal. And right lines being divided into two equal parts, perpendicular lines drawn from the points of the division to the circumferences shallbe equal. As if the two sections ABC and DEF, being set upon equal right lines AC & DF, be equal: then if each of the two lines AC & DF be divided into two equal parts in the points G and H, & from the said points be drawn to the circumferences two perpendicular lines BG and EH, the said perpendicular lines shallbe equal.   

The 1. Problem. The 1. Proposition. To find out the centre of a circle given.  
SVppose that there be a circle given ABC. It is required to find out the centre of the circle ABC. Draw in it a right line at all adventures, and let the same be AB. 
Constuction.  And (by the 10. of the first) divide the line AB into two equal parts in the point D. And (by the 11. of the same) from the point D raise up unto AB a perpendicular line DC, & (by the second petition) extend DC unto the point E. And (by the 10. of the first) divide the line CE into two equal parts in the point F. Then I say that the point F is the centre of the circle ABC. 
Demonstration leading to an impossibility.  For if it be not, let some other point, namely G, be the centre. And (by the first petition) draw these right lines GA, GD, and GB. And for as much as AD is equal unto DB, and DG is common unto them both, therefore these two lines AD and DG are equal to these two lines GD and DB, the one to the other, and (by the 15. definition of the first) the base GA is equal to the base GB. For they are both drawn from the centre G to the circumference: therefore (by the 8. of the first) the angle ADG is equal to the angle BDG. But when a right line standing upon a right line maketh the angles on each side equal the one to the other, either of those angles (by the 10. definition of the first) is a right angle. Wherefore the angle BDG is a right angle: but the angle FDB is also a right angle by construction. Wherefore (by the 4. petition) the angle FDB is equal to the angle BDG, the greater to the less, which is impossible. Wherefore the point G is not the centre of the circle ABC. In like wise may we prove that no other point besides F is the centre of the circle ABC. Wherefore the point F is the centre of the circle ABC: which was required to be done.  

Correlary.  

Correlary.  Hereby it is manifest, that if in a circle a right line do divide a right line into two equal parts, and make right angles on each side: in that right line which divideth the other line into two equal parts is the centre of the circle.    

The 1. Theorem. The 2. Proposition. If in the circumference of a circle be taken two points at all adventures: a right line drawn from the one point to the other shall fall within the circle.  
SVppose that there be a circle ABC. And in the circumference thereof, let there be taken at all adventures these two points A & B. Then I say that a right line drawn from A to B shall fall within the circle ABC. 
Demonstration leading to an impossibility.  For if it do not, let it fall without the circle, as the line AEB doth, which if it be possible imagine to be a right line. And (by the Proposition going before) take the centre of the circle, and let the same be D. And (by the first petition) draw lines from D to A, and from D to B. And extend DF to E. And for as much as (by the 15. definition of the first) DA is equal unto DB. Therefore the angle DAE is equal to the angle DBE. And for as much as one of the sides of the triangle DAE, namely the side AEB is produced, therefore (by the 16. of the first) the angle DEB, is greater than the angle DAE. But the angle DAE is equal unto the angle DBE. Wherefore the angle DEB is greater than the angle DBE. But (by the 18. of the first) unto the greater angle is subtended the greater side. Wherefore the side DB is greater than the side DE. But (by the 15. definition of the first) the line DB is equal unto the line DF. Wherefore the line DF is greater than the line DE, namely, the less greater than the greater: which is impossible. Wherefore a right line drawn from A to B falleth not without the circle. In like sort also may we prove that it falleth not in the circumference. Wherefore it falleth within the circle. If therefore in the circumference of a circle be taken two points at all adventures: a right line drawn from the one point to the other shall fall within the circle: which was required to be proved.   

The 2. Theorem. The 3. Proposition. If in a circle a right line passing by the centre do divide an other right line not passing by the centre into two equal parts: it shall divide it by right angles. And if it divide the line by right angles, it shall also divide the same line into two equal parts.  
SVppose that there be a circle ABC, 
The first para of this Proposition.  and let there be in it drawn a right line passing by the centre, and let the same be CD, dividing an other right line AB not passing by the centre into two equal parts in the point F. Then I say that the angles at the point of the division are right angles. 
Construction.  Take (by the first of the third) the centre of the circle ABC, and let the same be E. And (by the first petition) draw lines from E to A & from E to B. 
Demonstration.  And for as much as the line AF is equal unto the line FB, and the line FE is common to them both, therefore these two lines EF and FA are equal unto these two lines EF & FB. And the base EA is equal unto the base EB (by the 15. definition of the first). Wherefore (by the 8. of the first) the angle AFE is equal to the angle BFE. But when a right line standing upon a right line doth make the angles on each side equal the one to the other, either of those angles is (by the 10. definition of the first) a right angle. Wherefore either of these angles AFE, & BFE is a right angle. Wherefore the line CD passing by the centre, and dividing the line AB not passing by the centre into two equal parts, maketh at the point of the division right angles.  
But now suppose that the line CD do divide the line AB in such sort that it maketh right angles. 
The second part conversed of the first.  Then I say that it divideth it into two equal parts, that is, that the line AF is equal unto the line FB. 
Demonstration.  For the same order of construction remaining, for as much as the line EA is equal unto the line EB (by the 15. definition of the first). Therefore the angle EAF is equal unto the angle EBF (by the 5. of the first). And the right angle AFE is (by the 4. petition) equal to the right angle BFE. Wherefore there are two triangles EAF, & EBF having two angles equal to two angles, & one side equal to one side, namely the side EF which is common to them both, and subtendeth one of the equal angles, wherefore (by the 26. of the first) the sides remaining of the one, are equal unto the sides remaining of the other. Wherefore the line AF is equal unto the line FB. If therefore in a circle a right line passing by the centre do divide an other right line not passing by the centre into two equal parts, it shall divide it by right angles. And if it divide the line by right angles it shall also divide the same line into two equal parts: which was required to be demonstrated.   

The 3. Theorem. The 4. Proposition. If in a circle two right lines not passing by the centre, divide the one the other: they shall not divide each one the other into two equal parts.  
SVppose that there be a circle ABCD, and let there be in it drawn two right lines not passing by the centre and dividing the one the other, and let the same be AC and BD, which let divide the one the other in the point E. Then I say that they divide not each the one the other into two equal parts. 
Demonstration leading to an impossibility.  For if it be possible let them divide each the one the other into two equal parts, so that let AE be equal unto EC, & BE unto ED. And take the centre of the circle ABCD, which let be F. And (by the first petition) draw a line from F to E. Now for as much as a certain right line FE passing by the centre divideth an other line AC not passing by the centre into two equal parts, it maketh where the division is right angles (by the 3. of the third). Wherefore the angle FEA is a right angle. Again for as much as the right line FE, passing by the centre, divideth the right line BD not passing by the centre into two equal parts, therefore (by the same) it maketh where the division is right angles. Wherefore the angle FEB is a right angle. And it is proved that the angle FEA is a right angle. Wherefore (by the 4. petition) the angle FEA is equal unto the angle FEB, namely the less angle unto the greater: which is impossible. Wherefore the right lines AC and BD divide not each one the other into two equal parts. If therefore in a circle two right lines not passing by the centre, divide the one the other, they shall not divide each one the other into two equal parts: which was required to be demonstrated.  
In this Proposition are two cases. 
Two cases in this Proposition.  For the lines cutting the one the other, do either, neither of them pass by the centre, or the one of them doth pass by the centre, & the other not. The first is declared by the author. The second is thus proved.   

The 4. Theorem. The 5. Proposition. If two circles cut the one the other, they have not one and the same centre.  
SVppose that these two circles ABC, 
Construction.  and CBG do cut the one the other in the points C and B. Then I say that they have not one & the same centre. 
Demonstration leading to an impossibility.  For if it be possible let E be centre to them both. And (by the first petition) draw a line from E to C. And draw an other right line EFG at all adventures. And for as much as the point E is the centre of the circle ABC, therefore (by the 15. definition of the first) the line EC is equal unto the line EF. Again for as much as the point E is the centre of the circle CBG, therefore (by the same definition) the line EC is equal unto the line EG. And it is proved that the line EC is equal unto the line EF: wherefore the line EF also is equal unto the line EG, namely the less unto the greater: which is impossible. Wherefore the point E is not the centre of both the circles ABC, & CBG. I● like sort also may we prove that no other point is the centre of both the said circles. If therefore two circles cut the one the other, they have not one and the same centre: which was required to be proved.   

The 5. Theorem. The 6. Proposition. If two circles touch the one the other, they have not one and the same centre.  
SVppose that these two circles ABC, & CDE do touch the one the other in the point C. 
Demonstration leading to an impossibility.  Then I say that they have not one and the same centre. For if it be possible let the point F be centre unto them both. And (by the first petition) draw a line from F to C: and draw the line FEB at all adventures. And for as much as the point F is the centre of the circle ABC, therefore (by the 15. definition of the first) the line FC is equal unto the line FB. Again forasmuch as the point F is the centre of the circle CDE, therefore (by the same definition) the line FC is equal unto the line FE. And it is proved, that the line FC is equal unto the line FB, wherefore the line FE also is equal unto the line FB, namely the less unto the greater: which is impossible. Wherefore the point F is not the centre of both the circles ABC and CDE. In like sort also may we prove that no other point is the centre of both the said circles. If therefore two circles touch the one the other: they have not one and the same centre: which was required to be demonstrated.  

Two case● in this Proposition.  In this Proposition are two cases: for the circles touching the one the other, may touch either within or without. If they touch the one the other within, then is it by the former demonstration manifest, that they have not both one and the self same centre. It is also manifest if they touch the one the other without: for that every centre is in the midst of his circle.   

The 6. Theorem. The 7. Proposition. If in the diameter of a circle be taken any point, which is not the centre of the circle, and from that point be drawn unto the circumference certain right lines: the greatest of those lines shall be that line wherein is the centre, and the jest shall be the residue of the same line. And of all the other lines, that which is nigher to the line which passeth by the centre is greater than that which is more distant. And from that point can fall within the circle on each side of the lest line only two equal right lines.  
SVppose that there be a circle ABCD: and let the diameter thereof be AD. And take in it any point besides the centre of the circle, and let the same be F. 
Construction.  And let the centre of the circle (by the 1. of the third) be the point E. And from the point F let there be drawn unto the circumference ABCD these right lines FD, FC, and FG. Then I say that the line FA is the greatest: and the line FD is the jest. And of the other lines, the line FB is greater than the line FC, and the line FC is greater than the line FG. 
The first part of this Proposition.  Draw (by the first petition) these right lines BE, CE, and GOE And for as much as (by the 20. of the first) in every triangle two sides are greater than the third, 
Demonstration.  therefore the lines EB and EF are greater than the residue, namely then the line FB. But the line AE is equal unto the line BE (by the 15. definition of the first). Wherefore the lines BE and EF are equal unto the line AF. Wherefore the line AF is greater than then the line BF. Again for as much as the line BE is equal unto CE (by the 15. definition of the first) and the line FE is common unto them both, therefore these two lines BE and EF are equal unto these two CE and EF. But the angle BEF is greater than the angle CEF. Wherefore (by the 24. of the first) the base BF is greater than the base CF: and by the same reason the line CF is greater than the line FG. Again for as much as the lines GF and FE are greater than the line EG (by the 20. of the first). 
Second part.  But (by the 15. definition of the first) the line EG is equal unto the line ED: Wherefore the lines GF and FE are greater than the line ED, take away EF, which is common to them both, wherefore the residue GF is greater than the residue FD. Wherefore the line FA is the greatest, and the line FD is the jest, and the line FB is greater than the line FC, and the line FC is greater than the line FG. 
Third part.  Now also I say that from the point F there can be drawn only two equal right lines into the circle ABCD on each side of the lest line, namely FD. For (by the 23. of the first) upon the right line given EF and to the point in it, namely E, make unto the angle GEF an equal angle FEH: and (by the first petition) draw a line from F to H. Now for as much as (by the 15. definition of the first) the line EG is equal unto the line EH, and the line EF is common unto them both, therefore these two lines GE and EF are equal unto these two lines HE and EF, and (by construction) the angle GEF is equal unto the angle HEF. Wherefore (by the 4. of the first) the base FG is equal unto the base FH. 
This demonstrated by an argument leading to an impossibilie.  I say moreover that from the point F can be drawn into the circle no other right line equal unto the line FG. For if it possible let the line● FK be equal unto the line FG. And for as much as FK is equal unto FG. But the line FH is equal unto the line FG, therefore the line FK is equal unto the line FH. Wherefore the line which is nigher to the line which passeth by the centre is equal to that which is farther of, which we have before proved to be impossible.  

another demonstration of the latter part of the Proposition leading also to an impossibility.  Or else it may thus be demonstrated. Draw (by the first petition) a line from E to K: and for as much as (by the 15. definition of the first) the line GE is equal unto the line EK, and the line FE is common to them both, and the base GF is equal unto the base FK, therefore (by the 8. of the first) the angle GEF is equal to the angle KEF. But the angle GEF is equal to the angle HEF. Wherefore (by the first common sentence) the angle HEF is equal to the angle KEF the less unto the greater: which is impossible. Wherefore from the point F there can be drawn into the circle no other right line equal unto the line GF. Wherefore but one only. If therefore in the diameter of a circle be taken any point, which is not the centre of the circle, and from that point be drawn unto the circumference certain right lines: the greatest of those right lines shall be that wherein is the centre: and the lest shall be the residue. And of all the other lines, that which is nigher to the line which passeth by the centre is greater than that which is more distant. And from that point can fall within the circle on each side of the lest line only two equal right lines: which was required to be proved.  

¶ A Corollary.  

A Corollary.  Hereby it is manifest, that two right lines being drawn from any one point of the diameter, the one of one side, and the other of the other side, if with the diameter they make equal angles, the said two right lines are equal. As in this place are the two lines FG and FH.    

The 7. Theorem. The 8. Proposition. If without a circle be taken any point, and from that point be drawn into the circle unto the circumference certain right lines, of which let one be drawn by the centre and let the rest be drawn at all adventures: the greatest of those lines which fall in the concavity or hollowness of the circumference of the circle, is that which passeth by the centre: and of all the other lines that line which is nigher to the line which passeth by the centre is greater than that which is more distant. But of those right lines which end in the convexe part of the circumference, that is the least which is drawn from the point to the diameter: and of the other lines that which is nigher to the lest is always less than that which is more distant. And from that point can be drawn unto the circumference on each side of the least only two equal right lines.  

Third part.  Now also I say that from the point D can be drawn unto the circumference on each side of DG the least only two equal right lines. Upon the right line MD, and unto the point in it M make (by the 23. of the first) unto the angle KMD an equal angle DMB. And (by the first petition) draw a line from D to B. And for as much as (by the 15. definition of the first) the line MB is equal unto the line MK put the line MD common to the both, wherefore these two lines MK and MD are equal to these two lines BM and MD the one to the other, and the angle KMD is (by the 23. of the first) equal to the angle BMD: Wherefore (by the 4. of the first) the base DK is equal to the base DB.  
Or it may thus be demonstrated. 
another demonstration of the latter part, leading also to an impossibility.  Draw (by the first petition) a line from M to N. And for as much as (by the 15. definition of the first) the line KM is equal unto the line MN, and the line MD is common to them both. And the base KD is equal to the base DN (by supposition) therefore (by the 8. of the first) the angle KMD is equal to the angle DMN. But the angle KMD is equal to the angle BMD. Wherefore the angle BMD is equal to the angle NMD, the less unto the greater: which is impossible. Wherefore from the point D can not be drawn unto the circumference ABC on each side of DG the jest, more than two equal right lines. If therefore without a circle be taken any point and from that point be drawn into the circle unto the circumference certain right lines, of which let one be drawn by the centre, and let the rest be drawn at all adventures: the greatest of those right lines which fall in the concavity or hollowness of the circumference of the circle is that which passeth by the centre. And of all the other lines, that line which is nigher to the line which passeth by the centre, is greater than that which is more distant. But of those right lines which end in the convexe part of the circumference, that line is the jest which is drawn from the point to the dimetient: and of the other lines that which is nigher to the lest is always less than that which is more distant. And from that point can be drawn unto the circumference on each side of the jest only two equal right lines: which was required to be proved.  
This Proposition is called commonly in old books amongst the barbarous, Ca●d● Panonis, 
This Proposion is commonly called Ca●d● Panonis.  that is, the peacocks tail.  

¶ A Corollary.  
Hereby it is manifest, 
A Corollary.  that the right lines, which being drawn from the point given without the circle, and fall within the circle, are equally distant from the lest, or from the greatest (which is drawn by the centre) are equal the one to the other: but contrariwise if they be unequally distant, whether they light upon the concave or convexe circumference of the circle, they are unequal.    

The 8. Theorem. The 9 Proposition. If within a circle be taken a point, and from that point be drawn unto the circumference more than two equal right lines, the point taken is the centre of the circle.  
SVppose that the circle be ABC, and within it let there be taken the point D. And from D let there be drawn unto the circumference ABC more than two equal right lines, 
Construction.  that is, DA, DB, and DC. Then I say that the point D is the centre of the circle ABC. Draw (by the first petition) these right lines AB and BC: 
Demonstration.  and (by the 10. of the first) divide them into two equal parts in the points E and F: namely, the line AB in the point E, and the line BC in the point F. And draw the lines ED and FD, and (by the second petition) extend the lines ED and FD on each side to the points K, G, and H, L. And for as much as the line AE is equal unto the line EB, and the line ED is common to them both, therefore these two sides AE and ED are equal unto these two sides BE, and ED: and (by supposition) the base DA is equal to the base DB. Wherefore (by the 8. of the first) the angle AED is equal to the angle BED. Wherefore either of these angles AED and BED is a right angle. Wherefore the line GK divideth the line AB into two equal parts and maketh right angles. And for as much as, if in a circle a right line divide an other right line into two equal parts in such sort that it maketh also right angles, in the line that divideth is the centre of the circle (by the corollary of the first of the third). Therefore (by the same corollary) in the line GK is the centre of the circle ABC. And (by the same reason) may we prove that in the line HL is the centre of the circle ABC, and the right lines GK, and HL have no other point common to them both beside the point D. Wherefore the point D is the centre of the circle ABC. If therefore within a circle be taken a point, and from that point be drawn unto the circumference more than two equal right lines, the point taken is the centre of the circle: which was required to be proved.  

¶ An other demonstration.  
Let there be taken within the circle ABC the point D. 
another demonstration of the same leading also to an impossibility.  And from the point D let there be drawn unto the circumference more than two equal right lines, namely, DA, DB, and DC. Then I say that the point D is the centre of the circle. For if not, then if it be possible let the point E be the centre: and draw a line from D to E, and extend DE to the points F and G. Wherefore the line FG is the diameter of the circle ABC. And for as much as in FG the diameter of the circle ABC is taken a point, namely D, which is not the centre of that circle, therefore (by the 7. of the third) the line DG is the greatest, and the line DC is greater than the line DB, and the line DB is greate● than the line DA. But the lines DC, DB, DA, are also equal (by supposition): which is impossible. Wherefore the point E is not the centre of the circle ABC. And in like sort may we prove that no other point besides D. Wherefore the point D is the centre of the circle ABC: which was required to be proved.    

The 9 Theorem. The 10. Proposition. A circle cutteth not a circle in more points than two.  
FOr if it be possible let the circle ABC cut the circle DEF in more points than two, 
Demonstration leading to an impossibility.  that is, in B, G, H, & F. And draw lines from B to G, and from B to H. And (by the 10. of the first) divide either of the lines BG & BH into two equal parts, in the points K and L. And (by the 11. of the first) from the point K raise up unto the line BH a perpendicular line KC, and likewise from the point L raise up unto the line BG a perpendicular line LM, and extend the line CK to the point A, and LNM to the points X and E. And for as much as in the circle ABC, the right line AC divideth the right line BH into two equal parts and maketh right angles, therefore (by the 3. of the third) in the line AC is the centre of the circle ABC. Again, for as much as in the self same circle ABC the right line NX, that is, the line ME divideth the right line BG into two equal parts and maketh right angles, therefore (by the third of the third) in the line NX is the centre of the circle ABC. And it is proved that it is also in the line AC. And these two right lines AC and NX meet together in no other point besides O. Wherefore the point O is the centre of the circle ABC. And in like sort may we prove that the point O is the centre of the circle DEF. Wherefore the two circles ABC and DEF dividing the one the other have one and the same centre: which (by the 5. of the third) is impossible. A circle therefore cutteth not a circle in more points than two: which was required to be proved.  

another demonstration to prove the same.  
Suppose that the circle ABC do cut the circle DGF in more points than two, 
another demonstration of the same leading also to an impossibility.  that is, in B, G, F, and H. And (by the first of the third) take the centre of the circle ABC and let the same be the point K. And draw these right lines KB, KG, and KF. Now for as much as within the circle DEF is taken a certain point K, and from that point are drawn unto the circumference more than two equal right lines, namely, KB, KG, and KF: therefore (by the 9 of the third) K is the centre of the circle DEF. And the point K is the centre of the circle ABC. Wherefore two circles cutting the one the other have one and the same centre: which (by the 5. of the third) is impossible. A circle therefore cutteth not a circle in more points than two: which was required to be demonstrated.    

The 10. Theorem. The 11. Proposition. If two circles touch the one the other inwardly, their centres being given: a right line joining together their centres and produced, will fall upon the touch of the circles.  
SVppose that these two circles ABC, and ADE do touch the one the other in the point A. 
Construction.  And (by the first of the third) take the centre of the circle ABC, and let the same be F: and likewise the centre of the circle ADE, and let the same be G. Then I say that a right line drawn from F to G and being produced, will fall upon the point A. For if not, 
Demonstration leading to an impossibility.  then if it be possible let it fall as the line FGDH doth. And draw these right lines AF, & AG. Now for as much as the lines AG and GF are (by the 20. of the first) greater then the line FA, that is, than the line FH, take away the line GF which is common to them both. Wherefore the residue AG is greater than the residue GH. But the line DG is equal unto the line GA (by the 15. definition of the first). Wherefore the line GD is greater than the line GH: the less than the greater: which is impossible. Wherefore a right line drawn from the point F to the point G and produced, falleth not besides the point A, which is the point of the touch. Wherefore it falls upon the touch. If therefore two circles touch the one the other inwardly, their centres being given, a right line joining together their centres and produced, will fall upon the touch of the circles: which was required to be proved.  

another demonstration to prove the same.  
But now let it fall as GFC falleth, and extend the line GFC to the point H: 
another demonstration of the same leading also to an impossibility.  and draw these right lines AG and AF. And for as much as the lines AG and GF are (by the 20. of the first) greater then the line AF. But the line AF is equal unto the line CF, that is, unto the line FH. Take away the line FG common to them both. Wherefore the residue AG is greater than the residue GH, that is, the line GD is greater than the line GH: the less greater than the greater: which is impossible.  
Which thing may also be proved by the 7. Proposition of this book. For for as much as the line HC is the diameter of the circle ABC, 
The same again demonstrated by an argument leading to an absurdititie.  & in it is taken a point which is not the centre, namely, the point G, therefore the line GA is greater than the line GH by the said 7. Proposition. But the line GD is equal to the line GA (by the definition of a circle). Wherefore the line GD is greater than the line GH, namely, the part greater than the whole: which is impossible.    

The 11. Theorem. The 12. Proposition. If two circles touch the one the other outwardly, a right line drawn by their centres shall pass by the touch.  
SVppose that these two circles ABC and ADE do touch the one the other outwardly in the point A. And (by the third of the third) take the centre of the circle ABC, 
Demonstrati● leading to an impossibility.  and let the same be the point F: and likewise the centre of the circle ADE, and let the same be the point G. Then I say that a right line drawn from the point F to the point G shall pass by the point of the touch, namely, by the point A. For if not, then if it be possible, let it pass as the right line FCDG doth. And draw these right lines AF & AG. And for as much as the point F is the centre of the circle ABC, therefore the line FA is equal unto the line FC. Again for as much as the point G is the centre of the circle ADE, therefore the line GA is equal to the line GD. And And it is proved that the line FA is equal to the line FC. Wherefore the lines FA and AG are equal unto the lines FC and GD: Wherefore the whole line FG is greater than the lines FA and AG. But it is also less (by the 20. of the first): which is impossible. Wherefore a right line drawn from the point F to the point G shall pass by the point of the touch, namely, by the point A. If therefore two circles touch the one the other outwardly, a right line drawn by their centres, shall pass by the touch: which was required to be demonstrated.  

¶ An other demonstration after Pelitarius.  

another demonstration after Pelitarius leading also to an absurdity.  Suppose that the two circles ABC and DEF do touch the one the other outwardly in the point A: And let G be the centre of the circle ABC: From which point produce by the touch of the circles the line GA to the point F of the circumference DEF. Which for as much as it passeth not by the centre of the circle DEF (as the adversary affirmeth) draw from the same centre G an other right line GK, which if it be possible let pass by the centre of the circle DEF, namely, by the point H: cutting the circumference ABC in the point B, & the circumference DEF in the point D, & let the opposite point thereof be in the point K. And for as much as from the point G taken without the circle DEF is drawn the line GK passing by the centre H, and from the same point is● drawn also an other line not passing by the country, namely, the line GF. Therefore (by the 8. of this book) the outward part GD. of the line GK shall be less than the outward part GA of the line GF ● But the line GA is equal to the line GB. Wherefore the line GD is less than the line GB ● namely, the whole less than the part● which is absurd.    

The 12. Theorem. The 13. Proposition. A circle can not touch an other circle in more points than one, whether they touch within or without.  
FOr if it be possible, let the circle ABCD touch the circle EBFD first inwardly in more points than one, that is, in D and B. 
Of circles which touch the one the other inwardly.  Take (by the first of the third) the centre of the circle ABCD, and let the same be the point G: and likewise the centre of the circle EBFD, and let the same be the point H. Wherefore (by the 11. of the same) a right line drawn from the point G to the point H and produced, will fall upon the points B and D: let it so fall as the line BGHD doth. And for as much as the point G is the centre of the circle ABCD, therefore (by the 15. definition of the first) the line BG is equal to the line DG. Wherefore the line BG is greater than then the line HD: Wherefore the line BH is much greater than the line HD. Again for as much as the point H is the centre of the circle EBFD, therefore (by the same definition) the line BH is equal to the line HD: and it is proved that it is m●ch greater than it: which is impossible. A circle therefore can not touch a circle inwardly in more points than one.  
Now I say that neither outwardly also a circle toucheth a circle in more points than one. 
Of circles which touch the one the other outwardly.  For if it be possible, let the circle ACK touch the circle ABCD outwardly in more points then one, that is, in A and C: And (by the first petition) draw a line from the point A to the point C. Now for as much as in the circumference of either of the circles ABCD, and ACK, are taken two points at all adventures, namely, A and C, therefore (by the second of the third) a right line joining together those points shall fall within both the circles. But it falleth within the circle ABCD, & without the circle ACK: which is absurd. Wherefore a circle shall not touch a circle outwardly in more points than one, and it is proved that neither also inwardly. Wherefore a circle can not touch an other circle in more points than one, whether they touch within or without: which was required to be demonstrated.  

¶ An other demonstration after Pelitarius and Flussates.  

another demonstration after Pelitarius & Flussates, of circles which touch the one the other outwardly.  Suppose that there be two circles ABG and ADG, which if it be possible, let touch the one the other outwardly in more points than one, namely, in A and G. Let the centre of the circle ABG be the point I, and let the centre of the circle ADG be the point K. And draw a right line from the point I to the point K, which (by the 12. of this book) shall pass both by the point A and by the point G: which is not possible: for then two right lines should include a superficies, contrary to the last common sentence. It may also be thus demonstrated. Draw a line from the centre I to the centre K, which shall pass by one of the touches, as for example by the point A. And draw these right lines GK and GI', and so shall be made a triangle, whose two sides GK and GI' shall not be greater than the side IK: which i● contrary to the 20. of the first.  
But now if it be possible, let the foresaid circle ADG touch the circle ABC inwardly in more points than one, 
Of circles which touch the one the other inwardly.  namely, in the points A and G: and let the centre of the circle ABC be the point I, as before: and let the centre of the circle ADG be the point K, as also before. And extend a line from the point I to the point K, which shall fall upon the touch (by the 11. of this book). Draw also these lines KG, and IG. And for as much as the line KG is equal to the line KA (by the 15. definition of the first) add the line KI common to them both. Wherefore the whole line AI is equal to the two lines KG and KI: but unto the line AI is equal the line IG (by the definition of a circle). Wherefore in the triangle IKG the side IG is not less than the two sides IK and KG: which is contrary to the 20. of the first.    

The 13. Theorem. The 14. Proposition. In a circle, equal right lines, are equally distant from the centre. And lines equally distant from the centre, are equal the one to the other.  

The first part of this Theorem.  SVppose that there be a circle ABCD, and let there be in it drawn these equal right lines AB and CD. Then I say that they are equally distant from the centre. 
Construction.  Take (by the first of the third) the centre of the circle ABCD, and let the same be the point E. And (by the 12. of the first) from the point E draw unto the lines AB & CD perpendicular lines EF and EG. And (by the first petition) draw these right lines AE and CE. 
Demonstration.  Now for as much as a certain right line EF drawn by the centre cutteth a certain other right line AB not drawn by the centre, in such sort that it maketh right angles, therefore (by the third of the third) it divideth it into two equal parts. Wherefore the line AF is equal to the line FB. Wherefore the line AB is double to the line AF: and by the same reason also the line CD is double to the line CG. But the line AB is equal to the line CD. Wherefore the line AF is also equal to the line CG. And for as much as (by the 15. definition of the first) the line AE is equal to the line EC, therefore the square of the line EC is equal to the square of the line AE. But unto the square of the line AE, are equal (by the 47. of the first) the squares of the lines AF & FE: for the angle at the point F is a right angle. And (by the self same) to the square of the line EC are equal the squares of the lines EG and GC: for the angle at the point G is a right angle. Wherefore the squares of the lines AF and FE are equal to the squares of the lines CG and GE: of which the square of the line AF is equal to the square of the line CG: for the line AF is equal to the line CG. Wherefore (by the third common sentence) the square remaining, namely, the square of the line FE, is equal to the square remaining, namely, to the square of the line EG. Wherefore the line EF is equal to the line EG. 
Demonstration.  But right lines are said to be equally distant from the centre, when perpendicular lines drawn from the centre to those lines, are equal (by the 4. definition of the third). Wherefore the lines AB and CD are equally distant from the centre.  
But now suppose that the right lines AB and CD be equally distant from the centre, 
The second part which is the converse of the first.  that is, let the perpendicular line EF be equal to the perpendicular line EG. Then I say that the line AB is equal to the line CD. For the same order of construction remaining, we may in like sort prove that the line AB is double to the line AF, and that the line CD is double to the line CG. And for as much as the line AE is equal to the line CE, for they are drawn from the centre to the circumference, therefore the square of the line AE is equal to the square of the line CE. But (by the 47. of the first) to the square of the line AE are equal the squares of the lines EF and FA. And (by the self same) to the square of the line CE are equal the squares of the lines EG and GC. Wherefore the squares of the lines EF and FA are equal to the squares of the lines EG and GC. Of which the square of the line EG is equal to the square of the line EF, for the line EF is equal to the line EG. Wherefore (by the third common sentence) the square remaining, namely, the square of the line AF, is equal to the square of the line CG. Wherefore the line AC is equal unto the line CG. But the line AB is double to the line AF, and the line CD is double to the line CG. Wherefore the line AB is equal to the line CD. Wherefore in a circle equal right lines are equally distant from the centre. And lines equally distant from the centre, are equal the one to the other: which was required to be proved.  

¶ An other demonstration for the first part after Campane.  

another demonstration of the first part after Campane.  Suppose that there be a circle ABDC, whose centre let be the point E. And draw in it two equal lines AB and CD. Then I say that they are equally distant from the centre. Draw from the centre unto the lines AB and CD, these perpendicular lines EF and EG. And (by the 2. part of the 3. of this book) the line AB shall be equally divided in the point F. and the line CD shall be equally divided in the point G. And draw these right lines EA, EB, EC, and ED. And for as much as in the triangle AEB the two sides AB and AE are equal to the two sides CD and CE of the triangle CED, & the base EB is equal to the base ED. therefore (by the 8. of the first) the angle at the point A shall be equal to the angle at the point C. And for as much as in the triangle AEF the two sides AE and AF are equal to the two sides CE and CG of the triangle CEG, and the angle EAF is equal to the angle CEG, therefore (by the 4. of the first) the base EF i● equal to the base EG: which for as much as they are perpendicular lines, therefore the lines AB & CD are equally distant from the centre, by the 4. definition of this book.    

The 14. Theorem. The 15. Proposition. In a circle, the greatest line is the diameter, and of all other lines that line which is nigher to the centre is always greater than that line which is more distant.  
SVppose that there be a circle ABCD, and let the diameter thereof be the line AD, and let the centre thereof be the point E. And unto the diameter. AD let the line BC be nigher than the line FG. Then I say that the line AD is the greatest, and the line BC is greater than the line FG. Draw (by the 12. of the first) from the centre E to the lines BC and FG perpendicular lines EH and EK. 
Construction.  And for as much as the line BC is nigher unto the centre than the line FG, therefore (by the 4. definition of the third) the line EK is greater than the line EH. And (by the third of the first) put unto the line EH an equal line EL. And (by the 11. of the first) from the point L raise up unto the line EK a perpendicular line LM: and extend the line LM to the point N. And (by the first petition) draw these right lines, EM, EN, EF, and EG. And for as much as the line EH is equal to the line EL, therefore (by the 14. of the third, 
Demonstration.  and by the 4. definition of the same) the line BC is equal to the line MN. Again for as much as the line AE is equal to the line EM, and the line ED to the line EN, therefore the line AD is equal to the lines ME and EN. But the lines ME and EN are (by the 20. of the first) greater then the line MN. Wherefore the line AD is greater than the line MN. And for as much as these two lines ME and EN are equal to these two lines FE and EG (by the 15. definition of the first) for they are drawn from the centre to the circumference, and the angle MEN is greater than the angle FEG, therefore (by the 24. of the first) the base MN is greater than the base FG. But it is proved that the line MN is equal to the line BC: Wherefore the line BC also is greater than the line FG. Wherefore the diameter AD is the greatest, and the line BC is greater than the line FG. Wherefore in a circle, the greatest line is the diameter, and of all the other lines, that line which is nigher to the centre is always greater than that line which is more distant: which was required to be proved.  

¶ An other demonstration after Campane.  
In the circle ABCD, whose centre let be the point E, draw these lines, AB, AC, AD, FG, and HK, of which let the line AD be the diameter of the circle. 
another demonstration after Campane.  Then I say that the line AD is the greatest of all the lines. And the other lines each of the one is so much greater than each of the other, how much nigher it is unto the centre. join together the ends of all these lines with the centre, by drawing these right lines EB, EC, EG, EK, EH, and EF. And (by the 20. of the first) the two sides EF and EG of the triangle EFG, shall be greater than the third side FG. And for as much as the said sides EF & EG are equal to the line AD (by the definition of a circle) therefore the line AD is greater than the line FG. And by the same reason it is greater than every one of the rest of the lines, if they be put to be bases of triangles: for that every two sides drawn from the centre are equal to the line AD. Which is the first part of the Proposition. Again, for as much as the two sides EF and EG of the triangle EFG, are equal to the two sides EH and EK of the triangle EHK, and the angle FEG is greater than the angle HEK, therefore (by the 24. of the first) the base FG is greater than the base HK. And by the same reason may it be proved, that the line AC is greater than the line AB. And so is manifest the whole Proposition.    

The 15. Theorem. The 16. Proposition. If from the end of the diameter of a circle be drawn a right line making right angles: it shall fall without the circle: and between that right line and the circumference can not be drawn an other right line: and the angle of the semicircle is greater than any acute angle made of right lines, but the other angle is less than any acute angle made of right lines.  
SVppose that there be a circle ABC: whose centre let be the point D, and let the diameter thereof be AB. Then I say that a right line drawn from the point A, making with the diameter AB right angles, shall fall without the circle. 
The first part of this Theorem.  For if it do not, then if it be possible, let it fall within the circle as the line AC doth, and draw a line from the point D to the point C. 
Demonstration leading to an absurdity.  And for as much as (by the 15. definition of the first) the line DA is equal to the line DC, for they are drawn from the centre to the circumference, therefore the angle DAC is equal to the angle ACD. But the angle DAC is (by supposition) a right angle: Wherefore also the angle ACD is a right angle. Wherefore the angles DAC and ACD, are equal to two right angles: which (by the 17. of the first) is impossible. Wherefore a right line drawn from the point A, making with the diameter AB right angles, shall not fall within the circle. In like sort also may we prove, that it falleth not in the circumference. Wherefore it falleth without, as the line AE doth.  
I say also, 
Second part.  that between the right line AE, and the circumference ACB, can not be drawn an other right line. For if it be possible, let the line AF so be drawn. And (by the 12. of the first) from the point D draw unto the line FA a perpendicular line DG: And for as much as AGD is a right angle, but DAG is less than a right angle, therefore (by the 19 of the first) the side AD is greater than the side DG. But the line DA is equal to the line DH, for they are drawn from the centre to the circumference. Wherefore the line DH is greater than the line DG: namely, the less greater than the greater: which is impossible. Wherefore between the right line AE and the circumference ACB, can not be drawn an other right line.  
I say moreover, 
Third part.  that the angle of the semicircle contained under the right line AB and the circumference CHA, is greater than any acute angle made of right lines. And the angle remaining contained under the circumference CHA and the right line AE, is less than any acute angle made of right lines. For if there be any angle made of right lines greater than that angle which is contained under the right line BA and the circumference CHA, or less than that which is contained under the circumference CHA and the right line AE, then between the circumference CHA and the right line AE, there shall fall a right line, which maketh the angle contained under the right lines, greater than that angle which is contained under the right line BA and the circumference CHA, and less than the angle which is contained under the circumference CHA and the right line AE. But there can fall no such line, as it hath before been proved. Wherefore no acute angle contained under right lines, is greater than the angle contained under the right line BA and the circumference CHA, nor also less than the angle contained under the circumference CHA and the line AE.  

Correlary.  
Hereby it is manifest that a perpendicular line drawn from the end of the diameter of a circle toucheth the circle: and that a right line toucheth a circle in one point only. For it was proved (by the 2. of the third) that a right line drawn from two points taken in the circumference of a circle, shall fall within the circle. Which was required to be demonstrated.    

The 2. Problem. The 17. Proposition. From a point given, to draw a right line which shall touch a circle given.  

Construction.  SVppose that the point given be A, and let the circle given be BCD. It is required from the point A to draw a right line which shall touch the circle BCD. Take (by the first of the third) the centre of the circle, and let the same be E. And (by the first petition) draw the right line ADE. And making the centre E, and the space AE, describe (by the third petition) a circle AFG. And from the point D raise up (by the 11. of the first) unto the line EA a perpendicular line DF. And (by the first petition) draw these lines EBF and AB. Then I say, that from the point A is drawn to the circle BCD a touch line AB. 
Demonstration.  For for as much as the point E is the centre of the circle BCD, and also of the circle AFG, therefore the line EA is equal to the line EF, and the line ED to the line EB, for they are drawn from the centre to the circumference. Wherefore to these two lines AE and EB, are equal these two lines EF & ED, and the angle at the point E is common to them both: Wherefore (by the 4. of the first) the base DF is equal to the base AB, and the triangle DEF is equal to the triangle EBA, and the rest of the angles remaining to the rest of the angles remaining. Wherefore the angle EDF is equal to the angle EBA. But the angle EDF is a right angle: Wherefore also the angle EBA is a right angle, and the line EB is drawn from the centre. But a perpendicular line drawn from the end of the diameter of a circle, toucheth the circle (by the Corellary of the 16. of the third). Wherefore the line AB toucheth the circle BCD. Wherefore from the point given, namely, A, is drawn unto the circle given BCD, a touch line AB: which was required to be done.  

¶ An addition of Pelitarius.  

An addition of Pelitarius.  Unto a right line which cutteth a circle, to draw a parallel line which shall touch the circle.  
Suppose that the right line AB do out the circle ABC in the points A and B. It is required to draw unto the line AB a parallel line which shall touch the circle. Let the centre of the circle be the point D. And divide the line AB into two equal parts in the point E. And by the point E and by the centre D, draw the diameter CDEF. And from the point F (which is the end of the diameter) raise up (by the 11. of the first) unto the diameter CF a perpendicular line GFH. Then I say that the line GFH (which by the corollary of the 16. of this book toucheth the circle) is a parallel unto the line AB. 
This Problem commodious for the inscribing and circumscribing of figures in or abou● circles.  For forasmuch as the right line CF falling upon either of these lines AB & GH maketh all the angles at the point ● right angles (by the 3. of this book) and the two angles at the point Far supposed to be right angles: therefore (by the 29. of the first) the lines AB and GH are parallels: which was required to be done. And this Problem is very commodious for the inscribing or circumscribing of figures in or about circles.    

The 16. Theorem. The 18. Proposition. If a right line touch a circle, and from the centre to the touch be drawn a right line, that right line so drawn shallbe a perpendicular line to the touch line.  
SVppose that the right line DE do touch the circle ABC in the point C. And take the centre of the circle ABC, and let the same be F. 
Demonstration leading to an impossibility.  And (by the first petition) from the point F to the point C draw a right line FC. Then I say, that CF is a perpendicular line to DE. For if not, draw (by the 12. of the first) from the point F to the line DE a perpendicular line FG. And for as much as the angle FGC is a right angle, therefore the angle GCF is an acute angle: Wherefore the angle FGC is greater than the angle FCG, but unto the greater angle is subtended the greater side (by the 19 of the first). Wherefore the line FC is greater than the line FG. But the line FC is equal to the line FB, for they are drawn from the centre to the circumference: Wherefore the line FB also is greater than the line FG, namely, the less than the greater: which is impossible. Wherefore the line FG is not a perpendicular line unto the line DE. And in like sort may we prove, that no other line is a perpendicular line unto the line DE besides the line FC: Wherefore the line FC is a perpendicular line to DE. If therefore a right line touch a circle, & from the centre to the touch be drawn a right line, that right line so drawn shall be a perpendicular line to the touch line: which was required to be proved.  

¶ An other demonstration after Orontius.  
Suppose that the circle given be ABC, which let the right line DE touch in the point C. 
another demonstration after Orontius.  And let the centre of the circle be the point F. And draw a right line from F to C. Then I say that the line FC is perpendicular unto the line DE. For if the line FC be not a perpendiculer unto the line DE, then, by the converse of the x. definition of the first book, the angles DCF & FCE shall be unequal: & therefore the one is greater than a right angle, and the other is less than a right angle. (For the angles DCF and FCE are by the 13. of the first equal to two right angles) Let the angle FCE, if it be possible, be greater than a right angle, that is, let it be an obtuse angle. Wherefore the angle DCF ●hal be an acute angle. And forasmuch as by supposition the right line DE toucheth the circle ABC, therefore it cutteth not the circle. Wherefore the circumference BC falleth between the right lines DC & CF: & therefore the acute and rectiline angle DCF shall be greater than the angle of the semicircle BCF which is contained under the circumference BC & the right line CF. And so shall there be given a rectiline & acute angle greater than the angle of a semicircle: which is contrary to the 16. proposition of this book. Wherefore the angle DCF is not less than a right angle. In like sort also may we prove that it is not greater than a right angle. Wherefore it is a right angle, and therefore also the angle FCE is a right angle. Wherefore the right line FC is a perpendicular unto the right line DE by the 10. definition of the first● which was required to be proved.    

The 17. Theorem. The 19 Proposition. If a right line do touch a circle, and from the point of the touch be raised up unto the touch line a perpendicular line, in that line so raised up is the centre of the circle.  
SVppose that the right line DE do touch the circle ABC in the point C. And from C raise up (by the 11. of the first) unto the line DE a perpendicular line CA Then I say, that in the line CA is the centre of the circle. For if not, then if it be possible, let the centre be without the line CA, as in the point F. And (by the first petition) draw a right line from C to F. 
Demonstration leading to an impossibility.  And for as much as a certain right line DE toucheth the circle ABC, and from the centre to the touch is drawn a right line CF, therefore (by the 18. of the third) FC is a perpendicular line to DE. Wherefore the angle FCE is a right angle. But the angle ACE is also a right angle: Wherefore the angle FCE is equal to the angle ACE, namely, the less unto the greater: which is impossible● Wherefore the point F is not the centre of the circle ABC. And in like sort may we prove, that it is no other where but in the line AC. If therefore a right line do touch a circle, and from the point of the touch be raised up unto the touch line a perpendicular line, in that line so raised up is the centre of the circle: which was required to be proved.   

The 18. Theorem. The 20. Proposition. In a circle an angle set at the centre, is double to an angle set at the circumference, so that both the angles have to their base one and the same circumference.  
SVppose that there be a circle ABC, and at the centre thereof, 
Two cases in this Proposition the one when the angle set at the circumference includeth the centre.  namely, the point E, let the angle BEC be set, & at the circumference let there be set the angle BAC, and let them both have one and the same base, namely, the circumference BC. Then I say, that the angle BEC is double to the angle BAC. Draw the right line AE, and (by the second petition) extend it to the point F. Now for as much as the line AE is equal to the line EB, 
Demonstration.  for they are drawn from the centre unto the circumference, the angle EAB is equal to the angle EBA (by the 5. of the first). Wherefore the angles EAB and EBA are double to the angle EAB. But (by the 32. of the same) the angle BEF is equal to the angles EAB and EBA: Wherefore the angle BEF is double to the angle EAB. And by the same reason the angle FEC is double to the angle EAC. Wherefore the whole angle BEC is double to the whole angle BAC.  
Again, suppose that there be set an other angle at the circumference, 
The other when the same angle set at the circumference includeth not the centre.  and let the same be BDC. And (by the ●irst petition) draw a line from D to E. And (by the second petition) extend the line DE unto the point G. And in like sort may we prove, that the angle GEC is double to the angle EDC. Of which the angle GEB is double to the angle EDB. Wherefore the angle remaining BEC is double to the angle remaining BDC. Wherefore in a circle an angle set at the centre, is double to an angle set at the circumference, so that both the angles have to their base one and the same circumference: which was required to be demonstrated.   

The 19 Theorem. The 21. Proposition. In a circle the angles which consist in one and the self same section or segment, are equal the one to the other.  
SVppose that there be a circle ABCD, & in the segment thereof BAED, let there consist these angles BAD and BED. Then I say, that the angles BAD and BED are equal the one to the other. Take (by the first of the third) the centre of the circle ABCD, 
Construction.  and let the same be the point F. And (by the first petition) draw these lines BF and FD. 
Demonstration.  Now for as much as the angle BFD is set at the centre, and the angle BAD at the circumference, and they have both one and the same base, namely, the circumference BCD, therefore the angle BFD is (by the Proposition going before) double to the angle BAD: and by the same reason the angle BFD is also double to the angle BED. Wherefore (by the 7. common sentence) the angle BAD is equal to the angle BED. Wherefore in a circle the angles which consist in one and the self same segment, are equal the one to the other: which was required to be proved.  


Three cases in this Proposition.  
The first case.  
The second case.   In this proposition are three cases. For the angles consisting in one and the self same segment, the segment may either be greater than a semicircle, or less than a semicircle, or else just a semicircle. For the first case the demonstration before put serveth.  
The self same construction and demonstration will also serve, 
The third case.  if the angles were set in a semicircle as it is pla●ne to see, in the figure here set.   

The 20. Theorem. The 22. Proposition. If within a circle be described a figure of four sides, the angles thereof which are opposite the one to the other, are equal to two right angles.  
SVppose that there be a circle ABCD, and let there be described in it a figure of four sides, namely, ABCD. Then I say, that the angles thereof which are opposite the one to the other, are equal to two right angles. Draw (by the first petition) these right lines AC and BD. 
Construction.  Now for as much as (by the 32. of the first) the three angles of every triangle are equal to two right angles: therefore the three angles of the triangle ABC, namely, 
Demonstration.  CAB, ABC, and BCA, are equal to two right angles. But (by the 21. of the third) the angle CAB is equal to the angle BDC, for they consist in one and the self same segment, namely, BADC. And (by the same Proposition) the angle ACB is equal to the angle ADB, for they consist in one and the same segment ADCB. Wherefore the whol● angle ADC is equal to the angles BAC and ACB: put the angle ABC common to them both. Wherefore the angles ABC, BAC, and ACB, are equal to the angles ABC and ADC. But the angles ABC, BAC, and ACB, are equal to two right angles. Wherefore the angles ABC and ADC are equal to two right angles. And in like sort also may we prove, that the angles BAD and DCB are equal to two right angles. If therefore within a circle be described a figure of four sides, the angles thereof which are opposite the one to the other, are equal to two right angles● which was required to be proved.   

The 21. Theorem. The 23. Proposition. Upon one and the self same right line can not be described two like and unequal segments of circles, falling both on one and the self same side of the line.  

Demonstration leading to an impossibility.  FOr if it be possible, let there be described upon the right line AB two like & unequal sections of circles, namely, ACB & ADB, falling both on one and the self same side of the line AB. And (by the first petition) draw the right line ACD, and (by the third petition) draw right lines from C to B, and from D to B. And for as much as the segment ACB is like to the segment ADB: and like segments of circles are they which have equal angles (by the 10. definition of the third). Wherefore the angle ACB is equal to the angle ADB, namely, the outward angle of the triangle CDB to the inward angle: which (by the 16. of the first) is impossible. Wherefore upon one and the self same right line can not be described two like & unequal segments of circles, falling both on one & the self same side of the line: which was required to be demonstrated.  

An addition of Campane demonstrated by Pelitari●s.  Here Campane addeth that upon one and the self same right line cannot be described two like and unequal sections neither on one and the self same side of the line, nor on the opposite side. That they can not be described on one and the self same side, hath been before demonstrated, and that neither also on the opposite side, Pelitarius thus demonstrateth.   

The 22. Theorme. The 24. Proposition. Like segments of circles described upon equal right lines, are equal the one to the other.  
SVppose that upon these equal right lines AB and CD be described these like segments of circles, namely, AEB and CFD. Then I say, that the segment AEB is equal to the segment CFD. For putting the segment AEB upon the segment CFD, and the point A upon the point C, and the right line AB upon the right line CD, the point B also shall fall upon the point D, for the line AB is equal to the line CD. And the right line AB exactly agreeing with the right line CD, the segment also AEB shall exactly agreed with the segment CFD. 
Demonstration leading to an impossibility.  For if the right line AB do exactly agreed with the right line CD, and the segment AEB do not exactly agreed with the segment CFD, but differeth as the segment CGD doth: Now (by the 13. of the third) a circle cutteth not a circle in more points than two, but the circle CGD cutteth the circle CFD in more points than two, that is, in the points C, G, and D: which is (by the same) impossible. Wherefore the right line AB exactly agreeing with the right line CD, the segment AEB shall not but exactly agreed with the segment CFD: Wherefore it exactly agreeth with it, and is equal unto it. Wherefore like segments of circles described upon equal right lines, are equal the one to the other: which was required to be proved.  
This Proposition may also be demonstrated by the former proposition For if the sections AEB and CFD being like and set upon equal right lines AB and CD, 
another demonstration.  should be unequal, than the one being put upon the other, the greater shall exceed the less: but the line AB is one line with the line CD: so that thereby shall follow the contrary of the former Proposition.   

The 3. Problem. The 25. Proposition. A segment of a circle being given to describe the whole circle of the same segment.  
SVppose that the segment given be ABC. It is required to describe the whole circle of the same segment ABC. 
Construction.  Divide (by the 10. of the first) the line AC into two equal parts in the point D. And (by the 11. of the same) from the point D raise up unto the line AC a perpendicular line BD. And (by the first petition) draw a right line from A to B. Now then the angle ABD being compared to the angle BAD, is either greater than it, or equal unto it, or less than it.  


Three cases in this Proposition.  
The first case.   First let it be greater. And (by the 23. of the same) upon the right line BA, and unto the point in it A, make unto the angle ABD an equal angle BAE. And (by the second petition) extend the line BD unto the point E. And (by the first petition) draw a line from E to C. Now for as much as the angle ABE is equal to the angle BAE, 
Demonstration.  therefore (by the 6. of the first) the right line EB is equal to the right line AE. And for as much as the line AD is equal to the line DC, and the line DE is common to them both: therefore these two lines AD ●nd DE, are equal to these t●o lines CD and DE the one to the other. And the angle ADE ●● (by the 4. petition) equal to the angle CDE, for either of them is a right angle● Wherefor● (by the 4. of the first) the base AE is equal to the base CE. But it is pro●ed, that the line AE is squall to the line BE. Wherefore the line BE also is equal to the line CE. Wherefore these three lines AE, EB, and EC, are equal the one to the other. Wherefore making the centre E, and the space either AE, or EB, or EC, describe (by the third petition) a circle, and it shall pass by the points A, B, C. Wherefore there is described the whole circle of the segment given, And it is manifest, that the segment ABC is les●e than a semicircle, for the centre E falleth without it.  
The like demonstration also will serve if the angle ABD be equal to the angle BAD. 
The second case.  For the line AD being equal to either of these lines, BD, and DC, there are three lines● DA, DB, and DC, equal the one to the other. So that the point D shall be the centre of the circle being complete, and ABC shall be a semicircle.  
But if the angle ABD be less than the angle BAD, 
The third case.  then (by the 23. of the first) upon the right line BA, and unto the point in it A, make unto the angle ABD an equal angle within the segment ABC. And so the centre of the circle shall fall in the line DB, and it shall be the point E: and the segment ABC shall be greater than a semicircle, Wherefore a segment being given, there is described the whole circle of the same segment: ●hich was required to be done.  

¶ A Corollary. Hereby it is manifest, that in a semicircle the angle BAD is equal to the angle DBA: but in a section less than a semicircle, it is less: in a section greater than a semicircle, it is greater.  
And by this last general way, 
An addition.  if there be given three points, set howsoever, so that they be not all three in one right line, a man may describe a circle which shall pass by all the said three points. For as in the example before put, if you suppose only the 3. points A, B, C, to be given and not the circumference ABC to be drawn, yet following the self ●ame order you did before, that is, draw a right line from A to B and a other from B to C and divide the said right lines into two equal parts, in the points D and E, and erect the perpendicular lines DF and EF cutting the one the other in the point F, and draw a ●ight line from F to B: and making the centre the point F, and the space FB describe a circle, and it shall pass by the points A & C: which may be proved by drawing right lines from A to F, and from F to C. For forasmuch as the two si●es AD and DF of the triangle ADF are equal to the two sides BD and DF of the triangle BDF (for by supposition the line AD is equal to the line DB, and the line DF is common to them bot●) and the angle ADF is equal to the angle BDF (for they are both right angles) therefore (by the 4. of the first) the base AF is equal to the base BF. And by the same reason the line FC is equal to the line FB. Wherefore these three lines FA, FB and FC are equal the one to the other. Wherefore making the centre the point F and the space FB, it shall also pass by the points A and C. Which was required to be done. This proposition is very necessary for many things as you shall afterward see.  
Pelitarius here addeth a brief way how to find out the centre of a circle, which is commonly used of Artificers.    

The 23. Theorem. The 26. Proposition. Equal angles in equal circles consist in equal circumferences, whether the angles be drawn from the centres, or from the circumferences.  
SVppose that these circle's ABC and DEF, be equal. And from their centres, namely, the points G and H, let there be drawn these equal angles BGC and EHF: and likewise from their circumferences these equal angles BAC and EDF. Then I say, that the circumference BKC is equal to the circumference ELF. Draw (by the first petition) right lines from B to C, 
Construction.  and from E to F. And for as much as the circles ABC and DEF are equal, the right lines also drawn from their centres to their circumferences, 
Demonstration.  are (by the first definition of the third) equal the one to the other. Wherefore these two lines BG and GC, are equal to these two lines EH and HF. And the angle at the point G is equal to the angle at the point H: Wherefore (by the 4. of the first) the base BC is equal to the base EF. And for as much as the angle at the point A is equal to the angle at the point D, therefore the segment BAC is like to the segment EDF. And they are described upon equal right lines BC and EF. But like segments of circles described upon equal right lines, are (by the 24. of the third) equal the one to the other. Wherefore the segment BAC is equal to the segment EDF. And the whole circle ABC is equal to the whole circle DEF. Wherefore (by the third common sentence) the circumference remaining BKC is equal to the circumference remaining ELF. Wherefore equal angles in equal circles consist in equal circumferences, whether the angles be drawn from the centres or from the circumferences: which was required to be demonstrated.   

The 24. Theorem. The 27. Proposition. In equal circles the angles which consist in equal circumferences, are equal the one to the other, whether the angles be drawn from the centres, or from the circumferences.  
SVppose that these circle's ABC, and DEF, be equal. And upon these equal circumferences of the same circles, namely, upon BC and EF, let there consist these angles BGC and EHF drawn from the centres, and also these angles BAC and EDF drawn from the circumferences. Then I say, that the angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF. If the angle BGC be equal to the angle EHF, than it is manifest, that the angle BAC is equal to the angle EDF (by the 20. of the third). But if the angle BGC be not equal to the angle EHF, 
Demonstration leading to an impossibility.  then is the one of them greater than the other. Let the angle BGC be greater And (by the 23. of the first) upon the right line BG, and unto the point given in it G, make unto the angle EHF an equal angle BGK. But (by the 26. of the third) equal angles in equal circles consist upon equal circumferences, whether they be drawn from the centres or from the circumferences. Wherefore the circumference BK is equal to the circumference EF. But the circumference EF is equal to the circumference BC: Wherefore the circumference BK also is equal to the circumference BC, the less to the greater: which is impossible. Wherefore the angle BGC is not unequal to the angle EHF: Wherefore it is equal. And (by the 20. of the third) the angle at the point A is the half of the angle BGC: and (by the same) the angle at the point D is the half of the angle EHF. Wherefore the angle at the point A is equal to the angle at the point D. Wherefore in equal circles, the angles which consist in equal circumferences, are equal the one to the other, whether the angles be drawn from the centres or from the circumferences: which was required to be proved.   

The 25. Theorem. The 28. Proposition. In equal circles, equal right lines do cut away equal circumferences, the greater equal to the greater, and the less equal to the less.  
SVppose that these circle's ABC, and DEF, be equal. And in them let there be drawn these equal right lines BC and EF, which let cut away these circumferences BAC and DEF being the greater, & also these circumferences BGC and EHF being the less. Then I say, that the greater circumference BAC is equal to the greater circumference EDF: and the less circumference BGC is equal to the less circumference EHF. 
Construction.  Take (by the first of the third) the centres of the circles, and let the same be the points K and L. And draw these right lines, KB, KC, LE, and LF. And for as much as the circles are equal, 
Demonstration.  therefore (by the first definition of the third) the lines which are drawn from the centres are equal. Wherefore these two lines BK and KC, are equal to these two lines EL and LF. And (by supposition) the base BC is equal to the base EF. Wherefore (by the 8. of the first) the angle BKC is equal to the angle ELF. But (by the 26. of the third) equal angles drawn from the centres, consist upon equal circumferences. Wherefore the circumference BGC is equal to the circumference EHF: and the whole circle ABC is equal to the whole circle DEF. Wherefore the circumference remaining BAC, is (by the third common sentence) equal to the circumference remaining EDF. Wherefore in circles, equal right lines do cut away equal circumferences, the greater equal to the greater, and the less equal to the less: which was required to be proved.   

The 26. Theorem. The 29. Proposition. In equal circles, under equal circumferences are subtended equal right lines.  

The converse of the former Proposition.  SVppose that these circle's ABC and DEF, be equal. And in them let there be taken these equal circumferences, BGC and EHF: and draw these right lines BC and EF. Then I say, that the right line BC is equal to the right line EF. Take (by the first of the third) the centres of the circles, and let them be the points K and L, 
Construction.  and draw these right lines KB, KC, LE, LF. And for as much as the circumference BGC is equal to the circumference EHF, 
Demonstration.  the angle BKC is equal to the angle ELF (by the 27. of the third). And for as much as the circles ABC and DEF are equal the one to the other, therefore (by the first definition of the third) the lines which are drawn from the centres are equal. Wherefore these y. lines BK and KC, are equal to these y. lines LE and LF, & they comprehend equal angles. Wherefore (by the 4. of the first) the base BC is equal to the base EF. Wherefore in equal circles under equal circumferences, are subtended equal right lines: which was required to be demonstrated.   

The 4. Problem. The 30. Proposition. To divide a circumference given into two equal parts.  
SVppose that the circumference given be ADB. It is required to divide the circumference ADB into two equal parts. 
Construction.  Draw a right line from A to B. And (by the 10. of the first) divide the line AB into two equal parts in the point C. And (by the 11. of the first) from the point C raise up unto AB a perpendicular line CD. And draw these right lines AD and DB. And forasmuch as the line AC is equal to the line CB, 
Demonstration.  & the line CD is common to them both, therefore these two lines AC and CD are equal to these two lines BC and CD. And (by the 4. petition) the angle ACD is equal to the angle BCD, for either of them is a right right angle. Wherefore (by the 4. of the first) the base AD is equal to the base DB. But equal right lines do cut away equal circumferences, the greater equal to the greater, & the less equal to the less (by the 28. of the third) And either of these circumferences AD and DB is less than a semicircle. Wherefore the circumference AD is equal to the circumference DB. Wherefore the circumference given is divided into two equal parts: Which was required to be done.   

The 27. Theorem. The 31. Proposition. In a circle an angle made in the semicircle is a right angle: but an angle made in the segment greater than the semicircle is less than a right angle, and an angle made in the segment less than the semicircle, is greater than a right angle. And moreover the angle of the greater segment is greater than a right angle: and the angle of the less segment is less than a right angle.  

Second part.  And forasmuch as (by the 17. of the first) the two angles of the triangle ABC, namely, ABC and BAC are less than two right angles, and the angle BAC is a right angle. Therefore the angle ABC is less than a right angle, and it is in the segment ABC which is greater than the semicircle.  

Thir● part.  And forasmuch as in the circle there is a figure of four sides, namely, ABCD. But if within a circle be described a figure of four sides, the angles thereof which are opposite the one to the other are equal to two right angles (by the 22. of the third) Wherefore (by the same) the angles ABC and ADC are equal to two right angles. But the angle ABC is less than a right angle. Wherefore the angle remaining ADC is greater than a right angle, and it is in a segment which is less than the semicircle.  
Again forasmuch as the angle comprehended under the right lines AC and AF is a right angle, 
The fift and last part.  therefore the angle comprehended under the right line CA and the circumference ADC is less than a right angle. Wherefore in a circle an angle made in the semicircle is a right angle, but an angle made in the segment greater than the semicircle is less than a right angle, and an angle made in the segment less than the semicircle, is greater than a right angle. And moreover the angle of the greater segment is greater than a right angle: & the angle of the less segment is less than a right angle: which was required to be demonstrated.  
another demonstration to prove that the angle BAC is a right angle. 
another Demonstration to prove that the ang●e in a semicircle is a right angle.  Forasmuch as the angle AEC is double to the angle BAE (by the 32. of the first) for it is equal to the two inward angles which are opposite. But the inward angles are (by the 5. of the first) equal the one to the other, and the angle AEB is double to the angle EAC. Wherefore the angles AEB and AEC are double to the angle BAC. But the angles AEB and AEC are equal to two right angles: Wherefore the angle BAC is a right angle. Which was required to be demonstrated.  

Correlary.  
Hereby it is manifest, 
A Corollary.  that if in a triangle one angle be equal to the two other angles remaining the same angle is a right angle: for that the side angle to that one angle (namely, the angle which is made of the side produced without the triangle) is equal to the same angles, but when the side angles are equal the one to the other, they are also right angles.  

¶ An addition of Pelitarius.  
If in a circle be inscribed a rectangle triangle, the side opposite unto the right angle shall be the diameter of the circle.  

An addition of P●litarius.  Suppose that in the circle ABC be inscribed a rectangle triangle ABC, whose angle at the point B let be a right angle. Then I say, that the side AC is the diameter of the circle. For if not, then shall the centre be without the line AC, as in the point E. 
Demonstration lea●ing to an absurdity.  And draw a line from the point A to the point E, & produce it to the circumference to the point D: and let AED be the diameter: and draw a line from the point B to the point D. Now (by this 31. Proposition) the angle ABD shall be a right angle, and therefore shall be equal to the right angle ABC, namely, the part to the whole: which is absurd. Even so may we prove, that the centre is in no other where but in the line AC. Wherefore AC is the diameter of the circle: which was required to be proved.   

¶ An addition of Campane.  

An addition of Campane.  By this 31. Proposition, and by the 16. Proposition of this book, it is manifest, that although in mixed angles, which are contained under a right line and the circumference of a circle, there may be given an angle less & greater than a right angle, yet can there never be given an angle equal to a right angle. For every section of a circle is either a semicircle, or greater than a semicircle, or less, but the angle of a semicircle is by the 16. of this book, less than a right angle, and so also is the angle of a less section by this 31. Proposition: Likewise the angle of a greater section, is greater than a right angle, as it hath in this Proposition been proved.     

The 28. Theorem. The 32. Proposition. If a right line touch a circle, and from the touch be drawn a right line cutting the circle: the angles which that line and the touch line make, are equal to the angles which consist in the alternate segments of the circle.  
SVppose that the right line EF do touch the circle ABCD in the point B: and from the point B let there be drawn into the circle ABCD a right line cutting the circle, and let the same be BD. Then I say, that the angles which the line BD together with the touch line EF do make, are equal to the angles which are in the alternate segments of the circle, that is, the angle FBD is equal to the angle which consisteth in the segment BAD, and the angle EBD is equal to the angle which consisteth in the segment BCD. 
Construction.  Raise up (by the 11. of the first) from the point B unto the right line EF a perpendicular line BA. And in the circumference BD take a point at all adventures, and let the same be C. And draw these right lines AD, DC, and CB. 
Demonstration.  And for as much as a certain right line EF toucheth the circle ABC in the point B, and from the point B where the touch is, is raised up unto the touch line a perpendicular BA. Therefore (by the 19 of the third) in the line BA is the centre of the circle ABCD. Wherefore the angle ADB being in the semicircle, is (by the 31. of the third) a right angle. Wherefore the angles remaining BAD and ABD, are equal to one right angle. But the angle ABF is a right angle. Wherefore the angle ABF is equal to the angles BAD and ABD. Take away the angle ABD which is common to them both. Wherefore the angle remaining DBF, is equal to the angle remaining BAD, which is in the alternate segment of the circle. And for as much as in the circle is a figure of four sides, namely, ABCD, therefore (by the 22. of the third) the angles which are opposite the one to the other, are equal to two right angles. Wherefore the angles BAD and BCD, are equal to two right angles. But the angles DBF and DBE, are also equal to two right angles. Wherefore the angles DBF and DBE, are equal to the angles BAD and BCD. Of which we have proved that the angle BAD is equal to the angle DBF. Wherefore the angle remaining DBE, is equal to the angle remaining DCB, which is in the alternate segment of the circle, namely, in the segment DCB. If therefore a right line touch a circle, and from the touch be drawn a right line cutting the circle: the angles which that line and the touch line make, are equal to the angles which consist in the alternate segments of the circle: which was required to be proved.  
In this Proposition may be two cases. 
Two cases in this Proposition.  For the line drawn from the touch and cutting the circle, may either pass by the centre or not. If it pass by the centre, then is it manifest (by the 18. of this book) that it falleth perpendicularly upon the touch line, and divideth the circle into two equal parts, so that all the angles in each semicircle, are by the former Proposition, right angles, and therefore equal to the alternate angles made by the said perpendicular line and the touch line. If it pass not by the centre, then follow the construction and demonstration before put.   

The 5. Problem. The 33. Proposition. Upon a right line given to describe a segment of a circle, which shall contain an angle equal to a rectiline angle given.  
SVppose that the right line given be AB, and let the rectiline angle ge●●n be C. It is required upon the right line given AB to describe a segment of a circle which shall contain an angle equal to the angle C. Now the angle C is either an acute angle, or a right angle, or an obtuse angle.  


Three cases in this Proposition.  
The first case.   First, let it be an acute angle as in the first description. And (by the 23 of the first) upon the right line AB and to the point in it A describe an angle equal to the angle C, 
Construction.  and let the same be DAB. Wherefore the angle DAB is an acute angle. From the point A raise up (by the 11. of the first) unto the line AD a perpendicular line AF. And (by the 10. of the first) divide the line AB into two equal parts in the point F. And (by the 11. of the same) from the point F raise up unto the line AB a perpendicular line FG, and draw a line from G to B. 
Demonstration.  And forasmuch as the line AF is equal to the line FB and the line FG is common to them both, therefore these two lines AF and FG are equal to these two lines FB and FG: and the angle AFG is (by the 4. petition) equal to the angle GFB. Wherefore (by the 4. of the same) the base AG is equal to the base GB. Wherefore making the centre G and the space GA describe (by the 3. petition) a circle and it shall pass by the point B: describe such a circle & let the same be ABE: And draw a line from E to B. Now forasmuch as from the end of the diameter AE, namely, from the point A is 〈◊〉 a right line AD making together with the right line AE a right angle, therefore (by the corollary of the 16. of the third) the line AD toucheth the circle ABE. And forasmuch as a certain right line AD toucheth the circle ABE, & from the point A where the touch is, is drawn into the circled a certain right line AB: therefore (by the 32. of the third) the angle DAB is equal to the angle AEB which is in the alternate segment of the circle. But the angle DAB is equal to the angle C, wherefore the angle C is equal to the angle AEB. Wherefore upon the right line given AB is described a segment of a circle which containeth the angle AEB, which is equal to the angle given, namely, to C.  

The second case.  But now suppose that the angle C be a right angle. It is again required upon the right line AB to describe a segment of a circle, which shall contain an angle equal to the right angle C. 
Construction.  Describe again upon the right line AB and to the point in it A an angle BAD equal to the rectiline angle given C (by the 23. of the first) as it is set forth in the second description. And (by the 10. of the first) divide the line AB into two equal parts in the point F. And making the centre the point F and the space FA or FB describe (by the 3. petition) the circle AEB. 
Demonstration.  Wherefore the right line AD toucheth the circle AEB: for that the angle BAD is a right angle. Wherefore the angle BAD is equal to the angle which is in the segment AEB, for the angle which is in a semicircle is a right angle (by the 31. of the third) But the angle BAD is equal to the angle C. Wherefore there is again described upon the line AB a segment of a circle, namely, AEB, which containeth an angle equal to the angle given namely, to C.  
But now suppose that the angle C be an obtuse angle. 
The third case.  Upon the right line AB and to the point in it A describe (by the 23. of the first) an angle BAD equal to the angle C: as it is in the third description. 
Construction.  And from the point A raise up unto the line AD a perpendicular line AE (by the 11. of the first) And again by the 10. of the first) divide the line AB into two equal parts in the point F. And from the point F. ra●se up unto the line AB a perpendicular line FG (by the 11. of the same) & draw a line from G to B. 
Demonstration.  And now forasmuch as the line AF is equal to the line FB, and the line FG is common to them both, therefore these two lines AF and FG are equal to these two lines BF and FG: and the angle AFG is (by the 4. petition) equal to the angle BFG: wherefore (by the 4. of the same) the base AG is equal to the base GB. Wherefore making the centre G, and the space GA describe (by the 3. petition) a circle and it shall pass by the point B: let it be described as the circle AEB is. And forasmuch as from the end of the diameter AE is drawn a perpendicular line AD, therefore (by the corollary of the 16. of the third) the line AD toucheth the circle AEB & from the point of the touch, namely; A, is extended the line AB. Wherefore (by the 32. of the third) the angle BAD is equal to the angle AHB which is in the alternate segment of the circle. But the angle BAD is equal to the angle C. Wherefore the angle which is in the segment AHB is equal to the angle C. Wherefore upon the right line given AB, is described a segment of a circle AHB, which containeth an angle equal to the angle given, namely, C: which was required to be done.   

The 6. Problem. The 34. Proposition. From a circle given to cut away a section which shall contain an angle equal to a rectiline angle given.  
SVppose that the circle given be AC and let the rectiline angle given be D. It is required from the circle ABC to cut away a segment which shall contain an angle equal to the angle D. 
Construction.  Draw (by the 17. of the third) a line touching the circle, and let the same be EF: and let it touch in the point B. And (by the 23. of the first) upon the right line EF and to the point in it B describe the angle FBC equal to the angle D. 
Demonstration.  Now forasmuch as a certain right line EF toucheth the circle ABC in the point B: and frora the point of the touch, namely, B, is drawn into the circle a certain right line BC, therefore (by the 32. of the third) the angle FBC is equal to the angle BAC which is in the alternate segment. But the angle FBC is equal to the angle D. Wherefore the angle BAC which consisteth in the segment BAC is equal to the angle D. Wherefore from the circle given ABC is cut away a segment BAC, which containeth an angle equal to the rectiline angle given: which was required to be done.   

The 29. Theorem. The 35. Proposition. If in a circle two right lines do cut the one the other, the rectangle parallelogram comprehended under the segments or parts of the one line is equal to the rectangle parallelogram comprehended under the segment or parts of the other line.  
LEt the circle be ABCD, and in it let these two right lines AC and BD c●t the one the other in the point E. Then I say that the rectangle parallelogram contained under the parts AE and EC is equal to the rectangle parallelogram contained under the parts DE and EB. 

Two cases in this Proposition.  
First case.   For if the line AC and BD be drawn by the centre, then is it manifest, that for as much as the lines AE and EC are equal to the lines DE and EB by the definition of a circle, 
Demonstration.  the rectangle parallelogram also contained under the lines AE and EC is equal to the rectangle parallelogram contained under the lines DE and EB.  
But now suppose that the lines AC and DB be not extended by the centre, 
The second c●se.  and take (by the 1. of the third) the centre of the circle ABCD, and let the same be the point F, 
Construction.  and from the point F draw to the right lines AC and DB perpendicular lines FG and FH (by the 12. of the first) and draw these right lines FB, FC, and FE. And forasmuch as a certain right line FG drawn by the centre, 
Demonstration.  cutteth a certain right line AC not drawn by the centre in such sort that it maketh right angles, it therefore divideth the line AC into two equal parts (by the 3. of the third). Wherefore the line AG is equal to the line GC. and forasmuch as the right line AC is divided into two equal parts in the point G, and into two unequal parts in the point E: therefore (by the 5. of the second) the rectangle parallelogram contained under the lines. AE and EC together with the square of the line EG is equal to the square of the line GC. Put the square of the line GF common to them both, wherefore that which is contained under the lines AE & EC together with the squares of the lines EG and GF is equal to the squares of the lines GF & GC. But unto the squares of the lines EG & GF is equal the square of the line FE (by the 47. of the first): and to the squares of the lines GC and GF is equal the square of the line FC (by the same) Wherefore that which is contained under the lines AE and EC, together with the square of the line FE is equal to the square of the line FC. But the line FC is equal to the line FB. For they are drawn from the centre to the circumference. Wherefore that which is contained under the lines AE and E● together with the square of the line FE is equal to the square of the line FB. And by the same demonstration that which is contained under the lines DE and EB together with the square of the line FE is equal to the square of the line FB. Wherefore that which is contained under the lines AE and EC together with the square of the line EF is equal to that which is contained under the lines DE and EB together with the square of the line EF. Take away the square of the line FE which is common to them both. Wherefore the rectangle parallelogram remaining which is contained under the lines AE and EC is equal to the rectangle parallelogram remaining, which is contained under the lines DE and EB. If therefore in a circle two right lines do cut the one the other: the rectangle parallelogram comprehended under the segments or parts of the one line is equal to the rectangle parallelogram comprehended under the segments or parts of the other line: which was required to be demonstrated.  
In this Proposition are three cases: 
Three cases in this proposition.  For either both the lines pass by the centre, or neither of them passeth by the centre: or the one passeth by the centre and the other not. The two first cases are before demonstrated.  
Amongst all the Propositions in this third book, doubtless this is one of the chiefest. For it setteth forth unto us the wonderful nature of a circle. So that by it may be done many goodly conclusions in Geometry, as shall afterward be declared when occasion shall serve.   

The 30. Theorem. The 36. Proposition. If without a circle be taken a certain point, and from that point be drawn to the circle two right lines, so that the one of them do cut the circle, and the other do touch the circle: the rectangle parallelogram which is comprehended under the whole right line which cutteth the circle, and that portion of the same line that lieth between the point and the utter circumference of the circle, is equal to the square made of the line that toucheth the circle.  
SVppose that the circle be ABC: and without the same circle take any point at all adventures, and let the same be D. 
Construction.  And from the point D let there be drawn to the circle two right lines DCA and DB, and let the right line DCA cut the circle ACB in the point C, and let the right line BD touch the same. Then I say, that the rectangle parallelogram contained under the lines AD and DC, 
Two cases in this Proposition.  is equal to the square of the line BD. Now the line DCA is either drawn by the centre, or not.  
First let it be drawn by the centre. 
The first case.  And (by the first of the third) let the point F be the centre of the circle ABC, and draw a line from F to B. Wherefore the angle FBD is a right angle. 
Demonstration.  And for as much as the right line AC is divided into two equal parts in the point F, and unto it is added directly a right line CD, therefore (by the 6. of the second) that which is contained under the lines AD and DC together with the square of the line CF, is equal to the square of the line FD. But the line FC is equal to the line FB, for they are drawn from the centre to the circumference: Wherefore that which is contained under the lines AD and DC together with the square of the line FB, is equal to the square of the line FD. But the square of the line FD, is (by the 47. of the first) equal to the squares of the lines FB and BD (for the angle FBD is a right angle). Wherefore that which is contained under the lines AD and DC together with the square of the line FB, is equal to the squares of the lines FB and BD. Take away the square of the line FB which is common to them both. Wherefore that which remaineth, namely, that which is contained under the lines AD and DC, is equal to the square made of the line DB which toucheth the circle.  
But now suppose that the right line DCA be not drawn by the centre of the circle ABC. 
The second case.  And (by the first of the third) let the point E be the centre of the circle ABC. 
Construction.  And from the point E, draw (by the 12. of the first) unto the line AC a perpendicular line EF, and draw these right lines EB, EC, and ED. 
Demonstration.  Now the angle EFD is a right angle. And ●or as much as a certain right line EF drawn by the centre, cutteth a certain other right line AC not drawn by the centre, in such sort that it maketh right angles, it divideth it (by the third of the third) into two equal parts. Wherefore the line AF is equal to the line FC. And for as much as the right line AC is divided into two equal parts in the point F, & unto it is added directly an other right line making both one right line, therefore (by the 6. of the second) that which is contained under the lines DA and DC together with the square of the line FC, is equal to the square of the line FD: put the square of the line FE common to them both. Wherefore that which is contained under the lines DA and DC together with the squares of the lines CF and FE, is equal to the squares of the lines FD and FE. But to the squares of the lines FD and FE, is equal the square of the line DE (by the 47. of the first) for the angle EFD is a right angle. And to the squares of the lines CF and FE, is equal the square of the line CE (by the same). Wherefore that which is contained under the lines AD and DC together with the square of the line EC, is equal to the square of the line ED. But the line EC is equal to the line EB: for they are drawn from the centre to the circumference. Wherefore that which is contained under the lines AD and DC together with the square of the line EB, is equal to the square of the line ED. But to the square of the line ED, are equal the squares of the lines EB and BD (by the 47. of the first) for the angle EBD is a right angle: Wherefore that which is contained under the lines AD and DC together with the square of the line EB, is equal to the squares of the lines EB and BD. Take away the square of the line EB which is common to them both: Wherefore the residue, namely, that which is contained under the lines AD and DC, is equal to the square of the line DB. If therefore without a circle be taken a certain point, and from that point be drawn to the circle two right lines, so that the one of them do cut the circle, and the other do ●ouch the circle: the rectangle parallelogram which is comprehended under the whole right line which cutteth the circle and that portion of the same line that lieth between the point and the utter circumference of the circle, is equal to the square made of the line that toucheth the circle: which was required to be demonstrated.  

¶ Two Corollaries out of Campane.  
If from 〈◊〉 and the self same point taken without a circle be drawn into the circle lines how many soever: 
First Corollary.  the rectangle Parallelogrammes contained under every one of them and his outward par●, are equal the one to the other.  
And this is hereby manifest, for that every one of those rectangle parallelograms are equal to the square of the line which is drawn from that point and toucheth the circle by this 36. Proposition. Hereunto he addeth.  
If two lines drawn from one and the self same point do touch a circle, 
Second Corollary.  they are equal the one to the other.  
Which although it need no demonstration, for that the square of either of them is equal to that which is contained under the line drawn from the same point and his outward part: yet he thus proveth it.  
The same may be proved an other way: Draw a line from B to D. And (by the 5. of the first) the angle EBD is equal to the angle EDB. And for as much as the two angles ABE and ADE are equal, namely, for that they are right angles: if you take from them the equal angles EBD & EDB, the two other angles remaining, namely, the angles ABD and ADB shall be equal. Wherefore (by the 6. of the first) the line AB is equal to the line AD.   

¶ Hereunto also Pelitarius addeth this Corollary.  
From a point given without ● circle, can be drawn unto a circle only two touch lines. 
Third Corollary.   
The former description remaining, I say that from the point A can be drawn unto the circle BCD no more touch lines, but the two lines AB and AD. For if it be possible, let AF also be in the former figure a touch line, touching the circle in the point F. And prawe a line from E to F. And the angle at the point F shall be a right angle, by the 18. of this book: Wherefore it is equal to the angle EBA, which is contrary to the 20. of the first.  
This may also be thus proved. For as much as all the lines drawn from one and the self same point & touching a circle are equal, as we have before proved, but the lines AB and AF can not be equal, by the 8. Proposition of this book, therefore the line AF can not touch the circle BCD.    

The 31. Theorem. The 37. Proposition. If without a circle be taken a certain point, and from that point be drawn to the circle two right lines, of which, the one doth cut the circle and the other falleth upon the circle, and that in such sort, that the rectangle parallelogram which is contained under the whole right line which cutteth the circle, and that portion of the same line that lieth between the point and the utter circumference of the circle, is equal to the square made of the line that falleth upon the circle: then that line that so falleth upon the circle shall touch the circle.  
LEt the circle be ABC: and without the same circle take a point, and let the same be D, 
This proposition is the converse of the former.  & from the point D let there be drawn to the circle ABC two right lines DCA and DB: and let DCA cut the circle, and DB fall upon the circle. And that in such sort, that that which is contained under the lines AD and DC, be equal to the square of the line DB. Then I say, that the line DB toucheth the circle ABC. Draw (by the 17. of the third) from the point D a right line touching the circle ABC, and let the same be DE. 
Construction.  And (by the first of the same) let the point F be the centre of the circle ABC: and draw these right lines FE, FB, and FD. Wherefore the angle FED is a right angle. 
Demonstration.  And for as much as the right line DE toucheth the circle ABC, and the right line DCA cutteth the same, therefore (by the Proposition going before) that which is contained under the lines AD and DC, is equal to the square of the line DE. But that which is contained under the lines AD and DC, is supposed to be equal to the square of the line DB. Wherefore the square of the line DE is equal to the square of the line DB. Wherefore also the line DE is equal to the line DB. And the line FE is equal to the line FB, for they are drawn from the centre to the circumference. Now therefore these two lines DE and EF are equal to these two lines DB and BF, and FD is a common base to them both. Wherefore (by the 8. of the first) the angle DEF is equal to the angle DBF. But the angle DEF is a right angle. Wherefore also the angle DBF is a right angle. And the line FB being produced, shall be the diameter of the circle. But if from the end of the diameter of a circle be drawn a right line making right angles, the right line so drawn toucheth the circle (by the corollary of the 16. of the third). Wherefore the right line DB toucheth the circle ABC. And the like demonstration will serve if the centre be in the line AC. If therefore without a circle be taken a certain point, and from that point be drawn to the circle two right lines, of which the one doth cut the circle, and the other falleth upon the circle, and that in such sort, that the rectangle parallelogram which is contained under the whole right line which cutteth the circle, and that portion of the same line that lieth between the point and the utter circumference of the circle, is equal to the square made of the line that falleth upon the circle: then the line that so falleth upon the circle shall touch the circle: which was required to be proved.  

¶ An other demonstration after Pelitarius.  
Suppose that there be a circle BCD, whose centre let be E: 
another demonstration after Pelitarius.  and take a point without it, namely, A: And from the point A draw two right lines ABD, and AC: of which let ABD cut the circle in the point B, & let the other fall upon it. And let that which is contained under the lines AD and AB, be equal to the square of the line AC. Then I say, that the line AC toucheth the circle. For first if the line ABD do pass by the centre, draw the right line CE. And (by the 6. of the second) that which is contained under the lines AD and AB together with the square of the line EB, that is, with the square of the line EC (for the lines EB and EC are equal) is equal to the square of the line AE. But that which is contained under the lines AD and AB, is supposed to be equal to the square of the line AC: Wherefore the square of the line AC together with the square of the line CE, is equal to the square of the line AE. Wherefore (by the last of the first) the angle at the point C is a right angle. Wherefore (by the 18. of this book) the line AC toucheth the circle.  
But if the line ABD do not pass by the centre, draw from the point A the line AD, in which let be the centre E. And forasmuch as that which is contained under this whole line and his outward part, is equal to that which is contained under the lines AD and AB by the first Corollary before put, therefore the same is equal to the square of the line AC, wherefore the angle ECA is a right angle as hath before been proved in the first part of this Proposition. And therefore the line AC toucheth the circle: Which was required to be proved.    
The end of the third book of Euclides Elements.   

¶ The fourth book of Euclides Elements.  
THIS FOURTH BOOK entreateth of the inscription & circumscription of rectiline figures: 
The argument of this book.  how one right lined figure may be inscribed within an other right lined figure, and how a right lined figure may be circumscribed about an other right lined figure, in such as may be inscribed and circumscribed within or about the other. For all right lined figures cannot so be inscribed or circumscribed within or about the other. Also it teacheth how a triangle, a square, and certain other rectiline figures being regular may be inscribed within a circle. Also how they may be circumscribed about a circle. Likewise how a circle may be inscribed within them. And how it may be circumscribed about them. And because the manner of entreaty in this book is divers from the entreaty of the former books, he useth in this other words and terms than he used in them. The definitions of which in order here after follow.  

Definitions.  

A rectiline figure is said to be inscribed in a rectiline figure, 
First definition.  when every one of the angles of the inscribed figure toucheth every one of the sides of the figure wherein it is inscribed.  
As the triangle ABC is inscribed in the triangle DEF, because that every angle of the triangle inscribed, namely, the triangle ABC toucheth every side of the triangle within which it is described, namely, of the triangle DEF. As the angle CAB toucheth the side ED the angle ABC toucheth the side DF, and the angle ACB toucheth the side EF. So likewise the square ABCD is said to be inscribed within the square EFGH. for every angle of it toucheth some one side of the other. So also the Pentagon or five angled figure ABCDE is inscribed within the Pentagon or five angled figure FGHIK ● As you see in the figure●.   

Likewise a rectiline figure is said to be circumscribed about a rectiline figure, 
Second definition.  when every one of the sides of the figure circumscribed, toucheth every one of the angles of the figure about which it is circumscribed.  
As in the former descriptions the triangle DEF is said to be circumscribed about the triangle ABC, for that every side of the figure circumscribed, namely, of the triangle DEF toucheth every angle of the figure whereabout it is circumscribed. As the side DF of the triangle DEF circumscribed, toucheth the angle ABC of the triangle ABC about which it is circumscribed: and the side EF toucheth the angle BCA, and the side CD toucheth the angle CAB. Likewise understand you of the square EFGH which is circumscribed about the square ABCD: for every side of the one toucheth some one side of the other. Even so by the same reason the Pentagon FGHIK is circumscribed about the Pentagon ABCDE, as you see in the figure on the other side. And thus may you of other rectiline figures consider.  

The inscriptition and circumscription of rectiline figure's pertaineth only to regular figures.  By these two definitions it is manifest, that the inscription and circumscription of rectiline figures here spoken of, pertain to such rectiline figures only, which have equal sides and equal angles, which are commonly called regular. It is also to be noted that rectiline figures only of one kind or form can be inscribed or circumscribed the one within or about the other. As a triangle within or about a triangle: A square within or about a square: and so a Pentagon within or about a Pentagon, & likewise of others of one form. But a triangle can not be inscribed or circumscribed within or about a square: nor a square within or about a Pentagon. And so of others of divers kinds. For every plain rectiline figure hath so many angles as it hath sides. Wherefore the figure inscrided must have so many angles as the figure in which it is inscribed hath sides: and the angles of the one (as is said) must touch the sides of the other. And contrariwise in circumscription of figures, the sides of the figure circumscribed must touch the angles of the figure about which it is circumscribed.   


The third definition.  A rectiline figure is said to be inscribed in a circle, when every one of the angles of the inscribed figure toucheth the circumference of the circle.  
A circle by reason of his uniform and regular distance which it hath from the centre to the circumference may easily touch all the angles of any regular rectiline figure within it: and also all the sides of any figure without it. And therefore any regular rectiline figure may be inscribed within it, and also be circumscribed about it. And again a circle may be both inscribed within any regular rectiline figure, and also be circumscribed about it.  
As the triangle ABC is inscribed in the circle ABC ● for that every angle toucheth some one point of the circumference of the circle. As the angle CAB of the triangle ABC toucheth the point A of the circumference of the circle. And the angle ABC of the triangle toucheth the point B of the circumference of the circle. And also the angle ACB of the triangle 〈…〉 the point ● of the circumference of the circle. In like manner the square ADEF is inscribed in the same circle ABC: for that every angle of the square inscribed, toucheth some one point of the circle in which it is inscribed. And so imagine you of rectilined figures.   

A circle is said to be circumscribed about a rectiline figure, 
The fourth definition.  when the circumference of the circle toucheth every one of the angles of the figure about which it is circumscribed.  
As in the former example of the third definition. The circle ADEF is circumscribed about the triangle ABC, because the circumference of the circle which is circumscribed toucheth every angle of the triangle about which it is circumscribed● namely, the angles CAB, ABC, and BCA. Likewise the same circle ADEF is circumscribed about the square ADEF by the same definition as you may see.   

A circle is said to be inscribed in a rectiline figure, 
The fift definition.  when the circumference of the circle toucheth every one of the sides of the figure within which it is inscribed.  
As the circle ABCD is inscribed within the triangle EFG, because the circumference of the circle toucheth every side of the triangle in which it is inscribed● namely the side EF in the point B, and the side GF in the point C, and the side GE in the point D. Likewise by the same reason the same circle is inscribed within the square HIKL. And so may you judge of other rectiline figures.   

A rectilined figure is said to be circumscribed about a circle, 
The sixth devition.  when every one of the sides of the figure circumscribed toucheth the circumference of the circle.  
As in the former figure of the fift definition, the triangle EFG is circumscribed about the circle ABCD, for that every side of the same triangle being circumscribed toucheth the circumference of the circle, about which it is circumscribed. As the side EG of the triangle EFG toucheth the circumference of the circle in the point D: and the side EF toucheth it in the point B: and the side GF in the point C. Likewise also the square HIKL is circumscribed about the circle ABCD, for every one of his sides toucheth the circumference of the circle, namely, in the points A, B, C, D. And thus consider of all other regular right lined figures (for of them only are understanded these definitions) to be circumscribed about a circle, or to be inscribed within a circle: or of a circle to be circumscribed or inscribed about or within any of them.   


Seventh definition.  A right line is said to be coapted or applied in a circle, when the extremes or ends thereof, fall upon the circumference of the circle.  
As the line BC is said to be coapted or to be applied to the circle ABC, for that both his extremes fall upon the circumference of the circle in the points B and C. Likewise the line DE. This definition is very necessary, and is properly to be taken of any line given to be coapted and applied into a circle, so ●hat it exceed not the diameter of the circle given.    

The 1. Problem. The 1. Proposition. In a circle given, to apply a right line equal unto a right line given, which exceedeth not the diameter of a circle.  
SVppose that the circle given be ABC, and let the right line given, exceeding not the diameter of the same circle, be D. Now it is required in the circle given ABC, to apply a right line equal unto the right line D. 
Construction.  Draw the diameter of the circle ABC, and let the same be BC. Now if the line BC be equal unto the line D, then is that done which was required. For in the circle given ABC is applied a right line BC equal unto the right line D. 
Two cases in this Proposition.  But if not, 
First case●  then is the line BC greater than the line D. 
Second case.  And (by the third of the first) put unto the line D an equal line CE. And making the centre C, and the space CE, describe (by the third petition) a circle EGF, cutting the circle ABC in the point F, & draw a line from C to F. And for as much as the point C is the centre of the circle EGF, 
Demonstration.  therefore (by the 13. definition of the first) the line CF is equal unto the line CE. But the line CE is equal unto the line D. Wherefore (by the first common sentence) the line CF also is equal unto the line D. Wherefore in the circle given ABC, is applied a right line CA equal unto the right line given D: which was required to be done.   

The 2. Problem. The 2. Proposition. In a circle given, to describe a triangle equiangle unto a triangle given.  
SVppose that the circle given be ABC: and let the triangle given be DEF. Now it is required in the circle given ABC to describe a triangle equiangle unto the triangle given DEF. 
Construction.  Draw (by the 17. of the third) a right line touching the circle ABC, and let the same be GAH, and let it touch in the point A. And (by the 23. of the first) unto the right line AH, and unto the point in it A, describe an angle HAC equal unto the angle DEF. And (by the self same) unto the right line AG, and unto the point in it A, make an angle GAB equal unto the angle DFE. And draw a right line from B to C. And for as much as a certain right line GAH toucheth the circle ABC, 
Demonstration.  and from the point where it toucheth, namely, A, is drawn into the circle a right line AC; therefore (by the 31. of the third) the angle HAC is equal unto the angle ABC which is in the alternate segment of the circle. But the angle HAC is equal to the angle DEF. Wherefore the angle ABC is equal to the angle DEF. And by the same reason, the angle ACB is equal to the angle DFE. Wherefore the angle remaining, BAC, is equal unto the angle remaining, EDF. Wherefore the triangle ABC is equiangle unto the triangle DEF. And it is described in the circle given ABC. Wherefore in a circle given, is described a triangle equiangle unto a triangle given: which was required to be done.   

The 3. Problem. The 3. Proposition. About a circle given, to describe a triangle equiangle unto a triangle given.  
SVppose that the circle given be ABC, and let the triangle given be DEF. It is required about the circle ABC to describe a triangle equiangle unto the triangle DEF. 
Construction.  Extend the line EF on each side to the points G and H. And (by the first of the third) take the centre of the circle ABC, and let the same be the point K. And then draw a right line KB. And (by the 23. of the first) unto the right line KB, and unto the point in it K, make an angle BKA equal unto the angle DEG, and likewise make the angle BKC equal unto the angle DFH. And (by the 17. of the third) draw right lines touching the circle A, B, C, in the points A, B, C. And let the same be LAM, MBN, and NCL. 
Demonstration.  And for as much as the right lines LM, MN, & NL, do touch the circle ABC in the points A, B, C, and from the centre K unto the points A, B, C, are drawn right lines KA, KB, and KC, therefore the angles which are at the points A, B, C, are right angles (by the 18. of the third). And for as much as the four angles of the four sided figure AMBK, are equal unto four right angles: whose angles KAM, & KBM, are two right angles: therefore the angles remaining AKB, and AMB, are equal to two right angles. And the angles DEG, & DEF, are (by the 13. of the first) equal to two right angles. Wherefore the angles AKB, and AMB, are equal unto the angles DEG, and DEF: of which two angles the angle AKB is equal unto the angle DEG: Wherefore the angle remaining, AMB, is equal unto the angle remaining, DEF. In like sort may it be proved, that the angle LNM, is equal to the angle DFE. Wherefore the angle remaining MLN, is equal unto the angle remaining EDF. Wherefore the triangle LMN, is equiangle unto the triangle DEF: and it is described about the circle ABC. Wherefore about a circle given is described a triangle equiangle unto a triangle given: which was required to be done.  

¶ An other way after Pelitarius.  

another way after Peli●arius.  In the circle ABC inscribe a triangle GHK equiangle to the triangle EDF (by the former Proposition): so that let the angle at the point G be equal to the angle D, and let the angle at the point H be equal to the angle E: and let also the angle at the point K be equal to the angle F. Then draw the line LM parallel to the line GH, which let touch the circle in the point A (which may be done by the Proposition added of the said Pelitarius after the 17. Proposition). 
Construction.  Draw likewise the line MN parallel unto the line HK and touching the circle in the point B: And also draw the line LN parallel unto the line GK and touching the circle in the point C. And these three lines shall undoubtedly concur, as in the points L, M, and N, 
Demonstration.  which may easily be proved, if you produce on either side the lines GH, GK, and HK, until they cut the lines LM, LN, and MN, in the points O, P, Q, R, S, T. Now I say, that the triangle LMN circumscribed about the circle ABC, is equiangle to the triangle DEF. For it is manifest, that it is equiangle unto the triangle GHK, by the propriety of parallel lines. For the angle MTQ is equal to the angle at the point G of the triangle GHK (by the 29. of the first) and therefore also the angle at the point L, is equal to the self same angle at the point G (for the angle at the point L, is by the same 29. Proposition, equal to the angle MTQ). And by the same reason the angle at the point M, is equal to the angle at the point H of the self same triangle: and the angle at the point N, to the angle at the point K. Wherefore the whole triangle LMN, is equiangle to the whole triangle GHK: Wherefore also it is equiangle to the triangle DEF: which was required to be done.    

The 4. Problem. The 4. Proposition. In a triangle given, to describe a circle.  
SVppose that the triangle given be ABC. It is required to describe a circle in the triangle ABC. 
Construction.  Divide (by the 9 of the first) the angles ABC, and ACB into two equal par●es by two right lines BD and CD. And let these right lines meet together in the point D. And (by the 12. of the first) from the point D draw unto the right lines AB, BC and CA perpendicular lines, namely, DE, DF, and DG. 
Demonstration.  And forasmuch as the angle ABD is equal to though angle CBD, and the right angle BED is equal unto the right angle BFD. Now then there are two triangles EBD and FBD having two angles equal to two angles, and one side equal to one side, namely, BD which is common to them both, and subtendeth one of the equal angles. Wherefore (by the 26. of the first) the rest of the sides are equal unto the rest of the sides. Wherefore the line DE is equal unto the line DF: and by the same reason also the ●ne DG is equal unto the line DF. Wherefore these three right lines DE, DF, & DG, are equal the one to the other (by the first common sentence). Wherefore making the centre the point D, and the space DE, or DF, or DG, describe a circle and it will pass through the points E, F, G, and will touch the right lines AB, BC, and CA For the angles made at the points E, F, G, are right angles. For if the circle cut those right lines, then from the end of the diameter of the circle shall be drawn a right line making two right angles, 
Demonstration leading to an impossibility.  & falling within the circle: which is impossible, as it was manifest (by the 16. of the third). Wherefore the circle described, D being the centre thereof, and the space thereof being either DE, or DF, or DG, cutteth not these right lines AB, BC, & CA Wherefore (by the Corollary of the same) it toucheth them, and the circle is described in the triangle ABC. Wherefore in the triangle given ABC, is described a circle EFG: which was required to be done.   

The 5. Problem. The 5. Proposition. About a triangle given, to describe a circle.  
SVppose that the triangle given be ABC. It is required about the triangle ABC to describe a circle. Divide (by the 10. of the first) the right l●nes AB and AC into two equal parts in the points D and E. And from the points D and E (by the 11. of the first) raise up unto the lines AB & AC two perpendicular lines DF and EF. 
Three cases in this Proposition.  Now these perpendicular lines meet together either within the triangle ABC, or in the right line BC, or else without the right line BC.  
But now suppose that the right lines DF and EF do meet together without the triangle ABC in the point F. 
The third case.  Again as it is in the third description draw right lines from F to A, from F to B, and from F to C. And forasmuch as the line AD is equal unto the line DB, and the line DF is common unto them both, and maketh a right angle on each side of him, wherefore (by the 4. of the first) the base AF is equal unto the base BF. And in like sort may we prove that th● line CF is equal unto the line AF. Wherefore again making F the centre, and the space FA, or FB, or FC, describe a circle and it shall pass by the points A, B, C, and so is there a circle described about the triangle ABC, as ye see it i● in the third description. Wherefore about a triangle given is described a circle: which was required to be done.  

Correlary.  
Hereby it is manifest, that when the centre of the circle falleth within the triangle, the angle BAC being in a greater segment of a circle is less than a right angle. But when it falleth upon the right line BC the angle BAC being in a semicircle is a right angle. But when the centre falleth without the right line BC, the angle BAC being in a less segment of a circle, is greater than a right angle. Wherefore also when the angle given is less than a right angle, the right lines DF and EF will meet together within the said triangle. But when it is a right angle they will meet together upon the line BC. But when it is greater than a right angle, they will meet together without the right line BC.    

The 6. Problem. The 6. Proposition. In a circle given, to describe a square.  
SVppose that the circle given be ABCD. It is required in the circle ABCD to describe a square. Draw in the circle ABCD two diameters making right angles, 
Construction.  and let the same be AC and BD, and draw right lines from A to B, from B to C, from C to D, and from D to A. And forasmuch as the line BE is equal unto the line ED (by the 15. definition of the first) for the point E is the centre. 
Demonstration.  And the line EA is common to them both, making on each side a right angle: therefore (by the 4. of the first) the base AB is equal unto the base AD. And by the same reason also either of these lines BC and CD is equal to either of these lines AB and AD: wherefore ABCD is a figure of four equal sides. I say also that it is a rectangle figure. For forasmuch as the right line BD is the diameter of the circle ABCD, therefore the angle BAD being in the semicircle is a right angle (by the 31. of the third) And by the same reason every one of these angles ABC, BCD and CDA is a right angle. Wherefore the four sided figure ABCD is a rectangle figure, and it is proved that it consisteth of equal sides. Wherefore (by the 30. definition of the first) it is a square, and it is described in the circle ABCD: which was required to be done.   

The 7. Problem. The 7. Proposition. About a circle given, to describe a square.  
SVppose that the circle given be ABCD. It is required about the circle ABCD to describe a square. Draw in the circle ABCD two diameters making right angles, 
Construction.  where they cut the one the other, and let the same be AC and BD. And by the points A, B, C, D, draw (by the 17. of the third) right lines touching the circle ABCD, and let the same be FG, GH, HK, and KF. 
Demonstration.  Now for as much as the right line FG toucheth the circle ABCD in the point A, and from the centre E to the point A where the touch is, is drawn a right line EA, therefore (by the 18. of the third) the angles at the point A are right angles, and by the same reason the angles which are at the points B, C, D, are also right angles. And forasmuch as the angle AEB is a right angle, & the angle EBG is also a right angle, therefore (by the 28. of the first) the line GH is a parallel unto the line AC: and by the same reason the line AC is a parallel unto the line FK. In like sort also may we prove that either of these lines GF and HK is a parallel unto the line BED. Wherefore these figures GK, GC, AK, FB, and BK are parallelogrames● Wherefore (by the 34. of the first) the line GF is equal unto the line HK, and the line GH is equal unto the line FK. And forasmuch as the line AC is equal unto the line BD, but the line AC is equal unto either of these lines GH and FK: an● the line BD is equal to either of these lines GF and HK. Wherefore either of these lines GH and FK is equal to either of these lines GF and HK. Wherefore the figure FGHK consisteth of four equal sides. I say also that it is a rectangle figure. For forasmuch as GBEA is a parallelogram, and the angle AE● is a right angle: therefore (by the 34. of the first) the angle AGB is a right angle. In like sort may we prove that the angles at the points H, K, and F are right angles. Wherefore FGHK is a rectangle four sided figure, and it is proved that it consisteth of equal sides: wherefore it is a square, and it is described about the circle ABCD. Wherefore about a circle given is described a square: which was required to be done.   

The 8. Problem. The 8. Proposition. In a square given, to describe a circle.  
SVppose that the square given be ABCD. It is required in the square ABCD to describe a circle. 
Construction.  Divide (by the 10. of the first) either of these lines AB and AD into two equal parts in the points E and F. And by the point E (by the 31. of the first) draw a line EH parallel unto either of these lines AB and DC: 
Demonstration.  and (by the same) by the point F draw a line FK parallel unto either of these lines AD and BC. Wherefore every one of these figures AK, KB, AH, HD, AG, GC, BG, and GD is a parallelogram, and the sides which are opposite the one to the other, are (by the 34. of the first) equal the one to the other. And forasmuch as the line AD is equal unto the line AB, and the half of the line AD is the line AE, and the half of the line AB, is the line AF, therefore the line AE is equal unto the line AF: wherefore (by the same) the sides which are opposite are equal. Wherefore the line FG is equal unto the line EG. In like sort may we prove that either of these lines GH, and GK is equal to either of these lines FG and GOE Wherefore (by the first common sentence) these four lines GE, GF, GH, and GK are equal the one to the other. Wherefore making the centre G, and the space either GE, or GF, GH, or GK, describe a circle and it will pass by the points E, F, H, K, and will touch the right lines AB, BC, CD, and DA. For the angles at the points E, F, H, K, are right angles. For if the circle do cut the right lines AB, BC, CD, and DA, than the line which is drawn by the end of the diameter of the circle making right angles should fall within the circle, which is impossible (by the 16. of the third) Wherefore the centre being the point G and the space being GE, or GF, or GH, or GK if a circle be described, it shall not cut the rig●t lines AB, BC, CD, and DA. Wherefore it shall touch them. And it is described in the square ABCD: wherefore in a square given is described a circle: which was required to be done.   

The 9 Problem. The 9 Proposition. About a square given, to describe a circle.  
SVppose that the square given be AB CD. It is required about the square ABCD to describe a circle. Draw right lines from A to C, and from D to B, & let them cut the one the other in the point E. 
Construction.  And forasmuch as the line DA is equal unto the line AB, 
Demonstration.  and the line AC is common unto them both, therefore these two lines DA and AC are equal unto these two lines BA and AC, the one to the other. And the base DC is equal unto the base BC. Wherefore (by the 8. of the first) the angle DAC is equal unto the angle BAC. Wherefore the angle DAB is divided into two equal parts by the line AC. And in like sor● may we prove that every one of these angles ABC, BCD, and CDA is divided into two equal parts by the right lines AC and DB. And forasmuch as the angle DAB is equal unto the angle ABC, and of the angle DAB the angle EAB is the half; and of the angle ABC the angle EBA is the half: Therefore the angle EAB is equal unto the angle EBA: wherefore (by the 6. of the first) the side EA is equal unto the side EB. In like sort may we prove that either of these right lines EA and EB is equal unto either of these lines EC and ED. Wherefore these four lines EA, EB, EC, and ED are equal the one to the other. Wherefore making the centre E, and the space any of these lines EA, EB, EC, or ED. Describe a circle and it will pass by the points A, B, C, D, and shall be described about the square ABCD, as it is evident in the figure ABCD. Wherefore about a square given is described a circle: which was required to be done.  

¶ A Proposition added by Pelitarius.  
A square circumscribed about a circle, is double to the square inscribed in the same circle.  
This may also be demonstrated by the equality of the triangles and squares contained in the great squares.    

The 10. Problem. The 10. Proposition. To make a triangle of two equal sides called Isosceles, which shall have either of the angles at the base double to the other angle.  
TAke a right line at all adventures which let be AB, & (by the 11. of the second) let it be so divided in the point C, 
Construction.  that the rectangle figure comprehended under the lines AB and BC be equal unto the square which is made of the line AC. And making the centre the point A, & the space AB, describe (by the 3. petition) a circle BDE, and (by the 1. of the fourth) into the circle BDE apply a right line BD equal to the right line AC which is not greater than the diameter of the circle BDE. And draw lines from A to D and from D to C. And (by the 5. of the fourth) about the triangle ACD describe a circle ACDF. 
Demonstration.  And forasmuch as the rectangle figure contained under the lines AB and BC is equal to the square which is made of the line AC: (For that is by supposition) But the line AC is equal unto the line BD. Wherefore that which is contained under the lines AB and BC is equal to the square which is made of the line BD. And forasmuch as without the circle ACDF is taken a point B, and from B unto the circle ACDF are drawn two right lines BCA, and BD, in such sort that the one of them cutteth the circle, and the other endeth at the circumference, and that which is contained under the lines AB and BC is equal to the square which is made of the line BD, therefore (by the 17. of the third) the line BD toucheth the circle ACDF. And forasmuch as the line BD toucheth in the point D, and from D where the touch is, is drawn a right line DC, therefore (by the 32. of the same) the angle BDC is equal unto the angle DAC, which is in the alternate segment of the circle. And forasmuch as the angle BDC is equal unto the angle DAC, put the angle CDA common unto them both. Wherefore the whole angle BDA is equal to these two angles CDA, & DAC. But unto the angles CDA, & DAC is equal the outward angle BCD (by the 32. of the 1. ) Wherefore the angle BDA is equal unto the angle BCD. But the angle BDA is (by the 5. of the first) equal unto the angle CBD, for (by the 15. definition of the first) the side AD is equal unto the side AB: wherefore (by the 1. common sentence) the angle DBA is equal unto the angle BCD. Wherefore these three angles BDA, DBA, and BCD are equal the one to the other. And forasmuch as the angle DBC is equal unto the angle BCD, the side therefore BD is equal unto the side DC. But the line BD is by supposition equal unto the line CA two right angles. And either of the Angles at the base● is two ●ift parts of two right angles, or four fift parts of one right angle. Which shall manifestly appear, if we divide two right angles into five parts. For then in this kind of triangle, the angle at the top shall be one fift part, and either of the two angles at the base shall be two fift parts.  
This also is to be noted, that the line AC is the side of an equilater Pentagon to be inscribed in the circle ACD. For by the latter construction it is manifest, that the three arks AC, CD, and DE, of the less circle, are equal. And forasmuch as by the same it is manifest that the two lines AD and AE are equal, the ark also AE shall be equal to the ark AD (by the 20. of the third). Wherefore their halves also are equal. If therefore the ark AE be (by the 30. of the third) divided into two equal parts, the whole circumference ACDEA shall be divided into five equal arcs. And forasmuch as the lines subtending the said equal arks are (by the 2●. of the same) equal, therefore every one of the said sides shall be the side of an equilater Pentagon● which was required to be proved. And the same line AC shall be the side of an equilater ten angled figure to be inscribed in the circle BDE: the demonstration whereof I omit, for that it is demonstrated by Propositions following.  


A Proposition added by Petarilius.  ¶ A Proposition added by Pelitarius.  
Upon a right line given being finite, to describe an equilater and equiangle Pentagon figure.  
If we consider well this demonstration of Pelitarius, it will not be hard for us, upon a right line given to describe the rest of the figures whose inscriptions hereafter follow. 
Note.     

The 11. Problem. The 11. Proposition. In a circle given to describe a pentagon figure aequilater and equiangle.  
SVppose that the circle given be ABCDE. 
Construction.  It is required in the circle ABCDE to inscribe a figure of five angles of equal sides and of equal angles. Take (by the proposition going before) an Isosceles triangle FGH having either of the angles at the base GH double to the other angle, namely, unto the angle F. And (by the 2. of the fourth) in the circle ABCDE inscribe a triangle ACD equiangle unto the triangle FGH. So that let the angle CAD be equal to the angle F, & the angle ACD unto the angle G, and likewise the angle CDA to the angle H. Wherefore either of these angles ACD, and CDA is double to the angle CAD. Divide (by the 9 of the first) either of these angles ACD, and CDA into two equal parts by the right lines CE and DB. and draw right lines from A to B, from B to C, from C to D, from D to E, and from E to A. 
Demonstration.  And forasmuch as either of these angles ACD, and CDA is double to the angle CAD: and they are divided into two equal parts by the right lines CE and DB, therefore the five angles DAC, ACE, ECD, CDB, and BDA are equal the one to the other. But equal angles (by the 26. of the third) subtend equal circumferences. Wherefore the five circumferences AB, BC, CD, DE, and EA are equal the one to the other. And (by the 29, of the same) unto equal circumferences are subtended equal right lines: wherefore the five right lines AB, BC, CD, DE, & EA are equal the one to the other. Wherefore the figure ABCDE having five angles is equilater.  
Now also I say that it is equiangle. For forasmuch as the circumference AB is equal to the circumference DE, put the circumference BCD common unto them both. Wherefore the whole circumference ABCD is equal to the whole circumference EDCB: and upon the circumference ABCD consisteth the angle AED, and upon the circumference EDCB, consisteth the angle BAE. Wherefore the angle BAE is equal to the angle AED (by the 27. of the third) and by the same reason every one of these angles ABC, and BCD, and CDE, is equal to every one of these angles BAE and AED. Wherefore the five angled figure ABCDE, is equiangle, and it is proved, that is also equilater. Wherefore in a circle given is described a figure of five angles equilater and equiangle: which was required to be done.  

¶ An other way to do the same after Pelitarius.  
Here it is a pleasant thing to behold the variety of triangles: for in the triangle ABC either of the angles at the point A is one fift part of a right angle. Whereby is produced the side of a ten angled figure to be inscribed in the self-same circle: Which is manifest if we imagine the lines BL and LC to be drawn. For the ark BC is divided into two equal parts in the point L (by the 26. of the third). So then by the inscription of an equilater triangle, is known how to inscribe an Hexagon figure, namely, by dividing each of the arcs subtended under the sides of the triangle into two equal parts. And so always by the simple number of the side, is known the double thereof: as by a square is known an eight angled figure: and by an eight angled figure a sixteen angled figure. And so continually in the rest.  
Pelitarius teacheth yet an other way how to inscribe a Pentagon. 
another way also after Pelitarius.  Take the same circle that was before, namely, ABC, and the same triangle also DEF ● And (by the 17. of the third) draw the line MAN touching the circle in the point A. And upon the line AM and to the point A, describe (by the 23. of the first) the angle MAB equal to one of these two angles E or F (either of which, as it is manifest, is less than a right angle) by drawing the right line AB: which let out the circumference in the point B. Again, upon the line AN and to the point in it A, describe the angle NAC equal to the angle MAB, by drawing the right line AC: which let cut the circumference in the point C. And draw a right line from B to C. Then I say, that BC is the side of a Pentagon figure to be inscribed in the circle ABC. Which is manifest, if we divide the ark AB into two equal parts in the point H, and draw these right lines AH and BH, and if also we divide the ark AC into two equal parts in the point G, and draw these right lines AG and CG. For taking the quadrangle figure ABCG, it is manifest (by the 32. of the third) that the angle ABC is equal to the alternate angle NAC, and therefore is equal to the angle E. Likewise taking the quadrangle figure ACBH, the angle ACB shall be equal to the alternate angle MAB, and therefore is equal to the angle F. Wherefore (by the 32. of the first) as before, the triangle ABC is equiangle to the triangle DEF: And now may you proceed in the demonstration as you did in the former.    

The 12. Problem. The 12. Proposition. About a circle given, to describe an equilater and aquiangle pentagon.  
SVppose that the circle given be ABCDE. It is required about the circle ABCDE to describe a figure of five angles consisting of equal sides and of equal angles. 
Construction.  Take the points of the angles of a five angled figure described (by the 11. of the fourth) so that by the proposition going before, let the circumferences AB, BC, CD, DE, and EA be equal the one to the other. And by the points A, B, C, D, E, draw (by the 17. of the third) right lines touching the circle, and let the same be GH, HK, KL, LM, and MG. And (by the 1. of the third) take the centre of the circle & let the same be F. And draw right lines from F to B, from F to K, from F to C, from F to L, and from F to D. 
Demonstration.  And forasmuch as the right line KL, toucheth the circle ABCDE in the point C, and from the centre F unto the point C where the touch is, is drawn a right line FC: therefore (by the 18. of the third) FC is a perpendicular line vn●o KL. Wherefore either of the angles which are at the point C is a right angle, and by the same reason the angles which are at the points D and B are right angles. And forasmuch as the angle FCK is a right angle, therefore the square which is made of FK is (by the 47. of the first) equal to the squares which are made of FC and CK. And by the same reason also the square which is made of FK is equal to the squares which are made of FB and BK. Wherefore the squares which are made of FC and CK are equal to the squares which are made of FB and BK, of which the square which is made of FC is equal to the square which is made of FB. Wherefore the square which is made of CK is equal to the square which is made of BK. Wherefore the line BK is equal unto the line CK. And forasmuch as FB is equal unto FC, and FK is common to them both, therefore these two BF and FK are equal to these two CF & FK. And the base BK is equal unto the base CK. Wherefore (by the 8. of the first) the angle BFK is equal unto the angle KFC: and the angle BKF to the angle FKC. Wherefore the angle BFC is double to the angle KFC. And the angle BKC is double to the angle FKC. And by the same reason the angle CFD is double to the angle CFL, and the angle DLC is double to the angle FLC. And forasmuch as the circumference BC is equal unto the circumference CD, therefore (by the 27. of the third) the angle BFC is equal to the angle CFD. And the angle BFC is double to the angle KFC, and the angle DFC is double to the angle LFC, wherefore the angle KFC is equal unto the angle LFC. Now then there are two triangles FKC, and FLC, having two angles equal to two angles, and one side equal to one side, namely, FC, which is common to them both. Wherefore (by the 26. of the third, the other sides remaining are equal unto the sides remaining, and the angle remaining unto the angle remaining. Wherefore the right line KC is equal to the right line CL, and the angle FKC to the angle FLC. And forasmuch as KC is equal to CL: therefore KL is double to KC, and by the same reason also may it be proved that HK is double to BK. And forasmuch as it is proved that BK is equal unto KC, and KL is double to KC, and H K double to B K, therefore H K is equal unto KL. In like sort may we prove that every one of these lines HG, GM, & ML is equal unto either of these lines H K and KL. Wherefore the five angled figure GH KLM is of equal sides. I say also that it is of equal angles. For forasmuch as the angle F KC is equal unto the angle FLC, and it is proved that the angle H KL is double to the angle F KC, and the angle KLM is double to the angle FLC, therefore the angle H KL is equal to the angle KLM. In like sort may it be proved that every one of these angles KHG, HGM, and GML, is equal to either of these angles H KL, and KLM. Wherefore the five angles GH K, H KL, KLM, LMG, and MGH are equal the one to the other. Wherefore the five angled figure GH KLM is equiangle, and it is also proved that it is equilater, and it is described about the circle ABCDE: which was required to be done.  

¶ An other way to do the same after Pelitarius, by parallel lines.  
Suppose that the circle given be ABC, 
another way to do the sam● after Pelitarius.  whose centre let be the point F: and in it (by the former Proposition) inscribe an equilater and equiangle Pentagon ABCDE● by whose five angles draw from the centre beyond the circumference, five lines, FG, FH, FK, FL, and FM. And it is manifest, that the five angles at the centre F, are equal, when as the five sides of the triangles within are equal, and also their bases. It is manifest also, that th● five angles of the Pentagon which are at the circumference, are divided into ●●n equal ●ngles (by the 4. of the fir●●): Now then between the two lines FG and FH, draw the line GH parallel to the side AB, and touching the circle ABC (which is done by a Proposition added by Pelitarius after the 17. of the third). And so likewise draw these lines HK, KL, and LM, parallel to each of these sides BC, CD, and DE, and touching the circle. And for as much as the lines FG and FH fall upon the two parallel lines AB and GH, 
Demonstration.  the two angles FGH, & FHG, are equal to the two angles FAB and FBA, the one to the other (by the 29. of the first). Wherefore (by the sixth of the same) the two lines FG and FH are equal. And by the same reason, the two angles FHK & FKH, are equal to the two angles FGH and FHG the one to the other: and the line FK is equal to the line FH, and therefore is equal to the line FG. And forasmuch as the angles at the point F are equal, therefore (by the 4. of the first) the base HK is equal to the base GH. In like sort may we prove, that the three lines FK, FL, and FM, are equal to the two lines FG and FH. And also that the two bases KL and LM, are equal to the two bases GH and HK: and that the angles which they make with the lines FK, FL, and FM, are equal the one to the other. Now then draw the fift line MG: which shall be equal to the four former lines (by the 4. of the first) for that as we have proved, the two lines FG & FM, are equal, & the angle GFM is equal to every one of the angles at the point F. This line also MG toucheth the circle. For unto the point where the line LM toucheth the circle which let be N, draw the line FN. And it is manifest (by the 18. of the third) that either of the angle● at the point N, is a right angle. Wher●fore for as much as the angle L of the triangle FLN, is equal to the angle M of the triangle FMN, & the angle N of the one, is equal to the angl● N of the other: and the line FN is common to them both, the line NL shall (by the 26. of the first) be equal to the line NM. And so is the line ML divided equally in the point N. And forasmuch as the three sides of the triangle FGP are equal to the three sides of the triangle FMP, the angle P of the one shall be equal to the angle P of the other (by the 8. of the first). Wherefore either of them is a right angle (by the 13. of the same). And forasmuch as the two angles FMP and FPM of the triangle FMP, are equal to the two angles FMN & FNM of the triangle FMN, and the side FM is common to them both, therefore the line FP is equal to the line FN. But the line FN is drawn from the centre to the circumference. Wherefore also the line FP is drawn from the centre to the circumference. And forasmuch as the line MG is perpendicular to the line FP, therefore (by the Corollary of the 16. of the third) it toucheth the circle. Wherefore the Pentagon GHKLM circumscribed about the circle is equilater: it is also equiangle, a● it is easy to prove by the equality of the halves: which was required to be done.    

The 13. Problem. The 13. Proposition. An equilater and equiangle pentagon figure being given, to describe in it a circle.  
SVppose that the equilater & equiangle Pentagon figure given be ABCDE. It is required in the said five angled figure ABCDE to describe a circle. Divide (by the the 9 of the first) either of these angles BCD, and CDE into two equal parts, by these right lines CF and FD, 
Demonstration.  and from the point F where th●se right lines CF and DF meet. Draw these right lines FB, FA, & F●. And forasmuch as BC is equal unto CD ●nd CF is common to them both. Wherefore these two lines BC and CF are equal to these two lines DC and CF: and the angle BCF is equal to the angle DCF. Wherefore (by the 4. of the first) the base BF is equal to the base DF, and the triangle BCF is equal to the triangle DCF, and the angles remaining are equal unto the angles remaining the one to the other, under which are subtended equal sides. Wherefore the angle CBF is equal to the angle CDF. And forasmuch as the angle CDE is double to the angle CDF. But the angle CDE is equal to the angle ABC, and the angle CDF is equal to the angle CBF. Wherefore the angle CBA is double to the angle CBF. Wherefore the angle ABF is equal to the angle FBC. Wherefore the angle ABC is divided into two equal parts by the right line BF. In like sort also may it be proued● that either of these angles BAE, and AED are divided into two equal parts by either of these lines FA and EF. Draw (by the 12. of the first) from the point E to the right lines AB, BC, CD, and EA perpendicular line● FG, FH, FK, FL, and FM. And forasmuch as the angle HCF is equal 〈◊〉 the angl● KCF, and the right angle FHC is equal to the right angle FKC: now then there are two triangles FHC, and FKC● having two angles equal to two angles the one to the other, and one side equal to one side: for FC is commo●, unto them both, and is subtended under one of the equal angles. Wherefore (by the 2●. of the first) the sides remay●●ng are equal unto the sides remaining. Wherefore the perpendicular FH is equal unto the perpendicular FK. And in like sort also may it be prou●d that every one of these lines FL, FM, and FG is equal to every one of th●●● lines FH and FK. Wherefore these five right lines FG, FH, FK, FL, and FM are equal the one to the other. Wherefore making the centre F and the space FG, or FH, or FK, or FL, or FM. Describe a circle, and it will pass by the points G, H, K, L, M. And shall touch the right lines AB, BC, CD, DE, and EA (by the corollary of the 16. of the third) for the angles which are at the points G, H, K, L, M, are right angles. 
Demonstration leading to an absurdity.  For if it do not touch them but cut, then from the end of the diameter of the circle, shall be drawn a right line making two right angles, and falling with in the circle: which is (by the 16. of the third) proved to be impossible Wherefore making F the centre, and the space one of these lines, FG, FH, FK, FL, FM, describe a circle and it shall not cut the right lines AB, BC, CD, DE, and EA. Wherefore (by the corollary of the 16. of the third) it shall touch them as it is manifest in the circle GHKLM. Wherefore in the equilater and equiangle pentagon figure given, is described a circle: which was required to be done●  
By this proposition and the former, it is manifest that perpendicular lines drawn from the middle points of the sides of an equilater and equiangle Pentagon figure, 
A Corollary.  and produced, shall pass by the centre of the circle, in which the said Pentagon figure is inscribed, and shall also divide the opposite angles equally, as here, AK is one right line, and divideth the side CD and the angle A equally. And so of the rest which is thus proved. The space about the centre F is equal to four right angles, which are divided into ten equal angles by ten right lines meeting together in the point F. Wherefore the five angles AFM, MFE, EFL, LFD, and DFK are equal to two right angles. Wherefore (by the 14. of the first) the lines AF and FK make one right line. The like proof also will serve touching the re●t of the lines. And this is always true in all equilater figures which consist of uneven sides.   

The 14. Problem. The 14. Proposition. About a pentagon or figure of five angles given being equilater and equiangle, to describe a circle.  
SVppose that the Pentagon or figure of five angles given, being of equal sides and of equal angles, be ABCDE. It is required about the said Pentagon ABCDE, to describe a circle. Divide (by the 9 of the first) either of these angles BCD, 
Construction.  & CDE into two equal parts by either of these right lines CF, and DF. And from the point F where those right lines meet, draw unto the points B, A, E, these right lines FB, FA, FE. 
Demonstration.  And in like sort (by the Proposition going before) may it be proved, that every one of these angles CBA, BAE, and AED is divided into two equal parts, by these right lines FB, FA, & FE. And for as much as the angle BCD is equal to the angle CDE, & the half of the angle BCD is the angle FCD, and likewise the half of the angle CDE is the angle CDF. Wherefore the angle FCD is equal to the angle FDC. Wherefore the side FC is equal to the side FD. In like sort also may it be proved, that every one of these lines FB, FA, and FE is equal to every one of these lines FC and FD. Wherefore these five right lines FA, FB, FC, FD, and FE are equal the one to the other. Wherefore making the centre F; and the space FA, or FB, or FC, or FD, or FE. Describe a circle and it will pass by the points A, B, C, D, E, and shall be described ●bout the five angled figure ABCDE which is equiangle and equilater. Let this circle be described and let the same be ABCDE. Wherefore about the Pentagon given be●●g both equiangle and equilater, is described a circle: which was required to be ●one.   

The 15. Problem. The 15. Proposition. In a circle given to describe an hexagon or figure of six angles equilater and equiangle.  
SVppose that the circle given be ABCDEF. It is required in the circle given ABCDE to describe a figure of six angles of equal sides and of equal angles. D●aw the diameter of the circle ABCDEF, and let the same be AD. And (by the first of the third) take the c●n●re of the circled and let the same be G. And making the centre D, and the space DG, describe (by the third petition) a circle CGEH: and drawing right lines from E to G, and from G to C, extend them to the points B and F of the circumference of the circle given. And draw these right lines AB, BC, CD, DE, EF, and FA. Then I say, that ABCDEF is an Hexagon figure of equal sides and of equal angles. 
Demonstration.  For forasmuch as the point G is the centre of the circle ABCDEF, therefore (by the 15. definition of the first) the line GE is equal unto the line GD. Again forasmuch as the point D is the centre of the circle CGEH (therefore (by the self same) the line DE is equal unto the line DG: And it is proved that the line GE is equal unto the line GD. Wherefore the line GE is equal unto the line ED (by the first common sentence) wherefore the triangle EGD is equilater, (and bi● three angles, namely, EGD, GDE, DEG, are equal the one to the other. And forasmuch as (by the 5. of the first) in triangles of two equal sides colmonell called Isosceles, the angles at the base are equal the one to the other, and the three angles of a triangle are (by the 30. of the first) equal ●nto two right angles, therefore the angle EGD is the third part of two right angles. And in like sor●e may it be proved, that the angle DGC is the third part of two right angles. And forasmuch as the right line CG standing upon the right line EB doth (by the 13. of the fi●●t) make the two side angles ●GC, and CG● equal to two right angles, therefore the angle remaining CGB is the third part of two right angles. Wherefore the angles EGD, DGC, and CGB are equal the one to the other. Wherefore their head angles, that is, BGA, AGF, and FGE are (by the 15. of the first) equal to these angles EGD, DGC, and CGB. Wherefore these six angles EGD, DGC, CGB, BGA, AGF, and FGE are equal the one to the other. But equal angles consist upon equal circumferences (by 〈…〉 the third. Therefore these six circumferences AB● BE CD● DE● EF● and ●A are equal the one to the other. But under equal circumferences are ●●bt●●ded equal right lines (by the 29. of the same.) Wherefore these six right lines AB, BC, CD, DE, EF, and FA are equal the one to the other. Wherefore the Hexagon ABCDEF is equilater. I say also that it is equiangle. For forasmuch as the circumference AF is equal unto the circumference ED add the circumference ABCD common to them both. Wherefore the whole circumference FABCD is equal to the whole circumference EDCBA. And upon the circumference FABCD consisteth the angle FED: and upon the circumference EDCBA consisteth the angle AFE. Wherefore the angle AFE is equal to the angle DEF. In like sort also may it be proved that the rest of the angles of the Hexagon ABCDEF, that is every one of these angles FAB, ABC, BCD, and CDE is equal to every one of these angles AFE, and FED. Wherefore the Hexagon figure ABCDEF is equiangle, and it is proved that it is also equilater, and it is described in the circle ABCDEF. Wherefore in the circle given ABCDEF is described a figure of six angles of equal sides and of equal angles: Which was required to be done●  

¶ An other way to do the same after Orontius.  
Suppose that the circle given be ABCDEF in which first let there be described an equilater and equiangle triangle ACE (by the second of this book). 
An ●ther way to do the same after Orontius.  Wherefore the arks ABC, CDE, EFA are (by the 28. of the third) equal the one to the other. Divide every one of those three arks into two equal parts (by the 30. of the same) in the points B, D, & F. And draw these right lines AB● BC, CD, DE, EF, and FA. Now then by the 2. definition of this book there shall be described in the circle give an Hexagon figure ABCDEF, which must needs be equilater: for that every one of the arks which subtend the sides thereof are equal the one to the other. I say also that it is equiangle. For every angle of the Hexagon figure is set upon equal arks, namely, upon four such parts of the circumference whereof the whole circumference containeth six. Wherefore the angles of the Hexagon figure are equal the one to the other (by the 27. of the third). Wherefore in the circle given ABCDEF is inscribed an equilater and equiangle Hexagon figure: which was required to be done.   

¶ An other way to do the same after Pelitarius.  
Suppose that the circle in which is to be inscribed 〈◊〉 equilater & ●●●iangle Hexagon figure be ABCDE, 
another way after Pelitarius.  whose c●ntre let be F. And from the centre draw the semidiameter FA. And from the point A apply (by the first o●●hys book) the line ●● equal to the semidiameter. Which I say is the side of an equilater and equiangle Hexagon figure to be inscribed in the circle ABCDE. Draw ●●●ght ●ine from E to B ● And for as much as the line AB is equal to the line FA ● & it is also equal to the line FB, therefore the triangle AFB is equilater, and by the 5. of the first equiangle. Now then upon the centre F describe the angle BFC equal to the angle AF●, or to the angle FBA (which is all one) by drawing the right line FC. And draw a line from B to C. And for as much as the angle AFB is the third part of two right angles by the 5. and 32. of the first, the angle BFC also shall be the third part of two right angles. Wherefore either of the two angles remaining FBC and FCB, for as much as they are equal by the 5. of the first, shall be two third parts of two right angles (by the 32. of the same). Or (by the 4. of the first) forasmuch as the angle BFC is equal to the angle FBA, and the two sides FB and FC are equal to the two sides AB and BF, the base BC shall be equal to the base BF, and therefore is equal to the line FC. Wherefore the triangle FBC is equilater and equiangle. Lastly make the angle CFD equal to either of the angles at the point F, by drawing the line FD. And draw a line from C to D. Now then by the former reason the triangle FCD shall be equilater and equiangle. And for as much as the three angles at the point F are equal to two right angles (for each of them is the third part of two right angles) therefore (by the 14. of the first) AD is one right line: and for that cause is the diameter of the circle. Wherefore if the other semicircle AFD be divided into so many equal parts as the semicircle ABCD is divided into, it shall comprehend so many equal lines subtended unto it. Wherefore the line AB is the side of an equilater Hexagon figure to be inscribed in the circle: which Hexagon figure also shall be equiangle: For the half of the whole angle B is equal to the half of the whole angle C: which was required to be done.  
Now than if we draw from the centre F a perpendicular line unto AD, which let be FE, and draw also these right lines BE and CE: there shall be described a triangle BEC whose angle E which is at the top shall be the 6. part of two right angles by the 20. of the third. For the angle BFC is double unto it. And either of the two angles at the base, namely, the angles EBC, and ECB is dupla sesquialter to the angle E: that is, either of them containeth the angle E twice and half the angle E. And by this reason was found out the side of an Hexagon figure.  

Correlary.  
Hereby it is manifest, that the side of an Hexagon figure described in a circle is equal to a right line drawn from the centre of the said circle unto the circumference. And if by the points A, B, C, D, E, F, be drawn right lines touching the circle, then shall there be described about the circle an Hexagon figure equilater and equiangle, which may be demonstrated by that which hath been spoken of the describing of a Pentagon about a circle. And moreover by those things which have been spoken of Pentagons' we may in a Hexagon given either describe or circumscribe a circle: which was required to be done.     

The 16. Problem. The 16. Proposition. In a circle given to describe a quindecagon or figure of fifteen angles, equilater and equiangle.  
SVppose that the circle given be ABCD. It is required in the circle ABCD to describe a figure of fifteen angles consisting of equal sides and of equal angles. Describe in the circle ABCD the sides of an equilater triangle, and let the same be AC, and in the ark AC describe the side of an equilater pentagon and let the same be AB. 
Construction.  Now then of such equal parts whereof the whole circle ABCD containeth fifteen, of such parts I say, the circumference ABC being the third part of the circle shall contain five. And the circumference AB being the fift part of a circle shall contain three, wherefore the residue BC shall contain two. Divide (by the 30. of the first) the ark BC into two equal parts in the point E. 
Demonstration.  Wherefore either of these circumferences BE & EC is the fifteen part of the circle ABCD. If therefore there be drawn right lines from B to E, and from E to C, and then beginning at the point B or at the point C there be applied into the circle ABCD right lines equal unto EB or EC, and so continuing till ye come to the point C if you began at B, or to the point B if you began at C, and there shall be described in the circle ABCD a figure of fifteen angles equilater and equiangle: which was required to be done. And in like sort as in a pentagon, if by the points where the circle is divided, be drawn right lines touching the circle in the said points, there shall be described about the circle a figure of fifteen angles equilater & equiangle. And in like sort by the self same observations that were in Pentagons', we may in a figure of fifteen angles given being equilater and equiangle either inscribe, or circumscribe a circle.  


An addition of Flussates.  ¶ An addition of Flussates to find out infinite figures of many angles.  
If into a circle from one point be applied the sides of two 
* A Poligonon figure is a figure consisting of many sides.  Poligonon figures: the excess of the greater ark above the less, shall comprehend an ark containing so many sides of the Poligonon figure to be inscribed by how many unities the denomination of the Poligonon figure of the less side exceedeth the denomination of the Poligonon figure of the greater side: and the number of the sides of the Poligonon figure to be inscribed is produced of the multiplication of the denominations of the foresaid Poligonon figures the one into the other.  
As for example. Suppose that into the circle ABE be applied the side of an equilater and equiangle Hexagon figure (by the 15. of this book) which let be AB: and likewise the side of a Pentagon (by the 11. of this book) which let be AC: and the side of a square (by the 6. of this book) which let be AD: and the side of an equilater triangle (by the 2. of this book) which let be AE. Then I say, that the excess of the ark AD above the ark AB, which excess is the ark BD, containeth so many sides of the Poligonon figure to be inscribed, of how many unities the denominator of the Hexagon AB, which is six, exceedeth the denominator of the square AD, which is four. And forasmuch as that excess it two unities, therefore in BD there shall be two sides. And the denominator of the Poligonon figure which is to be inscribed shall be produced of the multiplication of the denominators of the foresaid Poligonon figures, namely, of the multiplication of 6. into 4. which maketh 24. which number is the denominator of the Poligonon figure, whose two sides shall subtend the ark BD. For of such equal parts whereof the whole circumference containeth 24, of such parts I say, the circumference AB containeth 4, and the circumference AD containeth 6. Wherefore if from AD which subtendeth 6. parts be taken away 4. which AB subtendeth, there shall remain unto BD two of such parts of which the whole containeth 24. Wherefore of an Hexagon and a square is made a Poligonon figure of 24. sides. Likewise of the Hexagon AB and of the Pentagon AC shall be made a Poligonon figure of 30. sides, one of whose sides shall subtend the ark BC. For the denomination of AB which is 6. exceedeth the denomination of AC which is 5. only by unity. So also forasmuch as the denomination of AB which is 6. exceedeth the denomination of AE which is 3. by 3. therefore the ark BE shall contain 3. sides of a Poligonon figure of .18. sides. And observing this self same method and order, a man may find out infinite sides of a Poligonon figure.    
The end of the fourth book of Euclides Elements.   

¶ The fifth book of Euclides Elements.  
THIS FIFTH BOOK of Euclid is of very great commodity and use in all Geometry, 
The argument of this fift book.  and much diligence aught to be bestowed therein. It aught of all other to be thoroughly and most perfectly and readily known. For nothing in the books following can be understand without it: the knowledge of them all depend of it. And not only they and other writings of Geometry, but all other Sciences also and arts: as Music, Astronomy, Perspective, Arithmetic, the art of accounts and reckoning, with other such like. This book therefore is as it were a chief treasure, and a peculiar jewel much to be accounted of. It entreateth of proportion and Analogy, or proportionality, which pertaineth not only unto lines, figures, and bodies in Geometry: but also unto sounds & voices, of which Music entreateth, as witnesseth Boetius and others which writ of Music. Also the whole art of Astronomy teacheth to measure proportions of times and movings. Archimedes' and jordan with other, writing of weights, affirm, that there is proportion between weight and weight, and also between place & place. You see therefore how large is the use of this fift book. Wherefore the definitions also thereof are common, although hereof Euclid they be accommodate and applied only to Geometry. The first author of this book was as it is affirmed of many, one Eudoxus who was Plato's scholar, 
The first author of this book Eudoxus.  but it was afterward framed and put in order by Euclid.  

Definitions.  


The first definition.  A part is a less magnitude in respect of a greater magnitude, when the less measureth the greater.  
As in the other books before, so in this, the author first setteth orderly the definitions and declarations of such terms and words which are necessarily required to the entreaty of the subject and matter thereof, which is proportion and comparison of proportions or proportionality. And first he showeth what a part is. Here is to be considered that all the definitions of this fifth book be general to Geometry and Arithmetic, and are true in both arts, even as proportion and proportionality are common to them both, and chief appertain to number, neither can they aptly be applied to matter of Geometry, but in respect of number and by number. Yet in this book, and in these definitions here set, Euclid seemeth to speak of them only Geometrically, as they are applied to quantity continual, as to lines, superficieces, and bodies: for that he yet continueth in Geometry. I will notwithstanding for facility and farther help of the reader, declare than both by example in number, and also in lines.  
For the clearer understanding of a part, it is to be noted, 
A part taken two manner of ways.  that a part is taken in the Mathematical Sciences two manner of ways. 
The fi●st way.  One way a part is a less quantity in respect of a greater, whether it measure the greater o● no. The second way, 
The second way.  a part is only that less quantity in respect of the greater, which measureth the greater. A less quantity is said to measure or number a greater quantity, 
How a less quantity is said to measure a greater.  when it, being oftentimes taken, maketh precisely the greater quantity without more or less, or being as oftentimes taken from the greater as it may, there remaineth nothing. As suppose the line AB to contain 3. and the line CD to contain 9 then doth the line AB measure the line CD: for that if it be take certain times, namely, 3. times, it maketh precisely the line CD, that is 9 without more or less. Again if the said less line AB be taken from the greater CD, as often as it may be, namely, 3. times, there shall remain nothing of the greater. So the number 3. is said to measure 12. for that being taken certain times, namely, four times, it maketh just 12. the greater quantity: and also being taken from 12. as often as it may, namely, 4. times, there shall remain nothing. And in this meaning and signification doth Euclid undoubtedly here in this define a part: 
In what signification Euclid here taketh a part.  saying, that it is a less magnitude in comparison of a greater, when the less measureth the greater. As the line AB before set, containing 3 is a less quantity in comparison of the line CD which containeth 9 and also measureth it. For it being certain times taken, namely, 3. times, precisely maketh it, or taken from it as often as it may, there remaineth nothing. Wherefore by this definition the line AB is a part of the line CD. Likewise in numbers, the number 5. is a part of the number 15. for it is a less number or quantity compared to the greater, and also it measureth the greater: for being taken certain times, namely, 3. times, it maketh 15. And this kind of part is called commonly pars metiens or mensurans, 
Par● metien● or mensuran●.  that is, a measuring part: some call it pars multiplicatina: 
Pars multiplicati●a.  and of the barbarous it is called pars aliquota, 
Pars aliquota.  that is an aliquote part. And this kind of part is commonly used in Arithmetic. 
This kind of part commonly used in Arithmetic.   
The other kind of a part, 
The other kind of part.  is any less quantity in comparison of a greater, whether it be in number or magnitude, and whether it measure or no. As suppose the line AB to be 17. and let it be divided into two parts in the point C, namely, into the line AC, & the line CB, and let the line AC the greater part contain 12. and let the line BC the less part contain 5. Now either of these lines by this definition is a part of the whole line AB. For either of them is a less magnitude or quantity in comparison of the whole line AB: but neither of them measureth the whole line AB: for the less line CB containing 5. taken as often as ye list, will never make precisely AB which containeth 17. If ye take it 3. times it maketh only 15. so lacketh it 2. of 17. which is to little. If ye take it 4. times, so maketh it 20. then are there three to much, so it never maketh precisely 17. but either to much or to little. Likewise the other part AC measureth not the whole line AB: for taken once, it maketh but 12. which is less than 17. and taken twice, it maketh 24. which are more than 17. by ●. So it never precisely maketh by taking thereof the whole AB, but either more or less. And this kind of part they commonly call pars constituens, or componens: 
Pars constit●ens, or componens.  Because that it with some other part or parts, maketh the whole. As the line CB together with the line AC maketh the whole line AB. Of the barbarous it is called pars aliquanta. 
Pars aliquanta.  In this signification it is taken in B●rla●● in the beginning of his book, in the definition of a part, when he saith: Every less number compared to a greater, is said to be a part of the greater, whether the less measure the greater, or measure it not.   

Multiplex is a greater magnitude in respect of the less, when the less measureth the greater. 
The second definition.   
As the line CD before set in the first example, is multiplex to the line AB. For that CD a line containing 9 is the greater magnitude, and is compared to the less, namely, to the line AB containing 3. and also the less line AB measureth the greater line CD: for taken 3. times, it maketh it, as was above said. So in numbers 12. is multiplex to 3: for 12 is the greater number, and is compared to the less, namely, to 3. which 3. also measureth it: 
Numbers very necessary for the understanding of this book and the other books following.  for 3 taken 4 times maketh 12. By this word multiplex which is a term proper to Arithmetic and number, it is easy to consider that there can be no exact knowledge of proportion and proportionality, and so of this fifth book with all the other books following, without the aid and knowledge of numbers.   

Proportion is a certain respect of two magnitudes of one kind, 
The t●ird definition.  according to quantity.  

Rational proportion divided ●●to two kinds.  Proportion rational is divided into two kinds, into proportion of equality, and into proportion of inequality. Proportion of equality is, when one quantity is referred to an other equal unto itself: as if ye compare 5 to 5, or 7 to 7, & so of other. And this proportion hath great use in the rule of Cosse. 
Proportion of equality.  For in it all the rules of equations tend to none other end but to find out and bring forth a number equal to the number supposed, which is to put the proportion of equality.  

Proportion of inequality.  Proportion of inequality is, when one unequal quantity is compared to an other, as the greater to the less, as 8. to 4: or 9 to 3: or the less to the greater as 4. to 8: or 3. to 9  

Proportion of the greater to the less.  Proportion of the greater to the less hath five kinds, namely, Multiplex, Superparticular, Superpartiens, Multiplex superperticular, and Multiplex superpartiens.  
Multiplex, is when the antecedent containeth in itself the consequent certain times without more or less: 
Multiplex.  as twice, thrice, four times, and so farther. And this proportion hath under it infinite kinds. For if the antecedent contain the consequent justly twice, it is called dupla proportion, 
Duple proportion.  as 4 to 2. If thrice tripla, 
Triple, quadruple. Quintuple.  as 9 to 3. If 4. times quadrupla as 12. to 3. If 5. times quintupla as 15. to 3. And so infinitely after the same manner.  

Superperticular.  Superperticular is, when the antecedent containeth the consequent only once, & moreover some one part thereof as an half, a third, or fourth, etc. This kind also hath under it infinite kinds. For if the antecedent contain the consequent once and an half, thereof it is called Sesquialtera, 
Sesquialtera.  as 6. to 4: if once and a third part Sesquitertia, 
Sesquitertia.  as 4. to 3: if once and a fourth part Sesquiquarta, 
Sesquiquarta.  as 5. to 4. And so in like manner infinitely.  

Superpartiens.  Superpartiens is, when the antecedent containeth the consequent only once, & moreover more parts than one of the same, as two thirds, three fourths, four fifthes and so forth. This also hath infinite kinds under it. For if the antecedent contain above the consequent two parts, it is called Superbipartiens, 
Superbipartiens.  as 7. to 5. If 3. parts Supertripartiens as 7. to 4. 
Supertripartiens.  If 4. parts Superquadripartiens, 
Superquadripartiens.  as 9 to 5. If 5. parts Superquintipartiens as 11. to 6. 
Superquintipartiens.  And so forth infinitely.  

Multiplex superperticular.  Multiplex Superperticular is when the antecedent containeth the consequent more than once, and moreover only one part of the same. This kind likewise hath infinite kinds under it. For if the antecedent contain the consequent twice and half thereof, it is called dupla Sesquialtera, 
Dupla Sesquialtera.  as 5. to 2. If twice and a third Dupla Sesquitertia as 7. to 3. 
Dupla sesquitertia.  If thrice and an half Tripla sesquialtera as 7. to 2. 
Tripla sesquialtera.  If four times and an half Quadr●pla Sesquialtera, as 9 to 2. And so going on infinitely.  

Multiplex superpartiens.  Multiplex Superpartient, is when the antecedent containeth the consequent more than once, and also more parts than one of the consequent. And this kind also hath infinite kinds under it. For if the antecedent contain the consequent twice, and two parts over, it is called dupla Superbipartiens as 8. to 3. 
Dupla superbipartiens.  If twice and three parts, dupla Supertripartiens as 11. to 4. 
Dupla supertripartiens.  If thrice and two parts, it is named Tripla Superbipartiens as 11. to 3. 
Tripla superbipartiens.  If three times and four parts Triple Superquadripartiens as 31. to 9 
Tripla superquad●ipartiens.  And so forth infinitely.  
Here is to be noted that the denomination of the proportion between any two numbers, is had by dividing of the greater by the less. For the quotient o● number produced of that division is even the denomination of the proportion. 
How to kno● the denomination of any proportion.  Which in the first kind of proportion, namely, multiplex, is ever a whole number, and in all other kinds of proportion it is a broken number.  
As if ye will know the denomination of the proportion between 9 and 3. Divide 9 by 3. so shall ye have in the quotient 3. which is a whole number, and is the denomination of the proportion: and showeth that the proportion between 9 & 3. is Tripla. So the proportion between 12. and 3. is quadrupla, for that 12. being divided by 3. the quotient is 4. and so of others in the kind of multiplex. And although in this kind the quotient be ever a whole number, yet properly it is referred to unity, and so is represented in manner of a broken number as 5/● and 4/● for unity is the denomination to a whole number.  
Likewise the denomination of the proportion between 4 and 3 is 1. 1/● for that 4 divided by 3. hath in the quotient 1 1/● one and a third part, of which third part, it is called sesquitercia: so the proportion between 7 and 6. is 1 ⅙ one and a sixth, of which fixed part it is called sesquisexta, and so of other of that kind. Also between, 7 and 5 the denomination of the proportion is 1 ●/● one and two fifthes, which denomination consisteth of two parts, namely, of the munerator and denominator of the quotient of 2. and 5: of which two fifthes it is called superbipartiens quintas: for 2 the numerator showeth the denomination of the number of the parts, and 5. the denominator, showeth the denomination, what parts they are, & so of others. Also the denomination between 5 and 2. is 2 ½ two and a half, which consisteth of a whole number and a broken, of 2. the whole number it is dupla, and of the half, it is called sesquialtera, so is the proportion dupla sesquialtera.  
Again the denomination of the proportion between 11. and 3. is 3 ●/● three and two thirds, consisting also of a whole number and a broken, of 3. the whole numbered it is called tripla, and of ●/● the broken number, it is called Superbipartiens tertias, so the proportion is tripla superbipartiens tertias. Thus much hitherto touching proportion of the greater quantity to the less.  
Proportion of the less quantity to the greater hath as many kinds, as that of the greater to the less, which kinds are in the same order: 
Proportion of the less in the greater.  and have also the self same names, but that to the names afore put ye must add here this word sub. As comparing the greater to the less, it was called multiplex, superparticular, superpartient, multiplex superparticular, and multiplex superpartient, now comparing the less quantity to the greater, it is called submultiplex, 
Submultiplex.  subsuperparticular, 
Subsuperparticular.  subsuperpartient, 
Subsuperpertient. etc.  submultiplex superparticular, and submultiplex superpartient. And so in like manner to all the inferior kinds of all sorts of proportion ye shall add that word sub. The examples of the former serve also here, only transposing the terms of the proportion making the antecedent consequent, and the consequent the antecedent. As 4. to 2. is dupla proportion: so 2. to 4. is subdupla. As 9 to 3. is tripla: so is 3. to ●. subtripla. And as 9 to 6. is sesquialtera, so 6. to 9 is subsesquialtera. As 7. to 5. is superbipartiens quintas, so is 5. to 7. subsuperbipartiens quintas. As 5. to 2. is dupla sesquialtera, so is 2. to 5. subdupla sesquialtera. And also as 8. to 3. is dupla superbipartiens tertias, so is 3. to 8. subdupla superbipartiens tertias. And so may ye proceed infinitely in all others. Thus much thought I good in this place for the ease of the beginner to be added touching proportion.   


The fourth definition.  proportionality, is a similitude of proportions.  
As in proportion are compared together two quantities, and proportion is nothing else but the respect and comparison of the one to the other, and these quantities are the terms of the proportion: so in proportionallitie are compared together two proportions. And proportionallitie is nothing else, but the respect & comparison of the one of them to the other. And these two proportions are the terms of this proportionallitie. He calleth it the similitude, that is, the likeness or idemptitie of proportions: 
Example of this definition in magnitudes.  As if ye will compare the proportion of the line A containing 2. to the line B containing 1, to the proportion of the line C containing 6. to the line D containing 3, either proportion is dupla. This likeness, idemptitie, or equality of proportion is called proportionallitie. 
Example thereof in numbers.  So in number 9 to 3. and 21. to 7. either proportion is tripla. Where note that proportions compared together, are said to be like the one to the other: 
Note.  but magnitudes compared together, are said to be equal the one to the other.   


The fifth definition.  Those magnitudes are said to have proportion the one to the other, which being multiplied may exceed the one the other.  
Before he showed and defined, what proportion was, now by this definition he declareth between what magnitudes proportion falleth, saying: That those quantities are said to have proportion the one to the other, which being multiplied, may exceed the one the other. 
An example of this definition in magnitudes.  As for that the line A being multiplied by what soever multiplication or number, as taken twice, thrice, or four, five, or more times, or once and half, or once and a third, & so of any other part, or parts, may exceed and become greater than the line B or contrariwise, than these two lines are said to have proportion the one to the other. And so ye may see that between any two quantities of one kind, there is a proportion. For the one remaining unmultiplied, & the other being certain times multiplied, shall be greater than it. As 3. to 24. hath a proportion, for leaving 24. unmultiplied, and multiplying 3. by 9, ye shall produce 27: which is greater than 24, and exceedeth it. Here is to be noted, that Euclid in defining what quantities have proportion, 
Why Euclid in defining of Proportion used multiplication.  was compelled to use multiplication, or else should not his definition be general to either kind of proportion: namely, to rational and irrational: to such proportion I say which may be expressed by number, and to such as cannot be expressed by any determinate number, but remaineth furred and innominable. In rational quantities which have one common measure, the excess of the one above the other is known, and by it is known the proportion, which may be expressed by some determinate number. But in irrational quantities which have no common measure, it is not so. For in them the excess of the one to the other is ever unknown, & therefore is furred, and innominable. As between the side of a square and the diameter therof● there is undoubtedly a proportion, for that the side certain times multiplied may exceed the diameter. Likewise between the diameter of a circle and the circumference thereof there is certainly, by this definition, a proportion, for that the diameter certain times multiplied may exceed the circumference of the circle: although neither of these proportions can be named & expressed by number. For this cause therefore used Euclid this manner of defining by multiplication.   

Magnitudes are said to be in one or the self same proportion, 
The sixth definition.  the first to the second, and the third to the fourth, when the equimultiplices of the first and of the third being compared with the equimultiplices of the second and of the fourth, according to any multiplication: either together exceed the one the other, or together are equal the one to the other, or together are less the one than other.  
In the definition last going before, he showed what magnitudes have proportion the one to the other, & now this definition showeth what magnitudes are in one and the self same proportion, 
An example of this definitiin in magnitudes.  and how to know whether they be in one and the self same proportion, or not. It is plain that every proportion hath two terms, so that when ye compare proportion to proportion, ye must of necessity, have 4. terms, that is, an antecedent and a consequent, to either of the proportions. As suppose A, B, C, D, to be four magnitudes, A the first, B the second, C the third, and D the fourth now if ye take the equimultiplices of A and C the first & the third, that is, if ye multiply A and C by one and the self same number, as let the multiplex of A be E, and let the equimultiplex of C be F. Likewise also if ye take the equimultiplices of B and D, the second and the fourth, that is if ye multiply them by any one number, whether it be by that number whereby ye multiplied A & C, or by any other number greater or less, as let the multiplex of B be G, and the equimultiplex of D be H: how it the equimultiplices of A and C be both greater hen the eqnimultiplices of B and D, that is if the multiplex of A be greater than the multiplex of B, and the multiplex of C be greater than the multiplex of D, or if they be both less than they: or both equal to them, 
An example in numbers.  then are the magnitudes A, B and ●, D in one and the self same proportion.  
Likewise in numbers 8. to 6. hath a proportion, also 4. to 3. hath a proportion: now to see whether they be in one and the self same proportion or not, set them in order as in the example here written, 8 the first, 6 the second, 4 the third, and 3. the fourth. Now take the equemultiplices of 8 and 4. the first, and the third, that is, multiply them by one and the self same number, suppose it be by 3. so the triple of 8 is 24. & the triple of 4. is 12: likewise take the equimultiplices of 6 and 3. the second and the fourth, multiplying them likewise by one and the self same number, suppose it be also by 3 as before ye did, the triple of 6 is 18. and the triple of 3 is 9 Now ye see that the triple of 8 the first, namely, 24. exceedeth the triple of 6. the second, namely, 18: likewise the triple of 4 the third number, namely, 12. exceedeth the triple of 3. the fourth, namely, 9 Wherefore by the first part of this definition, the numbers 8 to 6. and 4 to 3. are in one and the self same proportion, because that the equemultiplices of 8 and 4. the first & the third, do both exceed the equimultiplices of 6 and 3. the second and the fourth.  

another example in numbers.  Again, take the same numbers and try the same after this manner. Take the equimultiplices of 8. and 4. the first and the third, multiplying each by 3. as before ye did, so shall ye have 24 for the triple of 8. and 12 for the triple of 4. as ye had before. Then take the equimultiplices of 6 and 3. the second and the fourth, mutliplying them by some one number, but not by 3 as before ye did: but by 4. so for the quadruple of 6 the second number, shall ye have 24. and for the quadruple of 3 the fourth number, ye shall have 12. And now ye see that the equimultiplices of 8 and 4. the first and the third, namely, 24 and 12. are both equal to the multiplices of 6 and 3. the second and the fourth, namely, to 24 and 12. Wherefore the numbers given, are by the second part of this definition in one and the self same proportion, because the equimultiplices of 8 and 4 the first and the third, are both equal to the equimultiplices of 6 and 3. the second and the fourth.  

another example in numbers.  Again to show the same, and for the fullness of the definition, take the same numbers 8, 6, 4, 3. and take the equimultiplices of 8 and 4. the first and the third, multiplying each by 2. so have ye 16 for the duple of 8, the first number, and 8 for the duple of 4 the third number: then take also the equimultiplices of 6 and 3, the second and the fourth, mutliplying each by 3. so have ye 18 for the triple of 6 the second, and 9 for the triple of 3. the fourth number. And now ye see that the equimultiplices of 8 and 4. the first and the third, namely, 16. and 8 are both less than the equimultiplices of 6 and 3. the second & the fourth namely 18 and 9 For 16 are less than 18, and 8 are less than 9 Wherefore by the third part of this definition, the numbers proposed are in one and the self same proportion, for that the equimultiplices of 8 and 4 the first and the third are both less than the equimultiplices of 4 and 3 the second and the fourth.  
Farther in this definition, this particle (according to any multiplication) is most diligently to be considered, 
Note this particle according to any multiplication.  which signifies by any multiplication indifferently whatsoever. For whensoever the quantities be in one and the self same proportion, then by any multiplication whatsoever, the equimultiplices of the f●rst and the third, shall exceed the equimultiplices of the second and the fourth, or shall be equal unto them, or less than them. Yet it may so happen by some one multiplication; that the equimultiplices of the first and the third, do exceed the equimultiplices of the second and the fourth, and yet the quantities given shall not be in one and the self same proportion. As in this example here set, where the equimultiplices of 6 and 5, the first and the third, namely, 18. and 15. do both exceed the equimultiplices of 4 and 3. the second and the fourth, namely, 8 and 6. 
An example where the equimultiplices of the first and third exceed● the equimultiplices of the second and fourth, and yet the quantities given are not in one and the self same proportion.  yet are not the numbers given in one and the self same proportion. For 6 hath not that proportion to 4. which 5. hath to 3. In this example 6 and 5 the first and the third were multiplied by 3. which made their equimultiplices 18 and 15. which exceed the equimultiplices of 4 and 3, the second and the fourth being multiplied by 2. namely, 8 and 6: but if ye shall multiply 6 and 5 the first and the third by 2. ye shall produce 12 and 10 for their equimultiplices, and then if ye multiply 4 and 3. the second and the fourth by 3. so shall ye produce for their equimultiplices 12 and 9 Now ye see that by this multiplication the equimultiplices of the first and the third do not both exceed the equimultiplices of the second and fourth: for 12 the multiplex of 6 doth not exceed 12 the multiplex of 4. and therefore the numbers or quantities ar● not in one and the self same proportion, for that it holdeth not in all multiplications whatsoever.  
And because this definition requireth all manner of multiplications to bring forth the excesses, equalities, and wants of the antecedents above, to, or under the consequents; to avoid the tediousness and infinite labour thereof, I have set forth a rule much to be made of and esteemed, whereby ye may in any rational proportion produce equimultiplices of the first and the third equal to the equimultiplices of the second and the fourth. The rule is this, 
A rule to produce equimultiplices of the first and third equal to the equimultiplices of the second● and forth.  take two numbers whatsoever in that proportion in which your quantities are, & by the number which is antecedent multiply the consequents of your proportions, namely, the second and the fourth: and by the number which is the consequent multiply the antecedentes of your proportions, namely, the first and the third● then necessarily shallbe produced the equimultiplices of the first and the third equal to the equimultiplices of the second & the fourth. As by example, take 6 to 2. and 3 to 1, which are in one & the self same proportion, & taking these two numbers 9 & 3. which are in the same proportion, now by 9 the antecedent multiply the cons●quēts 2 & 1. and so shall ye have 18 & 9 for the equimultiplices of the second & the fourth, 
Example thereof.  then by 3 the consequent multiply the antedents 6 & 3, so shall ye have 18 & ● for the equimultiplices of the first & the third, which are equal to the former equimultiplices of the second & fourth. Whereof it followeth that if ye multiply 18 & 9 the equimultiplices of the first and the third by any number greater than 3 whereby they were now multiplied, they shall both ever exceed the equimultiplices of the second & the fourth: & if ye multiply them by any number less than 3. they shall ever both want of them. So that whatsoever multiplication it be, they shall ever both exceed, be equal, or want above, to, or from 18. ●nd 9 the equimultiplices of the second and fourth.   


The seventh definition  Magnitudes which are in one and the self same proportion, are called Proportional.  
As if the line A, have the same proportion to the line B, that the line C hath to the line D, then are the said four magnitudes A, B, C, D, called proportional. Also in numbers for that 9 to 3. hath that same proportion that 12 hath to 4: 
9 12  therefore these four numbers 9.3.12.4. 
3 4  are said to be proportional. Here is to be noted that this likeness or idemptitie of proportion which is called, as before was said proportionality, is of two sorts: the one is continual, the other is discontinuall. 
Proportionality of two sorts, continual and discontinuall.  Continual proportionality is, when the magnitudes set in like proportion, are so joined together, that the second which is consequent to the first, is antecedent to the third, and the fourth which is consequent to the third, is antecedent to the fift, and so continually forth. So every quantity or term in this proportionality, is both antecedent and consequent (consequent in respect of that which went before, 
An example of continual proportionality in numbers.  & antecedent in respect of that which followeth) except the first, which is only antecedent to that which followeth, and the last which is only consequent to that which went before. Take an example in these numbers, 16.8.4.2.1. 
16.8.4.2.1.  In what proportion 16. is to 8, in the same is 8. to 4, in the same also is 4. to 2, and likewise 2. to 1. For they all are in duple proportion: 16. the first is antecedent to 8, and 8. is consequent unto it: and the self same 8. is antecedent to 4: which 4 being consequent to 8. is antecedent to 2, which 2 likewise is consequent to 4. and antecedent to 1: which because he is the last, is only consequent, and antecedent to none, as 16. because it was the first, was antecedent only, and consequent to none. Also in this proportionality all the magnitudes must of necessity be of one kind, 
In coutinnall proportionality the quantities cannot be of one kind.  by reason of the continuation of the proportions in this proportionality, because there is no proportion between quantities of divers kinds. Discontinuall proportionality is, 
Discontinuall prop●rtionalitie.  when the magnitudes which are set in like proportion, are not continually set, as before they were, having one term referred both to that which went before, and to that which followeth, but have their terms distinct and severed asunder: as the first is antecedent to the second, so is the third antecedent to the fourth. Example in numbers, as 8 is to 4. 
Example of discontinual proportionality in numbers.  so is 6. to 3. for either proportion is duple. Where ye see, how each proportion hath his own antecedent and consequent distinct from the antecedent and consequent of the other, and no one number is antecedent and consequent in divers respects. And by reason of the discontinuance of the proportions in this proportionality, 
In discontinual proportionality the proportions may be of divers kinds.  the quantities compared, may be of divers kinds, because the consequent in the first proportion is not the antecedent in the second proportion. So that ye may compare superficies to superficies, or body to body in the self same proportion that ye do line to line.   

When the equemultiplices being taken, 
The eight definition.  the multiplex of the first exceedeth the multiplex of the second, & the multiplex of the third, exceedeth not the multiplex of the fourth: then hath the first to the second a greater proportion, then hath the third to the fourth.  
In the sixth definition was declared what magnitudes are said to be in ●●e and the same proportion: now he showeth in this definition what magnitudes are said to be in a greater proportion. And here is supposed the same order of multiplication, that there in that definition was used: namely, that the first and the third be equally multiplied, that is, by one & the self same number: and also that the second and the fourth be equally multiplied by the same or some other number: and then if the multiplex of the first, exceed the multiplex of the second: 
An example of this definition in magnitudes.  & the multiplex of the third; exceed not the multiplex of the fourth, the first hath a greater proportion to the second, then hath the third to the fourth. As suppose that there be four quantities, A, B, C, D: of which let A be the first, B the second, C the third, & D the fourth. And let A the first commine 6. and let B the second contain 2. & C the third 4. & D the fourth 3: Now take the equimultiplices of A and C the first & the third, which let be E and F, so that how multiplex E is to A so multiplex let F be to C: namely for example sake let either of them be triple: so have you 18. for the multiplex of A, and 12. for the multiplex of C. Likewise take the equimultiplices of B & D, the second & the fourth, multiplying them also by one and the self same number, as by 4: so have ye for the multiplex of B the second 8, namely, the line G, and for the multiplex of D the fourth 12, namely, the line H. Now because the line E multiplex to the first, namely, 18, exceedeth the line G multiplex to the second, namely, 8: And the line F multiplex to the third, namely, 12, exceedeth not the line H multiplex to the fourth, namely, 12 (for that they are equal) the proportion of A to B the first to the second, is greater than the proportion of C to D the third to the fourth. So likewise in numbers: take 11. to 2. & 7. to 3. and multiply 11. & 7. 
An example in numbers.  (the first, and the third) by 2, so shall ye have 22. for the multiplex of the first, and 14. for the multiplex of the third: and multiply 2. and 3. the second and the fourth by 6: so shall the multiplex of the second be 12. and the multiplex of the fourth be 18: Now ye see 22. the multiplex of the first, exceedeth 12, the multiplex of the second. But 14. the multiplex of the third, exceedeth not 18. the multiplex of the fourth: Wherefore the proportion of 11. to 2. the first to the second, is greater than the proportion of 7. to 3, the third to the fourth. And so of all other quantities and numbers, which are not in one and the self same proportion, ye may know when the first to the second hath a greater proportion than the third to the fourth.  

¶ An other example.  
This example have I set to declare that although the proportion of the first to the second be greater than the proportion of the third to the fourth, yet the multiplex of the first exceedeth not the multiplex of the second. Wherefore it is diligently to be noted, 
Note.  that it is sufficient to show that the proportion of the first to the second is greater than the proportion of the third to the fourth, if the want or lack of the multiplex of the first from the multiplex of the second, be less than the want or lack of the multiplex of the third to the multiplex of the fourth. As in this example 16. the multiplex of 8. the first, wanteth of 20. the multiplex of 4. the second, four: whereas 18. the multiplex of 9, the third, wanteth of 45, the multiplex of 9 the fourth, 27. And so of all others whereas (the proportions being divers) the equimultiplices of the first and the third are both less, than the equimultiplices of the second and the fourth. Likewise if the equimultiplices of the first and the third do both exceed the equimultiplices of the second & the first, them shall the excess of the multiplex of the first above the multiplex of the second, be greater than the excess of the multiplex of the third, above the multiplex of the fourth. As in these numbers here set, the equimultiplices of 6. and 4. the first and the third, namely, 12. and 8. do both exceed the equimultiplices of 2. and 3. the second and the fourth, namely, 4. and 6. But 12. the multiplex of the first exceedeth 4. the multiplex of the second by 4, and 8. the multiplex of the third exceedeth 6. the multiplex of the fourth by 2. but 8. is more than 2. Howbeit this is generally certain that when soever the proportion of the first to the second is greater than the proportion of the third to the fourth, there may be found some multiplication, that when the equimultiplices of the first and the third shall be compared to the equimultiplices of the second and the fourth, the multiplex of the first shall exceed the multiplex of the second, & the multiplex of the third shall not exceed the multiplex of the fourth, according to the plain words of the definition.  
In like manner when you have taken the equimultiplices of the first & the third, and also the equimultiplices of the second and the fourth, if the multiplex of the first exceed not the multiplex of the second, and the multiplex of the third exceed the multiplex of the fourth: then hath the first to the second a less proportion, then hath the third to the fourth. As in the example before, if ye change the terms, and make C the first, D the second, A the third, and B the fourth: then shall F, namely, 12. the multiplex of the first not exceed H, namely, 12. the multiplex of the second: but E, namely, 18. the multiplex of the third exceedeth G, namely, 8. the multiplex of the fourth. Wherefore the proportion of C to D, the first to the second, is less than the proportion of A to B, the third to the fourth.  
Even so in numbers. As in this example, 5. to 4. and 7. to 3. If ye multiply 5. and 7. the first and the third each by 3, ye shall for the multiplex of 5. the first have 15. and for the multiplex of 7. the third shall ye have 21: again if ye multiply 4. and 3. the second & the fourth by 6, for the multiplex of 4. the second ye shall have 24, and for the multiplex of 3. the fourth, ye shall have 18. So ye see that ●5. the multiplex of the first, is less than 24, the multiplex of the second: and 21. the multiplex of the third is greater than 18. the multiplex of the fourth. Wherefore the proportion of 5. to 4. the first to the second is less than the proportion of 7. to 3. the third to the fourth.    

Proportionallitie consisteth at the jest in three terms. 
The ninth definition.   
Before it was said, that proportionality is a likeness or an idemptitie of proportions. Wherefore of necessity in proportionality, there must be two proportions, and every proportion hath two terms, namely, his antecedent and consequent. Therefore in every proportionality th●re are four terms. But for that sometime, one term supplieth by divers relations, the room of two, for in respect to the first it is consequent, and in respect to that which followeth, it is antecedent: therefore three terms at lest and not under may suffice in proportionality, which three are in power four, and occupy the room of four, as is said. 
An example of this definition in magnitudes.  As suppose that A hath to B that proportion, that B hath to C: then are these three quantities A, B, C, set in the least number of proportionality. 
Example ●n numbers.  Likewise in numbers, as 8. 4. 2. and 9 6. 4.   

When there are three magnitudes in proportion, 
The tenth definition.  the first shall be unto the third in double proportion that it is to the second. But when there are four magnitudes in proportion the first shall be unto the fourth in triple proportion that it is to the second. And so always in order one more, as the proportion shall be extended.  

A rule to add proportions to proportions.  Multiply the antecedent of the one proportion by the antecedent of the other, and the number produced shall be the antecedent of the proportion which containeth them both. Likewise multiply the consequent of the one proportion by the consequent of the other, and the number produced shall be consequent of the proportion which shall contain them both.  
So if there be 4. quantities in continual proportion, the proportion of the first to the fourth, shall be triple to the proportion which is between the first and the second, that is, it shall contain it three times. As for example, Take 4. numbers in continual proportion 8. 4. 2. 1. 
8. 4. 2. 1.  You see that the proportion of 8 to 1. 
2 2 2  the first to the fourth, 
1 1 1  is octupla: the proportion of 8 to 4. the first to the second is dupla, now triple dupla proportion, that is, add 3. dupla proportions together, by the rule before given, as ye see in the example. Multiply all the antecedentes together 2. the antecedent of the first proportion, by 2. the antecedent of the second, so have ye 4: which 4. multiply by 2 the antecedent of the third proportion, so shall ye have 8 for a new antecedent. In like manner multiply all the consequentes together, 1. the consequent of the first proportion by 1. the consequent of the second proportion, so shall ye have 1, which 1. multiply again by 1. the consequent of the third proportion, so shall ye have again 1: which 1. let be consequent to your former antecedent 8: so have ye 8 to 1. which is octupla, which was also the proportion of the first to the fourth, which octupla is also brought fourth of the addition of three dupla proportions together, and containeth it three times, wherefore octupla is tripla to dupla, and therefore as the definition saith: the proportion of the first to the fourth is tripla to the proportion of the first to the second. And so consequently forth as long as the proportionality continueth according to the sentence of the definition, the terms of the proportions exceeding the number of proportions by one. As if ye have 5. terms in proportion, the proportion of the first to the fifth shall be quadrupla to the proportion of the first to the second, and if there be 6. terms, it shall be quintupla and so in order.   

Magnitudes of like proportion, are said to be antecedents to antecedentes, and consequentes to consequentes. 
The eleventh definition.   
For that before it was said, that proportion was a relation or a respect of one quantity to an other, now showeth he what magnitudes are said to be of like proportion, namely, these whose antecedents have like respect to their consequentes, and whose consequents receive like respects of their antecedents. 
Example of this definition in magnitude.  As putting 4. magnitudes A, B, CD. If A antecedent to B, be double to B, and C antecedent to D, be double also to D, then have the two antecedentes like respects to their consequents. Likewise if B the consequent be half of A, and also D the consequent be half of C, then the two consequentes B and D receive of their antecedentes like respects and relations. And by this definition, are these magnitudes A, B, C, D, of like proportion.  
Also in numbers, 9 3. 6. 2: because 9 the antecedent is triple to 3. his consequent, and the antecedent 6. is also triple to 2 his consequent: 
Example in numbers.  the two antecedents 9 and 6 have like respects to their consequentes, and because that 3 the consequent is the subtriple or third part of ●. his antecedent, and likewise 2 the consequent is the subtriple or third part of 6. his antecedent, the two consequentes 3 and 2 receive also like respects of their antecedentes, and therefore are numbers of like proportion.   

Proportion alternate, or proportion by permutation is, 
The twelf●h definition.  when the antecedent is compared to the antecedent, and the consequent to the consequent.  
The understanding of this definition & of all the definitions following, dependeth of the definition going before, and use it for a general supposition, namely, to have four quantities in proportion. Suppose four magnitudes A, B, C, D, to be in proportion, 
Example of this definition in magnituds.  namely, as A is to B, so let C be to D. Now if ye compare A the antecedent of the first proportion to C the antecedent of the second as to his consequent, & likewise if ye compare B the consequent of the first proportion as an antecedent to D the consequent of the second as to his consequent: then shall ye have the magnitudes in this sort: as A to C, antecedent to antecedent, 
Example in numbers.  so B to D, consequent to consequent, & this is called permutate proportion or alternate. In numbers as 12. to 6, so 8. to 4. either is dupla. Wherefore by permutation of proportion, as 12. to 8. antecedent to antecedent, so is 6. to 4. consequent to consequent, for either is sesquialtera.   


The thirteenth definition.  Converse proportion, or proportion by conversion is, when the consequent is taken as the antecedent, and so is compared to the antecedent as to the consequent.  
Suppose as before four magnitudes in proportion, A, B, C, D, as A to B, so C to D: 
Example of this definition in magnituds.  if ye refer B the consequent of the first proportion, as antecedent, to A the antecedent of the first, as to his consequent: and likewise if ye refer D the consequent of the second proportion as antecedent to C the antecedent of the second proportion, as to his consequent: them shall ye have the magnitudes in this order. As B to A consequent to antecedēt● so D to C consequent to antecedent. 
Example in numbers.  And this is called converse proportion. So also in numbers, 9 to 3, as 6. to 2, either is tripla, wherefore comparing 3. to 9, the consequent of the first to his antecedent 9, and also 2. the consequent of the second to his antecedent 6, by converse proportion it cometh to pass as 3. to 9, so 2. to 6: For either is subtripla.   


The fourteenth definition.  Proportion composed, or composition of proportion is, when the antecedent and the consequent are both as one compared unto the consequent.  
Suppose that in the former four magnituds in proportion, A, B, C, D, as A is to B, so is C to D: 
Example of this definition in magnituds.  if ye add A and B the antecedent and the consequent of the first proportion together, and compare them so added as one antecedent to B the consequent of the first proportion as to his consequent: and likewise if ye add together C and D the antecedent and the consequent of the second proportion, and so added, compare them as one antecedent to D the consequent of the second proportion, as to his consequent: then shall ye have the magnitudes in this order. As AB to B, so CD to D, for either of them is tripla. And this is called composed proportion, or composition of proportion. And so also in numbers. 
Example in numbers.  As 8. to 4, so 6. to 3: 8. and 4, the antecedent and consequent of the first proportion added together, make 12: which 12. as antecedent compare to 4. the consequent of the first proportion as to his consequent: so add together 6. and 3, the antecedent and consequent of the second proportion, they make 9: which 9 as antecedent compare to 3. the consequent of the second proportion, as to his consequent: so shall ye have by composition of proportion, as 12. to 4, so 9 to 3, for either of them is tripla.   

Proportion divided, or division of proportion is, 
The fi●t●ne definition.  when the excess wherein the antecedent exceedeth the consequent, is compared to the consequent.  
This definition is the converse of the definition going next before: 
This is the converse of the former definition.  in it was used composition, and in this is used division. As before so now suppose four magnitudes in proportion AB the first, B the second, CD the third, and D the fourth: as AB to B: so CD to D: AB, 
Example in magnitudes.  the antecedent of the first proportion exceedeth B the consequent of the first proportion by the magnitude A, wherefore A is the excess of the antecedent AB above the consequent B: so likewise CD the antecedent of the second proportion, exceedeth D the consequent of the same proportion, by the quantity C, wherefore C is the excess of the antecedent CD above the consequent D. Now if ye compare A the excess of AB the first antecedent, above the consequent B, as antecedent to B the consequent, as to his consequent: also if ye compare D the excess of the second antecedent CD, above the consequent D, as antecedent to D the consequent, as to his consequent: then shall your magnitudes be in this order. As A to B, so is C to D: which is called division of proportion, or proportion divided. 
Example in numbers.   
And so in numbers, as 9 to 6, so 12. to 8, either proportion is sesquialtera: the excess of 9 the antecedent of the first proportion above 6. the consequent of the same is 3●: the excess of 12. the antecedent of the second proportion above 8, the consequent of the same, is 4● then if ye compare 3. the excess of 9 the first antecedent above the consequent, as antecedent to 6, the consequent, as to his consequent: and also if ye compare .4 the excess of 12. the second antecedent above the consequent, as antecedent, to 8. the consequent, as to his consequent, ye shall have your numbers after this manner by division of proportion, as 3. to 6: so 4. to 8: for either proportion is subdupla.   

Conversion of proportion (which of the elders is commonly called everse proportion, 
The sixteen definition.  or eversion of proportion) is, when the antecedent is compared to the excess, wherein the antecedent exceedeth the consequent.  

An example of this definition in magnitudes.  Four magnitudes supposed as before, AB the first, B the second, CD the third, and D the fourth. As AB to A, so CD to C: AB the antecedent of the first proportion exceedeth B the consequent of the same by the magnitude A, wherefore A is the excess of the antecedent AB above the consequent B: so also the magnitude C is the excess of CD the antecedent of the second proportion above D the consequent of the same: now if ye refer AB the antecedent of the first proportion, as antecedent, to A the excess thereof above the consequent B, as to his consequent: if ye compare also CD the antecedent of the second proportion as antecedent to C the excess thereof above the consequent D, as to his consequent: then shall your magnitudes come to this order. As AB to A, so CD to C, and this is called conversion of proportion, and of some eversion of proportion. Likewise in numbers, as 9 to 6, so 12. to 8. 
An example in numbers.  either proportion is sesquialtera: the excess of 9 the antecedent of the first proportion above 6. the consequent of the same is 3: the excess of 12. the antecedent of the second proportion above 8. the consequent of the same, is 4: now compare the antecedent of the first proportion 9 as antecedent to 3. the excess thereof above 6. the consequent, as to his consequent, likewise compare 12. the antecedent of the second proportion as antecedent to 4. the excess thereof above 8. the consequent, as to his consequent: so shall your numbers be in this order by conversion of proportion: as 9 to 3: so 12. to 4: for either proportion is triple.   


The seventeenth definition.  Proportion of equality is, when there are taken a number of magnitudes in one order, and also as many other magnitudes in an other order, comparing two to two being in the same proportion, it cometh to pass, that as in the first order of magnitudes, the first is to the last, so in the second order of magnitudes is the first to the last. Or otherwise it is a comparison of extremes together, the middle magnitudes being taken away.  
To the declaration of this definition are required two orders of magnitudes equal in number, and in like proportion: 
An example of this definition in magnitudes.  As if there be taken in some determinate number certain magnitudes, namely, four, A, B, C, D. And also in the same number be taken other quantities, namely, four, E, F, G, H: then take the equal proportions by two and two: as A to B, so E to F: as B to C, so F to G: and as C to D, so G to H. Now according to the first definition, if A the first magnitude of the first order be to D the last magnitude of the same order, as E the first magnitude of the second order is to H the last magnitude the same, than it is called proportion of equality, or equal proportion.  
By the second definition, which is all one in substance with the first, ye leave the mean magnitudes in either order, namely, B, C, on the one side, and F, G on the other side, and only compare the extremes of each side together, which by this definition shall be in like proportion, as A is to D, so is E to H.  
Even so in numbers, take for example these two orders, 27. 9 12. 24. 25. and 9 3. 4. 8. 5. there are in each order as ye see, 
An example in numbers.  five numbers, then see that all the proportions taken by two & two be like: between 27 & 9, numbers of the first order, and between 9 and 3, numbers of the second order, there is one and the self same proportion, namely, tripla: also between ●, and 12, numbers of the first order, and 3. and 4, numbers of the second order, is like proportion, namely, subsesquitertia proportion: so between 12. and 24, numbers of the first order, and 4. and 8, numbers of the second order, is also like proportion, namely, subdupla: Last of all, between 24. and 15, numbers of the first row, and between 8. and 5, numbers of the second row, the proportion is one, namely, super●ripar●iens quintas. Wherefore by this definition, leaving out all the mean numbers of each side, ye may compare together only the extremes, and conclude that as 27. of the first row is to 15. the last of the same row, ●o is 9 the first of the second row to 5, the last of the same row: for the proportion of each is superquadripartiens quintas.  
Here is to be considered, 
Note.  that it is not of necessity that all the proportione in each row of numbers be set in like order, as in the one so in the other: but it shall be sufficient that the proportions be the same and in equal number in each row. Whether it be in the self same order, or in contrary, or inue●●ed order, it maketh no matter. As in these numbers● 12. 6. 2. in the ●●rst row● and ●●. 8● 4. in the second. As 12. is to 6, the firs● to the second of the first row, so is 8. to 4. the second to the third of the second row: either i● duple proportion. And as 6● to 2● the second to the third in the first order: so is 24. to 8. the first to the second in the second order. Where ye see that the proportions are not placed in one and the self same order, and course, and yet notwithstanding ye may conclude by equality of proportion, leaving the means 6. and 8: as 12. to 2. the first to the last of the first order, so 24. to 4. the first and last of the second order. And so of others whatsoever and how soever they be placed.   

An ordinate proportionality is, when as the antecedent is to the consequent, 
The eighttenth definition.  so is the antecedent to the consequent, and as the consequent is to another, so is the consequent to an other.  
For the declaration of this definition are also required two orders of magnitudes. 
An example of this definition in magnitudes.  Suppose in the first order, that the antecedent A, to his consequent B, have the same proportion that the antecedent D, hath to his consequent E in the second order: and make the consequent B antecedent to some other quantity, as to C. Also make the consequent E antecedent to an other quantity, as to F, so that there be the same proportion of B to C, which is of E to to F. And this disposition of proportions is called ordinate proportionality. Likewise in numbers, 18.9.3 and 6.3.1. 
Example in numbers.  As 18. to 9 antecedent to consequent, so is 6. to 3. antecedent to consequent: either is dupla proportion: and as 9 the consequent is to an other, namely, to the number 3, so is the consequent 3. to an other, namely, to unity. And this ordinate proportionality may be extended as far as ye li●t, as ye may see in the example of numbers in the definition next before.   


The nintenth definition.  An inordinate proportionality is, when as the antecedent is to the consequent, so is the antecedent to the consequent: and as the consequent is to an other, so is an other to the antecedent.  
This definition also as the other before, requireth two orders of magnitudes, Suppose in the first order that the antecedent A be to the consequent B, as the antecedent C, 
An example of this definition in magnitudes.  in the second order is to the consequent D, & let B the consequent of the first proportion be to some other, namely, to the magnitude E, as some other, namely, the magnitude F, is to the antecedent C of the second proportion: 
Example in numbers.  this kind of proportionality is called inordinate or perturbate. Take also an example in numbers, as 9 to 6. the antecedent to the consequent, so is 3 to 2 the antecedent to the consequent, either proportion is s●squ●ul tera, and as ● the consequent of the first proportion, is to an other, namely, to the number 3. so is another namely, the number 6. to 3. the antecedent of the second proportion, for either is dupla proportion.   

An extended proportionality is, when as the antecedent is to the consequent, so is the antecedent to the consequent, 
The 20. definition.  and as the consequent is to an other, so is the consequent to an other. Apertu●bate proportionality is, when, three magnitudes being compared to three other magnitudes, 
The 2●. definition.  it cometh to pass, that as in the first magnitudes the antecedent is to the consequent, so in the second is the antecedent to the consequent, & as in the first magnitudes the consequent is to an other magnitude, so in the second magnitudes is an other magnitude to the antecedent.  
These two last definitions here put by Zamberte seem all one with the other two last before set. 
These two last definitions not found in the greek examplers.  Wherefore it is not like that they were written and set here by Euclid, for that they seem no● necessary, but rather superfluous, neither are they found in the Greek examples commonly set forth in print, nor mentioned of any that hath written commentaries upon Euclid, old or new: Not of Campane, S●●ub●lius, Pellitarius, Orontius, nor Fl●ssates: wherefore it is not of necessity to add unto them any explanation or example either in magnitudes or in numbers. The examples of the two last definitions set before, may likewise serve for them also.    

The 1. Theorem. The 1. Proposition. If there be a number of magnitudes how many soever equemultiplices to a like number of magnitudes each to each: how multiplex on magnitude is to one, so multiplices are all the magnitudes to all.  
SVppose that there be a number of magnitudes, namely, AB, and DC equimultiplices. to a like number of magnitudes E and F each to each. Then I say, that how multiplex AB is to E, so multiplices are AB and DC to E and F. For forasmuch as how multiplex AB is to E, so multiplex is DC to F, therefore how many magnitudes there are in AB equal unto E so in any are there in DC equal unto F. 
Construction.  Divide AB into the magnitudes which are equal unto E, that is, into AG and GB: and likewise DC into the magnitudes which are equal unto F, that is, into DH and HC. 
Demonstration.  Now than the multitude of these DH & HC, is equal unto the multitude of these AG & GB. And forasmuch as AG is equal unto E, and DH unto F: therefore AG and DH, are equal unto E and F: and by the same reason forasmuch as GB is equal unto E, and HC unto F, GB also and HC are equal unto E and F. Wherefore how many magnitudes there are in AB equal unto E, so many are there in AB and DC equal unto E & F: Wherefore how multiplex AB is to E, so multiplices are AB and DC to E and F. If therefore there be a number of magnitudes how many soever equemultiplices to a like number of magnitudes each to each, how multiplex one magnitude is to one, so multiplices are all the magnitudes to all: which was required to be proved.   

The 2. Theorem. The 2. Proposition. If the first be equemultiplex to the second as the third is to the fourth, and if the fifth also be equemultiplex to the second as the sixth is to the fourth: then shall the first and the fifth composed together be equemultiplex to the second, as the third and the sixth composed together is to the fourth.  
SVppose that there be six quantities, of which let AB be the first, C the second, DE the third, F the fourth, BG the fifth, & EH the sixth: and suppose that the first, AB, be equemultiplex unto the second, C, as the third, DE, is to the fourth, F: and let the fift, BG, be equemultiplex unto the second, C, as the sixth, EH, is to the fourth, F. Then I say, that the first and the fifth composed together, which let be AG, is equemultiplex unto the second, C, as the third and sixth composed together, which let be DH, is to the fourth, F. For forasmuch as AB is equemultiplex to C, 
Demonstration●  as DE is to F, therefore how many magnitudes there are in AB equal unto C, so many magnitudes are there in DE equal unto F: and by the same reason how many there are in BG equal unto C, so many also are there in EH equal unto F. Wherefore how many there are in the whole AG equal unto C, so many are there in the whole DH equal unto F. Wherefore how multiplex AG is unto C, so multiplex is DH unto F. Wherefore the first and the fifth composed together, namely, AG is equemultiplex unto the second C, as the third and the sixth composed together, namely, DH, is to the fourth F. If therefore the first be equemultiplex to the second as the third is to the fourth, and if the fifth also be equemultiplex to the second as the sixth is to the fourth: then shall the first & the fifth composed together, be equemultiplex to the second, as the third and the sixth composed togethers is to the fourth: which was required to be proved.   

The 3. Theorem. The 3. Proposition. If the first be equemultiplex to the second, as the third is to the fourth, and if there be taken equemultiplices to the first & to the third: they shall be equemultiplices to them which were first taken, the one to the second, the other to the fourth.  
SVppose that there be four magnitudes, of which let A be the first, B the second, C the third, and D the fourth. And let the first, A, be equemultiplex to the second, B, as the third, C, is to the fourth, D. 
Construction.  And unto A and C take equemultiplices, which let be EF and GH, so that how multiplex EF is to A, so multiplex let HG be to C. Then I say, that EF is equemultiplex unto B, as GH is unto D. For forasmuch as EF is equemultiplex unto A, as GH is unto C, therefore how many magnitudes there are in EF equal unto A, so many magnitudes also are there in GH equal unto C. Let EF be divided into the magnitudes that are equal unto A, that is, into EK and KF. And likewise GH into the magnitudes equal unto C, that is, into GL and L H. Now then the multitude of these magnitudes EK and KF, 
Demonstration.  is equal unto the multitude of these magnitudes GL and LH. And forasmuch as A is equemultiplex to B, as C is to D: but EK is equal unto A, and GL unto C, therefore EK is equemultiplex unto B, as GL is unto D. And by the same reason KF is equemultiplex unto B, as L H is to D. Now then there are six magnitudes whereof EK is the first: B the second: GL the third: D the fourth: KF the fifth: LH the sixth. And forasmuch as the first EK is equemultiplex to the second B, as the third GL is to the fourth D: and the fift KF is equemultiplex to the second B, as the sixth LH is to the fourth D: therefore (by the second of the fift) the first is the fift composed together, namely, EF is equemultiplex unto the second B, as the third and the sixth composed together, namely, GH is to the fourth D. If therefore the first be equemultiplex to the second, as the third is to the fourth; and if there be taken the equemultiplices to the first and to the third, they shall be equemultiplices to them which were first taken, the one to the second, than other to the fourth: which was required to be proved.   

The 4. Theorem. The 4. Proposition. If the first be unto the second in the same proportion that the third is to the fourth: then also the equemultiplices of the first and of the third, unto the equemultiplices of the second and of the fourth, according to any mnltiplication, shall have the same proportion being compared together.  
SVppose that there be four magnitudes, of which let A be the first, ● the second, C the third, and D the fourth. And let the first A be unto the second B in the fame proportion, that the third C is unto the fourth D. And to A and C take equemultiplices E, and F, and likewise to B and D, 
Construction.  any other equimultiplices G & H. Then I say that as E is to G, so is F to H. Unto E & F take equemultiplices K & L, & unto G and H any other equemultiplices, that is, M and N. And forasmuch as E is equemultiplex unto A as F is unto C, 
Demonstration.  and unto E and F be taken the equemultiplices K & L, therefore (by the 3. of the fifth) K is equemultiplex to A, as L is to C: and by the same reason also M is equemultiplex unto B, as N is to D. And seeing that as A is to B, so is C to D, and of A and C are taken equemultiplices K and L, and likewise of B & D are taken other equemultiplices, namely, M and N, therefore if K exceed M, L also exceedeth N: and if it be equal, it is equal, and if it be lesse●●t is less (by the converse of the 6. definition of the fifth). And K and L be equemultiplices to E and F: and M and N are other equemultiplices to G and H. Wherefore as E is to G, so is F to H by the said sixth definition. If therefore the first be unto the second in the same proportion that the third is to the fourth: then also the equemultiplice● of the first and of the third, unto the equemultiplices of 〈◊〉 second & of the fourth, according to any multiplication, shall have the same proportion being compared together: which was required to 〈◊〉 proved.  
An Assumpt. 
ALemmae, or an assumpt.  Wherefore seeing it hath been proved that if K exceed M, L also exceedeth N, and if it be equal it is equally and if it be less, it is less: it is manifest that if M exceed K, N also exceedeth L: and if it be equal it is equal: and if it be less it is less: and by this reason as G is to E, so is H to F.  

A Corollary.  
Hereby it is manifest that if there be four magnitudes proportional, they shall also by conversion be proportional: that is, if the first be unto the second, 
A Corollary. Converse proportion.  as the thide is to the fourth: then by conversion as the second is to the first, so is the fourth to the third.    

The 5. Theorem. The 5. Proposition. If a magnitude be equemultiplex to a magnitude, as a part taken away of the one, is to a part taken away from the other: the residue also of the one, to the residue of the other, shall be equemultiplex, as the whole is to the whole.  
SVppose that the whole magnitude AB be unto the whole magnitude CD equemultiplex, as the part taken away of the one, namely, AE, is to the part taken away of the other, namely, CF. Then I say that the residuen of the one, namely, EB, is to the residue of the other, namely, DF equemultiplex as the whole AB is to the whole CD. How multiplex AE is to CF, so multiplex make EB to CG. 
Construction.  And forasmuch as (by the first of the fifth) AE is to GF equemultiplex, as AB is to GF: 
Demonstration.  but AE is to CF equemultiplex, as AB is to CD. Wherefore AB is equemultiplex to either of these GF and CD. Wherefore GF is equal unto CD. Take away CF which is common to them both. Wherefore that which remaineth namely, GC, is equal unto that which remaineth namely, DF. And forasmuch as AE is to CF equemultiplex as EB is to GC, but GC is equal unto DE, therefore AE is to CF equemultiplex as EB is to FD. But AE is put to be equemultiplex to CF, as AB is to CD, wherefore EB is to FD equemultiplex, as AB is to CD. Wherefore the residue EB is to the residue FD equemultiplex, as the whole AB is to the whole CD. If therefore a magnitude be equemultiplex to a magnitude, as a part taken away of the one is to a part taken away of the other: the residue of the one also to the residue of the other, shallbe equemultiplex as the whole is to the whole: which was required to be proved.   

The 6. Theorem. The 6. Proposition. If two magnitudes be ●quemultiplices to two magnitudes, & any par●es taken away of them also, be aequemultiplices to the same magnitudes: the residues also of them shall unto the same magnitudes be either equal, or equemultiplices.  
SVppose that there be two magnitudes AB and CD equemultiplices to two magnitudes E and F, and let the parts taken away of the magnitudes AB and CD, namely, AG and CH be equemultiplices to the same magnitudes E and F. 
Two cases in this Propotion.  Then I say that the residues GB and HD, are unto the self same magnitudes E and F either equal, or else equemultiplices.  
And in like sort may we prove, that if GB be multiplex to E, HD also shall be so multiplex unto F. If therefore there be two magnitudes equemultiplices to two magnitudes, 
The second.  and any parts taken away of them be also equemultiplices to the same magnitudes: the residues also of them shall unto the same magnitudes be either equal, or equemultiplices: which was required to be proved.   

The 7. Theorem. The 7. Proposition. Equal magnitudes have to one & the self same magnitude, one and the same proportion. And one and the same magnitude hath to equal magnitudes one and the self same proportion.  
SVppose that A and B be equal magnitudes, and take any other magnitude, namely, C. Then I say, that either of these A and B have unto C one and the same proportion: and that C also hath to either of these A and B one and the same proportion.  
I say moreover, that C hath to either of these A and B one and the same proportion. 
The second part demonstrated.  For the same order of construction remaining, we may in like sort prove, that D is equal unto E, & there is taken an other multiplex to C, namely, F. Wherefore if F exceed D, it also exceedeth E: and if it be equal it is equal: and if it be less it is less. But F is multiplex to C: and D & E are other equemultiplices to A and B. Wherefore as C is to A, so is C to B. Wherefore equal magnitudes have to one and the same magnitude, one and the same proportion: and one and the same magnitude hath to equal magnitudes one and the self same proportion: which was required to be demonstrated.   

The 8. Theorem. The 8. Proposition. Unequal magnitudes being taken, the greater hath to one and the same magnitude a greater proportion than hath the less. And that one and the same magnitude hath to the less a greater proportion than it hath to the greater.  
SVppose that AB and C be unequal magnitudes, of which let AB be the greater, and C the less. And let there be an other magnitude whatsoever, namely, D. Then I say that AB hath unto D a greater proportion than hath C to D: and also that D hath to C a greater proportion, than it hath to AB. For forasmuch as AB is greater than C, let there be taken a magnitude equal unto C, namely, BE.  
Now than the less of these two magnitudes AE and EB being multiplied will at the length be greater than D. 
The first part of this Proposition demonstrated.   

The second part of the proposition demonstrated.  I say moreover that D hath to C a greater proportion than D hath to AB. For the same order of construction still remaining, we may in like sort prove that N is greater than K, and that it is not greater than FH. And N is multiplex to D, and FH and K are certain other equemultiplices to AB and C. Wherefore D hath to C a greater proportion than D hath to AB.  

¶ For that Orontius seemeth to demonstrate this more plainly therefore I thought it not amiss here to set it.  
Suppose that there be two unequal magnitudes, of which le● A● b● the gre●ter, and C the less: and let there be a certain other magnitude, namely, D. Then I say first, that AB hath to D a greater proportion than hath C to D. For forasmuch as by supposition AB is greater than the magnitude C: therefore the magnitude AB containeth the same magnitude C, and an other magnitude beside. Let E● be equal unto C: and let AE be the part remaining of the same magnitude. 
First differ●c● of the first part.  Now AE and EB are either unequal or equal the one to the other. First let them be unequally and le● AE be less than EB. And unto AE the less take any multiplex whatso●u●r, so that it be greater than the magnitude D: and let the same be FG. And how multiplex FG is to AE, so multiplex let GH be to E● ● and K to C. Again take the duple of D ● which let be L, and then the triple, and le● the same be M. And so forward, always adding one: until there be produced su●h a multiplex to D which shall be n●xt greater than GH (that is, which amongst the 〈◊〉 o● D ● by the continual addition of one, doth first begin to exceade GH) and let th●●●me be N ● which let ●e quadruple to D. Now then the multiplex GH is the next multiplex less than N ● and therefore ●s not less than M, that is, is either equal unto it or greater than it. 
Demonstrati● of t●e same first difference.  And forasmuch ●s FG is equemultiplex to AE, as GH is to E●, therefore how multiplex F● is to AE, so multiplex is FH to AB (by the first of the fift). But how multiplex FG is to AE, so multiplex is K to C, therefore how multiplex FH is to AB, so multiplex is K to C. Moreover forasmuch as GH and K are equemultiplices unto EB and C ● and EB is by construction equal unto C, therefore (by the common sentence) GH is equal unto K. But GH is not less than M, as hath before been showed, and FG ● was put to be greater than D. Wherefore the whole FH is greater than these two D and M. But D and M are equal unto N. For N is quadruple to D. And M being triple to D, doth together with D make quadruple unto D. Wherefore FH is greater than N. Farther, K is proved to be equal to GH. Wherefore K is less than N. But FH and K are equemultiplices unto AB and C, unto the first magnitude, I say, and the third: and N is a certain other multiplex unto D, which representeth the second & the fourth magnitude. And the multiplex of the first exceedeth the multiplex of the second: but the multiplex of the third exceedeth not the multiplex of the fourth. Wherefore AB the first hath unto D the second a greater proportion, then hath C the third to D the fourth (by the 8. definition of this book).  
But if AE be greater than EB, let EB the less be multiplied until there be produced a multiplex greater than the magnitude D: which let be GH. And how multiplex GH is to EB, 
Second difference.  so multiplex let FG be to AE, and K also to C. Then take unto D such a multiplex as is next greater than FG: and again let the same be N, which let be quadruple to D. And in like sort as before may we prove, that the whole FH is unto AB equemultiplex as GH is to EB: and also that FH & K are equemultiplices unto AB ● & C: and finally that GH is equal unto K. And forasmuch as the multiplex N is next greater than FG: therefore FG is not less than M. But GH is greater than D by construction. Wherefore the whole FH is greater than D and M: and so consequently is greater than N. But K exceedeth not N: for K is equal to GH: for how multiplex K is to EB the less, so multiplex is FG to A● the greater. B●t those magnitudes which are equemultiplice● unto unequal magnitud●s are according to the same proportion unequal. Wherefore K is less than FG, and therefore i● much less than N. Wherefore again the multiplex of the first exceedeth the multiplex of the second: but the multiplex of the third exceedeth not th● multiplex of the fourth. Wherefore (by the 8. definition of the fift) A● the first, hath to D the second, a greater proportion, then hath C the third to D the fourth.  

Third difference.  But now if AE be equal unto EB, either of them shall be equal unto C. Wherefore unto either of thos● three magnitudes take equemultiplices greater than D. So that let FG be multiplex to AE, and GH unto EB, and K again to C: which (by the 6. common sentence) shall be equal the one to the other. Let N also be multiplex to D, and be next greater than every one of them, namely, let it be q●adrupl● to D. This construction finished, we may again prove that FH and K are equemultiplices to AB and C: and that FH the multiplex of th● first magnitude exceedeth N the multiplex of the second magnitude●●nd tha● K t●●●ultiplex of the third exceedeth not the multiplex of the fourth. Wherefore we may conclude that AB hath unto D a greater proportion, then hath C to D.  
Now also I say, that the self same magnitude D hath unto the less magnitude C a greater proportion, than it hath to the greater AB. 
The second part ●f this proposition.  And this may plainly be gathered by the foresaid discourse, without changing the order of the magnitudes & of the equemultiplices. For seeing that every way it is before proved, that FH exceedeth N, and K is exceeded of the self same N: therefore conversedly N exceedeth K, but doth not exceed FH. But N is multiplex to D that is to the first and third magnitude: and K is multiplex to the second, namely, to C ● and FH is multiplex to the fourth, namely, to AB ● Wherefore the multiplex of the first exceedeth the multiplex of the second: but the multiplex of the third exceedeth not the multiplex of the fourth. Wherefore (by the 8. definition of this fift book) D the first hath unto C the second a greater proportion, then hath D the third to AB the fourth: which was required to be proved.    

The 9 Theorem. The 9 Proposition. Magnitudes which have to one and the same magnitude one and the same proportion: are equal the one to the other. And those magnitudes unto whom one and the same magnitude hath one and the same proportion: are also equal.  
SVppose that either of these two magnitudes A and B have to C one and the same proportion. 
The first par● of this Proposition demonstrated.  Then I say that A is equal unto B. For if it be not, then either of these A and B should not have to C one & the same proportion (by the 8. of the fifth) but by supposition they have, wherefore A is equal unto B.  
Again, 
The second part proved.  suppose that the magnitude C have to either of these magnitudes A and B one and the same proportion. Then I say that A is equal unto B. For if it be not, C should not have to either of these A and B one and the same proportion (by the former proposition): but by supposition it hath, wherefore A is equal unto B. Wherefore magnitudes which have to one and the same magnitude one and the same proportion, are equal the one to the other. And thos● magnitudes unto whom one and the same magnitude hath one and the same proportion, are also equal: which was required to be proved.   

The 10. Theorem. The 10. Proposition. Of magnitudes compared to one and the same magnitude, that which hath the greater proportion, is the greater. And that magnitude whereunto one and the same magnitude hath the greater proportion, is the less.  
SVppose that A have to C a greater proportion than B hath to C. Then I say that A is greater than B. 
The first part of this proposition proved.  For if it be not, then either A is equal unto B or less than it. But A cannot be equal unto B, for then either of these A and B should have unto C one and the same proportion (by the 7 of the fifth): but by supposition they have not, wherefore A is not equal unto B. Neither also is A less than B, for than should A have to C a less proportion, then hath B to C (by the 8. of the fifth) but by supposition it hath not. Wherefore A is not less than B. And it is also proved that it is not equal, wherefore A is greater than B.  

The second part demonstrated.  Again suppose that C have to B a greater proportion than C hath to A. Then I say that B is less then A. For if it be not, then is it either equal unto it or else greater, but B cannot be equal unto A, for than should C have to either of these A and B one and the same proportion (by the 7. of the fifth) but by supposition it hath not, wherefore B is not equal unto A. Neither also is B greater than A, for than should C have to B a less proportion than it hath to A (by the 8. of the fifth) but by supposition it hath not: wherefore B is not greater than A. And it was proved that it is not equal unto A, wherefore B is less than A. Wherefore of magnitudes compared to one and the same magnitude, that which hath the greater proportion, is the greater. And that magnitude whereunto one and the same magnitude hath the greater proportion, is the less. Which was required to be proved.   

The 11. Theorem. The 11. Proposition. Proportions which are one and the self same to any one proportion, are also the self same the one to the other.  
SVpppose that as A is to B, so is C to D, and as C is to D, so is E to F. Then I say that as A is to B, so is E to F. 
Construction.  Take equemultiplices to A, C and E, which let be G, H, K. And likewise to B, D and F take any other equemultiplices, which let be L, M, and N. And because as A is to B, so is C to D: and to A and G are taken equemultiplices G & H; & to B and D are take certain other equemultiplices L & M. 
Demonstration●  If therefore G exceed L, then also H exceedeth M, and if it be equal it is equal, and if it be less it is less (by the converse of the 6● definition of the fifth). Again because that as C is to D, so is E to F: and to C and E are taken ●●●em●ltiplices H ●●d K: and likewise to D & F are taken certain other equemultiplices M & N. If therefore H exceed M, than also K exceedeth N: and if it be equal, it is equal, and if it be less, it is less (by the same converse) But if K exceed M, than also G exceedeth L: and if it be equal it is equal, and if it be less, it is less (by the same converse) Wherefore if G exceed L, then K also exceedeth N, and if it be equal it is equal, and if it be less, it is less. But G & K are equemultiplices of A & E. And L & N are certain other equemultiplices of B & F. Wherefore (by the 6. definition) as A is to B, so is E to F. Proportions therefore which are one and the self same to any one proportion, are also the self same one to the other: which was required to be proved.   

The 12. Theorem. The 12. Proposition. If there be a number of magnitudes how many soe●●r proportional: as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes.  
SVppose that there be a number of magnitudes how many soever, namely, A, B, C, D, E, F, in proportion: so that as A is to B, so let C be to D, and E to F. Then I say, that as A is to B, so 〈◊〉 A, C, E, to B, D, F. Take equemultiplices to A, C, and E. 
Constr●ction.  And let the same be G, H, K. And likewise to B, D, and F, ●ake any other equemultiplices, which to be L, M, N. And because that 〈◊〉 A is to B, so i● C to D, and E to F. 
Demonstration●  And to A, C, E, are taken ●quemultiplices G, H, K: and likewise to ●, D, F, are taken certain other equem●●tipli●●s L, M, N. If therefore G exceed L, H also exceedeth M, and KN: and if it be equal, it is equal, and if it be less, it is less (●y the converse of the six● definition of the fift). Wherefore if G exceed L, then G, H, K, also exceed L, M, N: and if they be equal, they are equal: and if they be less, they are less (by the same). But G, and G, H, K, are equemultiplices to the magnitude A, and to the magnitudes A, C, E. For (by the first of the fift) (if there be a number of magnitudes equemultiplices to a like number of magnitudes each to each, how multiplex one magnitude, is to one, so multiplices are all the magnitudes, to all). And by the same reason also L, and L, M, N, are equemultiplices to the magnitude B, and to the magnitudes B, D, F: Wherefore as A is to B, so is A, C, E, to B, D, F (by the sixth definition of the fift). If therefore there be a number of magnitudes how many soever, proportional: as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes: which was required to be proved.   

The 13. Theorem. The 13. Proposition. If the first have unto the second the self same proportion that the third hath to the fourth, and if the third have unto the fourth a greater proportion than the fifth hath to the sixth: them shall the first also have unto the second a greater proportion than hath the fifth to the sixth.  
SVppose that there be six magnitudes, of which let A be the first, B the second, C the third, D the fourth, E the fifth, and F the sixth. Suppose that A the first have unto B the second, the self same proportion that C the third hath to D the fourth. And let C the third have unto D the fourth, a greater proportion than hath E the fifth to F the sixth. Then I say that A the first hath to B the second a greater proportion, then hath E the fifth to F the sixth. 
Construction.  For forasmuch as C hath to D a greater proportion than hath E to F, therefore there are certain equemultiplices to C and E, and likewise any other equemultiplices whatsoever to D and F, which being compared together, the multiplex to C shall exceed the multiplex to D, but the multiplex to E shall not exceed the multiplex to F (by the converse of the eight definition of this book). Let those multiplices be taken, and suppose that the equemultiplices to C and E, be G and H: and likewise to D and F take any other equemultiplices whatsoever, and let the same be K and L, so that let G exceed K but let not H exceed L. And how multiplex G is to C, so multiplex let M be to A. And how multiplex K is to D, so multiplex also let N be to B. And because that as A is to B, so is C to D: and to A and C are taken equemultiplices M and G. 
Demonstration.  And likewise to B and D are taken certain other equemultiplices N & K: if therefore M exceed N, G also exceedeth K: and if it be equal it is equal, and if it be less it is less (by the conversion of the sixth definition of the fifth.) But by construction G excedet● K, wherefore M also exceedeth N, but H exceedeth not L. But M & H are equemultiplices to A & E: and N & L are certain other equemultiplices● whatsoever, to B and F. Wherefore A hath unto B a greater proportion than E hath to F (by the 8. definition.) If therefore the first have unto the second the self same proportion that the third hath to the fourth, and if the third have unto the fourth a greater proportion than the fifth hath to the sixth, then shall the firs● also have unto the second a greater proportion than hath the 〈◊〉 to the sixths' Which was required to be proved.  

¶ An addition of Campane.  
If there be four quantities, 
An addition of Campane  and if the first have unto the second a greater proportion than hath the third to the fourth: then shall there be some equemultiplices of the first and the third, which being compared to some equemultiplices of the second and the fourth, the multiplex of the first shall be greater than the multiplex of the second, but the multiplex of the third shall not be greater than the multiplex of the fourth.  
Although this proposition here put by Campane needeth no demonstration for that it is but the converse of the 8. definition of this book, yet thought I it not worthy to be omitted, for that it reacheth the way to find out such equemultiplices, that the multiplex of the first shall exceed the multiplex of the second, but the multiplex of the third shall not exceed the multiplex of the fourth.    

The 14. Theorem. The 14. Proposition. If the first have unto the second the self same proportion that the third hath unto the fourth: and if the first be greater than the third, the second also is greater than the fourth: and if it be equal it is equal: and if it be less it is less.  
SVppose that there be four magnitudes, of which let A be the first, B the second, C the third, and D the fourth: and let A the first, have unto B the second, the self same proportion that C the third hath unto D the fourth. And let A be greater than HUNDRED Then I say, that B also is greater than D. For forasmuch as A is greater than C, and there is a certain other magnitude, 
Demonstration.  namely, B, therefore (by the 8. of the fift) A hath unto B a greater proportion than C hath to B. And as A is to B, so is C to D. Wherefore C also hath unto D a greater proportion than C hath to B. But that magnitude whereunto one and the same magnitude hath the greater proportion, is the less (by the 10. of the fift). Wherefore D is less than B, & therefore B is greater than D. And in like sort may we prove, that if A be equal unto C, D shall also be equal unto D: and if A be less than C, B shall also be less then D. If therefore the first have unto the second the self same proportion that the third hath unto the fourth, and if the first be greater than the third, the second also is greater than the fourth, and if it be equal it is equal, and if it be less it is less: which was required to be proved.   

The 15. Theorem. The 15. Proposition. Like parts of multiplices, and also their multiplices compared together, have one and the same proportion.  
SVppose that AB be equemultiplex to C, as DE is to F. Then I say, that as C is to F, so is AB to DE. For forasmuch as how multiplex AB is to C, 
Construction.  so multiplex is DE to F: therefore how many magnitudes there are in AB equal unto C, so many are there in DE equal unto F. Divide AB into the magnitudes equal unto C, that is, into AG, GH, and HB: and likewise DE into the magnitudes equal unto E, that is, into DK KL, and LE. Now then the multitude of these AG, 
Demonstration.  GH, and HB, is equal to the multitude of these DK KL, and LE. And forasmuch as AG, GH, and HB, are equal the one to the other and likewise DK, KL, and LE, are also equal the one to the other: therefore as AG is to DK, so is GH to KL, and HB to LE. Wherefore (by the 12. of the fift) as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherefore as AG is to DK, so is AB to DE. But AG is equal unto C, and likewise DK to F. Wherefore as C is to F, so is AB to DE. Like parts therefore of multiplices and also their multiplices compared together, have one and the same proportion that their equemultiplices have: which was required to be demonstrated.   

The 16. Theorem. The 16. Proposition. If four magnitudes be proportional: then alternately also they are proportional.  
SVppose that there be four magnitudes proportional, namely, A, B, C, D, so that as A is to B, so let C be to D. Then I say, that alternately also they shall be in proportion, 
Demonstration of alternate proportion.  that is, as A is to C, so is B to D. Take equemultiplices unto A & B, and let the same be E & F. And likewise to C and D take any other equemultiplices what soever, 
Construction.  and let the same be G and H. 
Demonstration.  And forasmuch as how multiplex E is to A, so multiplex is F to B, but like parts of multiplices & also their multiplices have one and the self same proportion the one to the other (by the former Proposition). Wherefore as A is to B, so is E to F. But as A is to B, so is C to D● wherefore (by the 11. of the fift) as C is to D, so is E to F. Again, forasmuch as G and H are equemultiplices to C and D, but like parts of multiplices and also their multiplices have the one to the other one and the self same proportion (by the 15. of the fift). Wherefore as C is to D, so is G to H. But as C is to D, so is E to F. Wherefore as E is to F, so is G to H (by the 11. of the fift). But if there be four magnitudes in proportion, and if the first be greater than the third, the second also is greater than the fourth: and if it be equal it is equally and if it be less it is less (by the 14. of the fift). If therefore E exceed G, F shall also exceed H: and if it be equal, it is equal: and if it be less, it is less. But E and F are equemultiplices to A and B: and G and H are certain other equemultiplices to C & D. Wherefore (by the 6. definition of the fift) as A is to C, so is B to D. If therefore there be four magnitudes proportional, then alternately also they are proportional: which was required to be proved.   

The 17. Theorem. The 17. Proposition. If magnitudes composed be proportional, then also divided they shall be proportional.  
SVppose that the magnitudes composed being proportional, be AB, BE, CD, DF, so that as AB is to BE, so is CD to DF. Then I say, 
Demonstration of proportion by division.  that divided also they shall be proportional, as AE is to BE, so is CF to DF. Take equemultiplices unto AE, EB, CF, FD, and let the same be GH, HK, LM, and MN. 
Constructions  And likewise to EB, and FD, take any other equemultiplices what soever, and let the same be KO, and NP. And forasmuch as how multiplex GH is to AE, 
Demonstration.  so multiplex is KH to EB, therefore how multiplex GH is to AE, so multiplex is GK to AB (by the first of the fifth). But how multiplex GH is to AE, so multiplex is LM to CF: Wherefore how multiplex GK is to AB, so multiplex is LM to CF (by the 11. of the same). Again forasmuch as how multiplex LM is to CF, so multiplex is MN to DF, therefore how multiplex LM is to CF, so multiplex is LN to CD (by the first of the self same). But how multiplex LM is to CF so multiplex is GK to AB. Wherefore how multiplex GK is to AB, so multiplex is LN to CD. Wherefore GK and LN, are equemultiplices to● AB & CD. Again forasmuch as how multiplex HK the first is to EB the second, so multiplex is MN the third to FD the fourth. And how multiplex KO the fift is to EB the second, so multiplex is NP the sixth to FD the fourth. Wherefore (by the second of the same) how multiplex HO composed of the first and fift is to EB, so multiplex is MP composed of the third and sixth to FD. And for that as AB is to BE, so is CD to DF: and to AB & CD are taken equemultiplices GK and LN, and likewise to EB and FD are taken certain other equemultiplices, that is, HO, and MP. If therefore GK exceed HO, then LN also exceedeth MP: and if it be equal, it is equal: and if it be less, it is less (by the conversion of the sixth definition of the fift). Let GK exceed HO: Wherefore KH common to them both being taken away, the residue GH, shall exceed the residue KO. But if GK exceed HO, then doth LN exceed MP: Wherefore let LN exceed MP: and MN which is common to the both being taken away, the residue LM shall exceed the residue NP. Wherefore if GH exceed KO, then shall LM exceed NP. And in like sort may we prove, that if GH be equal unto KO, than LM shall be equal unto NP: and if it be less, it shall be less: but GH and LM are equemultiplices to AE and GF: and likewise KO and NP are certain other equemultiplices to EB, and FD. Wherefore as AE is to EB, so ●CF to FD (by the six● definition of the fift). If composed magnitudes therefore be proportional, then also divided they shall be proportional: which was required to be demonstrated.   

The 18. Theorem. The 18. Proposition. If magnitudes divided be proportional: then also composed they shall be proportional.  
SVppose that the magnitudes divided being proportional, be AE, EB, CF, & FD, so that as AE is to EB, so let CF be to FD. 
Demonstration of proportion by composition.  Then I say, that composed also they shall be proportional, that is, as AB is to BE, so is CD to DF. For if AB be not unto BE, as CD is to FD, then shall AB be unto BE, 
This proposition is the converse of the former.  as CD is either unto a magnitude less than FD, on unto a magnitude greater. Let it first be unto a less, namely, to DG. 
Demonstratiō●e●aing to an impossibility.  And forasmuch as, as AB is to BE, so is CD to DG: the composed magnitudes therefore are proportional, wherefore divided also they shall be proportional (by the 17. of the first). Wherefore as AE is to EB, so is CG to GD. But by supposition as AE is to EB, so is CF to FD. Wherefore (by the 11. of the fift) as CG is to GD, so is CF to FD. Now then there are four magnitudes, CG, GD, CF, and FD: of which the first CG is greater than the third CF. Wherefore (by the 14. of the fift) the second GD is greater than the fourth FD. But it is also put to be less than it: which is impossible. Wherefore it can not be that as AB is to BE, so is CD to a magnitude less than FD. In like sort may we prove, that it can not be so to a magnitude greater than FD. For by the same order of demonstration, it would follow that FD is greater than the said greater magnitude: which is impossible. Wherefore it must be to the self same. If therefore magnitudes divided be proportional, then also composed they shall be proportional: which was required to be proved.   

The 19 Theorem. The 19 Proposition. If the whole be to the whole, as the part taken away is to the part taken away: then shall the residue be unto the residue, as the whole is to the whole.  
SVppose that as the whole AB is to the whole CD, 
That which the fift of this book proved only touching multiplices, this proveth generally of all magnitudes.  so is the part taken away AE to the part taken away CF. Then I say, that the residue EB shall be unto the residue FD, as the whole AB is to the whole CD. For for that as the whole AB is to the whole CD, so is AE to CE, therefore alternately also (by the 16. of the fift) as AB is to AE, so is CD to CF. A●d for that when magnitudes composed are proportional, the same divided also are proportional (by the 17. of the fift) ● therefore as BE is to EA, so is DF to FC, Wherefore alternately also (by the 16. of the fift) as BE is to DF, so is EA to FC. But as AE is to CF, so (by supposition) is the whole AB to the whole CD. Wherefore the residue EB shall be unto the residue FD, as the whole AB is to the whole CD. If therefore the whole be to the whole, as the part taken away is to the part taken away, then shall the residue be unto the residue; as the whole is to the whole: which was required to be proved.  

¶ ALemma or Assumpt.  
And forasmuch as by supposition as AB is to CD, 
ALemma.  so is AE to CF: and alternately as AB is to AE, so is CD to CF. And now it is proved, that as, AB is to CD, so is EB to FD. Wherefore again alternately, as AB is to ●B, so is CD to FD. Wherefore it followeth, that as AB is to AE, so is CD to CF: and again, as the same AB is to EB, so is the same CD to DF.   

Corollary.  
And hereby it is manifest, 
A Corollary.  that if magnitudes composed be proportional, then also by conversion of Proportion (which of some is called Proportion by Eversion, 
Conversion of proportion.  and which is, as before it was defined, when the antecedent is compared to the excess, wherein the antecedent exceedeth the consequent) they shall be proportional.    

The 20. Theorem. The 20. Proposition. If there be three magnitudes in one order, and as many other magnitudes in an other order, which being taken two and two in each order, are in one and the same proportion, and if of equality in the first order the first be greater than the third, then in the second order the first also shall be greater than the third: and if it be equal, it shall be equal: and if it be less, it shall be less.  
SVppose that there be three magnitudes in one order, 
This proposition pertaineth to Proportion of equality inordinate proportionality.  namely, A, B, C, & let there be as many magnitudes in an other order, which let be D, E, F, which being taken two and two in each order, let be in one and the same proportion, that is, as A is to B, so let D be to E, and as B is to C, so let E be to F.  
But now if A be equal unto C, D also shall be equal unto F. For then A and C have unto B one and the same proportion (by the first part of the seventh of this book). 
The second difference.  And for that as A is to B, so is D to E, and as C is to B, so is F to E: therefore D and F have unto E one and the same proportion. Wherefore by the first part of the 9: of this book D is equal unto F.  
But now suppose that A be less then C. Then also shall D be less then F. For by the 8. of this book C shall have unto B a greater proportion than hath A to B. 
The third difference.  But as A is to B, so is D to E by supposition, and as C is to B, so have we proved is F to E. Wherefore F hath unto E a greater proportion than hath D to E. Wherefore by the first part of the 10. of this book F is greater than D. If therefore there be three magnitudes in one order, and as many other magn●tudes in an other order, which being taken two and two in each order, are in one and the same proportion, and if of equality in the first order the first be greater than the third, then in the second order also the first shall be greater than the third and if it be equal, it shall be equal: and if it be less, it shall be less: which was required to be proved.   

The 21. Theorem. The 21. Proposition. If there be three magnitudes in one order, and as many other magnitudes in an other order, which being taken two and two in each order are in one and the same proportion, and their proportion is perturbate: if of equality in the first order the first be greater than the third, than in the second order the first also shall be greater than the third: and if it be equal it shall be equal: and if it be less it shall be less.  
SVppose that there be three magnitudes in one order, 
Th●r proposition pertaineth to Proportion of equality in perturbate proportionality.  namely, A, B, C, and let there be as many other magnitudes in an other, which let be D, E, F: which being taken two & two in each order, let be in one and the same proportion, and let their proportion be perturbate. So that as A is to B, so let E be to F, & as B is to C, so let D be to E, and of equality let A be greater than G. Then I say, that D also is greater th●n E: and if it be equal it is equal: and if it be less it is less.  

The third difference.  Likewise, if A be less than C, D also is less than F. For then C shall have unto B a greater proportion, then hath A to B (by the 8. of the fift. Wherefore E also ●ath unto D a greater proportion than it ●ath to F. Wherefore by the second part of the 10. of this book D is less than F. If therefore there be three magnitudes in one order, & as many other magnitudes in an other order, which being taken two and two in each order are in one & the same proportion, & their proportion is perturbate, and if of equality in the first order the first be greater than the third, then in the second order the first also shall be greater than the third● and if it be equal it shall be equal: and if it be less it shall be less: which was required to be proved.   

The 22. Theorem. The 22. Proposition. If there be a number of magnitudes, how many soever in one order, and as many other magnitudes in an other order, which being taken two and two in each order are in one and the same proportion, they shall also of equality be in one and the same proportion.  

Proportion of equality in ordinate proportionality.  SVppose that there be a certain number of magnitudes in one order. As for example: A, B, C, and let there be as many other magnitudes in an other order, which let be D, E, F, which being taken two and two let be in one and the same proportion. So that as A is to B, so let D be to E, and as B is to C, so let E be to F. Then I say, that of equality they shall be in the same proportion, that is, as A is to C● so is D to F. 
Construction.  Take unto A and D equemultiplices G & H, and likewise to B & E take any other equemultiplices whatsoever, namely, K and L● and moreover unto C and F take any other equemultiplices also what soever, namely, M and N. And forasmuch as, 
Demonstration.  as A is to B, so is D to E: and unto A and D are taken equemultiplices G and H: and likewise unto B and E are taken certain other equem●ltiplices K and L: therefore (by the 4. of the fift) as G is to K● so is H to L. And (by the same reason) as K is to M, so is L to N. seeing therefore that there be in order three magnitudes G, K, M, & as many other magnitudes in an other order, namely, H, L, N, which being compared two to two are in one and the same proportion, therefore of equality (by the 20. of the fift) if N exceed M, then shall H exceed G: and if it be equal it shall be equal: and if it be less it shall be less. But G and H are equemultiplices unto A and D, and M and N are certain other equemultiplices unto C and F: therefore (by the 6. definition of the fift) as A is to C, so is D to F.  
So also if there be more magnitudes than three in ●ither order, 
When there are more than three magnitudes in either order.  the first of the one order shall be to the last, as the first of the other order is to the last. As if there were four in one order, namely, ABCD, and other four in the other order, namely, EFGH, 
A●CDE●GH  we may with three magnitudes A, B, C, and E, F, G, prove that as A is to C, so is E to G: And then leaving out in either order the second and taking the fourth, as leaving out B and F, and taking D and H, we may prove by these three and three A, C, D, and E, G, H, that as A is to D, so is E to H. And observing this order, this demonstration will serve how many soever the magnitudes be in either order. If therefore there be a number of magnitudes how many soever in one order, and as many other magnitudes in an other order, which being taken two and two in each order are in one and the same proportion, they shall also of equality be in one and the same proportion: which was required to be demonstrated.   

The 23. Theorem. The 23. Proposition. If there be three magnitudes in one order, and as many other magnitudes in an other order, which being taken two & two in each order are in one and the same proportion, and if also their proportion be perturbate: then of equality they shall be in one and the same proportion.  

Proportion of equality in perturbate proprotionalitie.  SVppose that there be in one order three magnitudes, namely, A, B, C, & let be taken in an other order as many other magnituds, which let be D, E, F which being taken two and two, in each order let be in one and the same proportion: and suppose that their proportion be perturbate. So that as A is to B, so let E be F, and as B is to C, so let D be to E. Then I say that as A is to C so is D to F. Take unto A, B, D, 
Construction.  equemultiplices, and let the same be GHK: and likewise unto C, E, F, take any other equemultiplices whatsoever, and let the same be LMN. And forasmuch as G and H are equemultiplices unto A and B, but the parts of equemultiplices are in the same proportion that their equemultiplices are (by the 15. of the fift) wherefore as A is to B, so is G to H. 
Demonstration.  And by the same reason also as E is to F, so is M to N. But as A is to B, so is E to F. Wherefore (by the 11. of the fift) as G is to H, so is M to N. And forasmuch as, as B is to C, so is D to E, and unto B & D are taken equemultiplices H & K: and likewise unto C and E are taken certain other equemultiplices L and M: therefore (by the 4. of the fifth) as H is to L, so is K to M, and alternately also (by the 16. of the fift) as B is to D, so is C to E. And forasmuch as H and K are the equemultiplices of B and D, but the parts of equemultiplices are in the same proportion that their equemultiplices are (by the 15. of the fift) therefore as B is to D, so is H to K. But as B is to D, so is C to E, therefore (by the 11. of the first) as H is to K, so is C to E. Again forasmuch as L and M are the equemultiplices of C and E, therefore as C is to E, so is L to M. But as C is to E, so is H to K: therefore as H it to K, so is L to M, and alternately (by the 16. of the fift) as H is to L, so is K to M. *  
(But as A is to B so is E to F (by supposition) wherefore as G is to H, so is E F, (by the 11. of the fift.) Again for as much as M and N are equemultiplices unto E a●d F, therefore again (by the 15. of the fifth) as ● is to ●, so ●● M to N. But as E is to F, so have we proved is G to H, wherefore as G is to H, so is M to N (by the 11. of the fift) And for that ●● ● is to ●, so is D to E (by supposition). And unto B and D are taken equemultiplices H and K: and unto C and E are taken certain other equemultiplices L and M, therefore as H is to L so is K to M, by the 4 of this book).  
But it is proved that as G is to H, so is M to N. Seeing therefore that there are in one orde three● magnitudes, namely, G, H, L, and as many other magnitudes in an other order, namely, K, M, N, which being taken two and two in each order, are in one and the same proportion, and their proportion is perturbate, therefore of equality (by the 21. of the fifth) if G exceed L, then shall K exceed N, and if it be equal, it shall be equal: and if it be less it shall be less. But G and K are equemultiplices unto A and D, and L and N are certain other equemultiplices unto C and F. Wherefore as A is to C, so is D to F (by the 6. definition of the fifth). If therefore there be three magnitudes in one order, and as many other magnitudes in an other order, which being taken two and two in each order, are in one and the same proportion, and if also their proportion be perturbate: then of equality they shall be in one and the same proportion: which was required to be proved.  
From this mark * first to the same mark again, you may if you will in stead of The●●● arguments which seem somewhat intricate, read those arguments following printed with an other letter, which are very perspicuous and brief, and followed of the most interpreters.  
And even as by the demonstration in three magnitudes is taken the proof in four magnitudes by leaving out one of 
Note.  the means: so by the demonstration in four magnitudes is taken the proof in five magnitudes by leaving out two of the means: and by the demonstration in five, the proof in six, by leaving out three means. And so forward continually, which is also to be understanded in the former kind of proportion of equality, which is in ordinate proportion.   

The 24. Theorem. The 24. Proposition. If the first have unto the second the same proportion that the third hath to the fourth, and if the fift have unto the second the same proportion that the sixth hath to the fourth: then also the first and fift composed together shall have unto the second the same proportion that the third and sixth composed together have unto the fourth.  
SVppose that there be six magnitudes AB, C, DE, F, BG, & EH: of which let AB be the first, C the second, DE the third, F the fourth, BG the fift, and EH the sixth. And suppose that AB the first, have unto C the second, the same proportion that DE the third hath to F the fourth, 
That which the second proposition of this book proved only touching multiplices, is here proved generally touching magnitudes.  and let BG the fift have unto C the second the same proportion that EH the sixth hath unto F the fourth. Then I say, that the first and fift composed together, namely, AG, hath unto C the second the same proportion, that the third and sixth composed together, namely, DH, hath unto F the fourth. For, for that as BG is to C, so is EH to F: then also by conversion (by the Corollary of the 4. of the fift) as C is to BG, so is F to EH. And for that as AB is to C, so is DE to F, but as C is to GB, so is F to EH: therefore of equality (by the 22. of the fift) as AB is to BG, so is DE to EH. And forasmuch as when magnitudes divided are proportional, they also composed are proportional (by the 18. of the fift): therefore as AG is to GB, so is DH to HE: but as BG is to C, so is EH to F: Wherefore again of equality (by the 22. of the fift) as AG is to C, so is DH to F. If therefore the first have unto the second the same proportion that the third hath to the fourth, and if the fift have unto the second the same proportion that the sixth hath to the fourth: then also the first and fift composed together shall have unto the second the same proportion that the third and sixth composed together have unto the fourth: which was required to be proved.   

The 25. Theorem. The 25. Proposition. If there be four magnitudes proportional: the greatest and the lest of them, shall be greater than the other remaining.  
SVppose that there be four magnitudes proportional AB, CD, E and F. So that as AB is to CD, so let E be to F. And let the greatest of them be AB, & the jest of them be F. Then I say, that these two magnitudes AB and F, are greater than the two magnitudes CD & E. Forasmuch as AB is supposed to be the greatest of all four, therefore it is greater than E. Therefore from the greater AB cut of (by the 3. of the first) unto E an equal magnitude AG. and likewise (by the same) from CD cut of unto F an equal magnitude CH. (Which may be done, for that the magnitude CD is greater than the magnitude E: for that as AB is to CD, so is E to F, therefore alternately as AB is to E, so is CD to F (by the 16. of the fift). But AB is greater than E: Wherefore also CD is greater than F: Which thing may also be proved by the 14. of the same.) Now for that as AB is to CD, so is E to F: but E is equal unto AG, and F is equal unto CH: therefore as AB is to CD, so is AG to CH: and forasmuch, as the whole AB is to the whole CD, so is the part taken away AG, to the part taken away CH: therefore the residue GB (by the 1●. of the fift) is unto the residue HD, as the whole AB is to the whole CD. But AB the first is greater than CD the third: Wherefore GB the second is greater than HD the fourth (by the 14. of the fift). And forasmuch as AG is equal unto E, & CH is equal unto F: therefore AG and F are equal unto CH and E. And forasmuch as if unto things unequal be added things equal, all shall be unequal (by the fourth common sentence): therefore seeing that GB and DH are unequal, and GB is the greater, if unto GB be added AG and F: and likewise if unto HD be added CH & E, there shall be produced AB and F greater than CD & E. If therefore there be four magnitudes proportional, the greatest and the lest of them, shall be greater than the other remaining: which was required to be demonstrated.  
Here follow certain propositions added by Campane, which are not to be contemned, and are cited even of the best learned, namely, of johannes Regio montanus, in the Epitome which he writeth upon Ptolemy.  

¶ The first Proposition.  
If there be four quantities, and if the proportion of the first to the second, be greater than the proportion of the third to the fourth: then contrariwise by conversion, the proportion of the second to the first, shall be less than the proportion of the fourth to the third.  

another demonstration of the same affirmatively.  It may also be demonstrated directly. For let E be unto B as C is to D. Then conversedly B is to E as D is to C. And forasmuch as A is greater than E by the first part of the tenth of this book, therefore by the second part of the 8 of the same B hath unto A a less proportion than hath B to E. Wherefore by the 13. of the same B hath unto A ● less proportion than hath D to C: which was required to be proved.   

¶ The second Proposition.  
If there be four quantities, and if the proportion of the first to the second be greater than the proportion of the third to the fourth, then alternately the proportion of the first to the third, shall be greater than the proportion of the second to the fourth.  

another demonstration of the same affirmatively.  This may also be demonstrated affirmatively, let E be unto B as C is to D. Now then by the first part of the tenth of this book, E is less than A: wherefore by the first part of the 8. of the same, the proportion of A to C is greater than the proportion of E to C. But alternately E is to C as B is to D. Wherefore (by the 13. of the same) A hath to C a 〈…〉   

¶ The third Proposition.  
If there be four quantities, and if the proportion of the first t● the second be greater, than the proportion of the third to the fourth: then by composition also the proportion of thee, f●●th and second to the second, shall be greater than the proportion of the third and fourth● to the fourth.  
This may also be demonstrated affirmatively. 
another demonstration of the same.  Forasmuch as the proportion of A to B is greater than the proportion of C to D: let E be unto B as C is to D. And so by the first part of the 10. of this book, E shall be less than A. And therefore by the common sentence EB shall be less than AB. Wherefore by the first part of the 8. of the same AB hath unto B a greater proportion than hath EB to B. But by composition EB is to B as CD is to D. For by supposition E is unto B as is to D. Wherefore (by the 12. of this book) AB hath to B a greater proportion than hath CD to D: which was required to be proved.   

¶ The fourth Proposition.  
If there be four quantities, and if the proportion of the first and the second to the second be greater than the proportion of the third and fourth to the fourth: then by division also the proportion of the first to the second, shall be greater th●n the proportion of the third to the fourth.  
Suppose that the proportion of AB to B be greater than the proportion of CD to D. 
Demonstration leading to an impossibility.  Then I say that by division also the proportion of A to B is greater than the proportion of C to D. For it cannot be the same. For then by composition AB should be to B as CD is to D. Neither also can it be less: for if the proportion of C to D be greater than the proportion of A to B, then by the former proposition, the proportion of CD to D should be greater than the proportion of AB to B: which is contrary also to the supposition. Wherefore the proportion of A to B is neither one and the same with the proportion of C to D, 〈…〉 it: Wherefore it is greater than it: which was required to be proved.  

another demonstration of the same affirmatively.  The same may also be proved affirmatively. Suppose that EB be unto B as CD is to D. Now then (by the first part of the 10. of the fifth) EB shall be less than AB: and therefore by the common sentence, E is less than A, wherefore by the first part of the 8. of this book, the proportion of E to B, is less than the proportion of A to B, but as E is to B, so is C to D: wherefore the proportion of C to D, is less than the proportion of A to B. Wherefore the proportion of A to B is greater than the proportion of C to D: which was required to be proved.   

¶ The fifth Proposition.  
If there be four quantities, and if the proportion of the first and the second to the second be greater than the proportion of the third and the fourth to the fourth: then by eversion the proportion of the first and second to the first, shall be less than the proportion of the third and fourth to the third.  
Suppose that the proportion of AB to B be greater than the proportion of CD to D. Then I say that by eversion the proportion of AB to A is less than the proportion of CD to HUNDRED 
Demonstration.  For by division by the former proposition the proportion of A to B is greater than the proportion of C to D. Wherefore by the first of these propositions conversedly, B hath unto A a less proportion than hath D to C. Wherefore by the 3. of the same by composition, the proportion of AB to A is less than the proportion of CD to C: which was required to be proved.   

¶ The sixth Proposition.  
If there be taken three quantities in one order, and as many in an other order, and if the proportion of the first to the second in the first order, be greater than the proportion of the first to the second in the latter order: then also the proportion of the first to the third in the first order, shall be greater than the proportion of the first to the third in the latter order.  
Suppose that there be three quantities in one order A, B, C, & as many other quantities in an other order D, E, F. And let the proportion of A to B in the first order be greater than the proportion of D to E in the second order, and let also the proportion of B to C in the first order, be greater than the proportion of E to F in the second order. Then I say that the proportion of A to C in the first order, is greater than the proportion of D to F in the second order. 
Demonstration●  For let G be unto C as E is to F. Now then by the first part of the 10 of this book G shall be less then B. And therefore by the second part of the 8. of the ●●me, the proportion of A to G i● greater th●n the proportion of ● to ●. Wherefore the proportion of A to G is much greater th●n the proportion of D to E. Now then let ● be 〈…〉 D is to E. Wherefore by the first part of the 100LS of the same, A is greater then H. And therefore by the first part of the 8. of the same, the proportion of A to C is greater than the proportion of H to C. But by proportion of equality H is to C as D is to F (for H is to G as D i● to E, and G is to C as E is to F. Wherefore by the 12. of the same A hath to C a greater proportion than hath D to F: which was required to be proved.   

¶ The seventh Proposition.  
If there be taken three quantities in one order, and as many other in an other order, and if the proportion of the second to the third in the first order be greater than the proportion of the first to the second in the latter order, if also the proportion of the first to the second in the first order be greater than the proportion of the second to the third in the latter order: then shall the proportion of the first to the third in the first order be greater, than the proportion of the first to the third in the latter order.  
Suppose that there be three quantities in one order A, B, C, and as many other in an other order D, E, F. And let the proportion of B to C in the first order, be greater than the proportion of D to E in the second order, and let also the proportion of A to B in the first order, be greater than the proportion of E to F in the second order. Then I say that A hath to C a greater proportion than hath D to F. This pertaineth to proportion of equality. For let G be unto C, as D is to E. And by the first part of the 10. of this boke● G shall be less than ●. And therefore by the second part of the 8. of the same, the proportion of A to G is greater than the proportion of A to B. Wherefore A hath unto G a much greater proportion than hath ● to F. Now then let H be unto G as E is to F. And by the first part of the 10. of the same, A shallbe greater than H. And by the first part of the 8. of the same, the proportion of A to C is greater than the proportion of H to C. But by the 23. of the same the proportion of H to C is as the proportion of D to F (for G is to C as D is to E, and H is to G as E is to F.) Wherefore (by the 12. of the same) the proportion of A to C is greater than the proportion of D to F, which was required to be proved.   

¶ The eight Proposition.  
If the proportion of the whole to the whole, be greater th●n the proportion of a part taken away, to a part taken away: them shall the proportion of the residue unto the residue be greater than the proportion of the whole to the whole.  
Suppose that there be two quantities AB & C D: from which let there be cut of these magnitudes AE and CF: and let the residue be EB and FD. And let the proportion of AB to CD be greater than the proportion of AE to CF. Then I say that the proportion of EB to FD is greater than the proportion of AB to CD. For (by the second of these propositions now added) alternately the proportion of A● to A● 〈◊〉 greater than the proportion of CD to CF. And therefore by eversion of proportion (by the 5. of the same) the proportion of AB to E● is less than the proportion of CD to FD. Wherefore again alternately the proportion of AB to CD is less than the proportion of EB to FD: which was required to be proved.   

¶ The ninth Proposition.  
If quantities how many soever in one order be compared to as many other in an other order, and if there be a greater proportion of every one that goeth before to that whereunto it is referred, then of any that followeth to that whereunto it is referred: the proportion of them all taken together unto all the other taken together, shall be greater, than the proportion of any that followeth to that whereunto it is compared, and also then the proportion of all them taken together to all the other taken together, but shall be less than the proportion of the first to the first.  
Suppose that there be three quantities in one order, A, B, C, & as many other in an other order D, E, F. And let the proportion of A to D be greater than the proportion of B to E, let also the proportion of B to E, be greater than the proportion of C to F. Then I say that the proportion of ABC taken all together, to DEF taken altogether, is greater than the proportion of B to E, and also then the proportion of C to F, & more over them the proportion of B & C taken together, to EF taken together, but is less than the proportion of A to D: 
Demonstration.  For forasmuch as A hath to D a greater proportion than hath B to E, therefore alternately A hath to B a greater proportion than hath D to E: wherefore by composition AB hath to B a greater proportion than hath DE to E. And again alternately AB hath to DE a greater proportion than hath B to E. Wherefore by the former proposition A hath to ● a greater proportion than hath AB to DE. And by the same reason may it be proved that hath to E a greater proportion than hath BC to EF. Wherefore A hath to D a greater proportion than hath BC to EF. Wherefore alternately A hath to BC a greater proportion than hath D to EF, wherefore by composition ABC hath to BC a greater proportion than hath DEF to EF. Wherefore again alternately ABC hath to DEF a greater proportion than hath BC to EF. Wherefore (by the former proposition) the proportion of A to D is greater than the proportion of ABC to DEF: Which was required to be proved.    
The end of the fifth book of Euclides Elements.   

¶ The sixth book of Euclides Elements.  
THIS SIXTH BOOK, is for use and practise, 
The argument of this sixth book.  a most special book. In it are taught the proportions of one figure to an other figure, & of their sides the one to the other, and of the sides of one to the sides of an other, likewise of the angles of the one to the angles of the other. Moreover it teacheth the description of figures like to ●igures given, and marvelous applications of figures to lines, evenly, or with decrease or excess, with many other Theorems, not only of the propo●tions of right lined figures, but also of sectors of circles, with their angles. On the Theorems and Problems of this Book depend for the most part, the compositions of all instruments of measuring length, breadth, or deepens, and also the reason of the use of the same instruments, as of the Geometrical ●quar●, 
This book necessary for the use of instruments of Geometry.  the Scale of the Astrolabe, the quadrant, the staff, and such other. The use of which instruments, besides all other mechanical instruments of raising up, of moving, and drawing huge things incredible to the ignorant, and infinite other gins (which likewise have their grounds out of this Book) are of wonderful and unspeakable profit, besides the inestimable pleasure which is in them.  

Definitions.  

1. Like rectiline figures are such, 
The first definition.  whose angles are equal the one to the other, and whose sides about the equal angles are proportional.  
As if ye take any two rectiline figures. As for example, two triangles ABC, and DEF: 〈…〉 of the one triangle be equal to the angles of the other, namely, if the angle A be equal to the angle D, and the angle B equal to the angle E, & also the angle C equal to the angle F. And moreover, i● the sides which contain the equal angles be proportional. As if the side AB have that proportion to the side BC, wh●ch the side DE hath to the side EF, and also if the side BC be unto the side CA, as ●he side EF is to the side FD, and moreover if the side CA be to the side AB, as the side FD is to the side DE, then are these two triangles said to be like: and so judge ye of any other kind of figures. As if in the parallelograms ABCD and EFGH, the angle A be equal to the angle E, and the angle B equal to the angle F, and the angle C equal to the angle G, and the angle D equal to the angle H. And furthermore, if the side AC have that proportion to the side CD which the side EG hath to the side GH, and if also the side CD be to the side DB as the side GH is to the side HF, and moreover, if the side DB be to the side BA as the side HF is to the side FE, and finally, if the side BA be to the side AC as the side FE is to the side EG, then are these parallelograms like.   


The second definition.  2. Reciprocal figures are those, when the terme● of proportion are both antecedentes and consequentes in either figure.  
As if ye have two parallelograms ABCD and EFGH. If the side AB to the side EF, an antecedent of the first figure to a consequent of the second figure, have mutually the same proportion, which the side EG hath to the side AC an antecedent of the second figure to a consequent of the first figure: then are these two figures Reciprocal. They are called of some, figures of mutual sides, and that undoubtedly not amiss nor unaptly. 
Reciprocal figures called mutual figures.  And to make this definition more plain, Campane and Pestitarius, and others● thus put it: Reciprocal figures, are when the sides of other 〈◊〉 mutually proportional, as in the example and declaration before given. Among the barbarous they are called Mutekesia, reserving still the Arabike word.   


The third definition.  3. A right line is said to be divided by an extreme and mean proportion, when the whole is to the greater part, as the greater part is to the less.  
As if the line AB, be so divided in the point C, that the whole line AB have the same proportion, to the greater part thereof, namely, to AC, which the same greater part AC hath to the less part thereof, namely, to CB, then is the line AB divided by an extreme and mean proportion. Commonly it is called a line divided by proportion ha●ing a mean and two extremes. How to divide a line in such sort was taught in the 11. Proposition of the second Book, but not under this form of proportion.   

4. The alitude of a figure is a perpendicular line drawn from the top to the base. 
The fourth definition.   
As the altitude or height of the triangle ABC, is the line AD being drawn perpendicularly from the point A, being the top or highest part of the triangle to the base thereof BC. So likewise in other figures as ye see in the examples here set. That which here ●ee calleth the altitude or height of a figure, in the first book in the 35. Proposition and certain other following, he taught to be contained within two equidistant lines: so that figures to have one altitude and to be contained within two equidistant lines, is all one. So in all these examples, if from the highest point of the figure ye draw an equidistant line to the base thereof, and then from that point draw a perpendicular to the same base that perpendicular is the altitude of the figure.   

5. A Proportion is said to be made of two proportions or more, when the quantities of the proportions multiplied the one into the other, produce an other quantity. 
The fifth definition.   
another example, where the greater inequality and the less inequality are mixed together 6. 4. 2. 3. the denomination of the proportion of 6. to 4, is 1 ●/●, 
another example.  of 4. to 2, is ●/●, and of 2. to 3, is ●/●: now if ye multiply as you aught, all these denominations together, ye shall produce 12. to 6, namely, dupla proportion.  
Forasmuch as so much hath hitherto been spoken of addition of proportions it shall not be unnecessary somewhat also to say of substraction of them. 
Of substraction of proportion.  Where it is to be noted, that as addition of them, is made by multiplication of their denominations the one into the other: so is the substraction of the one from the other done, by division of the denomination of the one by the denomination of the other. As if ye will from sextupla proportion subtrahe dupla proportion, take the denominations of them both. The denomination of sextupla proportion, is 6, the denomination of dupla proportion, is 2. Now divide 6. the denomination of the one by 2. the denomination of the other: the quotient shall be 3: which is the denomination of a new proportion, namely, tripla: so that when dupla proportion is subtrahed from sextupla, there shall remain tripla proportion. And thus may ye do in all others.   

6. A parallelogram applied to a right line, is said to want in form by a parallelogram like to one given: when the parallelogram applied wanteth to the filling of the whole line, by a parallelogram like to one given: 
The sixth definition.  and then is it said to exceed, when it exceedeth the line by a parallelogram like to that which was given.  
As let E be a Parallelogramme given, and let AB be a right line, to whom is applied the parallelogram ACDF. Now if it want of the filling of the line AB, by the parallelogram DFGB being like to the parallelogram given E, then is the parallelogram said to want in form by a parallelogram like unto a parallelogram given.  
Likewise if it exceed, as the parallelogram ACGD applied to the lin● AB● if it exceed it by the parallelogram FGBD being like to the parallelogram F which was given, then is the parallelogram ABGD, said to exceed in form by a parallelogram like to a parallelogram given.  
This definition is added by Flussates as it seemeth, it is not in any common Greek book abroad, nor in any Commentary. It is for many Theorems following very necessary.    

The 1. Theorem. The 1. Proposition. Triangles & parallelograms which are under one & the self same altitude: are in proportion as the base of the one is to the base of the other.  
And forasmuch as the lines CB, BG, and GH are equal the one to the other, 
Demonstration of the first part.  therefore the triangles also AHG, AGB and ABC, are (by the 38. of the first) equal the one to the other. Wherefore how multiplex the base HC is to the base BC, so multiplex also is the triangle AHC to the triangle ABC. And by the same reason also, how multiplex the base LC is to the base DC, so multiplex also is the triangle ALC to the triangle ADC. Wherefore if the base HC be equal unto the base CL, then (by the 38. of the first) the triangle AHC is equal unto the triangle ACL. And if the base HC exceed the base CL, then also the triangle AHC exceedeth the triangle ACL, and if the base be less, the triangle also shall be less. Now then there are four magnitudes, namely, the two bas●s BC and CD, and the two triangles ABC, and ACD, and to the base BC, and to the triangle ABC, namely, to the first and the third, are taken equemul●iplices, namely, the base HC and the triangle AHC, and likewise to the base CD and to the triangle ADC, namely, to the second and the fourth, are taken certain other equemultiplices, that is, the base CL, and the triangle ALC. And it hath been proved that if the multiplex of the first magnitude, that is, the base HC, do exceed the multiplex of the second, that is, the base CL, the multiplex also of the third, that is, the triangle AHC exceedeth the multiplex of the fourth● that is, the triangle ALC, and if the said base HC be equal to the said ba●● CL, the triangle also AHC is equal to the triangle ALC, and if it be less it i● less. Wherefore by the sixth definition of the fifth, as the first of the foresaid magnitudes is to the second, so is the third to the fourth. Wherefore as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.  
And because (by the 41. of the first) the parallelogram EC is double to the triangle ABC, 
Demonstration of the second part.  and (by the same) the parallelogram FC is double to the triangle ACD, therefore the parallelograms EC and FC are equemultiplices unto the triangles ABC and ACD. But the parts of equemultiplices (by the 15. of the fifth) have one and the same proportion with thei● equemultiplices. Wherefore as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram FC. And forasmuch as it hath been demonstrated, that as the base BC is to the base CD, so is the triangle ABC, to the triangle ACD, and as the triangle ABC is to the triangle ACD so is the parallelogram EC to the parallelogram FC. Wherefore (by the 11. of the fifth) as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. The parallelograms may also be demonstrated a part by themselves as the triangles are, if we describe upon the bases BG, GH, and DK & KL parallelograms under the self same altitude that the parallelogramme● given are. Wherefore triangles and parallelograms which are under one and the self same altitude, are in proportion, as the base of the one is to the base of the other: which was required to be demonstrated.  

Here Flussates addeth this Corollary.  
If two right lines being given, the one of them be divided how so ever: 
A Corollary added by Flussates.  the rectangle figures contained under the whole line undivided, and each of the segments of the line divided, are in proportion the one to the other, as the segments are the one to the other. For imagining the figures BA and AD in the former description, to be rectangled, the rectangle figures contained under the whole right line AC, and the segments of the right line BD, which is cu● in the point C, namely, the parallelograms BA and AD, are in proportion the one to the other, as the segments BC and CD are.    

The 2. Theorem. The 2. Proposition. If to any one of the sides of a triangle be drawn a parallel right line, it shall cut the sides of the same triangle proportionally. And if the sides of a triangle be cut proportionally, a right line drawn from section to section is a parallel to the other side of the triangle.  
SVppose that there be a triangle ABC, unto one of the sides whereof, namely, unto BC, let there be drawn a parallel line DE cutting the sides AC and AB in the points E and D. Then I say first that as BD is to DA, so is CE to EA. 
The first part of this Theorem.  Draw a line from B to E, & also from C to D. Wherefore (by the 37. of the first) the triangle BDE is equal unto the triangle CDE: for they are set upon one and the same base DE, and are contained within the self same parallels DE and BC. Consider also a certain other triangle ADE. Now things equal (by the 7. of the fifth) have to one self thing one and the same proportion. Wherefore as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. But as the triangle BDE is to the triangle ADE, so is the base BD to the base DA (by the first of this book.) For they are under one and the self same top, namely, E, and therefore are under one and the same altitude. And by the same reason as the triangle CDE is to the triangle ADE, so is the line CE to the line EA. Wherefore (by the 11. of the fifth) as the line BD is to the line DA, so is the line CE to the line EA.  
But now suppose that in the triangle ABC the sides AB & AC be cut proportionally so that as BD is to DA, so let CE be to EA, & draw a line from D to E. Then secondly I say that the line DE is a parallel to the line BC. 
Demonstration of the second part.  For the same order of construction being kept, for that as BD is to DA, so is CE to EA, but as BD is to DA, so is the triangle BDE to the triangle ADE (by the 1. of the sixth) & as CE is to EA, so (by the same) is the triangle CDE to the triangle ADE: therefore (by the 11. of the fifth) as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. Wherefore either of these triangles BDE and CDE have to the triangle ADE one and the same proportion. Wherefore (by the 9 of the fifth) the triangle BDE is equal unto the triangle CDE, and they are upon one and the self base, namely, DE. But triangles equal and set upon one base, are also contained within the same parallel lines (by the 39 of the first.) Wherefore the line DE is unto the line BC a parallel. If therefore to any one of the sides of a triangle be drawn a parallel line, it cutteth the other sides of the same triangle proportionally. And if the sides of a triangle be cut proportionally, a right line drawn from section to section, is parallel to the other side of the triangle: which thing was required to be demonstrated.  

¶ Here also Flussates addeth a Corollary.  
If a line parallel to one of the sides of a triangle do cut the triangle, it shall cut of from the whole triangle a triangle like to the whole triangle. 
A Corollary added by Flussates.  For as it hath been proved it divideth the sides proportionally. So that as EC is to EA, so is BD to DA, wherefore by the 18. of the fifth, as AC is to AE, so is AB to AD. Wherefore alternately by the 16. of the fifth as AC is to AB, so is AE to AD: wherefore in the two triangles EAD and CAB the sides about the common angle A are proportional. The said triangles also are equiangle. For forasmuch as the right lines AEC and ADB do fall upon the parallel lines ED and CB, therefore by the 29. of the firs● they make the angles AED and ADE in the triangle ADE equal to the angles ACB and ABC in the triangle ACB. Wherefore by the first definition of this book the whole triangle ABC is like unto the triangle cut of ADE.    

The 3. Theorem. The 3. Proposition. If an angle of a triangle be divided into two equal parts, and if the right line which divideth the angle divide also the base: the segments of the base shall be in the same proportion the one to the other, that the other sides of the triangle are. And if the segments of the base be in the same proportion that the other sides of the said triangle are: a right drawn from the top of the triangle unto the section, shall divide the angle of the triangle into two equal parts.  
SVppose that there be a triangle ABC, and (by the 9 of the first) let the angle BAC be divided into two equal parts by the right line AD, which let cut also the base BC in the point D. Then I say that as the segment BD is to the segment DC, so is the side BA to the side AC. 
Construction.  For by the point C (by the 31. of the first) draw unto the line DA a parallel line CE and extend the line BA till it concur with the line CE in the point E, and do make the triangle BEC. 
Demonstration of the first part.  But the line BA shall concur with the line CE (by the 5. petition) for that the angles EBC, and BCE are less than two right angles. For the angle ECB is equal to the outward and opposite angle ADB (by the 29. of the first.) And the two angles ADB and DBA of the triangle BAD are less than two right angles (by the 17. of the first) Now forasmuch as upon the parallels AD and EC falleth the right line AC, therefore by the 29. of the first) the angle ACE is equal unto the angle CAD. But unto the angle CAD is the angle BAD supposed to be equal. Wherefore the angle BAD is also equal unto the angle ACE. Again because upon the parallels AD and EC falleth the right line BAE, the outward angle BAD (by the 28. of the first) is equal unto the inward angle AEC. But before it was provell that the angle ACE is equal unto the angle BAD, wherefore the angle ACE is equal unto the angle AEC. Wherefore (by the 6. of the first) the side AE is equal unto the side AC. And because to one of the sides of the triangle BCE, namely, to EC is drawn a parallel line AD, therefore (by the 2. of the sixth) as BD is to DC, so is BA to AE. But AE is equal unto AC, therefore as BD is to DC, so is BA to AC. But now suppose that as the segment BD is to the segment DC, 
Demonstration of the second part, which is the converse of the first.  so is the side BA to the side AC, & draw a line from A to D. Then I say that the angle BAC is by the right line AD divided into two equal parts. For the same order of construction remaining, for that as BD is to DC, so is BA to AC, but as BD is to DC, so is BA to AE (by the 2. of the sixth) for unto one of the sides of the triangle BCE, namely, unto the side EC is drawn a parallel line AD. Wherefore also as BA is to AC, so is BA to AE (by the 11. of the fifth) Wherefore (by the 9 of the fifth) AC is equal unto AE. Wherefore also (by the 5. of the first) the angle AEC is equal unto the angle ACE, but the angle AEC (by the 29. of the first) is equal unto the outward angle BAD: and the angle ACE is equal unto the angle CAD which is alternate unto him: wherefore the angle BAD is equal unto the angle CAD. Wherefore the angle BAC is by the right line AD divided into two equal parts. Wherefore if an angle of a triangle be divided into two equal parts, and if the right line which divideth the angle cut also the base, the segments of the base shall be in the same proportion the one to the other, that the other sides of the said triangle are. And if the segments of the base be in the same proportion that the other sides of the said triangle are, a right line drawn from the top of the triangle unto the section divideth the angle of the triangle into two equal parts.  
This construction is the half part of that Gnomical figure described in the 43. proposition of the first book, which Gnomical figure is of great use in a manner in all Geometrical demonstrations.   

The 4. Theorem. The 4. Proposition. In equiangle triangles, the sides which contain the equal angles are proportional, and the sides which are subtended under the equal angles are of like proportion.  
SVppose that there be two equiangle triangles ABC and DCE: and let the angle ABC of the one triangle, be equal unto the angle DCE of the other triangle, and the angle BAC equal unto the angle CDE, and moreover, the angle ACB equal unto the angle DEC. Then I say, that those sides of the triangles ABC, & DCE, which include the equal angles, are proportional, and the side which are subtended under the equal angles are of like proportion. 
Construction.  For let two sides of the said triangles, namely, two of those sides which are subtended under equal angles: as for example the sides BC and CE, be so set that they both make one right line. And because the angles ABC & ACB are less than two right angles (by the 17. of the first): but the angle ACB is equal unto the angle DEC: therefore the angles ABC & DEC are less than two right angles. Wherefore the lines BA & ED being produced, will at the length meet together. Let them meet and join together in the point F. 
Demonstration.  And because by supposition the angle DCE is equal unto the angle ABC, therefore the line BF is (by the 28. of the first) a parallel unto t●e line CD. And forasmuch as by supposition the angle ACB is equal unto the angle DEC, therefore again (by the 28. of the first) the line AC is a parallel unto the line FE. Wherefore FADC is a parallelogram. Wherefore the side FA is equal unto the side DC: and the side AC unto the side FD (by the 34. of the first). And because unto one of the sides of the triangle BFE, namely, to FE is drawn a parallel line AC, therefore as BA is to AF, so is BC to CE (by the 2. of the sixth). But AF is equal unto CD. Wherefore (by the 11. of the fift) as BA is to CD, so is BC to CE, which are sides subtended under equal angles. Wherefore alternately (by the 16. of the fift) as AB is to BC, so is DC to CE. Again forasmuch as CD is a parallel unto BF, therefore again (by the 2. of the sixth) as BC is to CE, so is FD to DE. But FD is equal unto AC. Wherefore as BC is, to CE, so is AC to DE, which are also sides subtended under equal angles. Wherefore alternately (by the 16. of the fift) ●s BC is to CA, so is CE to ED● Wherefore forasmuch as it hath been demonstrated, that as AB is unto BE so is DC unto CE● but as DC is unto CA, so is CE unto ED● it followeth of equality (by the 22. of the fift) that ●s BA is unto AC so is CD unto DE● Wherefore in equiangle triangle●, the sides which include the equal angles are proportional: and the sides which are subtended under the equal angles are of like proportion ●hich was required to be demonstrated.   

The 5. Theorem. The 5. Proposition. If two triangles have their sides proportional, the triang●●s are equiangle, and those angles in them are equal, under which are subtended sides of like proportion.  
SVppose that there be two triangles ABC, & DEF, having their sides proportional, as AB is to BC, so let DE be to EF: 
This is the converse of the former proposition.  & as BC is to AC, so let EF be to DF: and moreover, as BA is to AC, so let ED be to DF. Then I say, that the triangle ABC is equiangle unto the triangle DEF: and those angles in them are equal under which are subtended sides of like proportion, that is, the angle ABC is equal unto the angle DEF: and the angle BCA unto the angle EFD: and moreover, the angle BAC to the angle EDF. Upon the right line EF, 
Construction  and unto the points in it E & F, describe (by the 23. of the first) angles equal unto the angles ABC & ACB, which let be FEG and EFG, namely, let the angle FEG be equal unto the angle ABC, and let the angle EFG be equal to the angle ACB. 
Demonstration.  And forasmuch as the angles ABC and ACB are less than two right angles (by the 17. of the first): therefore also the angles FEG and EFG are less than two right angles. Wherefore (by the 5. petition of the first) the right lines EG & FG shall at the length concur. Let them concur in the point G. Wherefore EFG is a triangle. Wherefore the angle remaining BAC is equal unto the angle remaining EGF (by the first Corollary of the 32. of the first). Wherefore the triangle ABC is equiangle unto the triangle GEF. Wherefore in the triangles ABC and EGF the sides, which include the equal angles (by the 4. of the sixth) are proportional, and the sides which are subtended under the equal angles are of like proportion. Wherefore as AB is to BC, so is GE to EF. But as AB is to BC, so by supposition is DE to EF. Wherefore as DE is to EF, so is GE to EF (by the 11. of the fift). Wherefore either of these DE and EG have to EF one and the same proportion. Wherefore (by the 9 of the fift) DE is equal unto EG. And by the same reason also DF is equal unto FG. Now forasmuch as DE is equal to EG and EF is common unto them both, therefore these two sides DE & EF, are equal unto these two sides GE and EF, and the base DF is equal unto the base FG. Wherefore the angle DEF (by the 8. of the first) is equal unto the angle GEF: and the triangle DEF (by the 4. of the first) is equal unto the triangle GEF: and the rest of the angles of the one triangle are equal unto the rest of the angles of the other triangle the one to the other, under which are subtended equal sides. Wherefore the angle DFE is equal unto the angle GFE: and the angle EDF unto the angle EGF. And because● the angle FED is equal unto the angle GEF: but the angle GEF is equal unto the angle ABC: therefore the angle ABC is also equal unto the angle FED. And by the same reason the angle ACB is equal unto the angle DFE● and moreover, the angle BAC unto the angle EDF. Wherefore the triangle ABC is equiangle unto the triangle DEF. If two triangles therefore have their sides proportional, the triangles shall be equiangle, & those angles in them shall be equal, under which are subtended sides of like proportion: which was required to be demonstrated.   

The 6. Theorem. The 6. Proposition. If there be two triangles whereof the one hath one angle equal to one angle of the other, & the sides including the equal angles be proportional: the triangles shall be equiangle, and those angles in them shall be equal, under which are subtended sides of like proportion.  
SVppose that there be two triangles ABC, and DEF, which let have the angle BAC of the one triangle equal unto the angle EDF of the other triangle, and let the sides including the equal angles be proportional, that is, as BA is to AC, so let ED be to DF. Then I say, that the triangle ABC is equiangle unto the triangle DEF: and the angle ABC is equal unto the angle DEF, and the angle ACB equal unto the angle DFE, which ●ngles are subtended to sides of like proportion. Unto the right line DF, and to the point in it D (by the 23. of the first) describe unto either of the angles BAC and EDF, 
Construction.  an equal angle FDG. And unto the right line DF, and unto the point in it F (by the same) describe unto the angle AC● an equal angle DFG. And forasmuch as the two angles BAC and ACB, are (by the 17. of the first) less than two right angles: therefore also the two angles edge and DFG, are less than two right angles. Wherefore the lines DG & FG being produced, shall concur (by the 5. petition). Let them concur in the point G. Wherefore DFG is a triangle. Wherefore the angle remaining ABC is equal unto the angle remaining DGF (by the 32. of the first). Wherefore the triangle ABC is equiangle unto the triangle DGF. Wherefore as BA is in proportion to AC, so is GD to DF (by the 4. of the sixth). But it is supposed, that as BA is to AC, so is ED to DF. Wherefore (by the 11. of the fift) as ED is to DF, so is GD to DF. Wherefore (by the 9 of the fift) ED is equal unto DG. And DF is common unto them both. Now then there are two sides ED and DF equal unto two sides GD and DF: and the angle EDF (by supposition) is equal unto the angle GDF. Wherefore (by the 4. of the first) the base EF is equal unto the base GF, and the triangle DEF is (by the same) equal unto the triangle GDF, and the other angles remaining in them are equal the one to the other, under which are subtended equal sides. Wherefore the angle DFG is equal unto the angle DFE: and the angle DGF unto the angle DEF. But the angle DFG is (by construction) equal unto the angle ACB. And the angle DGF is as it hath been proved, equal to the angle ABC. Wherefore also the angle ACB is equal unto the angle DFE. And the angle ABC is equal to the angle DEF. But by supposition the angle BAC is equal unto the angle EDF. Wherefore the triangle ABC is equiangle unto the triangle DEF. If therefore there be two triangles, whereof the one hath one angle equal to one angle of the other, and if also the sides including the equal angles be proportional: then shall the triangles also be equiangle, and those angles in them shallbe equal, under which are subtended sides of like proportion: which was required to be proved.   

The 7. Theorem. The 7. Proposition. If there be two triangles, whereof the one hath one angle equal to one angle of the other, and the sides which include the other angles, be proportional, and if either of the other angles remaining be either less or not less than a right angle: than shall the triangles be equiangle, and those angles in them shall be equal, which are contained under the sides proportional.  
SVppose that there be two triangles ABC and DEF, which let have one angle of the one, equal to one angle of the other, namely, the angle BAC equal unto the angle EDF. And let the sides which include the other angles, namely, the angles ABC and DEF be proportional, so that as AB is to BC, so let DE be to EF. And let the other angles remaining, namely, ACD and DFE be first either of them less than a right angle. Then I say that the triangle ABC is equiangle unto the triangle DEF. And that the angle ABC is equal unto the angle DEF, 
The first part of this proposition.  namely, the angles which are contained under the sides proportional, and that the angle remaining, namely, the angle C is equal unto the angle remaining, namely, to the angle F. For first the angle ABC is either equal to the angle DEF, or else unequal. If the angle ABC be equal to the angle DEF, 
Demonstration leading to an impossibility.  than the angle remaining, namely, ACB, shall be equal to the angle remaining DFE (by the corollary of the 32. of the first) And therefore the triangles ABC and DEF are equiangle. But if the angle ABC be unequal unto the angle DEF, then is the one of them greater than the other. Let the angle ABC be the greater, and unto the right line AB and unto the point in it B (by the 23. of the first) describe unto the angle DEF an equal angle ABG. And forasmuch as the angle A is equal unto the angle D, and the angle ABG is equal unto the angle DEF, therefore the angle remaining AGB is equal unto the angle remaining DFE (by the corollary of the 32. of the first. Wherefore the triangle ABG is equiangle unto the triangle DEF. Wherefore (by the 4. of the sixth) as the side AB is to the side BG, so is the side DE to the side EF. But by suppposition the side DE is to the side EF, as the side AB is to the side BC. Wherefore (by the 11. of the fifth) as the side AB is to the side BC so is the same side AB to the side BG. Wherefore AB hath to either of these BC and BG one and the same proportion, and therefore (by the 9 of the fifth) BC is equal unto BG. Wherefore (by the 5. of the first) the angle BGC is equal unto the angle BCG: but by supposition the angle BCG is less than a right angle. Wherefore the angle BGC is also less than a right angle. Wherefore (by the 13. of the first) the side angle unto it, namely, AGB is greater than a right angle, and it is already proved that the same angle is equal unto the angle F. Wherefore the angle F is also greater than a right angle. But it is supposed to be less which is absurd. Wherefore the angle ABC is not unequal unto the angle DEF, wherefore it is equal unto it. And the angle A is equal unto the angle D by supposition. Wherefore the angle remaining, namely, C, is equal unto the angle remaining, namely, to F (by the corollary of the 32. of the first) Wherefore the triangle ABC is equiangle unto the triangle DEF. 
The second part of this proposition.   
But now suppose that either of the angles ACB and DFE be not less than a right angle. That is, let either of them be a right angle, or either of them greater than a right angle. Then I say again that in that case also the triangle ABC is equiangle unto the triangle DEF. For if either of them be a right angle, forasmuch as all right angles are (by the 4. petition) equal the one to the other, strait way will follow the intent of the proposition. But if either of them be greater than a right angle, than the same order of construction that was before being kept, we may in like sort prove that the side BC is equal unto the side BG. Wherefore also the angle BCG is equal unto the angle BGC. But the angle BCG is greater than a right angle. Wherefore also the angle BGC is greater than a right angle. Wherefore two angles of the triangle BGC are greater than two right angles: which (by the 17. of the first) is impossible. Wherefore the angle ABC is not unequal unto the angle DEF. And therefore is it equal: but the angle A is equal unto the angle D (by supposition) Wherefore the angle remaining, namely, C is equal unto the angle remaining, namely, to F (by the corollary of the 32. of the first). Wherefore the triangle ABC is equiangle unto the triangle DEF. If therefore there be two triangles whero● the one hath one angle equal to one angle o● the other, and the sides which include the other angle be proportional, and if either of the other angles remaining be either less or not less than a right angle, the triangle shall be equiangle, and those angles in them shall be equal which are contained under sides proportional: which was required to be proved.   

The 8. Theorem. The 8. Proposition. If in a rectangle triangle be drawn from the right angle unto the base a perpendicular line, the perpendicular line shall divide the triangle into two triangles like unto the whole, and also like the one to the other.  
SVppose that there be a rectangle triangle ABC, whose right angle let be BAC: and (by the 12. of the first) from the point A to the line BC let there be drawn a perpendicular line AD, 
Construction.  which perpendicular line let divide the whole triangle ABC into these two triangles ABD and ADC. (Note that this perpendicular line AD, drawn from the right angle to the base, must needs fall within the triangle ABC, & so divide the triangle into two triangles. For if it should fall without, then producing the side BC unto the perpendicular line, there should be made a triangle, whose outward angle being an acute angle, should be less than the inward and opposite angle which is a right angle: which is contrary to the 16. of the first. Neither can it fall upon any of the sides AB or AC for th●n two angles of one and the self same triangle should not be less than too right angles, contrary to the self same 17. of the first. Wherefore it fall●●h within the triangle ABC) ● Then I say, that either of these triangles ABD and ADC, are like unto the whole triangle ABC, and moreover, that they are like the one to the other. First that the triangle ABD is like unto the whole triangle ABC is thus proved. 
Demonstration.  Forasmuch as (by the 4. petition) the angle BAC is equal unto the angle ADB, for either of them is a right angle. And in the two triangles ABC and ABD the angle B is common. Wherefore the angle remaining, namely, ACB is (by the Corollary of the 32. of the first) equal unto the angle remaining, namely, to BAD. Wherefore the triangle ABC is equiangle unto the triangle ABD. Wherefore the sides which contain the equal angles, are (by the 4. of the sixth) proportional. Wherefore as the side CB● which subtendeth the right angle of the triangle ABC, is unto the side BA which subtendeth the right angle of the triangle ABD, so is the same side AB which subtendeth the angle C of the triangle ABC, unto the side BD which subtendeth the angle BAD of the triangle ABD, which is equal unto the angle C: and moreover, the side AC unto the side AD which subtend the angle B common to both the triangles. Wherefore the triangle ABC is like unto the triangle ABD (by the 1. definition of the sixth). In like manner also may we prove, that the triangle ADC is like unto the triangle ABC. For the right angle ADC is equal to the right angle BAC, & the angle at the point C is common to either of those triangles. Wherefore the angle remaining, namely, DAC is equal to the angle remaining● namely, to ABC (by the Corollary of the 32. of the first). Wherefore the triangles ABC & ADC are equiangle. And therefore (by the 4. of the sixth) the sides which are about the equal angles are proportional. Wherefore as in the triangle ABC the side BC is to the side CA, so in the triangle ADC is the side AC to the side DC: and again, as in the triangle ABC the side CA is to the side AB, so in the triangle ADC is the side CD to the side DA. And moreover, as in the triangle ABC the side CB is to the side BA, so in the triangle ADC is the side CA to the side AD. Wherefore the triangle ADC is like unto the whole triangle ABC. Wherefore either of these triangles ABD, & ADC is like unto the whole triangle ABC.  
I say also, that the triangles ABD and ADC are like the one to the other. For forasmuch as the right angle BDA is equal unto the right angle ADC (by the 4. petition) and as it hath already been proved, the angle BAD is equal unto the angle C: therefore the angle remaining, namely, B is equal unto the angle remaining, namely, to DAC (by the Corollary of the 3●. of the first). Wherefore the triangle ABD is equiangle unto the triangle ADC. Wherefore as the side BD which in the triangle ABD subtendeth the angle BAD is unto the side DA which in the triangle ADC subtendeth the angle C which is equal unto the angle BAD, so is the side AD which in the triangle ABD subtendeth the angle B, unto the side DC which in the triangle ADC subtendeth the angle DAC which is equal unto the angle B: and moreover, so is the side BA unto the side AC which subtende the right angles. Wherefore the triangle ABD is like unto the triangle ADC. If therefore in a rectangle triangle be drawn from the right angle unto the base a perpendicular line, the perpendicular line shall divide the triangle into two triangles like unto the whole, and also like the one to the other: which was required to be proved.  

Corollary.  
Hereby it is manifest, that if in a rectangle triangle be drawn from the right angle unto the base a perpendicular line, the same line drawn is a mean proportional between the sections of the base: and moreover, between the whole base and either of the sections, the side annexed to the said section is the mean proportional. For is was proved, that as CD is to DA, so is DA to DB: and moreover, as CB is to BA, so is BA to BD: and finally, as BC is to CA, so is CA to CD.    

The 1. Problem. The 9 Proposition. A right line being given, to cut of from it any part appointed.  
LEt the right line given be AB. It is required that from the same line AB be cut of any part appointed. Suppose that a third part be appointed to be cut of. 
Construction.  From the point A draw a right line AC making with the line AB an angle: and in the line AC take a point at all adventures, which let be D. And beginning at D put unto AD two equal lines DE & EC (by the 2. of the first). And draw a right line from B to C: and by the point D (by the ●1. of the first) draw unto BC a parallel line DF. 
Demonstration.  Now forasmuch as unto one of the sides of the triangle ABC, namely, unto the side BC is drawn a parallel line FD, it followeth by the 2. of this book, that as CD is in proportion unto DA, so is BF to FA. But (by construction) CD is double to DA. Wherefore the line BF is also double to the line FA. Wherefore the line BA is triple unto the line AF. Wherefore from the right line● given AB, is cut of a third part appointed, namely, AF: which was required to be done.   

The 2. Problem. The 10. Proposition. To divide a right line given not divided, like unto a right line given being divided.  
SVppose that the right line given not divided be AB, and the right line given being divided, let be AC. It is required to divide the line AB which is not divided like unto the line AC which is divided. 
Construction.  Suppose the line AC be divided in the points D and E, & let the lines AB & AC so be put, that they make an angle at all adventures, and draw a line from B to C, and by the points D and E draw unto the line BC (by the 31. of the first) two parallel lines DF and EG: and by the point D unto the line AB (by the same) draw a parallel line DHK. Wherefore either of these figure● FH and HB are parallelograms. Wherefore the line DH is equal unto the line FG, and the line HK is equal unto the line GB. 
Demonstration.  And because to one of the sides of the triangle DKC, namely, to the side KC is drawn a parallel line HE, therefore the line CE (by th● 2. of the sixth) is in proportion unto the line ED as the line KH is to the line HD: but the line KH is equal unto the line BG, and the line HD is equal unto the line GF. Wherefore (by the 11. of the fift) as CE is unto ED, so is BG to GF. Again because to one of the sides of the triangle AGE, namely, to GE is drawn a parallel line FD, therefore the line ED (by the 2. of the sixth) is in proportion unto the line DA, as the line GF is to the line FA. And it is already proved that as CE is to ED, so is BG to GF. Wherefore as CE is to ED, so is BG to GF, and as ED is to DA, so is GF to FA. Wherefore the right line given not divided, namely, AB is divided like unto the right line given being divided, which is AC: which was required to be done.  

¶ A Corollary out of Flussates. 
A Corollary out of Flussates.   
By this Proposition we may divide any right line given, according to the proportion of any right lines given. For let those right lines having proportion be joined together directly, that they may make all one right line, and then join them to the line given anglewise. And so proceed as in the proposition, where you see that the right line given AB is divided into the right lines AF, FG and GB which have the self same proportion that the right lines AD, DE, and EC have.  
By this and the former proposition also may a right line given be easily divided into what parts so ever you will name. 
By this and the former proposition may a right line be divided into what parts soever you william.  As if you will divide the line AB into three equal parts, let the line DE be made equal to the line AD, and the line EC made equal to the same by the third of the first. And then using the self same manner of construction that was before: the line AB shall be divided into three equal parts. And so of any kind of parts whatsoever.    

The 3. Problem. The 11. Proposition. Unto two right lines given, to find a third in proportion with them.  
SVppose that there be two right lines given BA and AC, and let them be so put that they comprehend an angle howsoever it be. It is required to find unto BA and unto AC a third line in proportion. 
Construction.  Produce the lines AB and AC unto the points D and E. And unto the line AC (by the 2. of the first) put an equal line BD, and draw a line from B to C. And by the point D (by the 31. of the first) draw unto the line BC a parallel line DE, which let concur with the line AC in the point E. 
Demonstration.  Now forasmuch as unto one of the sides of the triangle ADE, namely, to DE is drawn a parallel line BC: therefore as AB is in proportion unto BD, so (by the 2. of the sixth) is AC unto CE. But the line BD is equal unto the line AC. Wherefore as the line AB is to the line AC, so is the line AC to the line CE. Wherefore unto the two right lines given AB and AC is found a third line CE in proportion with them: which was required to be done.  

¶ An other way after Pelitarius.  
Let the lines AB and BC be set directly in such sort that they both make one right line. 
another way after Pelitarius.  Then from the point A erect the line AD making with the line AB an angle at all adventures. And put the line AD equal to the line BC. And draw a right line from D to B which produce beyond the point B unto the point E. And by the point C draw unto the line DA a parallel line CE concurring with the line DE in the point E. Then I say that the line CE is the third line proportional with the lines AB and BC. For for●asmuch as by the 15. of the first, the angle B of the triangle ABD is equal to the angle B of the triangle CBE, and by the 29. of the same, the angle A is equal to the angle C, and the angle D to the angle E: therefore by the 4. of this book AB is to DA, as BC is to CE. Wherefore (by the 11. of the fifth) AB is to BC as BC is to CE: which was required to be done.   

¶ An other way also after Pelitarius.  
Let the lines AB and BC be so joined together, 
An ot●e● way after Pelitarius.  that they may make a right angle, namely, ABC. And draw a line from A to C, and from the point C draw unto the line AC a perpendicular CD (by the 11. of the first) And produce the line CD till it concur with the line AB produced unto the point D. Then I say that the line BD is a third line proportional with the lines AB and BC: which thing is manifest by the corollary of the 8. of this book.    

The 4. Problem. The 12. Proposition. Unto three right lines given to find a fourth in proportion with them.  
SVppose that the three right lines given be A, B, C. It is required to find unto A, B, C, a fourth line in proportion with them. Let there be taken two right lines DE & DF comprehending an angle as it shall happen, namely, EDF. 
Construction.  And (by the 2. of the first) unto the line A put an equal line DG. And unto the line B (by the same) put an equal line GOE And moreover, unto the line C put an equal line DH. Then draw a line from G to H. And by the point E (by the 31. of the first) draw unto the line GH a parallel line EF. Now forasmuch as unto one of the sides of the triangle DEF, 
Demonstration.  namely, unto the side EF is drawn a parallel line GH: therefore (by the 2. of the sixth) as the line DG is to the line GE, so is the line DH to the line HF. But the line DG is equal unto the line A, and the line GE is equal unto the line B, and the line DH unto the line C. Wherefore as the line A is unto the line B, so is the line C unto the line HF. Wherefore unto the three right lines given A, B, C, is found a fourth line HF in proportion with them: which was required to be done.  

¶ An other way after Campane.  
Suppose that there be three right lines AB, BC, and BD. It is required to add unto them a fourth line in proportion with them. 
another way after Campane.  join AB the first, with BD the third, in such sort that they both make one right line, namely, AD. And upon the said line AB erect from the point B the second line BC making an angle at all adventures. And draw a line from A to C. Then by the point D draw the line DE parallel to the line AC, which produce until it concur in the point E, with the line CB being likewise produced to the point E. Then I say that the line BE is the fourth line in proportion with the lines AB, BC, and BD: so that as AB is to BC, so is BD to BE. For forasmuch as by the 15 and 29. of the first the two triangles ABC and DBE are equiangle, therefore (by the 4. of this book) AB is to BC, as BD is to BE: which was required to be done.    

The 5. Problem. The 13. Proposition. Unto two right lines given, to find out a mean proportional.  
SVppose the two right lines given to be AB and BC. It is required between th●se two lines AB and BC to find out a mean line proportional. Let the lines AB and BC be so joined together that they both make one right line, namely, AC. 
Construction.  And upon the line AC describe a semicircle ADC, and from the point B raise up unto the line AC (by the 11. of the first) a perpendicular line BD, cutting the circumference in the point D: and draw a line from A to D, and a other from D to C. Now forasmuch as (by the 31. of the third) the angle in the semicircle ADC is a right angle, 
Demonstration.  and for that in the rectangle triangle ADC is drawn from the right angle unto the base a perpendicular line DB: therefore (by the Corollary of the 8. of the sixth) the line DB is a mean proportional between the segments of the base AB & BC. Wherefore between the two right lines given, AB & BC, is found a mean proportional DB: which was required to be done.  

¶ A Proposition added by Pelitarius.  
A mean proportional being given, to find out in a line given the two extremes. 
A proposition added by Pelitarius.  Now it behoveth that the mean given be not greater than the half of the line given.  
Flussates putteth this Proposition added by Pelitarius as a corollary following of this 13. proposition.    

The 9 Theorem. The 14. Proposition. In equal parallelograms which have one angle of the one equal unto one angle of the other, the sides shall be reciprocal, namely, those sides which contain the equal angles. And if parallelograms which having one angle of the one, equal unto one angle of the other, have also their sides reciprokal, namely, those which contain the equal angles, they shall also be equal.  
SVppose that there be two equal parallelograms AB and BC, having the angle B of the one equal unto the angle B of the other. 
The first part of this proposition.  And let the lines DB and DE be set directly in such sort that they both make one right line, namely, DE. And then (by the 14. of the first) shall the lines FB and BG be so set that they shall make also one right line, namely, GF. Then I say, that the sides of the parallelograms AB and BC, which contain the equal angles, are reciprocally proportional: that is, as BD is to BE, so is GB to BF. Make complete the parallelogram FE by producing the sides AF and CE, till they concur in the point H. Now forasmuch as the parallelogram AB is (by supposition) equal unto the parallelogram BC, and there is a certain other parallelogram FE: 
Demonstration of the of the same.  therefore (by the 7. of the fift) as the parallelogram AB is to the parallelogram FE, so is the parallelogram BC to the parallelogram FE. But as the parallelogram AB is to the parallelogram FE, so is the side DB to the side BE (by the first of this book). And (by the same) as the parallelogram BC is to the parallelogram FE, so is the side GB to the side BF. Wherefore also (by the 11. of the fift) as the side DB is to the side BE, so is the side GB to the side BF. Wherefore in the parallelograms AB and BC the sides which contain the equal angles, are reciprocally proportional: which was first required to be proved.  
But now suppose that the sides about the equal angles be reciprocally proportional so that as the side DB is to the side BE, 
The second part which is the converse of the first.  so let the side GB be to the side BF. Then I say, that the parallelogram AB is equal unto the parallelogram BC. For, for that as the side DB is to the side BE, so is the side GB to the side BF: but as the side DB is to the side BE, so (by the 1. of the sixth) is the parallelogram AB to the parallelogram FE: and as the side GB is to the side BF, so is the parallelogram BC to the parallelogram FE. Wherefore also (by the 11. of the fift) as the parallelogram AB is to the parallelogram FE, so is the parallelogram BC to the same parallelogram FE. Wherefore the parallelogram AB is equal unto the parallelogram BC (by the 9 of the fift). Wherefore in equal and equiangle parallelograms the sides which contain the equal angles are reciprocal: and if in equiangle parallelograms the sides which contain the equal angles be reciprocal, the parallelograms also shall be equal: which was required to be proved.   

The 10. Theorem. The 15. Proposition. In equal triangles which have one angle of the one equal unto one angle of the other, those sides are reciprokal, which include the equal angles. And those triangles which having one angle of the one equal unto one angle of the other, have also their sides which include the equal angles reciprokal, are also equal.  
SVppose that there be two equal triangles ABC, and ADE having one angle of the one equal unto one angle of the other, namely, the angle BAC equal unto the angle DAE. Then I say that in those triangles ABC and ADE, the sides which include the equal angles, are reciprokallie proportional, 
The first par● of this proposition.  that is, as the side CA is to the side AD, so is the side EA to the side AB. For let the lines CA and AD be so put, that they both make directly one right line. And so also the lines EA and AB shall both make one right line (by the 14. of the first) And draw a line from B to D. Now forasmuch as (by supposition) the triangle ABC is equal unto the triangle ADE. 
Demonstration of the same.  And there is a certain other triangle BAD, unto which the two equal triangles being compared, it will follow by the 7. of the fifth, that as the triangle ABC is unto the triangle BAD, so is the triangle EAD to the same triangle BAD. But as the triangle ABC is to the triangle BAD, so by the 1. of the sixth, is the base CA to the base AD: and as the triangle EAD is to the triangle BAD, so (by the same) is the base EA to the base AB. Wherefore (by the 11. of the fifth) as the side CA is to the side AD, so is the side EA to the side AB. Wherefore in the triangles ABC and ADE the sides which include the equal angles are reciprocally proportional.  
But now suppose that in the triangles ABC and ADE, the sides which include the equal angles, be reciprocally proportional, 
The second part which is the converse of the first.  so that as the side CA is to the side AD, so let the side EA be to the side AB. Then I say that the triangle ABC is equal unto the triangle ADE. For again draw a line from B to D. And for that as the line CA is to the line AD, so is the line EA to the line AB, but as the line CA is to the line AD, so is the triangle ABC to the triangle BAD, and as the line EA is to the line AB so is the triangle EAD to the triangle BAD. Wherefore as the triangle ABC is to the triangle BAD so is the triangle EAD to the same triangle BAD. Wherefore either of these triangles ABC and EAD have unto the triangle BAD one and the self same proportion. Wherefore (by the 9 of the fifth) the triangle ABC is equal unto the triangle EAD. If therefore there be taken equal triangles having one angle of the one equal unto one angle of the other, those sides in them shall be reciprokal, which include the equal angles. and those triangles which having one angle of the one equal unto one angle of the other, have also their sides which include the equal angles reciprokal, shall also be equal: which was required to be proved.   

The 11. Theorem. The 16. Proposition. If there be four right lines in proportion, the rectangle figure comprehended under the extremes: is equal to the rectangle figure contained under the means. And if the rectangle figure which is contained under the extremes, be equal unto the rectangle figure which is contained under the means: then are those four lines in proportion.  
SVppose that there b● four right lines in proportion, namely, AB, CD, E, and F: so that as the line AB is to the line CD, so let the line E be to the line F. Then I say, that the rectangle figure comprehended under the extremes AB and F, 
Demonstration of the first part.  is equal unto the rectangle figure contained under the means CD and E. From the point A (by the 11. of the first) raise up unto the right line AB a perpendicular line AG. And (by the same) from the point C unto the right line CD raise up a perpendicular line CH. And (by the 2. of the first) put the line AG equal unto the line F, and put also the line CH equal unto the line E, and make complete the parallelograms GB and HD. Now for that by supposition as the line AB is to the line CD, so is the line E to the line F. But the line E is equal unto the line CH, & the line F unto the line AG, therefore as the line AB is to the line CD, so is the line CH to the line AG. Wherefore in the parallelograms BG and DH the sides which include the equal angles, are reciprocally proportional. But equiangle parallelograms whose sides which include the equal angles, are reciprocal, are also equal (by the 14. of the sixth). Wherefore the parallelogram BG is equal unto the parallelogram DH. But the parallelogram BG is that which is contained under the lines AB and F, for the line AG is put equal unto the line F. And the parallelogram DH is that which is contained under the lines CD and E, for the line CH is put equal unto the line E. Wherefore the rectangle figure contained under the lines AB and F, is equal unto the rectangle figure contained under the lines CD and E. 
The second part which is the converse of the first.   
But now suppose that the rectangle figure comprehended under the lines AB and F, be equal unto the rectangle figure comprehended under the lines CD & E. Then I say, that the four right lines AB, CD, E and F, are proportional, that is, as the line AB is to the line CD, so is the line E to the line F. The same order of construction that was before being kept, forasmuch as that which is contained under the lines AB and F is equal unto that which is contained under the lines CD and E, but that which is contained under the lines AB and F is the parallelogram BG, for the line AG is equal unto the line F. And that also which is contained under the lines CD & E is the parallelogram DH, for the line CH is equal unto the line E. Wherefore the parallelogram BG is equal unto the parallelogram DH, & they are also equiangle. But in parallelograms equal & equiangle the sides which include the equal angles are reciprocal (by the 14. of the sixth). Wherefore as the line AB is to the line CD, so is the line CH to the line AG, but the line CH is equal unto the line E, and the line AG is equal unto the line F. Wherefore as the line AB is to the line CD, so is the line E to the line F. If therefore there be four right lines in proportion, the rectangle figure comprehended under the extremes, is equal to the rectangle figure contained under the means. And if the rectangle figure which is contained under the extremes, be equal unto the rectangle figure which is contained under the means, then are those four lines in proportioned which was required to be proved.   

The 12. Theorem. The 17. Proposition. If there be three right lines in proportion, the rectangle figure comprehended under the extremes, is equal unto the square that is made of the mean. And if the rectangle figure which is made of the extremes, be equal unto the square made of the mean, then are those three right lines proportional.  
SVppose that there be three lines in proportion A, B, C, so that as A is to B, so let B be to C. Then I say that the rectangle figure comprehended under the lines A and C is equal unto the square made of the line B. 
The first part of th●● Theorem.  Unto the line B (by the 2. of the first) put an equal line D. And because by supposition) as A is to B, so is B to C, but B is equal unto D, wherefore (by the 7. of the fifth) as A is to B, so is D to C, but if there be four right lines proportional, the rectangle figure comprehended under the extremes is equal unto the rectangle figure comprehended under the means (by the 16. of the sixth). Wherefore that which is contained under the lines A and C is equal unto that which is comprehended under the lines B and D. But that which is contained under the lines B and D, is the square of the line B, for the line B is equal unto the line D. Wherefore the rectangle figure comprehended under the lines A and C is equal unto the square made of the line B.  

The second part which is the converse of the first.  But now suppose that that which is comprehended under the lines A & C be equal unto the square made of the line B. Then also I say, that as the line A is to the line B, so is the line B to the line C. The same order of construction that was before, being kept, forasmuch as that which is contained under the lines A and C is equal unto the square which is made of the line B. But the square which is made of the line B is that which is contained under the lines B & D, for the line B is put equal unto the line D. Wherefore that which is contained under the lines A and C is equal unto that which is contained under the lines B and D. But if the rectangle figure comprehended under the extremes, be equal unto the rectangle figure comprehended under the mean lines, the four right lines shall be proportional (by the 16. of the sixth) Wherefore as the line A is to the line B, so is the line D to the line C. But the line B is equal unto the line D. Wherefore as the line A is to the line B, so is B to the line C. If therefore there be three right lines in proportion, the rectangle figure comprehended under the extremes, is equal unto the square that is made of the mean. And if the rectangle figure which is contained under the extremes, be equal unto the square made of the mean, then are those three right lines proportional: which was required to be demonstrated.  


A Corollary.  ¶ Corollary added by Flussates.  
Hereby we gather that every right line is a mean proportional between every two right lines which make a rectangle figure equal to the square of the same right line.    

The 6. Problem. The 18. Proposition. Upon a right line given, to describe a rectiline figure like, and in like sort situate unto a rectiline figure given.  
SVppose that the ri●ht line given be AB● and let the rectiline figure given be EG. It is required upon the right line given AB to describe a rectiline figure like, and in like sort situate unto the rectiline figure given GOE Draw a line from H to F, and unto the right line AB and to the p●i●t in it A, 
Description of the rectiline figure required.  make unto the angle E an equal angle DAB (by the 23. of the firs●) and unto the right line AB and unto the point in it B (by the same) make vn●o the angle EFH an equal angle ABD. Wherefore the angle remaining EHF is equal unto the angle remaining ADB. Wherefore the triangle HEF is equiangle unto the triangle DAB. Wherefore (by the 4. of the sixth) as the side HF is in proportion to the side DB, so is the side HE to the side DA, and the side EF to the side AB. Again (by the 23. of the first) unto the right line BD and unto the point in it D, make unto the angle FHG an equal angle BDC, and (by the same) unto the right line BD and unto the point in 〈◊〉 B, make unto the angle HFG an squall angle DBC. 
Demonstration.  Wherefore the angle remaining, namely, G, is equal unto the angle remaining, namely, to C. Wherefore the triangle HFG is equiangle unto the triangle DBC. Wherefore (by the 4. of the sixth) as the side HF is in proportion to the side DB, so is the side HG to the side DC, and the side GF to the side CB. And it is already proved that as HF is to DB, so is HE to DA, and EF to AB. Wherefore (by the 11. of the fift) as EH is to AD, so is EF to AB, and HG to DC, and moreover, GF to CB. And forasmuch as the angle EHF is equal unto the angle ADB, and the angle FHG is equal unto the angle BDC: therefore the whole angle EHG is equal unto the whole angle ADC, and by the same reason the angle EFG is equal unto the angle ABC. But (by construction) the angle E is equal unto the angle A, and the angle G is proved equal unto the angle C. Wherefore the figure AC is equiangle, unto the figure EG, and those sides which in it include the equal angles are proportional, as we have before proved. Wherefore the rectiline figure AC is (by the first definition of the sixth) like unto the rectiline figure given EG● Wherefore upon the right line given AB is described a rectiline figure AC like & in like sort ●●●uat● v●to the rectiline figure given EG: which was required to be done.   

The 13. Theorem. The 19 Proposition. Like triangles are one to the other in double proportion that the sides of like proportion are.  
SVppose the triangles like to be ABC and DEF, having the angle B of the one triangle, equal unto the angle E of the other triangle, & as AB is to BC, so let DE be to EF, so that let BC & EF be sides of like proportion. Then I say that the proportion of the triangle ABC unto the triangle DEF is double to the proportion of the side BC to the side EF. Unto the two lines BC and EF (by the 10. of the sixth) make a third line in proportion BG, so that as BC is to EF, so let EF be to BG, and draw a line from A to G. Now forasmuch as AB is to BC, as DE is to EF, therefore alternately (by the 16. of the fifth) as AB is to DE, so is BC to EF. 
Demonstration.  But as BC is to EF, so is EF to BG, wherefore also (by the 11. of the fifth) as AB is to DE, so is EF to BG. Wherefore the sides of the triangles ABG & DEF, which include the equal angles are reciprocally proportional. But if in triangles having one angle of the one equal to one angle of the other, the sides which include the equal angles, be reciprokal, the triangles also (by the 15. the sixth) shall be equal. Wherefore the triangle ABG is equal unto the triangle DEF. And for that as the line BC is to the line EF, so is the line EF to the line BG: but if there be three lines in proportion, the first shall have to the third double proportion that it hath to the second (by the 10. definition of the fifth) therefore the line BC hath unto the line BG double proportion that it hath to the line EF. But as BC is to BG, so (by the 1. of the sixth) is the triangle ABC to the triangle ABG. Wherefore the tiangle ABC is unto the triangle ABG in double proportion that the side BC is to the side EF. But the triangle ABG is equal to the triangle DEF. Wherefore also the triangle ABC is unto the triangle DEF in double proportion that the side BC is to the side EF. Wherefore like triangles are one to the other in double proportion that the sides of like proportion are: which was required to be proved.  

Corollary.  

A Corollary.  Hereby it is manifest that if there be three right lines in proportion, as the first is to the third, so is the triangle described upon the first, unto the triangle described upon the second, so that the said triangles be like, and in like ●ort described, for it hath been proved that as the line CB is to the line BG, so is the triangle ABC to the triangle DEF: which was required to be demonstrated.    

The 14. Theorem. The 20. Proposition. Like Poligonon figures, are divided into like triangles and equal in number, and of like proportion to the whole. And the one Poligonon figure is to the other Poligonon figure in double proportion that one of the sides of like proportion is to one of the sides of like proportion.  
SVppose that the like Poligonon figures be ABCDE, & FGHKL, having the angle at the point F equal to the angle at the point A, and the angle at the point ● equal to the angle at the point B, and the angle at the point H equal to the angle at the point C: and so of the rest. And moreover, as the side AB is to the side BC, so let the side FG be to the side GH, and as the side BC is to the side CD, so let the side GH be to the side HK and so forth. And let the sides AB & FG be sides of like proportion. Then I say first, 
The first par● of this Theorem.  that these Poligonon figures ABCDE & FGHKL, are divided into like triangles and equal in number. For draw these right lines, AC, AD, FH, & FK● And forasmuch as (by supposition, that is, by reason the figure ABCDE is like unto the figure FGHKL) the angle B is equal unto the angle G, and as the side AB is to the side BC, so is the side FG to the side GH, it followeth that the two triangles ABC and FGH have one angle of the one equal to one angle of the other, and have also the sides about the equal angles proportional. Wherefore (by the 6. of the sixth) the triangle ABC is equiangle unto the triangle FGH. And those angles in them are equal, under which are subtended sides of like proportion: namely, the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF. Wherefore (by the 4. of the sixth) the sides which are about the equal angles are proportional: and the sides which are subtended under the equal angles are of like proportion. Wherefore as AC is to BC, so is FH to GH. But by supposition as BC is to CD, so is GH to HK. Wherefore of equality (by the 22. of the fift) as AC is to CD, so is FH to HK. And forasmuch as by supposition the whole angle BCD is equal to the whole angle GHK, and it is proved that the angle BCA is e●●all to the angle GHF: therefore the angle remaining ACD is equal to the angle remaining FHK (by the 3. common sentence). Wherefore the ●ria●gl●s ACD, and FHK, have again one angle of the one equal to one angle of the other, and the sides which are about the equal sides are proportional. Wherefore (by the ●ame sixth of this book) the triangles ACD & FHK are equiangle. And (by the 4. of this book) the sides, which are about the equal angles 〈◊〉 proportional. And by the same reason may we prove that the triangle AD● is equiangle unto the triangle FKL. And that the sides which are about the equal angles are proportional. Wherefore the triangle ABC is like to the triangle FGH, and the triangle ACD to the triangle FHK, and also the triangle ADE to the triangle FKL (by the first definition of this sixth book). Wherefore the Poligonon figures given ABCDE, and FGHKL, are divided into triangles like and equal in number.  

The second part demonstrated.  I say moreover, that the triangles are the one to the other, and to the whole Poligonon figures proportional: that is, as the triangle ABC is to the triangle FGH, so is the triangle ACD to the triangle FHK, and the triangle ADE to the triangle FKL: and as the triangle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. For forasmuch as the triangle ABC is like to the triangle FGH, and AC and FH are sides of like proportion, therefore the proportion of the triangle ABC to the triangle FGH is double to the proportion of the side AC to the 〈◊〉 FH (by the former Proposition). And therefore also the proportion of the triangle ACD to the triangle FKH, is double to the proportion that the same side AC hath to the side FH. Wherefore (by the 11. of the fift) as the triangle ABC is to the triangle FGH, so is the triangle ACD to the triangle FHK. Again, forasmuch as the triangle ACD is like to the triangle FHK, and the sides AD & FK are of like proportion: therefore the proportion of the triangle ACD to the triangle FHK is double to the proportion of the side AD to the side FK (by the foresaid 19 of the sixth). And by the same reason the proportion of the triangle ADE to the triangle FKL, is double to the proportion of the same side AD to the side FK. Wherefore (by the 11. of the fift) as the triangle ACD is to the triangle FHK, so is the triangle ADE to the triangle FKL. But as the triangle ACD is to the triangle FHK, so is it proved that the triangle ABC is to the triangle FGH. Wherefore also (by the 11. of the fift) as the triangle ABC is to the triangle FGH, so is the triangle ADE to the triangle FKL. Wherefore the foresaid triangles are proportional: namely, as ABC is to FGH, so is ACD to FHK, and ADE to FKL. Wherefore (by the 12. of the fift) as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherefore as the triangle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. Wherefore the triangles are proportional both the one to the other, & also to the whole Poligonon figures.  
Lastly I say, 
The third part.  that the Poligonon figure ABCDE hath to the Poligonon figure FGHKL a double proportion to that which the side AB hath to the side FG: which are sides of like proportion. For it is proved, that as the triangle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. But the triangle ABC hath to the triangle FGH a double proportion to that which the side AB hath to the side FG (by the former 19 Proposition of this book): for it is proved, that the triangle ABC is like to the triangle FGH. Wherefore the proportion of the Poligonon figure ABCDE to the Poligonon figure FGHKL is double to the proportion of the side AB to the side FG: which are sides of like proportion. Wherefore like Poligonon figures are divided. etc. as before● which was required to be proved.  

The first Corollary.  
Hereby it is manifest, that all like rectiline figures what soever, are the one to the other in double proportion that the sides of like proportion are. 
The first Corollary.  For any like rectiline figures whatsoever are by this Proposition divided into like triangles and equal in number.   

The second Corollary.  
Hereby also it is manifest, that if there be three right lines proportional, 
The second Corollary.  as the first is to the third, so is the figure described upon the first to the figure described upon the second, so that the said figures be like and in like sort described. For it is proved, that the proportion of the Poligonon figure ABCDE to the Poligonon figure FGHKL is double to the proportion of the side AB to the side FG. And if (by the 11. of the sixth) unto the lines AB and FG we take a third line in proportion, namely, MN, the first line, namely, AB shall have unto the third line, namely, to MN, double proportion that it hath to the second line, namely, to FG (by the 10. definition of the fift). Wherefore as the line AB is to the line MN, so is the rectiline figure ABC to the rectiline figure FGH, the said rectiline figures being like & in like sort described.    

The 15. Theorem. The 21. Proposition. Rectiline figures which are like unto one and the same rectiline figure, are also like the one to the other.  
SVppose there be two rectiline figures A and B like unto the rectiline figure C. Then I say that the figure A is also like unto the figure B. For forasmuch as the figure A is like unto the figure C, 
Demonstration.  it is also equiangle unto it (by the conversion of the first definition of the sixth) & the sides including the equal angles shall be proportional. Again forasmuch as the figure B is like unto the figure C, it is also (by the same definition) equiangle unto it, and the sides about the equal angles are proportional. Wherefore both these figures A and B are equiangle unto the figure C, and the sides about the equal angles are proportional. Wherefore (by the first common sentence) the figure A is equiangle unto the figure B, and the sides about the equal angles are proportional, wherefore the figure B is like unto the figure A, which was required to be proved.   

The 16. Theorem. The 22. Proposition. If there be four right lines proportional, the rectiline figures also described upon them being like, and in like sort situate, shall be proportional. And if the rectiline figures upon them described be proportional, those right lines also shall be proportional.  
SVppose there be four right lines AB, CD, EF, and GH, and as AB is to CD, so let EF be to GH. And upon the lines AB and CD (by the 1●. of the sixth) let there be described two rectiline figures KAB, and LCD like the one to the other, and in like sort situate. And upon the lines EF and GH (by the same) let there be described also two rectiline figures MF and NH like the one to the other, and in like sort situate. 
The first part of this proposition.  Then I say that as the ●igure KAB is to the figure LCD, so is the figure MF to the figure NH. Unto the lines AB and CD (by the 11. of the sixth) make a third line in proportion, namely, O: and unto the lines EF and GH in like sort make a third line in a line proportion, namely, P. And for that as the line AB is to the line CD, so is the line EF to the line GH, but as the line CD is to the line O, so is the line GH to the line P. Wherefore of equality (by the 22. of the fifth) as the line AB is unto the line O, so is the line EF to the line P. But as the line AB is to the line O, so is the figure KAB to the figure LCD (by the second corollary of the 20. of the sixth). And as the line EF is to the line P, so is the figure M F to the figure NH. Wherefore (by the 11. of the fifth) as the figure KAB is to the figure LCD, so is the figure M F to the figure NH.  
But now suppose that as the figure KAB is to the figure LCD, 
The second part which is the converse of the first.  so is the figure M F to the figure NH, than I say that as the line AB is to the line CD, so is the line EF to the line GH. As the line AB is to the line CD, so (by the 1●. of the sixth) let the line EF be to the line QR, and upon the line QR (by the 18. of the sixth) describe unto either of these figures MF and NH a like figure, and in like sort situate SR. Now forasmuch as the line AB is to the line CD, so is the line EF to the line QR, and upon the lines AB and CD are described two figures like, and in like sort situate KAB and LCD, and upon the lines EF and QR are described also two figures like, and in like sort situate MF and SR, therefore as the figure KAB is to the figure LCD, so is the figure MF to the figure SR: wherefore also (by the 11. of the fifth) as the figure MF is to the figure SR, so is the figure MF to the figure NH, wherefore the figure M F hath to either of these figures NH, and SR one and the same proportion, wherefore by the 9 of the fifth, the figure NH is equal unto the figure SR. And it is unto it like, and in like sort situate. 
* Note that this is proved in the assumpt following.  But in like and equal rectiline figures being in like sort situate, the sides of like proportion on which they are described are equal. Wherefore the line GH is equal unto the line QR. And because as the line AB is to the line CD, so is the line EF to the line QR, but the line QR is equal unto the line GH, therefore as the line AB is to he line CD, so is the line EF to the line GH. If therefore there be four right lines proportional, the rectiline figures also described upon them being like and in like sort situate shall be proportional And if the rectiline figures upon them described being like and in like sort situate be proportional, those right lines also shall be proportional: which was required to be proved.  

An Assumpt.  
And now that in like and equal figures, being in like sort situate, the sides of like proportion are also equal (which thing was before in this proposition taken as granted) may thus be proved. 
An Assumpt.  Suppose that the rectiline figures NH and SR be equal and like, and as HG is to GN, so let RQ be to QS, and let GH and QR be sides of like proportion. Then I say that the side RQ is equal unto the side GH. For if they be unequal, the one of them is greater than the other, let the side RQ be greater than the side HG. And for that as the line RQ is to the line QS, so is the line HG to the line GN, and alternately also (by the 16. of the fifth) as the line RQ is to the line HG, so is the line QS, to the line GN, but the line RQ is greater than the line HG. Wherefore also the line QS is greater than the line GN. Wherefore also the figure RS is greater than the figure HN but (by supposition) it is equal unto it, which is impossible. Wherefore the line QR is not greater than the line GH. In like sort also may we prove that it is not less than it, wherefore it is equal unto it: which was required to be proved.  
Flussates demonstrateth this second part more briefly, 
another demonstration of the second part after Flussates.  by the first corollary of the ●0. of this book, thus. Forasmuch as the rectiline figures are by supposition in one and the same proportion, and the same proportion is double to the proportion of the sides AB to CD, and EF to GH (by the foresaid corollary) the proportion also of the sides shall be one and the self same (by the 7. common sentence) namely, the line AB shall be unto the line CD as the line EF is to the line GH.    

The 17. Theorem. The 23. Proposition. Equiangle Parallelogrammes have the one to the other that proportion which is composed of the sides.  
Flussates demonstrateth this Theorem without taking of these three lines, K, L, M, after this manner.  

another demonstration after flussates.  Forasmuch as (saith he) it hath been declared upon the 10. definition of the fift book, and ●ift definition of this book, that the proportions of the extremes consist of the proportions of the means, let us suppose two equiangle parallelograms ABGD, and GEZI, and let the angles at the point G in either be equal. And let the lines BG and GI' be set directly that they both make one right line, namely, BGI. Wherefore EGD also shall be one right line by the converse of the 15. of the first. Make complete the parallelogram GT. Then I say, that the proportion of the parallelograms AG & GZ is composed of the proportions of the sides BG to GI', and DG to GOE For forasmuch as that there are three magnitudes, AG, GT, and GZ, and GT is the mean of the said magnitudes: and the proportion of the extremes AG to GZ consisteth of the mean proportions (by the 5. definition of this book) namely, of the proportion of AG to GT, and of the proportion GT to GZ: But the proportion of AG to GT is one and the self same with the proportion of the sides BG to GI' (by the first of this book). And the proportion also of GT to GZ is one and the self same with the proportion of the other sides, namely, DG to GE (by the same Proposition). Wherefore the proportion of the parallelograms AG to GZ consisteth of the proportions of the sides BG to GI', and DC to GOE Wherefore equiangle parallelograms are the one to the other in that proportion which is composed of their sides: which was required to be proved.   

The 18. Theorem. The 24. Proposition. In every parallelogram, the parallelograms about the dimecient are like unto the whole, and also like the one to the other.  

Demonstration of this proposition wherein is first proved that the parallegramme EG is like to the whole parallelogram ABCD.  SVppose that there be a parallelogram ABCD, and let the dimecient thereof be AC: and let the parallelograms about the dimecient AC, be EG and HK. Then I say that either of these parallelograms EG and HK is like unto the whole parallelogram ABCD, and also are like the one to the other. For forasmuch as to one of the sides of the triangle ABC, namely, to BC is drawn a parallel line EF, therefore as BE is to EA, so (by the 2. of the sixth) is CF to FA. Again forasmuch as to one of the sides of the triangle ADC, namely, to CD is drawn a parallel line F●, therefore (by the same) as CF is to FA, so is DG to GA. But as CF is to FA, so is it proved that BE is to EA. Wherefore as BE is to EA, so (by the 11. of the fifth) ● is DG to GA. Wherefore by composition (by the 18. of the fifth) as BA is to AE● so is DA to AG. And alternately (by the 16. of the fifth) as BA is to AD, so is EA to AG. Wherefore in the parallelogrammes● ABCD and EG the sides which are about the common angle BAD are proportional. And because the line GF is a parallel unto the line DC● therefore the angle AGF (by the 29● of the● first) is equal unto the angle ADC, ● the angle GFA equal unto the angle DCA and the angle DAC is common to the two triangles ADC and AFG: Wherefore the triangle DAC is equiangle unto the triangle AGF. And by the same reason the triangle ABC is equiangle unto the triangle AEF. Wherefore the whole parallelogram ABCD is equiangle unto the parallelogram EG. Wherefore as AD is in proportion to DC, so (by the 4. of the sixth) is AG to GF, and as DC is to CA, so is GF to FA. And as AC is to CB, so is AF to FE. And moreover as CB is to BA, so is FE to EA. And forasmuch as it is proved that as D● is to CA, so is GF to FA: but as AC is to C●, so is AF to FE. Wherefore of equality (by the 22. of the fifth) as DC is to CB, so is GF to FE. Wherefore in the parallelograms ABCD and EG, the sides which include the equal angles are proportional. Wherefore the parallelogram ABCD is (by the first definition of the sixth) like unto the parallelogram EG.  
And by the same reason also the parallelogram ABCD is like to the parallelogram KH: 
That the parallelogram KH is like to the whole parallelogram ABCD.  wherefore either of these parallelograms EG and KH is like unto the parallelogram ABCD. But rectiline figures which are like to one and the same rectiline figure are also (by the 21. of the sixth) like the one to the other. Wherefore the parallelogram EG is like to the parallelogram HK. 
That the parallelograms EG and KH are like the one to the other.  Wherefore in every parallelogram, the parallelograms about the dimecient are like unto the whole, and also like the one to the other. Which was required to be proved.  

¶ An other more brief demonstration after Flussates.  
Suppose that there be a parallelogram ABCD, whose dimetient let b● A●, 
another Demonstration after Flussates.  about which let consist these parallelograms EK and TI, having the angles at the points ● and 〈…〉 with the whole parallelogram ABCD. Then I say, that those parallelograms EK and TI are like to the whole parallelogram DB and also al● like the one to the other. For forasmuch as BD, EK, and TI are parallelograms, therefore the right line AZG falling upon these parallel lines AEB, KZT, and DI G, or upon these parallel lines AKD, EZI, and BTG, maketh these angles equal the one to the other, namely, the angle EAZ to the angle KZA, & the angle EZA to the angle KAZ, and the angle TZG to the angle ZGI, and the angle TGZ to the angle IZG, and the angle BAG to the angle AGD: and finally, the angle BGA to the angle DAG. Wherefore (by the first Corollary of the 32. of the first, and by the 34. of the first) the angles remaining are equal the one to the other, namely, the angle B to the angle D, and the angle E to the angle K, and the angle T to the angle I Wherefore these triangles are equiangle and therefore like the one to the other, namely, the triangle ABG to the triangle GDA, and the triangle AEZ to the triangle ZKA, & the triangle ZTG to the triangle GIZ. Wherefore as the side AB is to the side BG, so is the side AE to the side EZ, and the side ZT to the side TG. Wherefore the parallelograms contained under those right lines, namely, the parallelograms ABGD, EK, & TI, are like the one to the other (by the first definition of this book). Wherefore in every parallelogram the parallelograms. etc. as before: which was required to be demonstrated.   

¶ A Problem added by Pelitarius.  
Two equiangle Parallelogrammes being given, so that they be not like, to cut of from one of them a parallelogram like unto the other.  

An addition of Pelitarius.  Suppose that the two equiangle parallelograms be ABCD and CEFG, which let not be like the one to the other. It is required from the parallelogram ABCD, to cut of a parallelogram like unto the parallelogram CEFG. Let the angle C of the one be equal to the angle C of the other. And let the two parallelograms be so 〈◊〉, that the lines BC & CG may make both one right line, namely, BG. Wherefore also the right lines DC and CE shall both make one right line, namely, DE. And draw a line from the point F to the point C, and produce the line FC till it concur with the line AD in the point H. And draw the line HK parallel to the line CD (by the 31. of the first). Then I say, that from the parallelogram AC is cut of the parallelogram CDHK, like unto the parallelogram EG. Which thing is manifest by this 24. Proposition. For that both the said parallelograms are described about one & the self same dimetient. And to the end it might the more plainly be seen, I have made complete the parallelogram ABGL.   

¶ An other Problem added by Pelitarius.  
Between two rectiline Superficieces, to find out a mean superficies proportional.  

Another addition of Pelitarius.  Suppose that the two superficieces be A and B, between which it is required to place a mean superficies proportional. Reduce the said two rectiline figures A and B unto two like parallelograms (by the 18. of this book) or if you think good reduce either of them to a square, (by the last of the second). And let the said two parallelograms like the one to the other and equal to the superficieces A and B, be CDEF and FGHK. And let the angles F in either of them be equal, which two angles let be placed in such sort, that the two parallelograms ED and HG may be about one and the self same dimetient CK (which is done by putting the right lines EF and FG in such sort that they both make one right line, namely, EG). And make complete the parallelogram CLK M. Then I say, that either of the supplements FL & FM is a mean proportional between the superficieces CF & FK, that is, between the superficieces A and B: namely, as the superficies HG is to the superficies FL, so is the same superficies FL to the superficies ED. For by this 24. Proposition the line HF is to the line FD, as the line GF is to the line FE. But (by the first of this book) as the line HF is to the line FD, so is the superficies HG to the superficies FL: and as the line GF is to the line FE, so also (by the same) is the superficies FL to the superficies ED. Wherefore (by the 11. of the fift) as the superficies HG is to the superficies FL, so is the same superficies FL to the superficies ED: which was required to be done.    

The 7. Problem. The 25. Proposition. Unto a rectiline figure given to describe an other figure like, which shall also be equal unto an other rectiline figure given.  
SVppose that the rectiline figure given, whereunto is required an other to be made like be ABC, and let the other rectiline figure whereunto the same is required to be made, equal be D. Now it is required to describe a rectiline figure like unto the figure ABC, and equal unto the figure D. 
Construction.  Upon the line BC describe (by the 44. of the first) a parallelogram BE equal unto the triangle ABC, and by the same upon the line CE, describe the parallelogram C M equal unto the rectiline figure D, and in the said parallelogram let the angle FCE, be equal unto the angle CBL. And forasmuch as the angle FCE is by construction equal to the angle CBL, 
Demonstration.  add the angle BCE common to them both. Wherefore the angles LBC and BCE are equal unto the angles BCE and ECF, but the angles LBC and BCE are equal to two right angles (by the 29. of the first) wherefore also the angles BCE and ECF are equal to two right angles. Wherefore the lines BC and CF (by the 14. of the first) make both one right line, namely, BF, and in like sort do the lines LE and EM make both one right line, namely, LM. Then (by the 13. of the sixth) take the mean proportional between the lines BC and CF, which let be GH. And (by the 18. of the sixth) upon the line GH, let there be described a rectiline figure KHG like unto the rectiline figure ABC, and in like sort situate. And for that as the line BC, is to the line GH, so is the line GH to the line CF: 
Demonstration.  but if there be three right lines proportional, as the first is to the third, so is the figure which is described of the first unto the figure which is described of the second, the said figures being like and in like sort situate (by the second corollary of the 20. of the sixth) wherefore as the line BC is to the line CF, so is the triangle ABC to the triangle KGH. But as the line BC is to the line CF, so is the parallelogram BE to the parallelogram EF (by the 1. of the sixth). Wherefore as the triangle A BC, is to the triangle KGH so is the parallelogram BE to the parallelogram EF. Wherefore alternately also (by the 16. of the fifth) as the triangle ABC is to the parallelogram BE, so is the triangle KGH, to the parallelogram EF: but the triangle ABC is equal unto the parallelogram BE, wherefore also the triangle KGH is equal unto the parallelogram EF: but the parallelogram FE is equal unto the rectiline figure D. Wherefore also the rectiline figure KGH is equal unto the rectiline figure D, and the rectiline figure KGH is by supposition like unto the rectiline figure ABC. Wherefore there is described a rectiline figure KGH like unto the rectiline figure given ABC, and equal unto the other rectiline figure given D: which was required to be done.   

The 19 Theorem. The 26. Proposition. If from a parallelogram be taken away a parallelogram like unto the whole and in like sort set, having also an angle common with it, then is the parallelogram about one and the self same dimecient with the whole.  
SVppose that there be a parallelogram ABCD, and from the parallelogram ABCD, take away a parallelogram AF like unto the parallelogram ABCD, and in like sort situate, having also the angle DAB common with it. Then I say, that the parallelograms ABCD and AF are both about one and the self same 
* 
By the dimetient is understand here the dimetient which is drawn from the angle which is common to them both to the opposite angle.  
Demonstration leading to an absurdity.   dimecient AFC, that is, that the dimecient AFC of the whole parallelogram ABCD passeth by the angle F of the parallelogram AF, and is common to either of the parallelograms. For if AC do not pass by the point F, then if it be possible let it pass by some other point, as AHC doth. Now than the dimetient AHC shall cut either the side GF or the side EF of the parallelogram AF. Let it cut the side GF in the point H. And (by the 31. of the first) by the point H let there be drawn to either of these lines AD and BC a parallel line HK wherefore GK is a parallelogram, and is about one and the self same dimetient with the parallelogram ABCD. And forasmuch as the parallelograms ABCD and GK are about one and the self same dimecient, therefore (by the 24. of the sixth) the parallelogram ABCD is like unto the parallelogram GK. Wherefore as the line DA is to the line AB so is the line GA to the line AK (by the conversion of the first definition of the sixth) And for that the parallelograms ABCD, and EG are (by supposition) like, therefore as the line DA is to the line AB so is the line GA to the line AE. Wherefore the line GA hath one and the self proportion to either of these lines AK and AE. Wherefore (by the 9 of the fifth) the line AK is equal unto the line AE, namely, the less to the greater, which is impossible. The self same inconvenience also will follow, if you put the dimetient AC to cut the side FE. Wherefore AC the dimetient of the whole parallelogram ABCD passeth by the angle and point F. And therefore the parallelogram AEFG is about one and the self same dimetient with the whole parallelogram ABCD. Wherefore if from a parallelogram be taken away a parallelogram like unto the whole, and in like sort situate, having also an angle common with it, then is that parallelogram about one and the self same dimetient with the whole: which was required to be proved.  

¶ An other demonstration after Flussates, which proveth this proposition affirmatively.  
From the parallelogram ABGD let there be taken away the parallelogram AEZK like and in like sort situate with the whole parallelogram ABGD, and having also the angle A common with the whole parallelogram. 
another way after Flussates.  Then I say that both their diameters, namely, AZ and AZG do make one and the self same right line. Divide the sides AB and B● into two equal parts in the points C and F (by the 10. of the first.) And draw a line from C to F. Wherefore the line CF is a parallel to the right line AG (by the corollary added by Campane after the 29. of the first) Wherefore the angles BAG and BCF are equal (by the 29. of the first): but the angle EAZ is equal unto the angle BAG (by reason the parallelograms are supposed to be like) wherefore the same angle EAZ is equal to the angle BCF, namely, the outward angle to the inward and opposite angle. Wherefore (by the 28. of the first) the lines AZ and CF are parallel lines. Now than the lines AZ and AG being parallels to one and the self same line, namely, to CF do concur in the point A. Wherefore they are set directly the one to the other, so that they both make one right line (by that which was added in the end of the 30. proposition of the first) wherefore the parallelograms ABGD, and AEZK are about one and the self fame dimetient: which was required to be proved.    

The 20. Theorem. The 27. Proposition. Of all parallelograms applied to a right line wanting in figure by parallelograms like and in like sort situate to that parallelogram which is described of the half line: the greatest parallelogram is that which is described of the half line being like unto the want.  


In this proposition are two cases, in the first the parallelogram compared to the parallelogram described of the half line is described upon a line greater than the half line: In the second upon a line less.  
The first case where the parellelogramme compared namely AF is described upon the line AK which is greater than the half line AC.   LEt there be a right line AB, and (by the 10. of the first) divide it in two equal parts in the point C. And unto the right line AB apply a parallelogram AD wanting in figure by the parallelogram DB, which let be like and in like sort described unto the parallelogram described of half the line AB, which is, BC. Then I say, that of all the parallelograms which may be applied unto the line AB and which want in figure by parallelograms like and in like sort situate unto the parallelogram DB the greatest is the parallelogram AD. For unto the right line AB let there be applied a parallelogram AF wanting in figure by the parallelogram FB, which let be like and in like sort situate unto the parallelogram DB. Then I say, that the parallelogram AD is greater than the parallelogram AF. For forasmuch as the parallelogram DB is like unto the parallelogram ●B, therefore (by the 26. of the sixth) they are about one and the self same dimetient. 
Demonstration of this case.  Let their dimetient be DB, and make complete the figure. Now forasmuch as (by the 43. of the first) the supplement FC is equal unto the supplement FE● add the figure FB common to them both. Wherefore the whole figure CR is equal unto the whole figure KE. But the figure CR is equal unto the figure CG (by the 36. of the first) for that the base AC is equal unto the base CB. Wherefore the figure GC is equal unto the figure KE. Add the figure CF common unto them both. Wherefore the whole figure AF is equal unto the whole Gnomon LMN. But the whole parallelogram DB is greater than the Gnomon LMN (by the 3. common sentence). Wherefore also it is greater than the parallelogram AF. But the parallelogram AD is equal unto the parallelogram DB (by the 36. of the first). Wherefore the parallelogram AD is greater than the parallelogram AF. Wherefore of all parallelograms applied to a right line wanting in figure by parallelograms like and in like sort situate, to that parallelogram which is described of the half line, the greatest parallelogram is that which is described of the half of the line, being like unto the want: which was required to be proved.  
Again, let AB be divided into two equal parts in the point C, and let the parallelogram applied upon the half line be AL, 
The second case where the parallelogram compared namely AE is described upon the line AD which is less than the line AC, Demonstration of the second case.  wanting in figure by the parallelogram LB, which let be like and in like sort situate unto the parallelogram AL. Again unto the line AB let there be applied an other parallelogram AE wanting in figure by the parallelogram EB being like and in like sort situate unto the parallelogram LB which is described upon half of the line AB. Then I say, that the parallelogram AL applied unto half the line is greater than the parallelogram AE. For forasmuch as the parallelogram EB is like unto the parallelogram LB, they are (by the 26. of the sixth) about one and the same dimetient. Let their dimetient be EB, and make complete the whole figure, and for that the figure LF is equal unto the figure LH (by the 36. of the first) for the base FG is equal unto the base GH, therefore the figure LF is greater than the figure KE. But the figure LF is equal unto the figure DL (by the 43. of the first). Wherefore the figure DL is greater than the figure KE: put the figure KD common to them both. Wherefore the whole parallelogram AL is greater than the whole parallelogram AE: which was required to be proved.   

The 8. Problem. The 28. Proposition. Upon a right line given, to apply a parallelogram equal to a rectiline figure given, & wanting in figure by a parallelogram like unto a parallelogram given. Now it behoveth that the rectiline figure given, whereunto the parallelogram applied must be equal, be not greater than that parallelogram, which so is applied upon the half line, that the defects shall be like, namely, the defect of the parallelogram applied upon the half line, and the defect of the parallelogram to be applied (whose defect is required to be like unto the parallelogram given).  
SVppose the right line given to be AB, and let the rectiline figure given whereunto is required to apply upon the right line AB an equal rectiline figure be C, which figure C, let not be greater than that parallelogram which is so applied upon the half line, that the defects shall be like, namely, the defect of the parallelogram applied upon the half line, and the defect of the parallelogram to be applied (whose defect is required to be like unto the parallelogram given). And let the figure whereunto the defect or want of the parallelogram is required to be like D. Now it is required upon the right line given AB, to describe unto the rectiline figure given C, an equal parallelogram wanting in figure by a parallelogram like unto D. 
Construction.  Let the line AB (by the 10. of the first) be divided into two equal parts in the point E. And (by the 18. of the sixth) upon the line ●B describe a rectiline figure EBFG like unto the parallelogram D and in like sort situate, which shall also be a parallelogram. And make complete the parallelogram AG. 
Two cases in this Proposition.  Now then the parallelogram AG is either equal unto the rectiline figure C, or greater than it by supposition. If the parallelogram AG be equal unto the rectiline figure C, then is that done which we ●ought for. 
The first case.  For then upon the right line AB is described unto the rectiline figure given C an equal parallelogram AG wanting in figure by the parallelogram GB, 
The second case.  which is like unto the parallelogram D. But if AG be not equal unto C then is AG greater than C, but AG is equal unto GB (by the first of the sixth). Wherefore also GB is greater than HUNDRED Take the excess of the rectiline figure BG above the rectiline figure C (by that which Pelitarius addeth after the 4●. of the first) And unto that excess (by the 15. of the sixth) describe an equal rectiline figure KLMN like and in like sort situate unto the rectiline figure D. But the rectiline figure D is like unto the rectiline GB, wherefore also the rectiline figure KLMN is like unto the rectiline figure GB (by the 25. of the sixth) Now then let the sides KL and GE be sides of like proportion, let also the sides LM and GF be sides of like proportion. And forasmuch as the parallelogram GB is equal unto the figures C and KM, therefore the parallelogram GB is greater than the parallelogram KM. Wherefore also the side GE is greater than the side KL, and the side GF is greater than the side LM, unto the side KL put an equal line GO (by the 2. of the first) and likewise unto the side LM put an equal line GP. And make perfect the parallelogram OGPX. Wherefore the parallelogram GX is equal & like unto the parallelogram KM. But the parallelogram KM is like unto the parallelogram GB. Wherefore also the parallelogram GX is like unto the parallelogram GB. Wherefore the parallelograms GX and GB are (by the 26. of the sixth) about one and the self same dimecient. Let their dimecient be GB, and make complete the figure. Now forasmuch as the parallelogram BG is equal unto the rectiline figure C, and unto the parallelogram KM, and the parallelogram GX, which is part of the parallelogram GB, is equal unto KM. Wherefore the Gnomon remaining YQV is equal unto the rectiline figure remaining, namely, to C. And forasmuch as the supplement PR is equal unto the supplement OS, put the parallelogram XB common unto them both. Wherefore the whole parallelogram PB is equal unto the whole parallelogram OB. But the parallelogram OB is equal unto the parallelogram TE by the 1. of the sixth, (for the side AE is equal unto the side EB) Wherefore the parallelogram TE is equal unto the parallelogram PB. Put the parallelogram OS common to them both. Wherefore the whole parallelogram TS is equal unto the whole gnomon YQV. But it is proved that the gnomon YQV is equal unto the rectiline figure C. Wherefore also the parallelogram TS is equal unto the rectiline figure C. Wherefore upon the right line given AB is applied a parallelogram TS equal unto the rectiline figure given C, and wanting in figure by a parallelogram XB which is like unto the parallelogram given D, for the parallelogram XB is like unto the parallelogram GX: which was required to be done.  

¶ A Corollary added by Flussates.  
Hereby it is manifest, that if upon a right line be applied a parallelogram wanting in figure by a square, 
A Corollary added by Flussates, and is put of Theon as an assumpt be●ore the 17. proposition of the tenth book: which ●or that it followeth of this proposition I thought it not amiss here to place.  the parallelogram applied shall be equal to the rectangle figure which is contained under the segments of the line given which are made by the application. For the rest of the line is equal to the other side of the parallelogram applied. For that they are sides of one & the self same square, as the parallelogram AG is contained under the lines AD and DB, or DG which is equal to DB.    

The 9 Problem. The 29. Proposition. Upon a right line given to apply a parallelogram equal unto a rectiline figure given, and exceeding in figure by a parallelogram like unto a parallelogram given.  
SVppose the right line given to be AB, and let the rectiline figure given whereunto is required upon the line AB to apply an equal parallelogram be C: let also the parallelogram whereunto the excess is required to be like, be D. Now it is required upon the right line AB to apply in parallelogram equal unto the rectiline figure C, and exceeding in figure by a parallelogram like unto the parallelogram D. 
Construction.  Let the line AB be (by the 10. of the first) divided into two equal parts in the point E. And upon the line EB (by the 18. of the sixth) describe a parallelogram BF like unto the figure D, and in like sort situate. And unto both these figures BF and C, describe an equal rectiline figure GH like unto the figure D and in like sort situate (by the 25. of the sixth). Wherefore the parallelogram GH is (by the 21. of the sixth) like unto the parallelogram BF. Let the sides KH and FL be sides of like proportion, and so also let the sides KG and FE be. And forasmuch as the parallelogram GH is (by construction) greater then the parallelogram FB, therefore the line KH is greater than the line FL, and the line KG is greater than the line FE. Extend the lines FL and FE to the points M & N, and unto the line KH put an equal line FLM, and likewise unto the line KG put an equal line FEN: and make perfect the figure MN. Wherefore the parallelogram MN is equal and like unto the parallelogram GH. But the parallelogram GH is like unto the parallelogram EL. Wherefore also the parallelogram MN is like unto the parallelogram EL. Wherefore the parallelograms EL and MN are (by the 26. of the sixth) about one and the same dimetient. Let the said dimetient be FO, and make perfect the figure. 
Demonstration.  Now forasmuch as the parallelogram GH is equal unto the figures EL and C. But by construction the parallelogram GH is equal unto the parallelogram MN. Wherefore the parallelogram MN is equal unto the figures EL and C. Take away the figure EL which is common to them both. Wherefore the Gnomon remaining, namely, VYX, is equal unto the rectiline figure C. And forasmuch as the line AE is equal unto the line EB, therefore the parallelogram A N is (by the 36. of the first) equal unto the parallelogram NB, that is, unto the parallelogram LP, which (by the 43. of the first) is equal unto the parallelogram NB. Add the parallelogram BO common to them both. Wherefore the whole parallelogram AO is equal unto the Gnomon VYX. But the Gnomon VYX is equal unto the rectiline figure C. Wherefore the parallelogram AO is equal unto the rectiline figure C. Wherefore upon the right line given AB is applied the parallelogram AO equal unto the rectiline figure given C, and exceeding in figure by the parallelogram QP which is like unto the parallelogram given D. For the parallelogram D is like unto the parallelogram BF: and the parallelogram BF is like unto the parallelogram PQ: for they are about one and the self same dimetient: which was required to be done.   

The 10. Problem. The 30. Proposition. To divide a right line given by an extreme and mean proportion.  
SVppose the right line given to be AB. It is required to divide the line AB by an extreme and mean proportion. 
Construction.  Upon the line AB describe (by the 46. of the first) a square BC. And upon the line AC (by the 29. of the sixth) apply a parallelogram CD equal unto the square BC, 
Demonstration.  and exceeding in figure by the figure AD like unto the figure BC. Now BC is a square. Wherefore also AD is a square. And forasmuch as BC is equal unto CD, take away the figure CE which is common to them both. Wherefore the figure remaining, namely, BF, is equal to the figure remaining, namely, to AD, and the angle E of the one is equal unto the angle E of the other. Wherefore (by the 2. definition of the sixth, and by the 14. of the sixth) the sides of the figures BF and DA, which contain the equal angles, are reciprocal. Wherefore as the side FE is to the side ED, so is the side AE to the side EB. But the side FE is equal unto the line AC, that is, unto the line AB, and the side ED is equal unto the line AE (by the 34. of the first). Wherefore as the line BA is to the line AE, so is the line AE to the line EB. But the line AB is greater than the line AE. Wherefore also the line AE is greater than the line EB. Wherefore the right line AB is divided by an extreme and mean proportion in the point E: and the greater segment thereof is AE: which was required to be done.  

another way.  

another way.  Suppose the right line given to be AB. It is required to divide the line AB by an extreme and mean proportion. Divide the line AB in the point C (by the 11. of the second) in such sort that the rectangle figure comprehended under the lines AB and BC may be equal unto the square described of the line CA And forasmuch as that which is comprehended under the lines AB & BC, is equal unto the square made of the line AC, therefore as the line BA is to the line AC, so (by the 17. of the sixth) is the line AC to the lin● CB. Wherefore the line AB is divided by an extreme & mean proportion in the point C. which was required to be done.    

The 21. Theorem. The 31. Proposition. In rectangle triangles the figure made of the side subtending the right angle, is equal unto the figures made of the sides comprehending the right angle, so that the said thr●e figures b● b● like and in like sort described.  
SVppose that there be a triangle ABC, whose angle BAC let be a right angle. Then I say that the figure which is described of the line BC is equal unto the two figures which are described of the lines BA & AC, the said three figures being like the one to the other, and in like sort described. From the point A (by the 12. of the first) let there be drawn unto the line BC a perpendicular line AD. 
Construction.  Now forasmuch as in the rectangle triangle ABC is drawn from the right angle A unto the base BC a perpendicular line AD, 
Demonstration.  therefore the triangles ABD, and ADC set upon the perpendicular line, are like unto the whole triangle ABC, and also like the one to the other (by the 8. of the sixth). And forasmuch as the triangle ABC is like unto the triangle ABD, therefore as the line CB is to the line BA, so is the line AB to the line BD. Now for that there are three right lines proportional, therefore (by the 2. corollary of the 20. of the sixth) as the first is to the third, so is the figure made of the first, to the figure made of the second, the said figures being like and in like sort described. Wherefore as the line BC is to the line BD, so is the figure made of the line BC to the figure made of the line BA, they being like and in like sort described. And by the same reason as the line BC is to the line CD, so is the figure made of the line BC, to the figure made of the line CA, they being like and in like sort described. Wherefore as the line BC is to the lines BD and DC, so is the figure made of the line BC to the figures made of the lines BA and AC, they being like and in like sort described. But the line BC is equal unto the lines BD and DC, wherefore the figure made of the line BC is equal unto the figures made of the lines BA and AC, they being like and in like sort described. Wherefore in rectangle triangles the figure made of the side subtending the right angle is equal unto the figures made of the sides comprehending the right angle, so that the said thr● figures be like and in like sort described: which was required to be proved.  

another way.  
Forasmuch as (by the first corollary of the 20. of the sixth) like rectiline figures are in double proportion to that that the sides of like proportion are, therefore the rectiline figure made of the line BC is unto t●e rectiline figure ●ade of the line BA in double proportion to that that the line CB is to the line BA, and (by the same) the square also made of the line BC is unto the square made of the line BA in double proportion to that that the line CB is unto the line BA. Wherefore also as the rectiline figure made of the line CB is to the rectiline figure made of the line BA, so is the square made of the line CB to the square made of the line BA. And by the same reason also, as the rectiline figure made of the line BC is to the rectiline figure made of the line CA, so is the square made of the line BC to the square made of the line CA Wherefore also as the rectiline figure made of the line BC is to the rectiline figures made of the lines BA and AC, so is the square made of the line BC to the squares made of the lines BA and AC. But the square made of the line BC is equal unto the squares made of the lines BA and AC (by the 47. of the first) Wherefore also the rectiline figure made of the line BC is equal unto the rectiline figures made of the lines BA and AC, the said three figures being like and in like sort described.  

The converse of this Proposition after Campane.  

The converse of the former proposition.  If the figure described of one of the sides of a triangle be equal to the figures which are described of the two other sides, the said figures being like and in like sort described, the triangle shall be a rectangle triangle.  
Suppose that ABC be a triangle, and let the figure described of the side BC be equal to the two figures described of the sides AB and AC, the said figures being like, and in like sort described. Then I say that the angle A is a right angle. Let the angle CAD be a right angle, and put the line AD equal to the line AB, and draw a line from D to C. Now then by this 31. proposition, the figure made of the line CD is equal to the two figures made of the lines AC and AD, the said figures being like and in like sort described. Wherefore also it is equal unto the figure made of the line BC, which is by supposition equal to the two figures made of the lines AC and AD (for the line AD is put equal to the line AB) wherefore the line DC is squall to the line BC. Wherefore (by the 8. of the first) the angle BAC is a right angle, which was required to be proved.     

The 22. Theorem. The 32. Proposition. If two triangles be set together at one angle, having two sides of the one proportional to two sides of the other, so that their sides of like proportion be also parallels: then the other sides remaining of those triangles shall be in one right line.  
SVppose the two triangles to be ABC, and DCE, and let two of their sides AC & DC make an angle ACD, and let the said triangles have two sides of the one, namely, BA and AC proportional to two sides of the other, 
Demonstration.  namely, to DC and DE, so that as AB is to AC, so let DC be to DE. And let AB be a parallel unto DC, and AC a parallel unto DE. Then I say, that the lines BC and CE are in one right line. For forasmuch as the line AB is a parallel unto the line DC, and upon them lighteth a right line AC: therefore (by the 29. of the first) the alternate angles BAC and ACD are equal the one to the other. And by the same reason the angle CDE is equal unto the same angle ACD. Wherefore the angle BAC is equal unto the angle CDE. And forasmuch as there are two triangles ABC and DCE, having the angle A of the one equal to the angle D of the other, and the sides about the equal angles are (by supposition) proportional, that is, as the line BA is to the line AC, so is the line CD to the line DE, therefore the triangle ABC is (by the 6. of the sixth) equiangle unto the triangle DCE. Wherefore the angle ABC is equal unto the angle DCE. And it is proved, that the angle ACD is equal unto the angle BAC. Wherefore the whole angle ACE is equal unto the two angles ABC and BAC. Put the angle ACB common to them both. Wherefore the angles ACE and ACB are equal unto the angles CAB, ACB, & CBA. But the angles CAB, ACB, and CBA, are (by the 32. of the first) equal unto two right angles. Wherefore also the angles ACE and ACB are equal to two right angles. Now then unto the right line AC, and unto the point in it C, are drawn two right lines BC and CE, not on one and the same side, making the side angles ACE & ACB equal to two right angles. Wherefore the lines BC and CE (by the 14. of the first) are set directly and do make one right line. If therefore two triangles be set together at one angle having two sides of the one proportional to two sides of the other, so that their sides of like proportion be also parallels: then the sides remaining of those triangles shall be in one right line: which was required to be proved.  
Although Euclid doth not distinctly set forth the manner of proportion of like rectiline figures, as he did of lines in the 10. Proposition of this Book, and in the 3. following it, yet as Flussates noteth, is that not hard to be done by the 22. of this Book. ●or two like rectiline figures being given to find out a third proportionally also between two rectiline superficieces given to find out a mean proportional (which we before taught to do by Pelitarius after the 24. Proposition of this book): and moreover three like rectiline figures being given to find out a fourth proportional like and in like sort described, and such kind of proportions, are easy to be found out by the proportions of lines. As thus. If unto two sides of like proportion we should found out a third proportional by the 11. of this boke● the rectiline figure described upon that line shall be the third rectiline figure proportional with the two first figures given by the 22. of this book. And if between two sides of like proportion be taken a mean proportional by the 13. of this Book: the rectiline ●igure described upon the said mean shall likewise be a mean proportional between the two rectiline figures given by the same 22. of the sixth. And so if unto three sides ge●en be found out the fourth side proportional (by the 12. of this book) the rectiline ●igure described upon the said fourth line shall be the fourth rectiline figure proportional. For if the right lines be proportional, the rectiline figures described upon them shall also be proportional, so that the said rectiline ●igures be like & in like sort described by the said 22. of the sixth.   

The 23. Theorem. The 33. Proposition. In equal circles, the angles have one and the self same proportion that the circumferences have, wherein they consist, whether the angles be set at the centres or at the circumferences. And in like sort are the sectors which are described upon the centres.  
SVppose the equal circles to be ABC, and DEF, whose centres let be G and H, and let the angles set at their centres G and H, be BGC, and EHF: and let the angles set at their circumferences be BAC and EDF. Then I say that as the circumference BC is to the circumference EF so is the angle BGC to the angle EHF: and the angle BAC to the angle EDF, and moreover the sector GBC to the sector HEF. Unto the circumference BC (by the 38. of the third) put as many equal circumferences in order as you will, namely, CK and KL, and unto the circumference EF put also as many equal circumferences in number as you will, namely, FM, and MN. And draw these right lines GK, 
That the angles at th● centre are in proportion the one to the other, as the circumferences whereon they are.  GL, HM, and HN. Now forasmuch as the circumferences BC, CK, and KL are equal the one to the other, the angles also BGC, and CGK, and KGL are (by the 27. of the third) equal the one to the other. Therefore how multiplex the circumference BL is to the circumference BC, so multiplex is the angle BGL to the angle BGC. And by the same reason also, how multiplex the circumference NE is to the circumference EF, so multiplex is the angle NHE to the angle EHF. Wherefore if the circumference BL be equal unto the circumference EN, the angle BGL is equal unto the angle EHN, and if the circumference BL be greater than the circumference EN, the angle BGL is greater than the angle NHE, and if the circumference be less, the angle is less. Now then there are four magnitudes, namely, the two circumferences BC and EF, and the two angles that is, BGC, and EHF, and to the circumference BC and to the angle BGC, that is, to the first and third are taken equemultiplices, namely, the circumference BL, and the angle BGL, and likewise to the circumference EF, and to the angle EHF, that is, to the second and fourth, are taken certain other equemultiplices, namely, the circumference EN and the angle EHN. And it is proved, that if the circumference BL exceed the circumference EN, the angle also BGL exceedeth the angle EHN. And if the circumference be equal, the angle is equal, and if the circumference be less, the angle also is less. Wherefore (by the 6. definition of the fifth) as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF●  
But as the angle BGC is to the angle EHF, 
That the angles at the circumferences are so also.  so is the angle BAC to the angle EDF, for the angle BGC is double to the angle BAC, and the angle EHF is also double to the angle EDF (by the 20. of the third) Wherefore as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. Wherefore in equal circles the angles are in one and the self same proportion that their circumferences are, whether the angles be set at the centres, or at the circumferences: which was required to be proved.  
I say moreover that as the circumference BC is to the circumference EF, 
That the sectors are so also.  so is the sector GBC to the sector HEF. Draw these lines BC and CK. And in the circumferences BC and CK take points at all adventures, namely, P and O. And draw lines from B to P, and from P to C, from C to O, and from O to K. And forasmuch as (by the 15. definition of the first) the two lines BG and GC are equal unto the two lines CG and GK, and they also comprehend equal angles, therefore (by the 4. of the first) the base BC is equal unto the base CK, & the triangle GBC is equal unto the triangle GCK. And seeing that the circumference BC is equal unto the circumference CK, therefore the circumference remaining of the whole circle ABC, namely, the circumference BLAKOC, is equal unto the circumference remaining of the self same circle ABC, namely, to the circumference CPBLAK. Wherefore the angle BPC is equal unto the angle COK (by the 27. of the third) Wherefore (by the 10. definition of the third) the segment BPC is like unto the segment COK, and they are set upon equal right lines BC and KC. But like segments of circles which consist upon equal right lines are also equal the one to the other (by the 24. of the third). Wherefore the segment BPC is equal unto the segment COK. And the triangle GBC is equal unto the triangle GCK. Wherefore the sector GBC is equal unto the sector GCK. And by the same reason also, the sector GKL is equal unto either of the sectors GBC and GCK. Wherefore the three sectors GBC and GCK, and GKL, are equal the one to the other. And by the same reason also the sectors HEF and HFM, and HMN are equal the one to the other. Wherefore how multiplex the circumference BL is to the circumference BC so multiplex is the sector GLB to the sector GBC. And by the same reason how multiplex the circumference NE is to the circumference EF, so multiplex is the sector HEN to the sector HEF. If therefore the circumference BL be equal unto the circumference EN, the sector also BGL is equal unto the sector EHN. And if the circumference BL exceed the circumference EN, the sector also BGL exceedeth the sector EHN. And if the circumference be less, the sector also is less. Now than there are four magnitudes, namely, the two circumferences BC and EF, and the two sectors GBC & HEF, and to the circumference BC, & to the sector GBC, namely, to the first and the third, are taken equemultiplices, that is, the circumference BL, and the sector GBL, and likewise to the circumference EF, and to the sector HEF, namely, to the second and fourth, are taken certain other equimultiplices, namely, the circumference EN and the sector HEN. And it is proved that if the circumference BL exceed the circumference EN, the sector also BGL exceedeth the sector EHN. And if the circumference be equal, the segment also is equal, and if the circumference be less, the segment also is less. Wherefore (by the conversion of the sixth definition of the fifth) as the circumference BC is to the circumference EF, so is the sector GBC unto the sector HEF: which is all that was required to be proved.  

Corollary. And hereby it is manifest, that as the sector is to the sector, so is angle to angle by the 11. of the fifth.  
Flussates here addeth five Propositions whereof one is a Problem having three Corollaryes following of it, and the rest are Theorems: which for that they are both witty, & also serve to great use, as we shall afterward see, I thought not good to omit, but have here placed them: but only that I have not put them to follow in order with the Propositions of Euclid as he hath done.  

¶ The first Proposition added by Flussates.  
To describe two rectiline figures equal and like unto a rectiline figure given and in like sort situate, which shall have also a proportion given.  
Suppose that the rectiline figure given be ABH. 
Construction of the Problem.  And let the proportion given be the proportion of the lines GC and CD. And (by the 10. of this book) divide the line AB like unto the line GD in the point E (so that as the line GC is to the line CD, so let the line AE be to the line EB). And upon the line AB describe a semicircle AFB. And from the point E erect (by the 11. of the first) unto the line AB a perpendicular line EF cutting the circumference in the point F. And draw these lines AF and FB. And upon either of these lines describe rectiline figures like unto the rectiline figure AHB and in like sort situate (by the 18. of the sixth): which let be AKF, & FIB. Then I say, that the rectiline figures AKF, and FIB, have the proportion given (namely, the proportion of the line GC to the line CD) and are equal to the rectiline figure given ABH unto which they are described like and in like sort situate. For forasmuch as AFB is a semicircle, therefore the angle AFB is a right angle (by the 31. of the third) and FE is a perpendicular line. 
Demonstartion of the same.  Wherefore (by the 8. of this book) the triangles AFE and FBE are like both to the whole triangle AFB and also the one to the other. Wherefore (by the 4. of this book) as the line AF is to the line FB, so is the line AE to the line EF, and the line EF to the line EB, which are sides containing equal angles. Wherefore (by the 22. of this book) as the rectiline figure described of the line AF is to the rectiline figure described of the line FB, so is the rectiline figure described of the line AE to the rectiline figure described of the line EF, the said rectiline figures being like and in like sort situate. But as the rectiline figure described of the line AE being the ●irst, is to the rectiline ●igure described of the line EF being the second, so is the line AE the first, 10. the line ●B the third (by the 2. Corollary of the 20. of this book). Wherefore the rectiline figure described of the line AF is to the rectiline figure described of the line FB, as the line A● is to the line EB. But the line AE is to, the line EB (by construction) as the line GC is to the line CD. Wherefore (by the 11. of the fift) as the line GC is to the line CD, so is the rectiline ●igure described of the line AF to the rectiline ●igure described 〈◊〉 the line ●B, the said rectiline figures being like and in like sort described. But the 〈…〉 described o● the lines AF and FB, are equal to the rectiline ●igure d●●●●●bed o● the line AB, unto which they are (by construction) described like and in like sort situate. Wherefore there are described two rectiline figures AKF and FIB equal and like unto the rectiline figure given ABH and in like sort situate, and they ha●e also the one to the other the proportion given, namely, the proportion of the line GC to the line CD: which was required to be done.   

¶ The first Corollary.  
To resolve a rectiline figure give into two like rectiline ●igures which shall hau● also a proporti● ge●en. For i● there be put three right lines in the proportion given, and if the line AB be cu● in the same proportion that the first line is to the third, 
The first Corollary.  the rectiline ●igures described of the line● A● and FB (which figures have the same proportion that the lines AE and EB have) shall be in double proportion to that which the lines AF and FB are (by the ●irs●●orollary o● the 20. o● this book). Wherefore the right lines AF and FB are the o●e to the other in the same proportion that the first of the three lines put is to the 〈◊〉. ●or t●e 〈◊〉 line to the third, namely, the line AE to the line EB is in double proportion that it is to the second, by the 10. definition of the fi●t.   

¶ The second Corollary.  
Hereby may we learn, how from a rectiline ●igure given to take away a part appointed, lea●ing, the rest of the rectiline ●igure like unto the whole. 
The second Corollary.  For if from the right line AB be cut of a part appointed, namely, EB (by the 9 of this book) as the line AE is to the line EB, so is the rectiline ●igure described of the line AF to the rectiline figure described of the line FB (the said figure's being supposed to be like both the one to the other and also to the rectiline ●igure described of the line AB, and being also in like sort situate). Wherefore taking away from the rectiline ●igure described of the line AB, the rectiline figure described of the line FB, the residue, namely, the rectiline figure described of the line AF shall be both like unto the whole rectiline ●igure given described of the line AB, and in like sort situate.   

¶ The third Corollary.  
To compose two like rectiline ●igures into one rectiline figure like and equal to the same figures. Let their sides of like proportion be set so that they make a right angle, 
The third Corollary.  as the lines AF and FB are. And upon the line subtending the said angle, namely, the line AB, describe a rectiline ●igure like unto the rectiline figures given and in like sort situate (by the 18. of this book) and the same shall be equal to the two rectiline figures given (by the 31. of this book).   

¶ The second Proposition.  
If two right lines cut the one the other obtuseangled wise, and from the ends of the lines which ●ut the one the other be drawn perpendicular lines to either line: the lines which are between the ends and the perpendicular lines are cut reciprocally.  
Suppose that there be two right lines AB and GD cutting the one the other in the point E, and making an obtuse angle in the section E. And from the ends of the lines, namely, A and G, let there be drawn to either line perpendicular lines, namely, from the point A to the line GD, which let be AD, and from the point G to the right line AB: which let be GB. Then I say, that the right lines AB and GD do, between the end A and the perpendicular B, and the end G and the perpendicular D, cut the one the other reciprocally in the point E: so that as the line AE is to the line ED, so is the line GE to the line EB. 
Demonstration of this proposition.  For forasmuch as the angles ADE and GBE, are right angles, therefore they are equal. But the angles AED and GEB are also equal (by the 15. of the first). Wherefore the angles remaining, namely, EAD, & EGB, are equal (by the Corollary of the 32. of the first). Wherefore the triangles AED and GEH, are equiangle. Wherefore the sides about the equal angles shall be proportional (by the 4. of the sixth). Wherefore as the line AE is to the line ED, so is the line GE to the line EB. If therefore two right lines cut the one the other obtuseangled wife. etc.: which was required to be proved.   

¶ The third Proposition.  
If two right lines make an acute angle, and from their ends be drawn to each line perpendicular lines cutting them: the two right lines given shall be reciprocally proportional as the segments which are about the angle.  
Suppose that there be two right lines AB and GB, making an acute angle ABG. And from the points A and G let there be drawn unto the lines AB and GB perpendicular lines AC and GE, cutting the lines AB and GB in the points E and ●. Then I say, that the lines, namely AB to GB, are reciprocally proportional, as the segments, namely, CB to EB which are about the acute angle B. 
Demonstration of this propositions  For forasmuch as th● right angles ACB and GER are equal, and the angle● ABG is common to the triangles ABC, and GBE ● therefore the angles remaining BAC and EGB are equal (by the Corollary of the 32. of the first). Wherefore the triangles ABC and GBE are equiangle. Wherefore the side● about the equal angles are proportional (by the 4. of the six) ● so that, as the line AB is to the line FC, so is the line GB to the line BE. Wherefore alternately as the line AB is to the line GB so is the line CB to the line BE. If therefore two right lines mak● a●●c●te angle● & c● which was required to be proved.   

● The fourth Propositions  
If in a circle be drawn two right lines cutting the one the other, the sections of the one to the sections of the other shall be reciprocally proportional.  
In the circle, AGB let these two right lines 〈…〉 one the other in the point E. 
Demonstration of this proposition.  Th●● I say, that reciprocally 〈◊〉 ●h● line AE is to the line ED, so is the line GE to the line EB. For forasmuch as (by the 35. of the third) the rectangle figure contained under the lines AE and EB is equal to the rectangle figure contained under the lines GE and ED, but in equal rectangle parallelograms the sides about the equal angles are reciprocal (by the 14. of the sixth). Therefore the line AE is to the line ED reciprocally as the line GE is to the line EB (by the second definition of the sixth). If therefore in a circle be drawn two right lines. etc.: which was required to be proved.   

¶ The fift Proposition.  
If from a point given be drawn in a plain superficies two right lines to the concave circumference of a circle: they shall be reciprocally proportional with their parts taken without the circle. And moreover a right line drawn from the said point & touching the circle, shall be a mean proportional between the whole line and the utter segment.  
Suppose that there be a circle ABD, and without it take a certain point, namely, G. And from the point G draw unto the concave circumference two right lines GB and GD, cutting the circle in the points C and E. And let the line GA touch the circle in the point A. Then I say, that the lines, namely, GB to GD are reciprocally as their parts taken without the circle, namely, as GC to GOE 
Demonstration of the first part of this proposition.  For forasmuch as (by the Corollary of the 36. of the third) the rectangle figure contained under the lines GB and GE is equal to the rectangle figure contained under the lines GD and GC, therefore (by the 14. of the sixth) reciprocally as the line GB is to the line GD, so is the line GC to the line GE, for they are sides containing equal angles. I say moreover, that between the lines GB and GE, or between the lines GD and GC the touch line GA is a mean proportional. 
Demonstration of the second part.  For forasmuch as the rectangle figure comprehended under the lines GB and GE is equal to the square made of the line AG (by the 36. of the third) it followeth that the touch line GA is a mean proportional between the extremes GB and GE (by the second part of the 17. of the sixth) for that by that Proposition the lines GB, GA, and GE are proportional. And by the same reason may it be proved that the line GA is a mean proportional between the lines GD and GC, and so of all others. If therefore from a point geuen● & c● which was required to be demonstrated.     
The end of the sixth book of Euclides Elements.   

¶ The seventh book of Euclides Elements.  
HITHERTO IN THE SIX books before hath Euclid passed through, and entreated of the Elements of Geometry without the aid and succour of number. But the matters which remain to be taught and to be spoken of in these his Geometrical books which follow as in the tenth, eleventh, and so forth, he could by no means fully and clearly make plain & demonstrate, without the help and aid of numbers. In the tenth is entreated of lines irrational and uncertain, and that of many & sundry kinds: and in the eleventh & the other following he teacheth the natures of bodies, and compareth their sides and lines together. All which for the most part are also irrational. And as rational quantities, and the comparisons and proportions of them, cannot be known, nor exactly tried, but by the mean of number, in which they are first seen and perceived: even so likewise cannot irrational quantities be known and found out without number. As straightness is the trial of crookedness, and inequality is tried by equality: so are quantities irrational perceived and known by quantities rational: which are ●irst and chiefly found among numbers. Wherefore in these three books following, being as it were in the midst of his Elements, he is compelled of necessity to entreat of numbers, 
Why Euclid in the midst of his works was compelled to add these three books of numbers.  although not so fully, as the nature of numbers requireth, yet so much as shall seem to be fit, and sufficiently to serve for his purpose. Whereby is seen the necessity, that the one Art, namely, Geometry, hath of the other, namely, of Arithmetic. And also of what excellency and worthiness Arithmetic is above Geometry: 
Arithmetic of more excellency than Geometry.  in that, Geometry borroweth of it principles, aid, and succour, and is as it were maimed with out it. Whereas Arithmetic is of itself, sufficient and needeth not at all any aid of Geometry, but is absolute and perfect in itself, and may well be taught and attained unto without it. Again the matter or subject where about Geometry is occupied, which are lines, figures, and bodies, are such as offer themselves to the senses, as triangles, squares, circles, cubes, and other are seen & judged to be such as they are, by the sight: but number, which is the subject and matter of Arithmetic, falleth under no sense, nor is represented by any shape, form, or figure: and therefore cannot be judged by any sense, but only by consideration of the mind, and understanding. Now things sensible are far under in degree then are things intellectual: 
Things intellectual of more worthiness the● things sensible.  and are of nature much more gross than they. Wherefore number, as being only intellectual, is more pure, more immaterial, and more subtle, far then is magnitude: and extendeth itself farther. For Arithmetic, not only aideth Geometry: but ministereth principles, and grounds to many other, nay rather to all other sciences and arts. 
Arithmetic ministereth prin●ciples and grounds in a manner to all sciences.  As to music, Astronomy, natural philosophy, perspective, with others. What other thing is in music entreated of, than number contracted to sound and voice? In Astronomy, who without the knowledge of number can do any thing, either in searching out of the motions of the heavens, and their courses, either in judging and foreshowing the effects of them? In natural philosophy, it is of no small force. The wisest and best learned philosophers that have been, as Pythagoras, Timeus, Plato, and their followers, found out & taught most pithily and purely, the secret and hidden knowledge of the nature and condition of all things, by numbers, and by the proprieties and passions of them. Of what force number is in perspective, let him declare and judge, who hath any thing traveled therein. Yea to be short, what can be worthily and with praise practised in common life of any man of any condition, without the knowledge of number. Yea it hath been taught of the chiefest amongst philosophers, that all natural things are framed, and have their constitution of number. Boetius saith Hoc fuit principals in anim● c●●ditoris exemplar: 
Boetius: Cap. 2. Lib. prim. Arithmeti.  Number (saith he) was the principal example and patron in the mind of the creator of the world. Doth not that great philosopher Timaus in his book, 
Timaus.  & also Plato in his Tim●●, following him, show how the soul is composed of harmonical numbers, and consonantes of music. Number compasseth all things, and is (after these men) the being and very essence of all things. And ministereth aid and help, as to all other knowledges, so also no small to Geometry. Which thing causeth Euclid● in the midst of his book of Geometry, to insert and place these three books of Arithmetic: as without the aid of which he could not well pass any father.  
In this seventh book, he ●irst placeth the general principles, and first grounds of Arithmetic, and setteth the definitions of number or multitude and the kinds thereof: as in the first book, he did of magnitude and the kinds and parts thereof. 
The argument of the seventh book.  After that he entreateth of numbers generally, and of their parts: and searcheth and demonstrateth in general the most common passions and proprieties of the same, and chiefly of numbers prime or incommensurable, and of numbers composed or commensurable and of their proprieties and partly also of the comparison o● proportion of one number to an other.  

¶ Definitions.  


The first definition.  1 Unity is that, whereby every thing that is, is said to be on.  
As a point in magnitude, is the lest thing in magnitude, and no magnitude at all, & yet the ground and beginning of all magnitudes: even so is unity in multitude or number, the lest thing in number, and no number at all, and yet the ground and beginning of all numbers. And therefore it is here in this place, of Euclid first defined: as in the first book, for the like reason and cause was a point first defined. Unity, saith Euclid, is that whereby every thing is said to be one: that is, unity is that, whereby every thing is divided and separated from an other, and remaineth on in itself pure and distinct from all others. Otherwise, were not this unity, 
Without unity should be confusion of things.  whereby all things are sejoined the on from the other, all things should suffer mixtion and be in confusion. And where confusion is, there is no order, nor any thing can be exactly known, either what it is, or what is the nature, and what are the properties thereof. Unity therefore is that which maketh every thing to be that which it is. Boetius saith very aptly: 
●oetius in his book d● unitate & uno.  unum quodque, idea est, quia unum numero est, that is every thing therefore is (that is, therefore hath his being in nature, and is that it is) for that it is on in number. According whereunto jordane (in that most excellent and absolute work of Arithmetic which he wrote) defineth unity after this manner.  
unitas, est res per se discretio: that is, unity is properly, and of itself the difference of any thing. That is, 
another desinition of unity.  unity is that whereby every thing doth properly and essentially differ, and is an other thing from all others. Certainly a very apt definition and it maketh plain the definition here set of Euclid.   


The second definition.  2 Number is a multitude composed of unities.  
As the number of three, is a multitude composed and made of three unities. Likewise the number of five is nothing else, but the composition & putting together of five unities. Although as was before said, between a point in magnitude, and unity in multitude, there is great agreement and many thi●●●● are common to them both, (for as a point is the beginning of magnitude, so is unity the beginning of number. And as a point in magnitude is indivisible, so is also unity in number indivisible) yet in this they differ and disagree. 
Difference between a point and unity.  There is no line or magnitude made of points, as of his parts. So that although a point be the beginning of a line, yet is it no part thereof. But unity, as it is the beginning of number, so is it also a part thereof, which is somewhat more manifestly set of Boetius in an other definition of number which he giveth in his Arithmetic, 
Boetius.  which is thus.  
Numerus, est quantitat● acernus ex unitatibus profusus, that is. Number is a mass or heap of quantities produced of unities: which definition in substance is all one with the first, wherein is said most plainly, that the heap or mass, 
another desinition of number.  that is, the whole substance of the quantity of number is produced & made of unities. So that unity is as it were the very matter of number. As four unities added together are the matter whereof the number 4. is made, & each of these unities is a part of the number four, namely, a fourth part, or a quarter. Unto this definition agreeth also the definition given of jordane, 
jordane.  which is thus. Number is a quantity which gathereth together things severed a sunder. 
another definition of numbers.  As five men being in themselves severed and distinct, are by the number five brought together, as it were into one mass, and so of others. And although unity be no number, yet it containeth in it the virtue and power of all numbers, and is set and taken for them. 
Unity hath in it the virtue and power of all numbers.   
In this place (for the Farther elucidation of things, partly before set, and chief hereafter to be set, because Euclid here doth make mention of divers kinds of numbers, and also defineth the same) is to be noted, that number may be considered three manner of ways. First, number may be considered absolutely, without comparing it to any other number, 
Number considered three manner of way●.  or without applying it to any other thing, only viewing ●nd paysing what it is in itself, and in his own nature only, and what parts it hath, and what proprieties and passions. As this number six, may be considered absolutely in his own nature, that it is an even number, and that it is a perfect number, and hath many more conditions and proprieties. And so conceive ye of all other numbers, whatsoever, of 9 12. and so forth.  
another way, number may be considered by way of comparison, and in respect of some other number either as equal to itself, or as greater than itself, or as less than itself. As 12. may be considered, as compared to 12. which is equal unto it, or as to 24. which is greater than it, for 12 is the half thereof, of as to 6. which is less than it, as being the double thereof. And of this consideration of numbers ariseth and springeth all kinds and varieties of proportion: as hath before been declared in the explanation of the principles of the fift book, so that of that matter it is needless any more to be said in this place.  
Thus much of this for the declaration of the things following.   

3 A part is a less number in comparison to the greater when the less measureth the greater. 
The third definition.   
As the number 3 compared to the number 12. is a part. For 3 is a less number than is 12. and moreover it measureth 12 the greater number. For 3 taken (or added to itself) certain times (namely, 4 times) maketh 12. For 3 four times is 12. Likewise is ● a part of 8: 2 is less than 8, and taken 4 times it maketh 8. For the better understanding of this definition, and how this word part, is diversly taken in Arithmetic and in Geometry, read the declaration of the first definition of the 5. book.   

4 Parts are a less number in respect of the greater, when the less measureth not the greater. 
The fourth definition.   
As the number 3 compared to 5, is parts of 5 and not a part. For the number 3 is less than the number 5, and doth not measure 5. For taken once it maketh but 3. once 3 is 3, which is less than 5. and 3 taken twice maketh 6, which is more than 5. Wherefore it is no part of 5 but parts, namely, three fifth parts of 5. For in the number 3 there are 3 unities, and every unity is the fifth part of 5. Wherefore 3 is three fifth parts of 5, and so of others.   

5 Multiplex is a greater number in comparison of the less, when the less measureth the greater. 
The fifth definition.   
As 9 compared to 3 is multiplex, the number 9 is greater than the number 3 And moreover 3 the less number measureth 9 the greater number. For 3 taken certain times, namely, 3 times maketh 9 three times three is 9 For the more ample and full knowledge of this definition, read what is said in the explanation of the second definition of the 5 book, where multiplex is sufficiently entreated of with all his kinds.   

6 An even number is that, which may be divided into two equal parts. 
The sixth definition.   
As the number 6 may be divided into 3 and 3 which are his parts, and they are equal, the one not exceeding the other. This definition of Euclid is to be understand of two such equal parts, which joined together, make the whole number: as 3 and 3 (the equal parts of 6) joined together, make 6, for otherwise many numbers both even and odd may be divided into many equal parts, as into 4. 5. 6● or more, and therefore into 2. As 9 may be divided into 3 and 3 which are his parts, and are also equal, for the one of them exceedeth not the other: yet is not therefore this number ● an even number, for that 3 and 3 (these equal parts of 9) added together make not 9 but only 6. Likewise taking the definition so generally, every number whatsoever should be an even number● for in that sort of understanding there is no number, but that it may be divided into two equal parts: as this number 7 may be divided into 3 parts namely, 3. 1. and 3. of which two, namely, 3 and 3 are equal, yet i● not 7 an even number, because 3 and 3 added together, make not 7. Boetius therefore in the first book of his Arithmetic, 
Boetius.  for the more plains, after this manner defineth an even number: Numerun pa● est, qui potest in ●qu●lia due dividi, una medi●●●n intercedenta, that is:  
An even number is that which may be divided into two equal parts, without an unity coming between them. 
another definition of an even number.  As 8 is divided into 4 and 4 two equal parts without an● unity coming between them, which added together, make 8, so that the sense of this definition is, that an even number is that which is divided into two such equal parts, which are his two half parts.  
Here is also to be noted, 
Note.  that a part is taken in this definition, and in certain definitions following, not in the signification as it was before defined, namely, for such a part as● measureth the whole number, but for any part which helpeth to the making of the whole, and into which the whole may be resolved, so are 3 and 5 parts of 8 in this sense but not in the other sense. For neither 3 not 5 measureth ●. Pythagoras and his scholars gave an other definition of an even number (which definition 〈◊〉 also hath) after this manner. 
Pithagora●.   
An even number is that which in one and the same division is divided into the greatest, and into the lest: into the greatest, as touching space, and into the lest as touching quantity. 
another definition.  As 10. is divided into 5 & 5. which 〈◊〉 his greatest parts (which greatness of parts he calleth space) and in the same division the number 10 is divided but into two parts: but into less than two parts nothing can be divided, which thing he calleth quantity, so that 10 divided into 5 and 5. in that one division, is divided into the greatest, namely, into his halues● and into the lest, namely, into two parts, and no more.  
There is also another definition more ancient, which is thus.  

another definition.  An even number is that which may be divided into two equal parts, and into two unequal parts, but in neither division (to the constitution of the whole) to the even part is added the odd, neither to the odd is added the even. As 8 may be divided diversly, partly into even parts, as into 4 and 4. likewise into 6 and 2. and partly into odd parts, as into 5 and 3. also into 7 and 1. In all which divisions, ye see no odd part joined to an even, nor an even part joined to an odd: but if the one be even, the other is even, and if the one be odd, the other is odd. In the two first divisions, both parts were even, and in the two last, both parts were odd. It is to be considered that the two parts added together must make the whole.  
This definition is general and common to all even numbers, except to the number 2. which can not be divided into two unequal parts, but only into two unities, which are equal.  

another definition.  There is yet given an other definition of an even number, namely, thus. An even number is that which only by an unity either above it or under it, differeth from an odd. As 8 being an even number differeth from 9 an odd number, being above it but by one. And also from 7. under it, it differeth likewise by one, and so o● others.   


The seventh definition.  7 An odd number is that which cannot be divided into two equal parts: or that which only by an unity differeth from an even number.  
As the number 5 can be by no means divided into two equal parts, namely, such two which added together, shall make 5. Or by the second definition, 5 an odd number, differeth from 6 an even number above ●t, by 1. And the same 5 differeth from 4 an even number under it likewise by 1.  
〈…〉 number after this manner.  
An odd number is that which cannot be divided into two equal parts, but that an unity shall be between them As if ye divide 5 into 2 and ●. which are two equal parts, 
another definition of an odd number.  there remaineth one or an unity between them to make the whole number 5.  
There is yet an other definition of an odd number. An odd number is that, which being divided into two unequal parts howsoever, the one is ever even, and the other odd. As if 9 be divided into two parts, which added together, maketh the whole, namely, into 4 and 5. which are unequal: 
another definition.  ye see the one is even, namely, 4 and the other is odd, namely, 5. so if ye divide 9 into 6 and 3. or into 8 and 1. the one part is ever even, and the other odd.   


The eight definition.  8 A number evenly even (called in latin pariter par) is that number, which an even number measureth by an even number.  
As 8 is a number evenly even. For 4 an even number measureth 8 by 2, which is also an even number. This definition hath much troubled many, and seemeth not a true definition, for that there are many numbers which even numbers do measure, and that by even numbers, which yet are not evenly even numbers, after most men's minds: as 24. which 6 an even number doth measure by four, which is also an even number, and yet as they think is not 24 an evenly even number, for that 8 an even number doth measure also ●4 by 3. an odd number. 
Campane.  Wherefore Campane to make his sentence plain, after this manner setteth forth this definition. Pariter par est quem cuncti pares ●um numerantes, paribus vicibu● numerane, that is:  

another definition of an evenly even number.  An evenly even number is, when all the even numbers which measure it, do measure it by even times, that is, by even numbers, as 16. All the even numbers whith measure 16, as are 8. 4. and 2. do measure it by even numbers. As 8 by 2, twice 8 is 16: 4 by 4, four times 4 is 16: and 2 by 8, 8 times 2 is 16. Which particle (all even number) added by Campane maketh 24 to be no evenly even number. For that some one even number measureth it by an odd number as 8 by 3. Flussates also is plainly of this mind, 
Flussates.  that Euclid gave not this definition in such manner as it is by Theon written, for the largeness & generality thereof, ●or that it extendeth it to infinite numbers which are not evenly even as he thinketh, for which cause in place thereof, he giveth this definition.  

another definition.  A number evenly even, is that which only even numbers do measure. As 16 is measured of none but of even numbers, and therefore is evenly even. There is also of Boetius given an other definition of more facility, 
Boetius.  including in it no doubt at all, which is most commonly used of all writers, and is thus.  

another definition.  A number evenly even is that which may be divided into two even parts, and that part again into two even parts, and so continually dividing without stay 〈◊〉 come to unity. As by example● 64. may be divided into 31 and. 32. And either of these parts may be divided into two even parts, for 32 may be divided into 16 and 16. Again, 16 may be divided into 8 and 8 which are even parts, and 8 into 4 and 4. Again 4 into ● and ●, and last of all may ● be divided into one and one.   

9 A number evenly odd (called in latin pariter impar) is that which an even number measureth by an odd number. 
The ninth definition.   
As the number 6 which 2 an even number measureth by 3 an odd number, three times 2 is 6. Likewise 10. which 2. an even number measureth by 5 an odd number. In this definition also is found by all the expositors of Euclid, the same want that was found in the definition next before. And for that it extendeth itself to large, for there are infinite numbers which even numbers do measure by odd numbers, which yet after their minds are not evenly odd numbers, as for example 12. For 4 an even number, measureth 12 by ● an odd number● three times 4 is 12. yet is not 12 as they think an evenly odd number. Wherefore Campane amendeth it after his thinking, 
Campane.  by adding of this word all, as he did in the first, and defineth it after this manner.  
A number evenly odd is, when all the even numbers which measure it, do measure it by uneven times, that is, by an odd number.  
As 10. is a number evenly odd, 
another definition.  for no even number but only 2 measureth 10. and that is by 5 an odd number. But not all the even numbers which measure 12. do measure it by odd numbers. For 6 an even number measureth 12 by 2 which is also even. Wherefore 12 is not by this definition a number evenly odd. Flussates also offended with the over large generality of this definition to make the definition agreed with the thing defined putteth it after this manner. 
Flussates.   
A number evenly odd, is that which an odd number doth measure only by an even number.  
As 14. which 7. an odd number doth measure only by 2. which is an even number. 
another definition.  There is also an other definition of this kind of number commonly given of more plains, which is this.  
A number evenly odd in that which may be divided into two equal parts, but that part cannot again be divided into two equal parts: as 6. may be divided into two equal parts into 3. and 3. but neither of them can be divided into two equal parts: for that 3. is an odd number and suffereth no such division. 
another definition.    

10 A number oddly even (called in latin in pariter par) is that which an odd number measureth by an even number. 
The tenth definition.   
As the number 12: for 3. an odd number measureth 12. by 4. which is an even number: three times 4. is 12.  
This definition is not found in the greek neither was it doubtless ever in this manner written by Euclid: 
This definition not found in the Greeks.  which thing the slenderness and the imperfection thereof and the absurdities following also of the same declare most manifestly. The definition next before given is in substance all one with this, For what number soever an euen● number doth measure by an odd, the self same number doth an odd number measure by an even. As 2. an even number measureth 6. by 3. an odd number. Wherefore 3. an odd number doth also measure the same number 6. by 2. an even number. Now if these two definitions be definitions of two distinct kinds of numbers, then is this number 6. both evenly even, and also evenly odd and so is contained under two divers kinds of numbers. Which is directly against the authority of Euclid who plainly. p●ouo●h here after in the 9 book, that every number whose half is an odd number, is a number evenly odd only. Flussates hath here very well noted, that these two evenly odd, and oddly even, were taken of Euclid for on and the self same kind of number. But the number which here aught to have been placed is called of the best interpreters of Euclid, numerus pariter par & nupar, that is a number evenly even, and evenly odd. Ye● and it is so called of Euclid himself in the 34. proposition of his 9 book: which kind of number Campanus and Flussates in stead of the insufficient and unapt definition before given, assign this definition.  
A number evenly even, and evenly ●dde, is that which an even number doth measure sometime by an even number, and sometime by an odd. 
another definition.   
As the number 12: for 2. an even number, measureth 12. by 6. an even number: two times 6. i● 12. Also 4. an even number measureth the same number 12. by 3. an odd number. Add therefore is 12. a number evenly even, and evenly odd, and so of such others.  
There is also an other definition given of this kind of number by Boetius and others commonly which is thus.  

Boe●ius definition of a number evenly even, and evenly ●d  A number eue●ly even and evenly odd is that, which may be divided into two equal parts, and each of them may a●ayne be divided into two equal parts: and so forth. But this division is at length stayed, and continueth not till it come to unity. As for example 48: which may be divided into two equal parts, namely, into 24. and 24. Again 24. which is on of the parts may be divided into two equal parts 12. and 12. Again 12. into 6. and 6. And again 6 may be divided into two equal parts, into 3. and 3: but 3 cannot be divided into two equal parts. Wherefore the division there stayeth: and continueth not till it come to unity as it did in these numbers which are evenly even only.   


The eleventh definition.  11 A number oddly odd is that, which an odd number doth measure by an odd number.  
As 25, which 5. an odd number, measureth by an odd number, namely, by 5. Five times five is 25: Likewise 21. whom 7. an odd number doth measure by 3, which is likewise an odd number. Three times 7. is 21.  

Flus●ates.  Flussatus giveth this definition following of this kind of number, which is all one in substance with the former definition.  
A number oddly odd, it that, which only an odd number doth measure.  

another definition.  As 15. for no number measureth 15. but only 5. and 3: also 25: none measureth it but only 5. which is an odd number, and so of others.   


The twelfth definition.  12 A prime (or first) number is that, which only unity doth measure.  
As 5.7.11.13. For no number measureth 5, but only unity. For u unities make the number 5. So no number measureth 7, but only unity .2. taken 3. times maketh 6. which is less than 7: and 2. taken 4. times is 8, which is more than 7. And so of 11.13. and such others. So that all prime numbers, 
Prime numbers called incomposed numbers.  which also are called first numbers, and numbers uncomposed, have no part to measure them, but only unity.   


The thirteenth definition.  13 Numbers prime the one to the other are they, which only unity doth measure, being a common measure to them.  
As 15. and 22. be numbers prime the one to the other .15. of itself is no prime number, for not only unity doth measure it, but also the numbe●● 5. and 3, for ●. times 5. is lx. Likewise 22. is of itself no prime number: for it is measured by 2. and 11, besides unity. For 11. twice, or 2. eleven times, make 22. So that although neither of these two numbers 15. and 22. be a prime or incomposed number, but either have parts of his own, whereby it may be measured beside unity: yet compared together, they are prime the one to the other: for no one number doth as a common measure, measure both of them but only unity, which is a common measure to all numbers. The numbers 5. and 3. which measure 15. will not measure 22: again, the numbers 2. and 11. which measure 22, do not measure 15.   

14 A number composed, is that which some one number measureth. 
The fourteenth definition.   
A number composed is not measured only by unity, as was a prime number, but hath some number which measureth it. As 15: the number 3. measureth 15, namely, taken 5. times. Also the number 5. measureth 15, namely, taken 3. times. 5. times ●, and 3. times 5, is 15. Likewise 18. is a composed number, it is measured by these numbers 6.3.9.2. and so of others. These numbers are also called commonly second numbers, as contrary to prime or first numbers.   

15 Numbers composed the one to the other, are they, which some one number, being a common measure to them both, doth measure. 
The fifteenth definition.   
As 12. and 8. are two composed numbers the one to the other. For the number 4, is a common measure to them both: 4. taken three times maketh 12: and the same 4. taken two times maketh 8. So are 9 and 15: 3. measureth them both. Also 10. and 25: for 5. measureth both of them: and so infinitely of others. In this do numbers composed the one to the other or second numbers, differre from numbers prime the one to the other: for that two numbers being composed the one to the other, each of them severally is of necessity a composed number. As in the examples before 8. and 12. are composed numbers: likewise 9 and 15: also 10, and 25: but if they be two numbers prime the one to the other, ●it is not of necessity, that each of them severally be a prime number. As 9 and 22. are two numbers prime the one to the other: no one number measureth both of them: and yet neither of them in itself and in his own nature is a prime number, but each of them is a composed number. For 3. measureth 9, and 11. and 2. measure 22.   

16 A number is said to multiply a number, when the number multiplied, 
The sixteenth definition.  is so oftentimes added to itself, as there are in the number multiplying unities: and an other number is produced.  
In multiplication are ever required two numbers, the one is whereby ye multiply, commonly called the multiplier or multiplicant, the other is that which is multiplied. 
Two numbers required in multiplication.  The number by which an other is multiplied, namely, the multiplyer, is said to multiply. As if ye will multiply 4. by 3, then is three said to multiply 4: therefore according to this definition, because in 3. there are three unities: add 4,3. times to itself, saying 3. times 4: so shall ye bring forth an other number, namely, 12, which is the sum produced of that multiplication: and so of all other multiplications.   

17 When two numbers multiplying themselves the one the other, produce an other: 
The seventeenth definition.  the number produced is called a plain or superficial number. And the numbers which muliply themselves the one by the other, are the sides of that number.  
As let these two numbers 3. and 6, multiply the one the other, saying, 3. times 6, or six times 3, they shall produce 18. This number 18. thus produced, is called a plain number, or a superficial number. And the two multiplying numbers which produced 12, namely, 3. and 6, are the sides of the same superficial or plain number, that is, the length and breadth thereof. Likewise if 9 multiply 11, or eleven nine, there shall be produced 99 a plain number, whose sides are the two numbers 9 and 11 ● as the length and breadth of the same. They are called plain and superficial numbers, 
Why they are called superficial numbers.  because being described by their unities on a plain superficies, they represent some superficial form or figure Geometrical, having length and breadth. As ye see of this example: and so of others. And all such plain or superficial numbers do ever represent right angled figures as appeareth in the example.   


The eighteenth definition.  18 When three numbers multiplied together the one into the other, produce any number, the number produced, is called a solid number: and the numbers multiplying themselves the one into the other, are the sides thereof.  
As taking these three numbers 3.4. 5. multiply the one into the other. First 4. into 5. saying, four times 5. is 20: then multiply that number produced, namely, 20. into 3: which is the third number, so shall ye produce 60. which is a solid number: and the three numbers which produced the number, namely, 3.4. and 5. are the sides of the same. And they are called solid numbers, 
Why they are called solid numbers.  because being described by their unities, they represent solid and bodylicke figures of Geometry, which have length, breadth, and thickness. As ye see this number 60. expressed here by his unities. Whose length is his side 5, his breadth is 3, and thickness 4. And thus may ye do of all other three numbers multiplying the one the other.   


The ninetenth definition.  19 A square number is that which is equally equal: or that which is contained under two equal numbers.  
As multiply two equal numbers the one into the other. As 9 by 9 ye shall produce 81, which is a square number. Euclid calleth it a number equally equal, because it is produced of the multiplication of two equal numbers the one into the other. Which numbers are also said in the second definition to contain a square number. As in the definitions of the second book two lines are said to contain a square or a parallelogram figure. It is called a square number, 
Why it is called a square number.  because being described by his unities it representeth the figure of a square in Geometry. As ye here see doth the number 81. whose sides, that is to say, whose length and breadth, are 9 and 9, equal numbers: which also are said to contain the square number 81: and so of others.   


The twentieth definition.  20 A cube number is that which is equally equal equally: or that which is contained under three equal numbers.  
As multiply three equal numbers the one into the other, as 9, 9, and 9: first 9 by 9, so shall ye have 81: which again multiply by 9, so shall ye produce 729. which is a cube number. And Euclid calleth it a number equally equal equally, because it is produced of the multiplication of three equal numbers the one into the other: which three numbers are said in the second definition (wherein he speaketh more applying to Geometry) to contain the cube number. It is therefore called a cube number, 
Why it is called a cube number.  because being described by his unities, it representeth the form of a cube in Geometry, whose sides, that is to say, whose length, breadth, and thickness, are the three equal numbers 9, 9, and 9, of which he was produced: which three sides also are said to contain the cube number 729: behold here the description thereof.   


The twenty one definition.  21 Numbers proportional are, when the first is to the second equemultiplex, as the third is to the fourth, or the self same part, or the self same parts.  
Here he defineth which numbers are called proportional, that is, what numbers have one and the self same proportion. For example 6. to 3: and 4. to 2, are numbers proportional, and have one and th● self same proportion: for 6. the first is to 3. the second equemultiplex, as 4. the third is to 2. the fourth: ●. is double to 3: and so is 4. double to 2. Likewise these four numbers are in like proportion 3. 9.4.1●. for what part 3. is of 9, such part is 4. of 12 ●3. of 9 is a third part, so is also 4. of 12. a third part. So are these four numbers also in proportion ●. 5: 4. 10: what parts 2. are of 5, such parts are 4. of 10: 2, of 5, are two ●ift parts, likewise 4. of 10 are two fift parts. Moreover, these numbers 8.6: 1●. 9: be in proportion, for what and how many parts 8. are of 6, such & so many parts are 12. of 9: 8. of 6, is four third parts, for one third part of 6. is 2, which taken four times maketh 8: so 12. of 9, is also four third parts: for one third part of 9 is 3, which taken four times make 1●. And so conceive ye of all other proportional numbers.  
In the 〈◊〉 definition of the u book, Euclid gave a ●arre other definition of magnitudes proportional, and much unlike to this which he here giveth of numbers proportional: 
Why the definition of proportional magnitudes is unlike to the definitio of proportional numbers.  the reason is as there also was partly noted, for that there he gave a definition common to all quantities discrete and continual, rational, and irrational: and therefore was constrained to give the definition by the excess, equality or want of their equemultiplices, and that generally only: for that irrational quantities have no certain part or common measure to be measured by or known, neither can they be expressed by any certain numbers. But here in this place because in numbers there are no irrational quantities, but all are certainly known, so that both they and the proportions between them may be expressed by numbers certain and known, by reason of their parts certain, and for that they have some common measure to measure them (at the least unity which is a common measure to all numbers) he giveth here this definition of proportional numbers, by that the one is like equemultiplex to the other, or the same part, or the same parts: which definition is much easier than was the other: and is not so large, as is the other, neither extendeth itself generally to all kind of quantities rational and irrational, but containeth itself within the limits and bonds of rational quantity and numbers.   

22 Like plain numbers, and like solid numbers, are such, 
The twenty two definition.  which have their sides proportional.  
Before he showed that a plain number hath two sides, and a solid number three sides. Now he showeth by this definition which plain numbers, and which solid numbers are like the one to the other. The likeness of which numbers dependeth altogether of the proportions of the sides of these numbers. So that if the two sides of one plain number, have the same proportion the one to the other, that the two sides of the other plain number have the one to the other, then are such two plain numbers like. For an example 6 and 24 be two plain numbers, the sides of 6 be 2 and 3, two times 3 make 6: the sides of 24 be 4 and 6, four times 6 makes 24. Again the same proportion that is between 3 and 2 the sides of 6. is also between 6 and 4 the sides of 24. Wherefore 24 and 6 be two like plain and superficial numbers. And so of other plain numbers. After the same manner is it in solid numbers. If three sides of the one be in like proportion together, as are the three sides of the other, then is the one solid number like to the other. As 24 and 192 be solid numbers, the sides of 24. are 2. 3. and 4, two times there taken 4 times are 24. the sides of 192 are 4.6. and 8: for four times 6. 8 times make 192. Again the proportion of 4 to 3 is sesquitercia, the proportion of 3 to 2 is sesquialtera, which are the proportions of the sides of the one solid number, namely, of 24: the proportion between 8 and 6 is sesqu●●ercia, the proportion between 6 and 4 is sesquialtera, which are the proportions of the sides of the other solid number, namely, of 191. And they are one and the same with the proportione of the 〈◊〉 of the other wherefore th●se two solid numbers 24 and 192 be like, and so of other solid numbers.   

23 A perfect number is that, which is equal to all his parts. 
The twenty three definition.   
As the parts of 6 are 1. 2. 3. three is the half of 6, two the third part, and 1. the sixth part, and more ports 6 hath not: which thr●● paytes 1. ●. 3. added together, make 6 the whole number, whose parts they are. Wherefore 6 is a perfect number. So likewise is 28 a perfect number, the parts whereof are these numbers 14. 7. 2. and 1: 14 is the half thereof, 7 is the quarter, 4 is the seventh part, 2 is a fourteenth part, and 1 an 28 part, and these are all the parts of 28. all which, namely, 1, 2, 3, 4, 7 and 14 added together, make justly without more or l●sse 28. Wherefore 〈…〉 a perfect number, and so of others the like. This kind of numbers is very rare and seldom found. 
Perfect numbers rare & of great use in magic & in secret philosophy.  From 1 to 10, there is but one perfect number, namely, 6. From 10 to an 100, there is also but one, that is, 28. Also from 100 to 1000 there is but one which is 496. From 1000 to 10000 likewise but one. So that between every s●ay in numbering, which is ever in the tenth place, there is found but one perfect number And for their rareness and great perfection, they are of marvelous use in magic, and in the secret part of philosophy.  
This kind of number is called perfect, 
In what respect a number is perfect.  in respect 〈…〉 numbers which are imperfect. For as the nature of a perfect number standeth in this, that all parts added together are equal to the whole: and make the whole: so in an imperfect number all the parts added together are not equal to the whole, nor make the whole, but make either more or less. Wherefore of imperfect numbers there are two kinds, 
Two kinds of imperfect numbers.  the one is called abundant or a building, the other 〈◊〉, or wanting.  
A number abunding is that whose parts being all added together make more than the w●●ds number whose parts they are, as 12. is an abundant number. For all the parte● of 12. namely, 6.4.3.2. and 1. added together make 16: which are more than 12. Likewise 18. is a number abunding, all his part● namely, 9.6.3.2. and 1. added together make 20. which are more than 18: and so of others.  

A ●●mber wan●●ng●  A number diminute, or wanting is that whose parts being all added together, make less than the whole, or number whose parts they are.  
As 9 is a diminute, or wanting number, for all his parts, namely 3. and 1. (more parts he hath not) added together make only 4: which are less than 9 Also 26. is a diminute number, all his parts, namely. 13.2.1. added together make only 16: which is a number much less than 26. And so of such like.  

Common sentences.  CAmpane and Flussates here add certain common sentences, some of which for that they are in these three books following sometimes alleged, I thought good here to annex.   


●irst common sentence.  1 The less part is that which hath the greater denomination: and the greater part is that, which hath the less denomination.  
As the numbers 6. and 8. are either of them a part of the number 24: 6. is a fourth part, 4. times 6. is 24: and 8. is a third part, 3. times 8. is 24. Now forasmuch as 4 (which denominateth what part 6. is of 24) is greater than 3. (which denominateth what part 8. is of 24.) therefore is 6. a less part of 24● than is 8. and so is 8. a greater part of 24. than 6. is. And so in others.   


●●cond ●ommon sentence.  2 Whatsoever numbers are equemultiplices to one & the self same number, or to equal numbers, are also equal the one to the other.  
As if unto the number 3 be taken two numbers containing the same number four times, that is being equemultiplices to the same number three: the said two numbers shallbe equal. For 4. times 3. will ever be 12. So also will it be if unto the two equal numbers 3. & 3. be taken two numbers, the one containing the one number 3. four times, the other containing the other number 3. also four times, that is, being equemultiplices to the equal numbers 3. and 3.   


Third common sentence.  3 Those numbers to whom one and the self same number is equmultiplex, or whose equemultiplices are equal: are also equal the on to the other.  
As if the number 18. be equemultiplex to any two numbers, that is, contain any two numbers twice, thrice, four times & c? As for example 3. times: then are the said two numbers equal. For 18. divided by 3. will ever bring forth 6. So that that division made twice will bring forth 6. and 6. two equal numbers. So also would it follow if the two numbers had equal equemultiplices, namely, if 18. and 18. which are equal numbers contained any two numbers 3. times.   

4 If a number measure the whole, and a part taken away: it shall also measure the residue. 
Forth common sentence.   
As if from 24. be taken away 9 there remaineth 15. And for as much as the number 3 measureth the whole number 24, & also the number taken away, namely, 9 it shall also measure the residue, which is 15● For 3. measureth 15 by five, five times 3. is 15. And so of others.   


●i●th common sentence.  5 If a number measure any number: it also measureth every number that the said number measureth.  
As the number 6. measuring the number 12. shall also measure all the numbers that 1●. measureth● as the numbers 24.36.48.60. and so forth: which the number 12. doth measure by the number● 2.3.4. and 5. And for as much as the number 12. doth measure the numbers 24.36.48. and 60. And the number 6, doth measure the number 12. (namely by 2.) It followeth by this common sentence, that the number 6. measureth each of th●se numbers 24. 36.48. and 60. And so of others.   

6 If a number measure two numbers, 
Sixth common sentence.  it shall also measure the number composed of them.  
As the number 3 measureth these two numbers 6. and 9● it measureth 6. by 2● and 9 by 3. And therefore by this common sentence it measureth the number 15. which is composed of the numbers 6. and 9: namely it measureth it by 5.   

7 If in numbers there be proportions how manysoever equal or the self same to one proportion: 
Seventh com●mon sentence.  they shall also be equal or the self same the one to the other.  
As if the proportion of the number 6. to the number 3. be as the proportion of the number 8. to the number 4, if also the proportion of the number 10. to the number 5. be as the proportion of the number 8. to the number 4: then shall the proportion of the number 6. to the number 3. be as the proportion of the number 10. is to the number 5: namely, each proportion is duple. And so of others. Euclid in his ●. book the 11. proposition demonstrated this also in continual quantity: which although as touching that kind of quantity it might have been put also as a principle (as in numbers he taketh it) yet for that in all magnitudes their proportion can not be expressed, (as hath before been noted & shallbe afterward in the tenth book more at large made manifest:) therefore he demonstrateth it there in that place, and proveth that it is true as touching all proportions generally whither they be rational or irrational.    

¶ The first Proposition. The first Theorem. If there be given two unequal numbers, and if in taking the less continually from the greater, the number remaining do not measure the number going before, until it shall come to unity: then are those numbers which were at the beginning given, prime the one to the other.  
SVppose that there be two unequal numbers AB the greater, and CD the less, and from AB the greater, take away CD the less as o●ten as you can leaving FA, 
Constr●ctio●.  and from CD take away FA as often as you can, leaving the number GC. And from FA take away GC as often as you can, and so do continually till there remain only unity, which let be HA. 
Demonstrati●● leading to an absurdity.  Then I say that no number measureth the numbers AB and CD. For if it be possible let some number measure them, and let the same be E. Now CD measuring AB leaveth a less number than itself, which let be FA ● And FA measuring DC leaveth also a less than itself, namely, GC. And GC measuring FA leaveth unity HA. And forasmuch as the number E measureth DC, and the number CD measureth the number BF, therefore the number E also measureth BF, and it measureth the whole number BA, wherefore it also measureth that which remaineth, namely, the number FA (by the 4. common sentence of the seventh). But the number AF measureth the number DG, wherefore E also measureth DG. And it measureth also the whole DC, wherefore it also measureth that which remaineth, namely, the number GC (by the same common sentence): but GC measureth the number FH, wherefore also E measureth FH, and it meas●reth the whole number FA, wherefore (by the former common sentence) it also measureth that which remaineth HA, which is unity, itself being a number, which is impossible. Wherefore no number doth measure the numbers AB and CD, wherefore the numbers AB and CD are prime numbers the one to the other: which was required to be proved.  

The converse of this proposition after Campane.  
And if the two numbers, namely AB and CD be prime the one to the other. Then the less being continually taken from the greater there shallbe no stay of that sustraction, till that you come to unity. For if in the continual substraction ther● be a stay before you come to unity. 
The converse of ●his proposition.  Suppose that HA be the number whereat the stay is made, which also being subtrahed out of GC leaveth nothing. Wherefore HA measureth GC wherefore also it measureth FH by the 5. common sentence of the seventh. And for as much as it also measureth i● self, therefore it also measureth the whole AF by the sixth common sentence of the seventh, wherefore also it measureth DG by the 5. common sentence. But it is before proved that it measureth GC, wherefore it measureth the whole CD, by the sixth common sentence of the seventh: wherefore also it measureth BF by the 5. common sentence of the seventh. And it is also proved that it measu●eth FA, wherefore also it measureth the whole number AB by the sixth common sentence of the seventh. Now for as much as the number HA measureth the numbers AB and CD, therefore the numbers AB and CD are numbers composed: wherefore they are not prime the one to the other: which is contra●y to the supposition.  
And by this proposition if there be two numbers given. It is easy to find out, whether they be prime the one to the other or no. 
How to ●now whether two numbers given be prime the one to the other.  For if by such continual substraction of the less from the greater, you come at the length to unity. Then are those numbers given prime the one to the other. But if there be a stay before you come to unity, then are the numbers given, numbers composed the one to the other.    

¶ The 1. Problem. The 2. Proposition. Two numbers being given not prime the one to the other, to find out their greatest common measure.  
SVppose the two numbers given not prime the one to the other to be AB and CD. It is required to find out the greatest common measure of the said numbers AB and CD. 
Two cases in this problem.  Now the numbe● CD either measureth the number AB or not. If CD measure AB it also measureth itself. 
The first case.  Wherefore CD is a common measure to the numbers CD and AB. And it is manifest also that it is the greatest common measure● for there is no number greater than CD that will measure CD.  

The second case.  But if CD do not measure AB, then if of the numbers AB and CD, the less be continually taken away from the greater, there will before you come to unity, be left a number, which will measure the number going before (by the 1. of the seventh). For if there should not, then should the numbers AB and CD be prime the one to the other, which is contrary to the supposition. Let the said number left by the continual substraction of the less number out of the greater be FC. So that le● the number CD measuring AB, and subtrahed out of it as often as you can le●e a less number than itself, namely AE. And let AE measuring CD, and subtrahed out of it as often as you can leave a lesse-then itself namely, CF. And suppos● th●t CF do so measure AE that there remain nothing. Then I say that CF is a common measure to the numbers AB and CD. For forasmuch as CF measureth AE, and AE measureth DF, 
Demonstration of the second case.  therefore CF also measureth DF (by the fifth common sentence of the seventh) and it likewise measureth itself, wherefore it also measureth the whole CD (by the sixth common sentence of the seuenth●) ● but CD measureth BE, wherefore CF also measureth BE (by the ●ifte common sentence of the seventh). And it measureth also EA: 
That CF is a common measure to the numbers AB and CD.  wherefore it also measureth the whole BA (by the sixth common sentence of the seventh): and it also measureth CD as we have before proved: wherefor● the number CF measureth the numbers AB & CD wherefore the number CF is a common measure to the numbers AB & CD.  
I say also that it is the greatest common measure. For if CF be not the greatest common measure to AB and CD, 
That CF is the greatest common measure to AB and CD.  let there be a number greater than CF, which measureth AB and CD: which let be G. And forasmuch as G measureth CD, and CD measureth BE, therefore G also measureth BE (by the ●●ft common sentence of the seventh). And it measureth the whole AB, wherefore also it measureth the residue, namely, AE (by the 4. common sentence of the se●enth). But AE measureth DF, wherefore G also measureth DF (by the foresaid 5. common sentence of the seventh). And it measureth the whole CD. Wherefore it also measureth the residue FC: namely, the greater number the less: which is impossible. No number therefore greater than CF shall measure those numbers AB and CD: wherefore CF is the greatest common measure to AB and CD: which was required to be done.  

Corrolary.  
Hereby it is manifest, that if a number measure two numbers it shall also measure their greatest common measure. For if it measure the whole & the part taken away, it shall always measure the residue also, which residue is at the length, the greatest common measure of the two numbers given.    

¶ The 2. Problem. Th' 3. Proposition. Three numbers being given, not prime the one to the other: to find out their greatest common measure.  

The second case.  But now suppose that D do not measure C. First I say that D & C are not prime numbers the one to the other. For forasmuch as the numbers A, B, C, are not prime the one to the other (by supposition) some one number will measure them: but that number that measureth the numbers A, B, C, shall also measure the numbers A, B, and shall likewise measure the greatest common measure of AB, namely, D (by the Corollary of the second of the seventh). And the said number measureth also C. Wherefore some one number measureth the numbers D and C. Wherefore D and C are not prime the one to the other.  
Now then let there be taken (by the 2. of the seventh) the greatest common measure unto the numbers D and C, which let be the number E. And forasmuch as E measureth D, and D measureth the numbers A, B, therefore E also measureth the numbers A, B (by the sixth common sentence): and it measureth also C. Wherefore E measureth the numbers A, B, C. Wherefore E is a common measure unto the numbers A, B, C. I say also that it is the greatest. For if E be not the greatest common measure unto the numbers A, B, C, let there be some number greater than E, which measureth the numbers A, B, C. And let the same number be ●. And forasmuch as F measureth the numbers A, B, C: it measureth also the numbers A, B. Wherefore also it measureth the greatest common measure of the numbers A, B (by the Corollary of the 2. of the seventh). But the greatest common measure of the numbers A, B, is D. Wherefore E measureth D. And it measureth also the number C. Wherefore F measureth the numbers D, C. Wherefore also (by the same Corollary) it measureth the greatest common measure of the numbers D, C. But the greatest common measure of the numbers D, C, is E. Wherefore F measureth E, namely, the greater number the less: which is impossible. Wherefore no number greater than E shall measure the numbers A, B, C. Wherefore E is the greatest common measure to the numbers A, B, C: which was required to be done.  

¶ Corollary.  
Wherefore it is manifest, that if a number measure three numbers, it shall also measure their greatest common measure. And in like sort more numbers being given not prime the one to the other may be found out their greatest common measure, and the Corollary will follow.    

¶ The 2. Theorem. The 4. Proposition. Every less number is of every greater number, either a part, or parts.  

Two cases in this Proposition.  SVppose there be two numbers A and BC. Of which let BC be the less. Then I say, that BC is either a part or parts of A. For the numbers A and BC are either prime the one to the other, or not. First let A and BC be prime the one to the other. 
The first case.  And divide the number BC into those unities which are in it. Now every one of the unities which 〈◊〉 in BC i● s●me certain part of A ● Wherefore BC are parts of A.  
But now suppose that the number● A and BC 〈◊〉 not prime the one to the other. Then BC either measureth A or not measureth it. 
The second case.  If BC measure: A, then is BC a part of A. But if not● take (by the 2. of the seventh) the greatest common ●eas●r● of A and BC, and let the same be D. And let BC be divided into as man● pa●te● as it hath equal unto D, that is into BE, EF ● and FC. And forasmuch as D measureth A, therefore D i● a part of A. But D is equal unto every one of these parts BE, EF, and FC. Wherefore also every one of these parts BE, EF, and FC, is a part of A. Wherefore the number BC is parts of A. Wherefore every less number is of every greater number, either a part or parts: which was required to 〈◊〉 proved.   

¶ The 3. Theorem. The 5. Proposition. If a number be a part of a number, and an other number the self same part of an other number, than both the numbers added together shall be the self same part of both the numbers added together, which one number was of one number.  
SVppose the number A to be a part of the number BC, and let an other number, namely, D, be the self same part of an other number, namely, of EF. Then I say, 
This proposition, and the 6. proposition in discrete quantity, answer to the first of the fifth in continual quantity.  that the numbers A and D added together, are the self same part of the numbers BC and EF added together, that A is of BC. For forasmuch as what part the number A is of the number BC, the self same part is the number D of the number ● F, therefore h●w many numbers there are in BC equal unto A, so many numbers are there in EF equal unto D. Divide I● say, BC into the numbers that are equal unto A, that is, into BG, and GC, and likewise divide EF into the numbers that are equal unto D, that is, into EH and HF ● Now then the multitude of these BG and GC is equal unto the multitude of these EH and HF. 
Demonstration.  And forasmuch as BG is equal unto A, and EH, unto D, therefore BG and EH are equal unto A & D. And by the same reason forasmuch as GC is equal unto A, and HF unto D: therefore GC and HF are also equal unto A and D. Wherefore how many numbers there are in BC equal unto A ● so many are there in BC and EF equal unto A and D. Wherefore how multiplex BC is to A, so multiplex are both the numbers BC and EF, to both the numbers A and D. Wherefore what part A is of BC, the self part also are A & D added together, of BC and EF added together: which was required to be proved.   

¶ The 4. Theorem. The 6. Proposition. If a number be parts of a number, and an other number the self same parts of an other number● than both numbers added together shall be of both numbers added together the self same parts, that one number was of one number.  
SVppose that the number AB be parts of the number C, and let an other number, namely, DE be the self same parts of an other number, namely, of F. Then I say, that the numbers AB and DE added together, are of the numbers C and F added together the self same parts, that AB is of C. 
Construction.  For forasmuch as what parts AB is of C, the self same parts is DE of F: therefore how many parts of C there are in AB, so many parts of F are there in DE. Divide AB into the parts of C, that is, into AG and GB, and likewise DE into the parts of F, that is, into DH and HERALD 
Demonstration.  Now then the multitude of these AG and GB is equal to the multitude of these DH and HERALD And forasmuch as what part AG is of C, the self same part is DH of F ● therefore what part AG is of C, the self same part is AG and DH added together of C and F added together. And by the same reason also what part GB is of C, the self same part is GB and HE added together of C and F added together. Wherefore what parts AB is of C, the self same parts are AB and DE added together of C and F added together: which was required to be demonstrated.   

¶ The 5. Theorem. The 7. Proposition. If a number be the self same part of a number, that a part taken away is of a part taken away: then shall the residue be the self same part of the residue, that the whole was of the whole.  

Thi● proposition and the next following in discreet quantity answereth to the fifth proposition of the fifth book in continual quantity.  SVppose that the number AB be of the number CD the self same part, that the part taken away AE is of the part taken away CF. Then I say, that the residue EB is of the residue FD the self same part that the whole AB is of the whole CD. What part AE is of CF, the self same part let EB be of GC. And for that what part AE is of CF, the same part is EB of CG, therefore what part AE is of CF, the same part (by the 5. of the seventh) is AB of FG. But what part AE is of CF, the same part (by supposition) is AB of CD. Wherefore what part AB is of FG, the self same part is AB of CD. 
Construction.  Wherefore AB is one & the self same part of both these numbers GF and CD. Wherefore GF is equal unto CD (by the second common sentence of the seventh). Take away CF which is common to them both. Wherefore the residue GC is equal unto the residue FD. 
Demonstration.  And forasmuch as what part AE is of CF, the same part is EB of GC: but GC is equal unto FD: therefore what part AE is of FC, the self same part is EB of FD. But what part AE is of CF, the same part is AB of CD. Wherefore what part EB is of FD, the same part is AB of CD. Wherefore the residue EB is of the residue FD the self same part that the whole AB is of the whole CD: which was required to be demonstrated.   

¶ The 6. Theorem. The 8. Proposition. If a number be of a number the self same parts, that a part taken away is of a part taken away, the residue also shall be of the residue the self same parts that the whole is of the whole.  
SVppose that the number AB be of the number CD the self same parts that the part taken away AE is of the part taken CF. Then I say, that the residue EB is of the residue FD the self same parts that the whole AB is of the whole CD. Unto AB put an equal number GH. 
Constu●ction.  Wherefore what parts GH is of CD, the self same parts is AE of CF. Divide GH into the parts of CD, that is, GK, and KH, and likewise AE into the parts of CF, that is, into AL and LE. Now then the multitude of these GK and KH, is equal unto the multitude of these AL and LE. 
Demonstration.  And forasmuch as what part GK is of CD, the self same part is AL of CF: but CD is greater than CF. Wherefore GK is greater than AL. Put unto AL an equal number MG. Wherefore what part GK is of CD, the same part is GM of CF. Wherefore the residue MK is (by the 7. of the seventh) of the residue FD, the self same part that the whole GK is of the whole CD. Again, forasmuch as what part KH is of CD, the self same part is EL of CF: but CD is greater than CF. Wherefore HK is greater than EL. Put unto EL an equal number KN. Wherefore what part KH is of CD, the self same part is KN of CF. Wherefore the residue also NH is (by the 7. of the seventh) of the residue FD, the self same part that the whole KH is of the whole DC. Wherefore both these MK and NH added together are (by the 5. of the seventh) of DF the self same parts that the whole HG is of the whole CD. But both these MK and NH added together, are equal unto EB. And HG is equal unto BA. Wherefore the residue EB is of the residue FD the self same parts that the whole AB is of the whole CD: which was required to be proved.  

¶ An other demonstration after Flussates.  
Suppose that the number AB be of the number CD the self same parts that the part taken away AE is of the part taken away CF. 
another demonstration after Flussates.  Then I say, that the residue EB is of the residue FD the self same parts that the whole AB is of the whole CD. Let EB be of CI the self same parts that AB is of CD, or AE of C●. Now forasmuch as EB is of CI the self same parts that AE is of CF: therefore both these AE and EB added together are of both these CF and CI added together (that is, the whole AB is of the whole FI) the self same parts that AE is of CF (by the sixth of this book). But what parts AE is of CF, the self same parts is the number AB of the number CD (by supposition). Wherefore what parts the number AB is of the number FI, though self same parts is the same number AB of the number CD. Wherefore the numbers FI and CD are equal. Take away the number CF which is common to them both. Wherefore the numbers remaining CI and ●D are equal. Wherefore what parts the number EB is of the number CI, the self same parts is the same number EB of the number FD. But what parts EB is of CI, the self same parts (by construction) is AB of CD. Wherefore what parts the residue EB is of the residue FD, the self same parts is the whole AB of the whole CD: which was required to be proved.    

¶ The 7. Theorem. The 9 Proposition. If a number be a part of a number, and if an other number be the self same part of an other number: then alternately what part or parts the first is of the third, the self same part or parts shall the second be of the fourth.  
SVppose that the number A be of the number BC the self same part, that an other number D is of an other number EF. And let A be less then D. Then I say, that alternately what part or parts A is of D, the self same part or parts is BC of EF. 
Construction.  For forasmuch as what part A is of BC, the self same part is D of EF, therefore how many numbers there are in BC equal unto A, so many are there in EF equal unto D. Divide BC into the numbers equal unto A, that is, into BG & GC: and likewise EF into the numbers equal unto D, that is, into EH and HF. Now then the multitude of these BG and GC, is equal unto the multitude of these EH & HF. 
Demonstration.  And forasmuch as the numbers BG and GC are equal the one to the other, the numbers also EH and HF are equal the one to the other: and the multitude of these BG & GC is equal unto the multitude of these EH and HF. Wherefore what part or parts BG is of EH, the self same part or parts is GC of HF. Wherefore what part or parts BG is of EH, the self same part or parts (by the fift & sixth of the seventh) are BG and GC added together, of EH and HF added together. But BG is equal unto A, and EH unto D. Wherefore what part or parts A is of D, the self same part or parts is BG of EF: which was required to be demonstrated.   

¶ The 8. Theorem. The 10. Proposition. If a number be parts of a number, and an other number the self same parts of an other number, then alternately what parts or part the first is of the third, the self same parts or part is the second of the fourth.  
SVppose that the number AB be of the number C the self same parts, that an other number DE is of an other number F, and let AB be less than DE. Then I say, that alternately also what part or parts AB is of DE, the self same parts or part is C of F. Forasmuch as what parts AB is of C, the self same parts is DE of F: 
Construction.  therefore how many parts of C there are in AB, so many parts of F also are there in DE. Divide AB into the parts of C, that is, into AG and GB. And likewise DE into the parts of F, that is, DH and HERALD Now then the multitude of these AG and GB, is equal unto the multitude of these DH and HERALD 
Demonstration.  And forasmuch as what part AG is of C, the self same part is DH of F, therefore alternately also (by the former) what part or parts AG is of DH, the self same part or parts is C of F. And by the same reason also what part or parts GB is of HE, the same part or parts is C of F. Wherefore what part or parts AG is of DH, the self same part or parts is AB of DE (by the 6. of the seventh). But what part or parts AG is of DH, the self same part or parts is it proved that C is of F. Wherefore what parts or part AB is of D E, the self same parts or part is C of F: which was required to be proved.   

¶ The 9 Theorem. The 11. Proposition. If the whole be to the whole, as a part taken away is to a part taken away: then shall the residue be unto the residue, as the whole is to the whole.  

This proposition i● discreet quantity answereth to the ninth proposition of the fifth book in continual quantity.  SVppose that the whole number AB be unto the whole number CD, as the part taken away AE, is to the part taken away CF. Then I say that the residue EB, is to the residue FD, as the whole AB is to the whole CD. For forasmuch as, AB is to CD, as AE is to CF: therefore what part or parts AB is of CD, the self same part or parts is AE of CF. Wherefore also the residue EB is of the residue FD (by the 8. of the seventh) the self same part o● parts that AB is of CD. 
Demonstration.  Wherefore also (by the 21. definition of this book) as EB is to FD, so is AB to CD: which was required to be proved.   

¶ The 10. Theorem. The 12. Proposition. If there be a multitude of numbers how many soever proportional: as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes.  
SVppose that there be a multitude of numbers how many soever proportional, namely, A, B, C, D, so that as A is to B, so let C be to D. 
This in discreet quantity answereth to the twelve proposition of the fifth in continual quantity.  Then I say that as one of the antecedentes, namely, A is to one of the consequentes, namely, to B, or as C is to D, so are all the antecedentes: namely, A and C to all the consequentes, namely, to B and D. For forasmuch as (by supposition) as A is to B, so is C to D, therefore what part or parts A is of B, the self same part or parts is C of D (by the 21. definition of this book) wherefore alternately what part or parts A is of C the self same part or parts is B of D (by the ninth and tenth of the seventh) wherefore both these numbers added together, 
Demonstration.  A and C, are of both these numbers B and D added together the self same part or parts that A is of B (by the 5. and 6. of the seventh) wherefore (by the 21. definition of the seventh) as one of the antecedents, namely, A, is to one of the consequentes, namely, to B, so are all the antecedentes A and C to all the consequentes B & D. Which was required to be proved.   

¶ The 11. Theorem. The 13. Proposition. If there be four numbers proportional: then alternately also they shall be proportional.  
SVppose that there be four numbers proportional, 
This in discrete quantity answereth to the sixteenth proposition of the fifth book in continual quantity.  A, B, C, D, so that as A is to B, so let C be to D. Then I say that alternately also they shallbe proportional, that is. as A is to C, so is B to D. For forasmuch as (by supposition) as A is to B, so is C to D, therefore (by the 21. definition of this book) what part or parts A is of B the self same part or parts is C of D. Therefore alternately what part or parts A is of C the self same part or parts is B of D (by the 9 of the seventh) & also (by the 10. of the same): wherefore as A is to C, so is B to D (by the 21. definition of this book): which was required to be proved.  
Here is to be noted, that although in the foresaid example and demonstration the number A be supposed to be less than the number B, and so the number C is less than the number D: 
Note.  yet will the same serve also though A be supposed to be greater than B, whereby also C shall be greater than D, as in th●s example here put. For for that (by supposition) as A is to B, so is C to D, and A is supposed to be greater than B, and C greater than D: therefore (by the 21. definition of this Book) how multiplex A is to B, so multiplex is C to D, and therefore what part or parts B is of A, the self same part or parts is D of C. Wherefore alternately what part or parts B is of D, the self same part or parts is A of C, and therefore by the same definition, B is to D, as A is to C. And so must you understand of the former Proposition next going before.   

¶ The 12. Theorem. The 14. Proposition. If there be a multitude of numbers how many soever, and also other numbers equal unto them in multitude, which being compared two and two are in one and the same proportion: they shall also of equality be in one and the same proportion.  

This in discrete quantity answereth to t●● twenty one proposition o● the fifth book in continual quantity.  SVppose that there be a multitude of numbers how many soever: namely, A, B, C, and let the other numbers equal unto them in multitude be D, E, F: which being compared two and two, let be in one and the same proportion: that is, as A to B, so let D be to E: and as B is to C, so let E be to F. Then I say, that of equality, as A is to C, so is D to F. For forasmuch as by supposition as A is to B, so is D to E: therefore alternately also (by the 13 of the seventh) as A is to D, so is B to E. Again, for that as B is to C, so is E to F: 
Demonstration.  therefore alternately also (by the self same) as B is to E, so is C to F. But as B is to E, so is A to D. Wherefore (by the seventh common sentence of the seventh) as A is to D, so is C to F. Wherefore alternately (by the 13. of the seventh) as A is to C, so is D to F: which was required to be demonstrated.  
After this Proposition, Campane demonstrateth in numbers these four kinds of proportionality, 
Certain additions of campane.  namely, proportion converse, composed, divided, and everse: which were in continual quantity, demonstrated in the 4. 17. 18. and 19 propositions of the fift book. And first he demonstrateth converse proportion in this manner.  

The second case.  But if A be greater than B, C also is greater than D: and what part or parts B is of A, the self same part or parts is D of C. Wherefore (by the same definition) as B is to A, so i● D to C: which was required to be proved.  

proportionality divided, is thus demonstrated.  
Suppose that the number AB be to the number B, as the number CD is to the number D. Then I say, 
Proportionality divided.  that divided also, as A is to B, so is C to D. For for that as AB is to B, so is CD to D: therefore alternately (by the 14. of this book) as AB is to CD, so is B to D. Wherefore (by the 11. of this book) as AB is to CD, so is A to C. Wherefore as B is to D, so is A to C: and for that as A is to C, so is B to D, therefore alternately as A is to B, so is C to D.   

proportionality composed, is thus demonstrated.  

Proportionality composed.  If A be unto B, as C is to D: then shall AB be to B, as CD is to D. For alternately A is to C, as B is to D. Wherefore (by the 13. of this book) as AB, namely, all the antecedentes are to CD, namely, to all the consequentes, so is B to D, namely, one of the antecedentes to one of the consequentes. Wherefore alternately as AB is to B, so is CD to D.   

Euerse proportionality, is thus proved.  

Euerse proportionality.  Suppos● that AB be to B, as CD is to D: then shall AB be to A, as CD is to C. For alternately AB is to CD, a● B is to D. Wheref●r● (by the 13. of this boo●●) A● is ● CD, as A is to C. Wherefore alternately AB i● to A, a● CD i● to C: whi●h was required to be proved.    

¶ A proportion here added by Campane.  
If the proportion of the first to the second, be as the proportion of the third to the fo●rth, and if the proportion of ●he fift to the second be as the proportion of the sixth to the fourth: then the proportion of the first and the fifth taken together, shall be to the second, as the proportion ●f the third and the sixth taken together to the fourth.  
And after the same manner may you prove the converse of this Proposition. If B be to A, as D is to C● and if also B be unto E, as D is to F: Then shall B be to AE, as D is to CF. For by converse proportionality, 
The conuers● of the same pr●position.  A is to B, as C is to D. Wherefore of equality, A is to E, as C is to F. Wherefore by composition A and E be to E, as C and F are to F. Wherefore conversedly, E is to A and E, as F is to C and F. But by supposition, B is to E, as D is to F. Wherefore again by Proportion of equality, B is to A and E, as D is to C and F: 
Demonstration.  which was required to be proved.  

A Corollary.  
By this also it is manifest that if the proportion of numbers how many soever unto the first, be as the proportion of as many other numbers unto the second, 
A Corollary following th●se propositions ad●ed by Campane.  then shall the proportion of the numbers composed of all the numbers that were antecedentes to the first, be to the first, as the number composed of all the numbers that were antecedentes to the second is to the second. And also conversedly if the proportion of the first to numbers how many soever, be as the proportion of the second to as many other numbers: then shall the proportion of the first to the number composed of all the numbers that were consequentes to itself, be as the proportion of the second to the number composed of all the numbers that were consequences to itself.    

¶ The 13. Theorem. The 15. Proposition. If unity measure any number, and an other number do so many times measure an other number: unity also shall alternately so many times measure the third number, as the second doth the fourth.  
SVppose that unity A do measure the number BC: and let an other number D so many times measure some other number, namely, EF. Then I say, that alternately, unity A shall so many times measure the number D, as the number BC doth measure the number EF. 
Co●str●ctio●.  For forasmuch as unity A doth so many times measure BC, as D doth EF: therefore how many unities there are in BC, so many numbers are there in EF equal unto D. Divide (I say) BC into the unities which are in it, that is, into BG, GH, and HC. And divide likewise EF into the numbers equal unto D, that is, into EK, KL, and LF. Now than the multitude of these BG, GH, and HC, is equal unto the multitude of these EK, KL, LF. 
Demonstration.  And forasmuch as these unities BG, GH, and HC, are equal the one to the other, and these numbers EK, KL, & LF, are also equal the one to the other, and the multitude of the unities BG, GH, and HC, are equal unto the multitude of the numbers EK, KL, & LF: therefore as unity BG is to the number EK, so is unity GH to the number KL, and also unity HC to the number LF. Wherefore (by the 12● of the seventh) as one of the antecede●t●●●s to one of the consequentes, so are all the anteceden●es to all the consequentes: Wherefore as unity BG is to the number EK, so is the number BC to the number EF. But unity BG is equal unto unity A, and the number EK to the number D. Wherefore (by the 7. common sentence) as unity A is to the number D, so is the number BC to the number EF. Wherefore unity A measureth the number D, so many times, as BC measureth EF (by the 21 definition of this book): which was required to be proved.   

¶ The 14. Theorem. The 16. Proposition. If two numbers multiplying themselves the one into the other, produce any numbers: the numbers produced shall be equal the one into the other.  
SVppose that there be two numbers A and B: and let A multiplying B produce C, and let B multiplying A produce D. Th●n I say, that the number C● equal unto the n●mber D. 
Demonstration.  Take any unity, namely E. And forasmuch as A multiplying B produced C, therefore B measureth C by the unities which are in A. And unity E measureth the number A by those unities which are in the number A. Wherefore unity E so many times measureth A, as B measureth C. Wherefore alternately (by the 15. of the seventh) unity E measureth the number B so many times as A measureth C. Again, for that B multiplying A produced D, therefore A measureth D by th● unities which are in B. And unity E measureth B by the unities which are in B. Wherefore unity E so many times measureth the number B, as A measureth D. But unity E so many times measureth the number B, as A measureth C. Wherefore A measureth either of these numbers C and D a like. Wherefore (by the 3. common sentence of this book) C is equal unto D: which was required to be demonstrated.   

The 15. Theorem. The 17. Proposition. If one number multiply two numbers, and produce other numbers, the numbers produced of them, shall be in the self same proportion, that the numbers multiplied are.  
SVppose that the number A multiplying two numbers B and C, do produce the numbers D and E. Then I say that as B is to C, so is D to E. Take unity, namely, F. And forasmuch as A multiplying B produced D, therefore B measureth D by those unities that are in A. And unity F measureth A by th●s● v●iti●● which are in A. 
●emonstra●ion.  Wherefore unity F so many times measureth the number A, as B measureth D. Wherefore as unity I is to the number A. so is the number B to the number D (by the 21 definition of this book) And by the same reason, as unity F is to the number A, so is the number C to the number E: wherefore also (by the 7. common sentence of this book) as B is to D, so is C to E. Wherefore alternately (by the 15. of the seventh) as B is to C, so is D to E. If therefore one number multiply two numbers● and produce other numbers: the numbers produced of them, shall be in the self same proportion, that the numbers multiplied are: which was required to be proved.  

Here Flu●●tes adds thi● Corollary.  
If two number● having one and the sam● proportion with two other numbers do multiply th● o●e the other alternately, 
A Corollary added by Fluss●tes.  and produce any numbers, the numbers produced of them shall be equal the one to the other.  
Suppose that there be two number● ● and B, and also two other numbers C and D, having th● same proportion that the numbers A and B have: and let the numbers A and B multiply the number● C & D alternately, that is, let A multiplying D produce F, and let B multiplying C produce E. Then I say that the numbers produced namely, E & F are equal. Let A and B multiply the one the other in such sort, that let A multiplying B produce G, and let B, multiplying A produce H, Now then the numbers G and H are equal by the 16. of this booke● And forasmuch as A multipli●ng the two numbers B and D, produced the numbers G and F, therefore G is to ●, as B is to D by this proposition. So likewise B multiplying the two numbers A and C produced the two numbers H and E. Wherefore by the same H is to E as A is to C. But alternately (by the 13. of this book) A is to C as B is to D, but as A is to C so is H to E, and as ● is to D, so is G to ●. Wherefore by the seventh common sentence, as H is to E ● so is G to F. Wherefore alternately (by the 13. of this book) H is to G as E is to F. But it is proved that G & H are equal: Wherefore E and F (which have the same proportion that A and B have) are equal. If therefore there be two numbers, etc. Which was required to be proved.    

¶ The 16. Theorem. The 18. Proposition. If two numbers multiply any number, & produce other numbers: the numbers of them produced, shall be in the same proportion that the numbers multiplying are.  
SVppose that two numbers A and B multiplying the number C, do produce the numbers D and E. 
Demonstration.  Then I say that as A is to B, so is D to E. For forasmuch as A multiplying C produced D, therefore C multiplying A produceth also D (by the 16. of this book.) And by the same reason C multiplying B produceth E. Now then one number C multiplying two numbers A and B, produceth the numbers D and E. Wherefore by the 17. of the seventh, as A is to B, so is D to E: which was required to be demonstrated.  
This Proposition, and the former touching two numbers, may be extended to numbers how many soever. 
This proposition and the former may be extended to numbers how many soever.  So that if one number multiply numbers how many soever, and produce any numbers, the proportion of the numbers produced, and of the numbers multiplied, shall be one and the self same. Likewise if numbers how many soever multiply one number, and produce any numbers, the proportion of the numbers produced● and of the numbers multiplying shall be one and the self same: which thing by this and the former proposition repeated as often as is needful, is not hard to prove.   

¶ The 17. Theorem. The 19 Proposition. If there be four numbers in proportion: the number produced of the first and the fourth, is equal to that number which is produced of the second and the third. And if the number which is produced of the first and the fourth be equal to that which is produced of the second & the third: those four numbers shall be in proportion.  
But now again, suppose that E be equal unto F. Then I say that as A is to B, so is C to D. 
The second part of this proposition which is the converse of the first.  For the same order of construction remaining still, forasmuch as A multiplying C & D produced G and E, therefore by the 17. of the seventh, as C is to D so is G to E, but E is equal unto F (But if two numbers be equal, one number shall have unto them on● and the same proportion) wherefore as G is to E, so is G to F. But as G is to E, so is C to D. Wherefore as C is to D, so is G to F, but as G is to F, so is A to B by the 18. of the seventh, wherefore as A is to B so is C to D: 
Demonstration.  which was required to be proved.  

An assumpt added by Campane.  Here Campane addeth, that it is needless to demonstrate, that if one number have to two numbers one and the same proportion, the said two numbers shall be equal: or that if they be equal, one number hath to them one and the same proportion. For (saith he) if G have unto E and F one and the same proportion, them either, what part or parts G is to E, the same part or parts is G also of F: or how multiplex G is to E, so multiplex is G to F (by the 21. definition) And therefore by the 2 and 3 common sentence, the said numbers shall be equal. And so conversedly, if the two numbers E and F be equal, then shall the numbers E and F be either the self same part or parts of the number G, or they shall be equemultiplices unto it. And therefore by the same definition the number G shall have to the numbers E and F one and the same proportion.   

¶ The 18. Theorem. The 20. Proposition. If there be three numbers in proportion, the number produced of the extremes, is equal to the square made of the middle number. And if that number which is produced of the extremes, be equal to the square made of the middle number, those three numbers shall be in proportion.  
SVppose there be three numbers in proportion, A, B, C, as A is to B, so let B be to C. 
This proposition in numbers demonstrateth that which the 17. of the sixth demonstrateth in lines.  Then I say that the number produced of A and C is equal to the square number which is made of B. Put unto B an equal number D. Wherefore as A is to B, so is D to C. Wherefore that which is produced of A into C, is equal unto that which is produced of B into D. But that which is produced of B into D is equal to that which is made of B (for B is equal unto D) wherefore that which is produced of A into C is equal to that which is made of B.  

Demonstration.  But now suppose that that which is produced of A into C, be equal to that which is made of B. 
The second part which is the converse of the first.  Then I say that as A is to B, so is B to C. For (the same order of construction remaining) forasmuch as that which is produced of A into C, is equal to that which is made of B, but that which is made of B, is equal to that which is produced of B into D. (For B and D are by supposition equal): therefore that which is produced of A into C is equal to that which is produced of ● into D: 
Demonstration  wherefore (by the second part of the former proposition) as A is to B, so is D to C: but D is equal to E. Wherefore as A is to B, so is the same B to C: which was required to be proved.   

¶ The 19 Theorem. The 21. Proposition. The left numbers in any proportion, measure any other numbers having the same proportion equally, the greater the greater, & the less the less.  
SVppose that CD & EF be the lest numbers that have one & the same proportion with the numbers A and B. 
Demonstration leading to an impossibility.  Then I say, that the number CD so many times measureth the number A, as the number EF measureth the number B. For forasmuch as by supposition CD is to EF, as A is to B, and CD and EF are also supposed to be less than A and B: therefore CD and EF are either part or parts of A and B (by the 4. of this book, and by the 21. definition of the same). But they are not parts. For if it be possible, let CD be parts of A. Wherefore EF is the self same parts of B, that CD is of A. Wherefore how many parts of A there are in CD, so many parts are there of B in EF. Divide CD into the parts of A, that is, into CG and GD. And likewise divide EF into the parts of B, that is, into EH and HF. Now then the multitude of these CG and GD, is equal unto the multitude of these EH & HF. And forasmuch as CG and GD are numbers equal the one to the other, and these numbers EH and HF are also equal the one to the other, and the multitude of these CG and GD, is equal to the multitude of these EH and HF: therefore as CG is to EH, so is GD to HF. Wherefore (by the 12. of the seventh) as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherefore as CG is to EH, so is CD to EF. Wherefore CG and DH ●re in the self same proportion that CD and EF are, being also less than CD and EF: which is impossible. For CD and EF are supposed to be the lest that have 〈◊〉 and the same proportion with them. Wherefore CD is not parts of A: wherefore it is a part. Wherefore EF is of B ●he self same part, that CD is of A. Wherefore CD so many 〈◊〉 measureth A as EF doth B● which was required to be demonstrated.   

¶ The 20. Theorem. The 22. Proposition. If there be three numbers, and other numbers equal unto than in multitude, which being compared two and two are in the self same proportion, and if also the proportion of them be perturbate, then of equality they shall be in one and the same proportion.  
SVppose that there be three numbers A, B, and C and let the other numbers equal ●nto them in multitude be D, E, and F. 
This proposition in discreet quantity answereth to the 23. proposition of the fifth book in continual quantity.  And le● two and two compared together be in one and the same proportioned and let the proportion of them be perturbate; so that 〈…〉 as B is to C, so let D be to E. 〈…〉 is to C, so is D to F. ●or, for 〈…〉 therefore that which is produced of A into F, is (by the 19 of the seventh) equal to that which is produced of B into E. Again for that as B is to C, so is D to E, therefore that which is produced of D into C, is equal to that which is produced of B into E. And it is proved, that that which is produced of A into F, is equal to that which is produced of B into E. Wherefore that which is produced of A into F, is equal to that which is produced of D into C. Wherefore (by the second part of the 19 of the seventh) as A is to C, so is D to F: which was required to be proved.  
The same may also be proved if in either order be more than three numbers: as it was proved in the 23. of the fift touching more magnitudes than three.   

¶ The 21. Theorem. The 23. Proposition. Numbers prime the one to the other: are the lest of any numbers, that have one and the same proportion with them.  

This and the eleven propositions following, declare the passion's and properties of● prime numbers.  SVppose that A and B be numbers prime the one to the other. Then I say that A and B are the lest of any numbers that have one and the same proportion with them. For if A and B be not the lest of any numbers that have one and the same proportion with them, then are there some numbers less than A and B, being in the self same proportion that A and B are. Let the same be C and D. Now forasmuch as the lest numbers in any proportion measure any other numbers having the same proportion equally, the greater the greater, and the less the less (by the 21. of the seventh) that is, the antecedent the antecedent, and the consequent the consequent: therefore C so many times measureth A, as D measureth B. How many times C measureth A, so many unities let there be in E. 
Demonstration leading to an impossibility.  Wherefore D measureth B by those unities which are in E. And forasmuch as C measureth A by those unities which are in E, therefore E also measureth A by those unities which are in C. And by the same reason E measureth B, by those unities which are in D. Wherefore E measureth A and B being prime numbers the one to the other which (by the 13. definition of the seventh) is impossible. Wherefore there are no other numbers less than A and B, which are in the self same proportion that A and B are. Wherefore A and B are the lest numbers that have one and the same proportion with them: which was required to be demonstrated.   

¶ The 22. Theorem. The 24. Proposition. The lest numbers that have one and the same proportion with them: are prince the one to the other.  
SVppose that A and B be the lest numbers that have one and the same properti●● with them. 
This is the converse of the former proposition.  Then I say that A and B be prime the one to the other. For if A & B be not prime the one to the other, then shall some one number measure A & B. Let the same be C. And how oftentimes C measureth A, so many unities let there be in D: and how oftentimes C measureth B, so many unities let there be in E. And forasmuch as C measureth A by those unities that are in D, therefore C multiplying D produceth A● And by the same reason C multiplying E produceth B. 
Demonstration leading to an absurdity.  Wherefore the number C multiplying two numbers D and E, produceth A & B. Wherefore (by the 17. of the seventh) as D is to E, so is A to B. But the numbers D and E are less than A and ●, and are also in the self same proportion with them, which is impossible. Wherefore no number measureth these numbers A and B. Wherefore A and B are prime the on to the other: which was required to be demonstrated.   

¶ The 23. Theorem. The 25. Proposition. If two numbers be prime the on to the other: any number measuring one of them shallbe prime to the other number.  
SVppose that A and B be two prime numbers the on to the other. And let some number, namely, C, measure A. Then I say that C and B are prime numbers the on to the other. For if C and B be not prime the one to the other, let some number measure C and B, and let the same be D. 
Demonstration leading to an absurdity.  And forasmuch as D measureth C, and C measureth A, therefore D also measureth A (by the fift common sentence) and D measureth B. Wherefore D measureth A and B being numbers prime the on to the other, which is impossible (by the 13. definition of the seventh) wherefore no number measureth these numbers B and C. Wherefore B and C be numbers prime the on to the other. Which was required to be proved.   

¶ The 24. Theorem. The 26. Proposition. If two numbers be prime to any one number, the number also produced of them shall be prime to the self same.  
SVppose that the two numbers A and B be prime to any one number, namely, to C, and let A multiplying B produce D. Then I say that C and D are prime numbers the one to the other. For if C and D be not prime the one to the other than some number shall measure them. Let there be a number that measureth them, and let the same be E. 
Demonstration leading to an absurdity.  And forasmuch as A and C are prime numbers the one to the other, and the number E measureth C. Therefore E and A are (by the 25. of the seventh) prime the one to the other. And forasmuch as E measureth D. How many times E measureth D, so many unities let there be in F. Wherefore F also measureth D by those unities which are in E. Wherefore E multiplying F produceth D. But A also multiplying B produced D. Wherefore that which is produced of E into F is equal to that which is produced of A into B. But if that which is produced of the extremes be equal to that which is produced of the means, then are those four numbers in proportion (by the 19 of the seventh) wherefore ●● E is to A, so is B to F. But A and E are prime the one to the other. Yea they are prime and the lest in the same proportion (by the 23 of the seventh.) But the lest numbers in any proportion measure any other numbers having the same proportion equally, the greater the greater, and the less the less, that is, the antecedent the antecedent, and the consequent the consequent by the 21. of the seventh wherefore E measureth B, and it also measureth C: wherefore E measureth C and B, which are by supposition numbers prime the one to the other, which is impossible by the 13. definition of the seventh. Wherefore no number measureth those numbers C and D. Wherefore C and D are prime the one to the other: which was required to be proved.   

¶ The 25. Theorem. The 27. Proposition. If two numbers be prime the one to the other, that which is produced of the one into himself, is prime to the other.  
SVppose that there be two numbers prime the one to the other, A and B, and let A multiplying himself produce HUNDRED Then I say that B and C are prime the one to the other. For put unto A an equal number, 
Demonstration.  namely, D. And forasmuch as A & B are prime the one to the other, and A is equal unto D: therefore D and B also are prime the one to the other. Wherefore either of these numbers D and A is prime to B. Wherefore that which is produced of D into A is prime unto B by the former proposition. But that number which is produced of D into A is the number C. Wherefore C and B are prime numbers the one to the other: which was required to be proved.   

¶ The 26. Theorem. The 28. Proposition. If two numbers be prime to two numbers, each to either of both: the numbers produced of them shall be prime the one to the other.  
SVppose that there be two numbers A and B prime to two numbers C and D, either of both, to either of both: namely, let either of these A and B be prime to C, and also to D. 
Demonstration.  And let A multiplying B produce E, and let C multiplying D produce F. Then I say that E and F are prime numbers the one to the other. For forasmuch as either of these A & B are prime unto C, therefore that which is produced of A into B is prime unto C by the 26. of the seventh. But that which is produced of A into B is the number E, therefore E and C are prime the one to the other. And by the same reason also E and D are prime the one to the other● Wherefore either of these numbers C and D are prime unto E: wherefore that also which is produced of C into D is prime unto E by the same. But that which is produced of C into D, is the number F. Wherefore E and F are numbers prime the one to the other: which was required to be demonstrated.   

¶ The 27. Theorem. The 29. Proposition If two numbers be prime the one to the other, and each multiplying himself bring forth certain numbers: the numbers of them produced shall be prime the one to the other. And if those numbers given at the beginning multiplying the said numbers produced, produce any numbers: they also shall be prime the one to the other: and so shall it be continuing infinitely.  
SVppose that there be two numbers A and B prime the one to the other. And let A multiplying himself produce C: and multiplying C let it produce E. Likewise let B multiplying himself produce D, and multiplying D let it produce F. Then I say what C and D are numbers prime the one to the other. And likewise that E and F are numbers prime the one to the other. For forasmuch as A and B, 
Demonstration.  are prime the one to the other, and A multiplying himself produced C, therefore C and B are prime the one to the other (by the 27. of the seventh). And by the same reason forasmuch as C and B are prime the one to the other, and B multiplying himself produced D, therefore C and D are prime the one to the other. Again forasmuch as A and B are prime the one to the other, and B multiplying himself produced D. Therefore (by the 27 of the seventh) A and D are prime the one to the other. Now then forasmuch as two numbers A and C are prime to two numbers B and D, either of both to either of both: therefore (by the 28. of the seventh) that which is produced of A into C is prime to that which is produced of B into D. But that which is produced of A into C is the number E, and that which is produced of B into D is the number F. Wherefore E and F are numbers prime the one to the other. And so always if A & B multiplying the numbers E and F do produce any numbers, the numbers produced, may by the former Proposition, be proved to be prime the one to the other: which was required to be proved.   

¶ The 28. Theorem. The 30. Proposition. If two numbers be prime the one to the other: then both of them added together, shall be prime to either of them. And if both of them added together be prime to any one of them, than also those numbers given at the beginning, are prime the one to the other.  
SVppose that these two numbers AB and BC being prime numbers be added together. Then I say, that both these added together, namely, the number ABC, is prime to either of these AB, and BC. For if CA and AB be not prime the one to the other, some number than shall measure them. 
Demonstration of the first part leading to an absurdity.  Let some number measure them, and let the same be D. Now then forasmuch as D measureth the whole CA and the part taken away AB, it measureth also the residue CB (by the 4. common sentence). And it measureth BA. Wherefore D measureth these numbers AB and BC, being prime the one to the other: which is impossible (by the 13. definition of the seventh). Wherefore no number measureth these numbers CA and AB. Wherefore CA and AB are prime the one to the other. And by the same reason also may it be proved, that CA and BC are prime the one to the other. Wherefore the number AC is to either of these numbers AB and BC, prime.  
But now suppose that the numbers CA and AB be prime the one to the other. Then I say, that the numbers AB and BC are also prime the one to the other, For if, AB & BC be not prime the one to the other: some one number measureth these numbers AB and BE Let some one number measure them, and let the same be D. And forasmuch as D measureth either of these numbers AB and BC, it shall also measure the whole CA (by the 6. common sentence). 
Demonstration of the second part which is the con●c●se of the first, leaning also to an absurdity.  And it also measureth AB. Wherefore D measureth these numbers CA and AB being prime the one to the other: which is impossible (by the 13. definition of the seventh). Wherefore no number measureth these numbers AB and BC. Wherefore AB and BC are prime the one to the other, which was required to be proved.   

¶ The 29. Theorem. The 31. Proposition. Every prime number is to every number which it measureth not, prime.  
SVppose that there be given a prime number, namely, A, and let B be an other number, which it measureth not. 
Demonstrasion leading to an absurdity.  Then I say, that the numbers B & A are prime the one to the other. For if A and B be not prime the one to the other, than some number measureth them. Let there be a number that measureth them, and let the same be C. Now C is no unity. And forasmuch as C measureth B, but A measureth not B, therefore C is not one and the same number with A. And forasmuch as C measureth A & B, it also measureth A being a prime number, and being not one and the same with it: which is impossible (by the 13. definition of the seventh). Wherefore no number measureth these numbers A and B. And therefore A and B are prime the one to the other: which was required to be proved.   

¶ The 30. Theorem. The 32. Proposition. If two numbers multiplying the one the other produce any number, and if also some prime number measure that which is produced of them: then shall it also measure one of those numbers which were put at the beginning.  
SVppose that two numbers A and B multiplying the one to the other do produce the number C: and let some prime number, namely, D measure C. Then I say that D measureth one of these numbers either A or B. Suppose that it measure not A: now D is a prime number. Wherefore A and D are prime the one to the other (by the proposition next going before). And how often D measureth C, so many unities let there be in E. 
Demonstrasion.  And forasmuch as D measureth C by those unities which are in E, therefore D multiplying E produceth C: but A also multiplying B produced C wherefore that which is produced of D into E is equal to that which is produced of A into B. Wherefore (by the 19 o● the seventh) as A is to D, so is E to B: but D and A are prime numbers: and therefore the lest numbers in that proportion: but the lest in any proportion measure the numbers having the self same proportion with them equally, the greater, the greater, and the less, the less, that is the antecedent, the antecedent, and the consequent, the consequent (by the 21. of the seventh). wherefore the consequent D measureth the consequent B. In like sort may we prove that if D measure not B it measureth A. Wherefore D measureth one of these numbers A or B: which was required to to be proved.  

A Corollary.  

A Corollary ●●ded by Campave.  Hereby it is manifest that if a number measure a number produced of two numbers multiplied the one into the other, or be commensurable to the same, it shall also either measure one of the two number● multiplied, or be comme●surable with one of them.    

The 31. Theorem. The 3●. Proposition. Every composed number, is measured by some prime number.  
SVppose that A be a composed number. Then I say that A is measured by some prime number. For forasmuch as A is a composed number, some number must needs measure it (by the 14. definition of the seventh). Let there be a number that measureth it, and let the same be B. Now if B be a prime number than is that manifest which we seek for: 
Demonstration l●ading to an impossibility.  but if it be composed number some number must needs measure it (by the sel●e same definition) Let there be a number that measureth it, and let the same be C. And forasmuch as C measureth B, and B measureth A: therefore C also measureth A (by the 5. common sentence): and if C be a prime number than is that manifest which we sought for. But if it be a composed number some number shall measure it: and the like consideration being had there shall at the length be found some prime number which measureth the number going before, which shall also measure A. For if there be not found any such prime number than shall infinite numbers decresing measure the said number A of which the one is less than the other, which is impossible in numbers. Wherefore some prime number shall at the length be found which shall measure the number going before and which shall also measure the number A (by the 5. common sentence). Every composed number therefore is measured by some prime number: which was required to be proved.  

¶ An other way.  
Suppose that A be a composed number. 
another demonstration.  Then I say that some prime number measureth it. For forasmuch as A is a composed number, some number shall measure it (by the 14. definition of the seventh) Let the lest number that measureth it be B. Then I say that B is a prime number. For if B be not a prime number some number shall measure it. Let C measure it. Wherefore C is less than B. And forasmuch as C measureth B, and B measureth A, therefore C also measureth A being less than B, which by supposition is the lest number that measureth A, which is absurd. Wherefore B is not a composed number, but a prime number, which was required to be proved.    

The 32. Theorem. The 34. Proposition. Every number is either a prime number, or else some prime number measureth it.  
SVppose that there be a number A. 
Demonstration.  Then I say that A is either a prime number, or else some prime number measureth it. For if A be a prime number than is that had which is required. But if it be a composed number, some prime number shall measure it (by the 33. of the seventh). Every number therefore is either a prime number, or else some prime number measureth it: which was required to be demonstrated.   

¶ The 3. Problem. The 35. Proposition. How many numbers soever being given, to found out the lest numbers that have one and the same proportion with them.  
SVppose that there be a multitude of numbers given, namely, A, B, and C. It is required to find out the lest numbers that have one and the same proportion with these numbers A, B, C. 
Two cases in this Proposition.  These numbers A, B, C, are either prime the one to the other, or not prime. If A, B, C, be prime the one to the other, then are they the lest that have one and the same proportion with them (by the 23. of the seventh). 
The first case.   

The second case.  But if they be not prime, take by the 3. of the seventh, unto A, B, C, the greatest common measure, which let be the number D. And how often D measureth every one of these A, B, C, 
Demonstration.  so many unities let there be in every one of these numbers EFG. Wherefore these numbers E, F, G, do measure these numbers A, B, C, by those unities which are in D. Wherefore these numbers E, F, G, measure these numbers A, B, C, equally. Wherefore E, F, G, are by the 18. of the seventh, in the self same proportion that A, B, C, are. Now than I say that they also are the lest. For if E, F, G, 
Demonstration leading to an absurdity.  be not the lest that have one and the same proportion, with A, B, C, there shall then be some numbers less than E, F, G, being in the self same proportion that A, B, C, are. Suppose that the same numbers be H, K, L, which shall measure the numbers A, B, C, equally. How many times H measureth A, so many unities let there be in M. Wherefore either of these K and L measureth either of these B and C by those unities which are in M (by the 21. of the seventh). And forasmuch as H measureth A by those unities which are in M, therefore M also measureth A by those unities which are in H. And by the same reason M measureth either of these B and C by those unities which are in either of these K and L. Wherefore M measureth these numbers A, B, C. And forasmuch as H measureth A by those unities which are in M, therefore H multiplying M, produceth A. And by the same reason E multiplying D, produceth A. Wherefore that which is produced of E into D is equal to that which is produced of H into M. Wherefore (by the 19 of the seventh) as E is to H, so is M to D. But E is greater than H, wherefore M also is greater than D, and it measureth these numbers A, B, C, which is impossible. For D is supposed to be the greatest common measure unto A, B, C. Wherefore there shall be no other numbers less than E, F, G, and in the self same proportion with A, B, C. Wherefore E, F, G, are the least numbers which have one and the same proportion with A, B, C: which was required to be done.  

A Corollary.  
Hereby it is manifest that the greatest common measure to numbers how many soever: 
A Corollary added by Campane.  measureth the said numbers by the numbers in the lest proportion that the numbers given are.    

¶ The 4. Problem. The 36. Proposition. Two numbers being given, to find out the least number which they measure.  
SVppose that the ●●o numbers ge●e● be A and B. It is required to find 〈◊〉 the jest number which they measure● No● A and B are ei●h●r prime ●he one to the other, or not. 
Two cases in this proposition.  Suppose first that A and ● be prime the one to the other: and let A multiplying B produce C: wherefore B, multiplying A produ●●● also C (by the 16. of the seventh.) 
The first case.  Wherefore A and B measure C. Now also I say, that C is the jest number which they measure● 
Demonstration leading to an absurdity.  For if it be not, those numbers A and ● measure some number less the● C: let them measure some number less than C, and let the same be D: and how often A measureth D, so many unities let there be in E● and how often ● measureth D, so many unities let there be in ●● Wherefore A multiplying E producent D, and B multiplying F pro●●●●●● also D. Wherefore that which is produced of A into ●, is equal to that which is produced of B into ●: wherefore (by the 19 of the seventh) as A is to B, so is F to E. But A and B are prime: yea they are prime and also the jest in that proportion (by the 23. of the seventh): but the jest numbers in any proportion measure those numbers that have one and the same proportion with them equally: the greater the greater: and the less the less (by the 21 of the seventh). Wherefore B measureth E, namely, the consequent, the consequent. And forasmuch as A multiplying B and E produced C and D: therefore (by the 17. of the seventh) as ● is to E, so is C to D. But B measureth E. Wherefore C also measureth D, the greater, the less which is impossible. Wherefore if those numbers A and B be prime, they shall measure no number less then C. Wherefore C is the jest number which A and B measure.  
But now suppose that A and B be not prime the one to the other, 
The second case●  and take (by the 35. of the seventh) the jest numbers that have one and the same proportion with A and B, and let the same be F and E. Wherefore that which is produced of A into E, is equal to that which is produced of B into F (by the 19 of the seventh). Let A multiplying E produce C● wherefore B multiplying F produceth also C. Wherefore A and B measure C. Then I say, that C is the jest number that they measure. For if it be not, those numbers A and B shall measure some number less than C: let them measure a number less than C, and let the same be D● And how often A measureth D, so many unities let there be in G, And how often ● measureth D● so many unities let there be in H. 
Demonstration leading to an absurdity.  Now then A multiplying G produceth D. And B multiplying H produceth also D. Wherefore that which is produced of A into G is equal to that which is produced of B into H. Wherefore (by the 19 of the seventh) as A is to B, so is H to G● But as A is to B, so is F to E. Wherefore as F is to E, so is H to G: but the jest numbers in any proportion measure the numbers that have the same proportion with them equally, the greater the greater, & the less the less (by the 21. of the seventh). Wherefore E measureth G. And forasmuch as A multiplying G and E produced C and D● therefore (by the 17. of the fift) as E is to G, so is C to D. But E measureth G. Wherefore C also measureth D, the greater the less: which is impossible. Wherefore those numbers. A and B do not measure any number less than ●. Wherefore ● is the jest number that is measured by A and B● whi●h was required to be done.   

The 33. Theorem. The 37. Proposition. If two numbers measure any number, the lest number also which they measure, measureth the self same number.  
SVppose that there be two numbers given A and B ● and let them measure the number CD: 
Demonstration leading to an impossibility.  and let the lest number that they measure be E. Then I say that E also measureth the number CD. For if E do not measure CD, let E measuring CD, that is subtrahed out of CD as often as you can, as for example, once, leave a less than itself, namely, CF. And let the number subtrahed which E measureth be FD, and forasmuch as A and B measure E, and E measureth DF, therefore A and B also measure DF. And they measure the whole CD, wherefore by the 4. common sentence of the seventh they measure also that which remaineth CF being less then E: which is impossible. Wherefore E of necessity measureth CD, which was required to be proved.   

¶ The 5. Problem. The 38. Proposition. Three numbers being given, to find out the lest number which they measure.  
SVppose that there be three numbers given A, B, C. It is required to find out the lest number which they measure. Take (by the 36. of the seventh) the least number which A and B measure, and let the same be D. 
Two cases in this proposition.  Now then C either measureth D or else measureth it not. 
The first case.  First let it measure it. And the numbers also A and B measure D: wherefore A, B, C, measure D. Now then I say that D is the lest number which they measure. For if not, let the numbers A, B, C, measure some number less than D, and let the same be E. 
Demonstration lea●i●g ●o an absurdity.  And forasmuch as A, B, C, measure E therefore also A and B measure E, wherefore (by the 37. of the seventh) the least number which those numbers A and B measure shall also measure E. But the lest number which A and B measure is D. Wherefore D measureth E, the greater the less: which is impossible. Wherefore these numbers A, B, C, shall not measure any number less then D. Wherefore D is the lest number that A, B, C, do measure.  
But now suppose that C measure not D. And take (by the 36. of the seventh) the least number which those two numbers C and D do measure, and let the same be E. 
The second case.  And forasmuch as A and B measure D, and D measureth E, therefore A and B also measure E, and C also measureth E, wherefore A, B, C, also measure E. I say moreover that E is the lest number which A, B, C measure. For if it be not, let there be some less number than E which they measure, 
Demonstration leading to an absurdity.  and let the same be F. And forasmuch as A, B, C, measure F. Therefore A and B also measure F: wherefore the lest number which these numbers A and B do measure doth also measure F (by the 37. of the seventh). But the lest number which A and B do measure is D. Wherefore D measureth F. And C also measureth F. Wherefore D and C measure F. Wherefore the lest number which C and D do measure, shall also (by the self same) measure F. But the lest ●●mbe● which C & D measure is E. Wherefore E measureth F, namely, the greater, the less, which is impossible. Wherefore these numbers A, B, C, do not measure any number less then E. Wherefore E is the lest number which A, B, C, do measure: which was required to be demonstrated.  
In like manner also how many numbers soever being given, may be found out the lest number which they ●easure. For if unto the three numbers A, B, C, be added a forth, then if the said forth number measure the number E, then is E the least number which the four numbers given measure. But if it do not measure E, then by the 37. of this book must you find but the lest number which E and the forth number measure. Which shall be the number sought for. And so likewise if there be five, six, or how manysoever given.  

Corollary.  
Hereby it is manifest that the least common measure to numbers howmanysoever, 
A Corollary.  measureth every number which the said numbers how many soever measure.    

¶ The 34. Theorem. The 39 Proposition. If a number measure any number: the number measured shall have a part after the denomination of the number measuring.  
SVppose that there be a number B, which let measure the number A. Then I say, that A hath a part taking his denomination of the number B. For how often B measureth A, so many unities let there be in C. And let D be unity. And forasmuch as B measureth A, by those unities which are in C, and unity D measureth C by those unities which are in C, therefore unity D, so many times measureth the number C, as B doth measure A. 
Demonstration.  Wherefore alternately (by the 15. of the seventh) unity D, so many times measureth B, as C doth measure A. Wherefore what part unity D is of the number B, the same part is C of A. But unity D is a part of B having his denomination of B. Wherefore C also is a part of A having his denomination of B. Wherefore A hath C as a part taking his denomination of B: which was required to be proved.  
The meaning of this Proposition is, that if three measure any number, that number hath a third part, and if four measure any number the said number hath a fourth part. And so forth.   

¶ The 35. Theorem. The 40. Proposition. If a number have any part: the number whereof the part taketh his denomination shall measure it.  
SVppose that the number A have a part, namely, B: and let the part B have his denomination of the number C. Then I say, that C measureth A. Let D be unity. And forasmuch as B is a part of A, having his denomination of C: 
The converse of the former proposition.  and D being unity is also a part of the number C, having his denomination of C: therefore what part unity D is of the number C, the same part is also B of A: wherefore unity D so many ●●●es measureth the number C, 
Demonstration.  as B measur●●● A. Wherefore alternately (by ●●e 15. ●f the s●●●nth) unity D so many ●●mes m●●●●reth the nu●●be● B, 〈◊〉 C meas●●eth A. Wheref●●● C measureth A: which was requi●●d to b● proved.  
This Proposition is the converse of the former: and the meaning thereof is, that every number having a third part is measured of three, and having a fourth part is measured of four. And so forth.   

¶ The 6. Problem. Th' 41. Proposition. To find out the lest number, that containeth the parts given.  
SVppose that the parts given be A, B, C, namely, let A be an half part, B a third part, & C a fourth part. 
Construction.  Now it is required to find out the lest number which containeth the parts A, B, C. Let the said parts A, B, C, have their denominations of the numbers D, E, F. And take (by the 38. of the seventh) ●he lest number which the numbers D, E, F, measure, and let the same be G. And forasmuch as the numbers D, E, F measure the number G, therefore the number G hath parts denominated of the numbers D, E, F (by the 39 of the seventh). But the parts A, B, C, have their denomination of the numbers D, E, F. Wherefore G hath those parts A, B, C. I say also that it is the lest number which hath these parts. 
Demonstration le●ding to an absurdity.  For if G be not the lest number which containeth those parts A, B, C, then let there be some number less than G which containeth the said parts A, B, C. And suppose the same to be the number H. And forasmuch as H hath the said parts A, B, C, therefore the numbers that the parts A, B, C, take their denominations of, shall measure H (by the 40. of the seventh) But the numbers whero● the parts A, B, C, take their denominations of, be D, E, F. Wherefore the numbers D, E, F, measure th● number H which is less than G, which is impossible. For G is supposed to be t●● l●●s● number that the numbers D, E, F, do measure. Wherefore there is no number less than G, which containeth these parts A, B, C: which was required to be done.  

Corrolary.  

A Corollary ad●ed by Campane.  Hereby it is manifest that if there be taken the lest number, that numbers how many soever do measure, the said number shall be the least which hath the parts denominated of the said numbers how many soever.  
Campane after he hath taught to find out the first lest number that containeth the parts given, teacheth also to find out the second lest number, 
How to ●inde out the second least number and the third, and so ●orth infinitely.  that is, which except the lest of all is less than all other, and also the third least, and the fourth etc. The second is found out by doubling the number G. For the numbers which measure the number G ●hall also measure the double thereof (by the 5. commo● sentence of the seventh). But there cannot be given a number greater than the number G, & less than the double thereof, whom the parts given shall ●●asure● For forasmuch as the parts given do ●easur● the whole, namely, which is less than the double, and they also measure the part taken away, namely, the number G, they should also measure the residue, namely, a number less than G, which is proved to be the least number that they do measure, which is impossible● wherefore the second number which the said parts given do measure● must, exceeding G, needs reach to the double of G, and the third to the triple, and the fourth to the quadruple, and so infinitely, for those parts can never measure any number less than the number G.  
By this Proposition also it is easy to found out the lest number containing the parts given of parts. As if we would find out the lest number which containeth one third 〈◊〉 ●● an hal●e part, 
How to si●● out the lest ●●m● a con●ay●●g ●●e pa●●s of parts.  and one fourth part of a third part, reduce the said d●●ers fraction into simple fraction (by the common 〈◊〉 of reducing of fr●●●ions) namely, the 〈◊〉 of an hal●e into a 〈◊〉 part of an wholey and ●he fourth of thi●d into a twelfth part of an● whole. And then by this Problem search out the lest number which containeth a six● part and a twelfth part, and so have you done.    
The end of the seventh book of Euclides Elements.   

¶ The eighth book of Euclides Elements.  
AFter that Euclid hath in the seventh book entreated of the proprieties of numbers in general, and of certain kinds thereof more specially, and of prime and composed numbers with others: now in this eight book he prosecuteth farther, and findeth out and demonstrateth the properties and passions of certain other kinds of numbers: 
The Argu●●●● of the eight books.  as of the lest numbers in proportion, and how such may be found out infinitely in whatsoever proportion: which thing is both delectable, and to great use. Also here is entreated of plain numbers, and solid: and of their sides, and proportion of them. Likewise of the passions of numbers square and cube, and of the natures and conditions of their sides, and of the mean proportional numbers of plain, solid, square, and cube numbers, with many other things very requisite and necessary to be known.  

¶ The first Theorem. The first Proposition. If there be numbers in continual proportion howmanysoever, and if their extremes be prime the one to the other: they are the lest of all numbers that have one and the same proportion with them.  
SVppose that the numbers in continual proportion be A, B, C, D. And let their extremes namely, A and D be prime the one to the other. Then I say that the numbers A, B, C, D, are the lest of all numbers that have one and the same proportion with them. For if they be not, let E, F; G, H being less numbers than A, B, C, D, be in the self same proportion that ABCD are. 
Demonstration leading to an absurdity.  And forasmuch as the numbers A, B, C, D, are in the self same proportion that the numbers E, F, G, H, are, & the multitude of these numbers E, F, G, H, is equal to the multitude of these ●●mbers A, B, C, D, therefore of equality (by the 14. of the seventh) as A is to D, so is E to H. But A and D are prime the one to the other, yea they are prime and the lest that ha●e the same proportion with them: But the lest numbers in any proportion measure the numbers that have the same proportion with them equally, the antecedent the antecedent, and the consequent the consequent (by the 21. of the seventh) wherefore A measureth E, the greater the less: which is impossible. Wherefore the numbers E, F, G, H, being less than A, B, C, D, are not in the same proportion that A, B, C, D, are, wherefore A, B, C, D, are the lest of all numbers which ha●e one and the same proportion with them● which was required to be demonstrated.   

¶ The 1. Problem. The 2. Proposition. To find out the lest numbers in continual proportion, as many as shall be required, in any proportion given.  
SVppose that the proportion given in the least numbers be A to B. It is required to find out the least numbers in continual proportion, as many as shall be required, in the same proportion that A is to B. 
Construction.  Let there be required four. And let A multiplying himself produce C: and multiplying B let it produce D: and likewise let B multiplying himself produce E. And moreover let A multiplying those numbers C, D, E, 
Demonstration.  produce F, G, H: and let B multiplying E produce K. And forasmuch as A multiplying himself produced C, and multiplying B produced D, now then the number A multiplying two numbers A and B produced C & D. Wherefore (by the 17. of the seventh) as A is to B, so is C to D. Again, forasmuch as A multiplying B produced D, and B multiplying himself produced E, therefore each of those numbers A and B multiplying B, bringeth forth these numbers D and E. Wherefore (by the 18. of the seventh) as A is to B, so D to E. But as A is to B, so is C to D. Wherefore as C is to D, so is D to E. And forasmuch as A multiplying C and D produced F and G, therefore (by the 17. of the seventh) as C is to D, so is F to G. But as C is to D, so is A to B. Wherefore as A is to B, so is F to G. Again forasmuch as A multiplying D and E produced G and H, therefore (by the 17. of the seventh) as D is to E, so is G to H. But as D is to E, so is A to B. Wherefore as A is to B, so is G to H. And forasmuch as those numbers A and B multiplying E produced H and K, therefore (by the 18. of the seventh) as A is to B, so is H to K. And it is proved, that as A is to B, so is ● to ●, and G to H: wherefore as F is to G, so is G to H, and H to K. Wherefore these numbers C, D, E, and F, G, H, K, are proportional in the same proportion, that A is to B. Now I say, that they are also the jest. For forasmuch as A and B are the jest of all numbers that have the same proportion with them: but the jest numbers that have one & the same proportion with them are prime the one to the other (by the 24. of the seventh): therefore A and B are prime the one to the other: and each of these numbers A & B multiplying himself produced these numbers C and E, and likewise multiplying each of these numbers C and E they produced F and K. Wherefore (by the 29. of the seventh) C, E, and F, K, are prime the one to the other. But if there be numbers in continual proportion how many soever, and if their extremes be prime the one to the other, they are the jest of all numbers that have the same proportion with them (by the first of the eight). Wherefore these numbers C, D, E, and F, G, H, K, are the jest of ●ll n●mber ●hat ●a●e ●he same proportion with A and B● And forasmuch 〈◊〉 (by the ●9. of the seventh) that always happeneth touching the extre●●●● namely, that A and B multiplying the numbers produced 〈◊〉 shall produce ●ther prime numbers, namely, the extremes of five numbers in continual proportion, therefore (by the first of this book) all five are the jest of that proportion. And so infinitely: which was required to be done.  

¶ Corollary.  
Hereby it is manifest, that if three numbers being in continual proportion, be the jest of all numbers that have the same proportion with them, their extremes are squares: and if there be four their extremes are cubes. For the extremes of three are produced of the multiplying of the numbers A and B into themselves. And the extremes of four are produced of the multiplying of the roots A and B into the squares C and E, whereby are made the cubes F and K.    

The 2. Theorem. The 3. Proposition. If there be numbers in continual proportion how many soever, and if they be the jest of all numbers that have one and the same proportion with them: their extremes shall be prime the one to the other.  
SVppose that the numbers in continual proportion being the lest of all numbers that have the same proportion with them be A, B, C, D. 
This proposition is the ●●uerse of the first.  Then I say that their extremes A and D are prime the one to the other. Take (by the 2. of the eight, or by the 35. of the seuen●h) the two least numbers that are in the same proportion that A, B, C, D, are, and let the same be the numbers E, F. And after that take three numbers G, H, K, and so always forward on more (by the former proposition) until the multitude ●aken be equal to the multitude of the numbers given A, B, C, D. And let those numbers be L, M, N, O. Wherefore (by the 29. of the seventh) their extremes L, O, 
Demonstration●  are prime the one to the other. For forasmuch as E and F are prime the one to the other, and each of them multiplying himself produced G and K, & likewise each of these, G & K multiplying himself produced L & O, therefore (by the 29 of the seventh) G & K are prime the 〈◊〉 to th● other, & so likewise are L and O prime the one to the other. And forasmuch as A, B, C, D, are the lest of all numbers that have the same proportion with them, and likewise L, M, N, O, are the lest of all numbers that are in the same proportion that A, B, C, D, are, and the multitude of these numbers A, B, C, D, is equal to the multitude of these L, M, N, O: therefore every one of these A, B, C, D, is equal unto every one of these L, M, N, O. Wherefore A is equal unto L, and D is equal unto O. And forasmuch as L and O are prime the one to the other, and L is equal unto A, and O is equal unto D: therefore A and D are prime the one to the other: which was required to be proved.   

The 2. Problem. The 4. Proposition. Proportions in the lest numbers how many soever being given, to find out the lest numbers in continual proportion in the said proportions given.  
SVppose that the proportions in the lest numbers given, be A to B, C to D, and E to F. It is required to find out the lest numbers in continual proportion, in the same proportion that A is to B, and that C is to D, and that E is to F. Take the lest number whom B and C do measure, and let the same be G. And how often B measureth G, so many times let A measure H. And how often C measureth G, so many times let D measure K. Now E either measureth K or measureth it not. 
Two cases in this proposition. The first case.  First let it measure it. And how often E measureth K, so many times let F measure L. And forasmuch as how often A measureth H, so many times doth B measure G: therefore by the 17. of the seventh, as A is to B, so is H to G. And by the same reason as C is to D● so is ● to K, and moreover as E is to F, so is K to L. Wherefore these numbers H, G, K, L, are in continual proportion, and in the same proportion that A is to B, and that C is to D, and moreover that E is to F. I say also that they are the lest in those proportions● 
Demonstration leading to an absurdity.  For if H, G, K, L, be not the lest numbers in continual proportion, and in the same proportions that A is to B, and C to D, and E to F, then are there some numbers less● then H, G, K, L, in the same proportions that A is to B and C to D, and E to F, let those numbers be N, X, M, O. And forasmuch as A is to B, so is N to X● and A and B are the jest, but the lest measure those numbers that have one and the same proportion with them equally, the greater, the greater, and the less the less, that is, the antecedent, the antecedent, & the consequent the consequent (by the 21. of the seventh) therefore B measureth X. And by the same reason C also measureth X, wherefore B and C measure X. Wherefore the lest number whom B and C measure, shall also by the 37 of the seventh measure X. But the lest number whom B and C measure is G. Wherefore G measureth X, the greater, the less, which is impossible. Wherefore there shall not be any less numbers than H, G, K, L, in continual proportion, and in the same proportions that A is to B, and C to D, and E to F.  
But now suppose that E measure not K. 
The second case.  And by the 36. of the seventh, take the lest number whom E and K measure, and let the same be M. And how often K measureth M, so often let either of these G and H measure either of these N and X. 
Demonstration.  And how often E measureth M, so often let F measure O. And forasmuch as how often G measureth N, so often doth H measure X, therefore as ● is to G, so is X to N. But as H is to G, so is A to B Wherefore as A is to B, so is X to N. And by the same reason as C is to D, so is N to M. Again, forasmuch as how often E measureth M, so often F measureth O, therefore as E is to F so is M to O. Wherefore X, N, M, O, are in continual proportion, and in the same proportions that A is to B, and C to D, and E to F. I say also that they are the lest in that proportion. For if X, N, M, O, be not the lest in continual proportion, and in the same proportions that A is to B, and C to D, and E to F, then shall there be some numbers less than X, N, M, O, in continual proportion, and in the same proportions that A is to B, and C to D, and E to F. Let the same be the numbers P, R, S, T. And for that as P is to R, so is A to B, and A and B are the lest, but the lest numbers measure those numbers that have one & the same proportion with them equally, the greater the greater, and the less the less, that is, the antecedent the antecedent, and the consequent the consequent by the 21. of the seventh, therefore B measureth R. And by the same reason also C measureth R. Wherefore B and C measure R. Wherefore the lest number wh●m B and C measure shall also measure R (by the 37. of the seventh). But the jest number whom B and C measure is G, wherefore G measureth R. And as G is to R, so is K to S. Wherefore K measureth S. And E also measureth S. Wherefore E and K measure S. Wherefore the lest number whom E and K measure, shall (by the self same) measure S. But the lest number whom E and K measures M. Wherefore M measureth S the greater the less, which is impossible. Wherefore there are no numbers less than X, N, M, O, in continual proportion, & in the same proportions that A is to B, and C to D, and E to F. Wherefore X, N, M, O are the lest numbers in continual proportion, and in the same proportions that A is to B, and C to D, and E to F: which was required to be done.   

¶ The 3. Theorem. The 5. Proposition. Plain or superficial numbers are in that proportion the one to the other which is composed of the sides.  

This proposition in numbers answereth to the of the sixth touching parellelogrammes.  SVppose that A and B be plain or superficial numbers, and let the sides of A be the numbers C and D, and let the sides of B, be the numbers E and F. Then I say that A is to B in that proportion that is composed of the sides. Take (by the fourth of the eight) the least numbers in continual proportion, and in the same proportions that C is to E, and D to F. 
Construction.  And let the same be the numbers G, H, K: 
Demonstration.  so that as C is to E, so let G be to H, & as D is to F, so let H be to K. Wherefore those numbers G, H, K, have the proportions of the sides: but the proportion of G to K is con posed of that which G hath to H and of that which H hath to K: wherefore G is unto K in that proportion which is composed of the sides. Now I say that as A is to B, so is G to K. For let D multiplying E produce L. And forasmuch as D multiplying C produced A, and multiplying E produced L: therefore (by the 17. of the seventh) as C is to E, so is A to L. But as C is to E, so is G to H, wherefore as G is to H, so is A to L. Again forasmuch as E multiplying D produced L, & multiplying F produced B: therefore (by the 17. of the seventh) as D is to F, so is L to B. But as D is to F, so is H to K, wherefore as H is to K, so is L to B. And it is proved that as G is H, so is A to L. Wherefore of equality (by the 14. of the seventh) as G is to K, so is A to B. But G is unto K in that proportion which is composed of the sides, wherefore A is unto B in that proportion which is composed of the sides: which was required to be demonstrated.  

¶ An other demonstration of the same after Campane.  
Suppose that A and B be plain numbers: and let the sides of A be the numbers C and D: and let the numbers E and F be the sides of the number B. And let D multiplying E produce the number G. Then I say that the proportion of A to B is composed of the proportions of C to E & D to F that is, 
another demonstration after Campane.  of the sides of the superficial number A to the sides of the superficial number B. For forasmuch as D multiplying E produced G, and multiplying C it produced A, therefore by (the 17. of the seventh) A is to G as C is to E: again forasmuch as E multiplying D produced G and multiplying F it produceth B, therefore by the same G is to B as D is to F. Wherefore the proportions of the sides namely, of C to E and of D to F are one and the same with the proportions of A to G and G to B. But (by the fifth definition of the sixth) the proportion of the extremes A to B is composed of the proportions of the means, namely, of A to G and G to B, which are proved to be one and the same with the proportions of the sides C to E, and D to F. Wherefore the proportion of the superficial numbers A to B is composed of the proportions of the sides C to E, and D to F. Wherefore pla●ne. etc. which was required to be proved.    

¶ The 4. Theorem. The 6. Proposition. If there be numbers in continual proportion how many soever, and if the first measure not the second, neither shall any one of the other measure any one of the other.  
SVppose that there be numbers how many soever in continual proportion, namely, f●ue, A, B, C, D, E. And suppose that A measure not B. Then I say, that neither shall any other of the numbers A, B, C, D, E, measure any one of the other. That A, B, C, D, E, do not in continual order measure one the other, it is manifest: for A measureth not B. Now I say, that neither shall any other of them measure any other of them. I say that A shall not measure C. For how many in multitude A, B, C, are, take so many of the jest numbers that have one and the same proportion with A, B, C, (by the 35. of the seventh) and let the same be F, G, H. 
Demonstration.  And forasmuch as F, G, H are in the self same proportion that A, B, C, are: and the multitude of these numbers A, B, C, is equal to the multitude of those numbers F, G, H, therefore of equality (by the 14. of the seventh) as A is to C, so is F to H. And for that as A is to B, so is F to G, but A measureth not B, therefore neither doth F measure G. Wherefore F is not unity. For if F were unity, it should measure any number. But F and H are prime the one to the other (by the 3. of the eight). Wherefore F measureth not H: & as F is to H, so is A to C, wherefore neither doth A measure C. In like sort may we prove that neither shall any other of the numbers A, B, C, D, E, measure any other of the numbers A, B, C, D, E: which was required to be demonstrated.   

¶ The 5. Theorem. The 7. Proposition. If there be numbers in continual proportion how many soever, and if the first measure the last, it shall also measure the second.  
SVppose that there be a multitude of numbers in continual proportion, namely, A, B, C, D. And let A the first measure D the last. 
Demonstration leading to an impossibility.  Then I say, that A the first measureth B the second. For if A do not measure B, neither shall any other measure any other (by the 7. of the eight): which (by supposition) is not true. For A is supposed to measure D. Now then A measuring D, shall also measure B: which was required to be proved.   

¶ The 6. Theorem. The 8. Proposition. If between two numbers there fall numbers in continual proportion: how many numbers fall between them, so many also shall fall in continual proportion between other numbers which have the self same proportion.  
SVppose that between the two numbers A and B, do fall in continual proportion the numbers C and D. And as A is to B, so let E be to F. Then I say, that how many numbers in continual proportion do fall between A and B, so many numbers also in continual proportion shall there fall between E and F. How many A, B, C, D, are in multitude, take (by the 35. of the seventh) so many of the lest numbers that have one and the same proportion with A, B, C, D, and let the same be G, H, K, L. Wherefore their extremes G and I are prime the one to the other (by the 3. of the eight). And forasmuch as A and C, and D and B, are in the self same proportion that G & H, and K and L are, 
Demonstration.  and the multitude of these numbers A, C, D, B, is equal to the multitude of these numbers G, H, K, L: therefore of equality (by the 14. of the seventh) as A is to B, so is G to L. But as A is to B, so is E to F. Wherefore as G is to L, so is E to F. But G and L are prime the one to the other: yea they are prime and the lest. But the lest numbers measure those numbers that have the same proportion with them equally, the greater the greater, and the less the less (by the 21. of the seventh) that is, the antecedent, the antecedent, & the consequent, the consequent. Wherefore how many times G measureth E, so many times L measureth F● How often G measureth E, so often let H measure M, and K measure N Wherefore these numbers G, H, K, L, equally measure these numbers E, M, N, F. Wherefore (by the 18. of the seventh) these numbers G, H, K, L, are in the self same proportion that E, M, N, F, are. But G, H, K, L, are in the self same proportion that A, C, D, B, are: wherefore those numbers A, C, D, B, are in the self same proportion that E, M, N, F, are● But A, C, D, B, are in continual proportion: wherefore also E, M, N, F, are in continual proportion. Wherefore how many numbers in continual proportion fall between A and B, so many also in continual proportion fall there between E and F: which was required to be demonstrated.  

A Corollary added by Flussates.  

A Corollary added by Flussates.  Between two numbers whose proportion is superparticular, or superbipartient, there falleth no mean● proportional. For the lest numbers of that proportion differ the one from the other only by unity or by two. But if between the greater numbers of that proportion there should fall a mean proportional then should there fall also a mean proportional between the lest numbers which have the same proportion by this Proposition. But between numbers differing only by unity or by two, there falleth no mean proportional.    

¶ The 7. Theorem. The 9 Proposition. If two numbers be prime the one to the other, and if between them shall fall numbers in continual proportion: how many numbers in continual proportion fall between them, so many also shall fall in continual proportion between either of those numbers and unity.  
SVppose that there be two numbers prime the one to the other A and B: and let there fall between them in continual proportion these numbers C and D: and let E be unity. Then I say, that how many numbers in continual proportion fall between A and B, so many also shall fall in continual proportion between A and unity E: 
Construction.  and likewise between B and unity E. Take (by the 35. of the seventh) the two least numbers that are in the same proportion that A, C, D, B, are: and let the same be F and G: and then take three of the lest numbers that are in the same proportion that A, C, D, B, are: and let the same be H, K, L: and so always in order one more, until the multitude of them be equal to the multitude of these numbers A, C, D, B: and those being so taken let them be M, N, X, O. Now it is manifest, that F multiplying himself produced H, and multiplying N produced M. 
Demonstration.  And G multiplying himself produced L, and multiplying L produced O. And forasmuch as M, N, X, O, are (by supposition) the least of all numbers that have the same proportion with G, F: and A, C, D, B, are (by the first of the eight) the least of all numbers that have the same proportion with G, F: and the multitude of these numbers M, N, X, O, is equal to the multitude of these numbers A, ●, D, B: therefore every one of these numbers M, N, X, O, is equal to every one of these numbers A, C, D, B. Wherefore M is equal unto A, and O is equal unto B. And forasmuch as F multiplying himself produced H: therefore ● measureth H by those unities which are in F: and unity E measureth F by those unities which are i● F: wherefore (by the 15. of the seventh) unity E, so many times measureth the number F, as F measureth H● wherefore as unity E is to the number F, so is F to H. Again forasmuch as F multiplying N produced M, therefore H measureth M by those unities which are in F. And unity E measureth F by th●se unities which are in F: wherefore (by the self same) unity E so many times measureth F, as H measureth M. Wherefore as unity E is to the numbers F, so is H to M. But it is proved, that as unity E is to the number F, so is F to H: wherefore as unity E is to the number F, so is F to H, and H to M. But M is equal unto A● wherefore as unity E is to the number F, so is F to H, & H to A. And by the same reason as unity E is to the number G, so is G to L and L to B. Wherefore how many numbers fall in continual proportion between A and B: so many numbers also in continual proportion fall there between unity E and the number A, and likewise between unity E and the number ●: which was required to be demonstrated.   

¶ The 8. Theorem. The 10. Proposition. If between two numbers and unity fall numbers in continual proportion: how many numbers in continual proportion fall between either of them & unity so many also shall there fall in continual proportion between them.  
SVppose that between the two numbers A, B, and unity C ● do fall these numbers in continual proportion D, E, and F, G. 
This proposition is the converse of the former.  Then I say that how many numbers in continual proportion there are between either of these A, B, and unity C, so many numbers also in continual proportion shall there fall between A and B. Let D multiplying F produce H, and let D multiplying H produce K, and like wise let F multiplying H produce L. 
Construction.  And for that by supposition as unity C is to the number D, so is D to E, therefore how many times unity C measureth the number D, 
Demonstration.  so many times doth D measure E. But unity C measureth D by those unities which are in D ● wherefore D measureth E by those unities which are in D. Wherefore D multiplying himself produceth E. Again for that as unity C is to the number D, so is E to A, therefore how many times unity C measureth the number D, so many times E measureth A. But unity C measureth D, by those unities which are in D, therefore E measureth A by those unities which are in D. Wherefore D multilying E produced A. And by the same reason F multiplying himself produced G, and multiplying G produced B. And forasmuch as D multiplying himself produced E, and multiplying F produced H, therefore (by the 17. of the seventh) as D is to F, so is E to H. And by the same reason as D is to F, so is H to G. Wherefore as E is to H, so is H to G. Again forasmuch as D multiplying E produced A, and multiplying H produced K, therefore (by the 17. of the seventh) as E is to H, so is A to K. But as E is to H, so is D to F, therefore as D is to F, so is A to K. Again forasmuch as D multiplying H produced K, and F multiplying H produced L, therefore (by the 17. of the seventh) as D is to F, so is K to L. But as D is to F, so is A to K, wherefore as A is to K, so is K to L. Again forasmuch as F multiplying H produced L and multiplying G produced B, therefore (by the 17. of the seventh) as H is to G, so is L to B. But as H is to G so is D to F, wherefore as D is to F so i● L to B. And it is proved that as D is to F, so is A to K, and K to L, and L to B. Wherefore the numbers A, K, L, B, are continual proportion. Wherefore how many numbers in continual proportion fall between either of these numbers A, B, & unity C, so many also in continual proportion fall there between the numbers A and B: which was required to be proved.   

¶ The 9 Theorem. The 11. Proposition. Between two square numbers there is one mean proportional number. And a square number to a square, is in double proportion of that which the side of the one is to the side of the other.  
SVppose that there be two square numbers A and B, and let the side of A be C, & let the side of B be D. Then I say that between these square numbers A and B, there is one mean proportional number, and also that A is unto B in double proportion of that which C is to D. Let C multiplying D produce E. 
The first part of this proposition demonstrated.  And forasmuch as A is a square number, & the side thereof is C, therefore C multiplying himself produced A. And by the same reason D multiplying himself produced B. Now forasmuch as C multiplying C produced A, and multiplying D produced E, therefore (by the 17. of the seventh) as C is to D, so is A to E. Again forasmuch as C multiplying D produced E, and D multiplying himself produced B, therefore these two numbers C and D multiplying one number, namely, D, produce E and B. Wherefore (by the 18. of the seventh) as C is to D, so is E to B. But as C is to D, so is A to E. Wherefore as A is to E, so is E to B. Wherefore between these square numbers A and B, there is one mean proportional number, namely, E. 
The second part demonstrated.  Now also I say that A is unto B in double proportion of that which C is to D. For forasmuch as there are three numbers in continual proportion, A, E, B, therefore (by the 10. definition of the fift) A is unto B in double proportion of that which A is to E. But as A is to E, so is C to D. Wherefore A is unto B in double proportion of that which the side C is unto the side D: which was required to be proved.   

¶ The 10. Theorem. The 12. Proposition. Between two cube numbers there are two mean proportional numbers. And the one cube is to the other cube in triple proportion of that which the side of the one is to the side of the other.  
●Vppose that there be two cube numbers A and B, and let the side of A be C, and let the side of B be D. Then I say that between those cube numbers A and B, there are two mean proportional numbers, and that A is unto B in triple proportion of that which C is to D. Let C multiplying himself produce E, and multiplying D let it produce F, 
Construction.  and let D multiplying himself produce G. And let C multiplying F produce H, and let D multiplying F produce K. 
The first part of this pr●position demonstrated.  And forasmuch as A is a cube number, and the side thereof is C, & C multiplying himself produceth E, therefore C multiplying E produceth A. And by the same reason for that D multiplying himself, produced G, therefore D multiplying G produceth B. And forasmuch as C multiplying C and D produced E and F: therefore by the 17. of the fift, as C is to D, so is E to F. And by the same reason also, as C is to D, so is F to G. Again forasmuch as C multiplying E and F produced A and H, therefore as E is to F, so is A to H. But as E is to F, so is C to D. Wherefore as C is to D: so is A to H. Again forasmuch as each of these numbers C and D multiplying F produced H and K, therefore (by the 18. of the seventh) as C is to D, so is H to K. Again forasmuch as D multiplying F and G produced K & B: therefore (by the 17. of the seventh) as F is to G, so is K to B. But as F is to G, so is C to D, wherefore as C is to D, so is K to B. And it is proved that as C is to D, so is A to H, and H to K, and K to B: wherefore between these cube numbers A and B, there are two mean proportional numbers, that is, H and K.  
Now also I say, that A is unto B in triple proportion of that which C is to D. 
The second part demonstrated.  For forasmuch as there are four numbers proportional A, H, K, B, therefore (by the 10. definition of the fift) A is unto B in triple proportion of that which A is unto H. But as A is unto H, so is C to D, wherefore A is unto B in triple proportion of that which C is to D: which was required to be proved.   

¶ The 11. Theorem. The 13. Proposition. If there be numbers in continual proportion how many so ever, and each multiplying himself produce certain numbers, the numbers of them produced shall be proportinall. And if those numbers given at the beginning multiplying the numbers produced, produce other numbers, they also shallbe proportional: and so shall it be continuing infinitely.  
SVppose that there be a multitude of numbers in continual proportion, namely, A, B, C, as A is to B, so let B be to C. And let A, B, C, multiplying each himself bring forth the numbers D, E, F, & multiplying the numbers D, E, F, let them bring forth the numbers G, H, K. Then I say that D, E, F, are in continual proportion, and also that. G, H, K, are in continual proportion. 
Construction.  For it is manifest that the numbers D, E, F, are square numbers, & that the numbers G, H, K, are cube numbers. Let A multiplying B produce L. And let A & B multiplying L produce M and N. And again let B multiplying C produce X: and let B and C multiplying X produce O and P. 
Demonstration.  Now by the discourse of the proposition going before we may prove that D, L, E, and also G, M, N, H, are in continual proportion and in the same proportion that A is to B and likewise that E, X, F, and also H, O, P, K, are in continual proportion and in the same proportion that B is to C. But as A is to B, so is B to C. Wherefore D, L, E, are in one and the same proportion with E, X, F and moreover G, M, N, H, are in one and the same proportion with H, O, P, K, and the multitude of these numbers D, L, E, is equal to the multitude of these numbers E, X, F, and likewise the multitude of these numbers G, M, N, H, is equal to the multitude of these numbers H, O, P, K, wherefore of equality (by the 14. of the seventh) as D is to E, so is E to F. And as G is to H, so is H to K: which was required to be proved.   

¶ The 12. Theorem. The 14. Proposition. If a square number measure a square number, the side also of the one shall measure the side of the other. And if the side of the one measure the side of the other, the square number also shall measure the square number.  
SVppose that there be two square numbers A and B, and let the sides of them be C and D● and let A measure B. Wherefore C also shall measure D. 
The first part of this proposition.  Let C multiplying D produce E. Wherefore (by the 17. and 18. of the seventh, and 13. of the eight) those numbers A, E, B, are in continual proportion, and are in the same proportion that C is to D. And forasmuch as A, E, B, are in continual proportion, and A measureth B; therefore (by the 7. of the eight) A measureth E. But as A is to E, so is C to D: wherefore C measureth D.  
But now suppose that the side C do measure the side D. 
The second part is the converse of the first.  Then I say, that the square number A also measureth the square number B. For the same order of construction remaining, we may in like sort prove, that the numbers A, E, B, are in continual proportion, & in the same proportion, that C is to D. And for that as C is to D, so is A to E, but C measureth D: therefore A measureth E: and A, E, B, are in continual proportion: wherefore A measureth B. If therefore a square number measure a square number, the side also of the one shall measure the side o● the other. And if the side of the one measure the side of the other, the square number also shall measure the square number: which was required to be demonstrated.   

¶ The 13. Theorem. The 15. Proposition. If a cube number measure a cube number, the side also of the one shall measure the side of the other. And if the side of the one measure the side of the other, the cube number also shall measure the cube number.  
SVppose that the cube number A do measure the cube number B, and let the side of A be C, and the side of B be D. Then I say, that C measureth D. Let C multiplying himself produce E, & multiplying D let it produce F. And let D multiplying himself produce G. And moreover let C and D multiplying F produce H and K. 
The first part of this proposition.  Now it is manifest (by the 17, and 18. of the seventh, and 12. of the eight) that those numbers E, F, G, and also A, H, K, B, are in continual proportion, & in the same proportion that C is to D. And forasmuch as A, H, K, B, are in continual proportion, and A measureth B, therefore (by the 7. of the eight) A also measureth H. But as A is to H, so is C to D. Wherefore C also measureth D.  

The second part is the converse of the first.  But now suppose that the side C do measure the side D. Then I say, that the cube number A also measureth the cube number B. For the same order of construction being kept, in like sort may we prove, that A, H, K, B, are in continual proportion, and in the same proportion that C is to D. And forasmuch as C measureth D, but as C is to D, so is A to H, therefore A measureth H: wherefore A also measureth B. If therefore a cube number measure a cube number, the side also of the one shall measure the side of the other. And if the side of the one measure the side of the other, the cube number also shall measure the cube number: which was required to be proved.   

¶ The 14. Theorem. The 16. Proposition. If a square number measure not a square number, neither shall the side of the one measure the side of the other. And if the side of the one measure not the side of the other, neither shall the square number measure the square number.  


A negative proportion.  
The first part of this proposition.   SVppose that A and B be two square numbers, and let the side of A be C: and let the side of B be D. And be it that A measureth not B. Then I say, that neither shall C measure D. For if C do measure D, then (by the 14. of the eight) A also measureth B. But A by supposition measureth not B: wherefore neither doth C measure D.  

The second part is the converse of the first.  But now again suppose that the side C measure not the side D. Then I say, that neither shall the square number A measure the square number B. For if A do measure B, then shall C (by the 14. of the eight) measure D. But (by supposition) C measureth not D. Wherefore neither doth A measure B: which was required to be proved.   

¶ The 15. Theorem. The 17. Proposition. If a cube number measure not a cube number, neither shall the side of the one measure the side of the other. And if the side of the one measure not the side of the other, neither shall the cube number measure the cube number.  
SVppose that the cube number A do not measure the cube number B: 
A negative proposition.  and let the side of A be C: and the side of B be D. Then I say, that C shall not measure D. For if C do measure D, then (by the 15. of the eight) A also shall measure B. 
The first part of this proposition.  But (by supposition) A measureth not B: wherefore neither shall C measure D.  
But now suppose that the side C measure not the side D. 
The second part is the converse of the first.  Then I say, that neither shall the cube number A measure the cube number B. For if A do measure B, then also (by the 15. of the eight) shall C measure D. But (by supposition) C measureth not D. Wherefore neither shall A measure B: which was required to be proved.   

¶ The 16. Theorem. The 18. Proposition. Between two like plain or superficial numbers there is one mean proportional number. And the one like plain number is to the other like plain number in double proportion of that which the side of like proportion, is to the side of like proportion.  
SVppose that there be two like plain or superficial numbers A & B. And let the sides of A be the numbers C, D: and the sides of B be the numbers E, F. And forasmuch as like plain numbers are those which have their sides proportional (by the 22. definition of the seventh) therefore as C is to D, so is E to F. 
Demonstration of the fi●st part of this proposition.  Then I say that between A and B there is one mean proportional number, and that A is unto B in double proportion of that which C is unto E, or of that which D is unto F, that is, of that which side of like proportion is to side of like proportion. For for that as C is to D, so is E to ●, therefore alternately (by the 13. of the seventh) as C is to E, so is D to F. And forasmuch as A is a plain or superficial number, and the sides thereof are C and D: therefore D multiplying C produced A. And by the same reason also E multiplying F produced B. Let D multiplying E produce G. And forasmuch as D multiplying C produced A, and multiplying E produced G, therefore (by the 17. of the seventh) as C is to E, so is A to G. But as C is to E, so is D to F, wherefore as D is to F, so is A to G. Again forasmuch as E multiplying D produced G, and multiplying ● produced B, therefore (by the 17. of the seventh) as D is to F, so is G to B. But it is proved that as D is to F, so is A to G: wherefore as A is to G, so is G to B. Wherefore these numbers A, G, B, are in continual prorortion. Wherefore between A and B there is one mean proportional number.  

Demonstration of the second part.  Now also I say that A is unto B in double proportion of that which side of like proportion is to side of like proportion, that is, of that which C is unto E, or of that which D is unto F. For forasmuch as A, G, B, are in continual proportion, therefore (by the 10. definition of the ●ift) A is unto B in double proportion of that which A is unto G. But as A is to G, so is C to E, and D to F: wherefore A is unto B in double proportion of that which C is to E, or D to F: which was required to be demonstrated.   

¶ The 17. Theorem. Th' 19 Proposition. Between two like solid numbers, there are two mean proportional numbers. And the one like solid number, is to the other like solid number in triple proportion of that which side of like proportion is to side of like proportion.  
SVppose that there be two like solid numbers A and B. And let the sides of the number A, be the numbers C, D, E. And let the sides of the number B, be the numbers F, G, H. And forasmuch as (by the 22. definition of the seventh) like solid numbers have their sides proportional, therefore as C is to D, so is F to G, and as D is to E, so is G to H. Then I say that between A and B, there are two mean proportional numbers. And that A is unto B in triple proportion of that which C is to F, or of that which D is to G, or moreover of that which E is unto H.  
For let C multiplying D produce K. And let F multiplying G, produce L. 
Demonstration of the first part of this proposition.  And forasmuch as C, D, are in the self same proportion that F, G, are, & of C & D is produced K, and of F and G is produced L, therefore K and L are like plain numbers. And therefore between those numbers K and L, there is one mean proportional number (by the 18. of the se●enth) Let the same be M. Wherefore M is produced of D and F, as it is manifest by the proposition going before. Wherefore as K is to M, so is M to L. And forasmuch as D multiplying C produced K, and multiplying F produced M: therefore (by the 17. of the seventh) as C is to F, so is K to M, but as K is to M, so is M to L. Wherefore these numbers K M, L, are in continual proportion and in the same proportion that C is to D. And for that as C is to D, so is F to G, therefore alternately (by the 13. of the seventh) as C is to F, so is D to G. Again, for that as D is to E, so is G to H, therefore alternately also as D is to G, so is E to H. Wherefore these numbers K, M, L, are in continual proportion, and in the same proportion that C is to F, and that D is to G, and moreover that E is to H. Now let E and H multiplying M produce N and X. And forasmuch as A is a solid number, and the sides thereof are C, D, E, therefore E multiplying that which is produced of C and D, produceth A. But that which is produced of C and D is K. Wherefore E multiplying K produceth A. And (by the same reason H multiplying that which is produced of F and G, that is multiplying L produceth B. And forasmuch as E multiplying K produced A and multiplying M produced N, therefore (by the 17. of the seventh), as K is to M, so is A to N. But as K is to M, so is C to F, & D to G. and moreover E to H, therefore as C is to F, and D to G, and E to H, so is A to N. Again, forasmuch as E multiplying M produced N, and H multiplying M, produced X, therefore (by the 18. of the seventh) as E is to H, so is N to X. But as E is to H, so is C to F, and D to ●. Wherefore as C is to F, and D to G, and E to H, so is A to N, and N to X. Again forasmuch as H multiplying M, produced X, and multiplying L produced B, therefore (by the 17. of the seventh) as M is to L, so is X to B. But as M is to L, so is C to F, and D to G, and E to H therefore as C is to F, and D to G, and E to H● so is not only X to B, but also A to N, and N to X. Wherefore these numbers A, N, X, B, are in continual proportion, and that in the proportions of the sides. 
The second part.  I say moreover that A is unto B in triple proportion of that, which side of like proportion, is to side of like proportion, that is, of that which the number C hath to the number F, or of that which D hath to G, or moreover of that which E hath to H. For forasmuch as there are four numbers in continual proportion, that is, A, N, X, B, therefore (by the 10. definition of the fift) A is unto B in triple proportion of that which A is unto N. But as A is to N, so is it proved that C is to F, and D to G, and moreover E to H. Wherefore A is unto B in triple proportion of that which side of like proportion is unto side of like proportion, that is of that which the number C is to the number F, and of that which D is to G, and moreover of that which E is ●o H: Which was required to be proved.   

¶ The 18. Theorem. The 20. Proposition. If between two numbers there be one mean proportional number: those numbers are like plain numbers.  
SVppose that between the two numbers A and B there be one mean proportional number, and let the same be C. 
This proposition is the converse of the 18. proposition.  Then I say, that those numbers A and B are like plain numbers. Take (by the 35. of the seventh) two of the lest numbers that have one & the same proportion with A, C, B: and let the same be the numbers D, E. 
Construction.  Wherefore as D is to E, so is A to C, but as A is to C, so is C to B, wherefore as D is to E, so is C to B. Wherefore how many times D measureth A, so many times doth E measure C. How many times D measureth A, so many unities let there be in F. Wherefore F multiplying D produceth A, and multiplying E it produceth C: wherefore A is a plain number: and the sides thereof are D and F (by the 17. definition of the seventh). Again forasmuch as D and E are the least numbers that have one & the same proportion with C, B, therefore (by the 21. of the seventh) how many times D measureth C, so many times doth E measure B. How often E measureth B, so many unities let there be in G. Wherefore E measureth B by those unities which are in G: wherefore G multiplying E produceth B: wherefore B is a plain number (by the 17. definition of the seventh). 
Demonstration.  And the sides thereof are E and G. Wherefore those two numbers A and B are two plain numbers. I say moreover that they are like. For forasmuch as F multiplying E produced C: and G multiplying E produced B: therefore (by the 17. of the seventh) as F is to G, so is C to B, but as C is to B, so is D to E, wherefore as D is to E, so is F to G. Wherefore A and B are like plain numbers, for their sides are proportional: which was required to be proved.   

¶ The 19 Theorem. The 21. Proposition. If between two numbers, there be two mean proportional numbers, those numbers are like solid numbers.  

This proposition is the converse of the 19 proposition.  SVppose that between two numbers A and B, there be two mean proportional numbers C, D. Then I say that A and B are like solid numbers. Take (by the 3● of the seventh, or 2. of the eight) three of the lest numbers that have one and the same proportion with A, C, D, B, and let the same be E, F, G. Wherefore (by the 3. of the eight) their extremes E, G are prime the one to the other. And forasmuch as between the numbers E and G there is one mean proportional number: 
Construction.  therefore (by the 20 of the eight) they are like plain numbers. Suppose that the sides of E, be H and K. And let the sides of G, be L and M. Now it is manifest that these numbers E, F, G, are in continual proportion, 
Demonstration.  and in the same proportion that H is to L, and that K is to M. And forasmuch a● E, F, G are the lest numbers that have one and the same proportion with A, C, D, therefore of equality (by the 14. of the seventh) as E is to G, so is A to D. But E, G, are (by the 3. of the eight) prime numbers, yea they are prime and the lest, but the lest numbers (by the 21. of the seventh) measure those numbers that have one & the same proportion with them equally, the greater the greater, and the less the less, that is, the antecedent the antecedent, & the consequent the consequent: therefore how many time● E measureth A, so many times G measureth D. How many times E measureth A, so many unities let there be in N. Wherefore N multiplying E, produceth A. But E is produced of the numbers H, K. Wherefore N multiplying that which is produced of H, K, produceth A. Wherefore A is a solid number, and the sides thereof are H, K, N. Again, forasmuch as E, F, G, are the lest numbers that have one and the same proportion with C, D, B, therefore how many times E measureth C, so many times G measureth B. How oftentimes G measureth B, so many unities let there be in X. Wherefore G measureth B by those unities which are in X. Wherefore X multiplying G produceth B. But G is produced of the numbers L, M. Wherefore X multiplying that number which is produced of L and M, produceth B. Wherefore B is a solid number, and the sides thereof are L, M X. Wherefore A, B are solid numbers. I say moreover that they are like solid numbers. For forasmuch as N and X multiplying E produced A and C: therefore by the 18. of the seventh, as N is to X, so i● A to C, that is E ●o F. But as E is to F, so is H to L, and K to M: therefore as H is to L, so is K to M, and N to X. And H, K, N, are the sides of A, and likewise L, M, X, a●● th● sides of B● wherefore A, B● are like solid numbers: which was required to be proved.   

¶ The 20. Theorem. The 22. Proposition. If three numbers be in continual proportion, and if the first be a square number, the third also shall be a square number.  
SVppose that there be three numbers in continual proportion A, B, C, and let the first be a square number. 
Demonstration.  Then I say that the third is also a square number. For forasmuch as between A and C there is one mean proportional number● namely B, therefore (by the 20. of the eight) A and C are like plain numbers. But A is a square number. Wherefore C also is a square number: which was required to be proved.   

¶ The 21. Theorem. The 23. Proposition. If four numbers be in continual proportion, and if the first be a cube number, the fourth also shall be a cube number.  
SVppose that there be four numbers in continual proportion A, B, C, D. 
Demonstration.  And let A be a cube number. Then I say that D also is a cube number. For forasmuch as between A and D there are two mean proportional numbers B● C. Therefore A, D are like solid numbers (by the 21. of this book) But A is a cube number, wherefore D also is a cube number● which was required to be demonstrated.   

¶ The 22. Theorem. The 24. Proposition. If two numbers be in the same proportion that a square number is to a square number, and if the first be a square number, the second also shall be a square number.  
SVppose that two numbers A and B be in the same proportion, that the square number C is unto the squ●●● number D. And let A be a square number. 
Demonstration.  Then I say that B also is a square number. For forasmuch as C and D are square numbers. Therefore G and D are like plain numbers. Wherefore (by the 18. of the eight) between C and D there is one mean proportional number. But as C is to D, so is A to B. Wherefore between A and B there is one mean proportional number (by the 8. of the eight) But A is a square number. Wherefore (by the 22. of the eight) B also is a square number which was required to be proved.   

¶ The 23. Theorem. The 25. Proposition. If two numbers be in the same proportion the one to the other, that a cube number is to a cube number, and if the first be a cube number, the second also shall be a cube number.  
SVppose that two numbers A and B be in the same proportion the one to the other, that the cube number C is unto the cube number D. And let A be a cube number. Then I say that B also is a cube number. For forasmuch as C, D, are cube numbers, therefore C, 
Demonstration.  D are like solid numbers, wherefore (by the 19 of the eight) between C and D there are two proportional numbers. But how many numbers fall in continual proportion between C and D, so many (by the 8. of the eight) fall there between the numbers that have the same proportion with them. Wherefore between A and B there are two mean proportional numbers which let be E and F. And forasmuch as there are four numbers in continual proportion, namely, A, E, F, B, and A is a cube number, therefore (by the 2●. of the eight) B also is a cube number: which was required to be demonstrated.  

A Corollary added by Flussates.  

A Corollary added by Flussates.  Between a square number and a number that is not a square number, falleth not the proportion of one square number to an other. For if the first be a square number, the second also should be a square number which is contrary to the supposition. Likewise between a cube number, and a number that is no cube number falleth not the proportion of one cube number to an other. For if the first be a cube number, the second also should be a cube number, which is contrary to the supposition, & therefore impossible.    

¶ The 24. Theorem. The 26. Proposition. Like plain numbers, are in the same proportion the one to the other, that a square number is to a square number.  
SVppose that A and B be like plain numbers. Then I say that A is unto B in the same proportion that a square number is to a square number. For forasmuch as A, B, are like plain numbers, 
Construction.  therefore between A and B there falleth one mean proportional number (by the 18. of the eight). Let there fall such a number, and let the same be C. And (by the 35. of the seventh) take the three lest numbers that have one and the same proportion with A, C, B, and let the same be D, E, F: wherefore (by the corollary of the 2 ● of the eight) their 〈◊〉 that is D, F, are square numbers. And for that as D is to F, so is A to B, (by the 14. of the seventh): and D, F, are square numbers. Therefore A is unto B in that proportion, that a square number is unto a square num●er: which was required to be proved.   

The 25. Theorem. The 27. Proposition. Like solid numbers are in the same proportion the one to the other, that a cube number is to a cube number.  
SVppose that a A and B be like solid numbers. Then I say that A is unto B, in the same proportion, that a cube numb is to to a cube number. For forasmuch as A, B, are like solid numbers. 
Construction.  Therefore (by the 19 of the eight) between A and B there fall two mean proportional numbers. Let there fall two such numbers, and let the same be C and D. And take (by the 35. of the seventh) the least numbers that have one and the same proportion with A, C, D, B, 
Demonstration.  and equal also with them in multitude, and let the same be E, F, G, H. Wherefore (by the corollary of the 2. of the eight) their extremes, that is, EH, are cube numbers. But as E is to H, so is A to B. Wherefore A is unto B in the same proportion, that a cube number is to a cube number: which was required to be demonstrated.  

¶ A Corollary added by Flussates.  
If two nnmbers be in the same proportion the one to the other that a square number is to a square number: those two numbers shall be like superficial numbers. 
A Corollary added by Flussates.  And if they be in the same proportion the one to the other that a cube number is to a cube number, they shall be like solid numbers.  
First let the number A have unto the number B the same proportion, that the square number C hath to the square number D. Then I say, that A and B are like superficial numbers. For forasmuch as between the square numbers C and D there falleth a mean proportional (by the 11. of this book) there shall also between A and B (which have one and the same proportion with C and D) fall a mean proportional (by the 8. of this book). Wherefore A and B are like superficial numbers (by the 20. of this book).  
But if A be unto B, as the cube number C, is to the cube number D. Then are A & B like solid numbers. For forasmuch as C and D be cube numbers, there falleth between them too mean proportional numbers (by the 12. of this book). And therefore (by the 8. of the same) between A and B (which are in the same proportion that C is to D) there falleth also two mean proportional numbers. Wherefore (by the 21. of this book) A and B are like solid numbers.   

another Corollary added also by Flussates.  
If a number multiplying a square number, produce not a square number: the said number multiplying shall b● no square number. 
Another Corollary added by Flussates.  For if it should be a square number, than should it and the number multiplied being like superficial numbers (by reason they are square numbers) have a mean proportional (by the 18. of this book). And the number produced of the said mean should be equal to the number contained under the extremes, which are square numbers (by the 20. of the seventh). Wherefore the number produced of the extremes being equal to the square number produced of the mean, should be a square number. But the said number by supposition, is no square number. Wherefore neither is the number multiplying the square number, a square number.  
The first part of the first Corollary is the converse of the 26. Proposition of this book, and hath some use in the tenth book. The second part of the same also is the converse of the 27. Proposition of the same.    
The end of the eighth book of Euclides Elements.   

¶ The ninth book of Euclides Elements.  
IN THIS NINTH BOOK Euclid continueth his purpose touching numbers: partly prosecuting things more fully, 
The Argument of the ni●th book.  which were before somewhat spoken of, as of square and cube numbers: and partly setting out the natures and proprieties of such kinds of number, as have not yet been entreated of: which yet are most necessary to be known. As are numbers even, and odd: whose passions and conditions are in this book largely taught, with their compositions, and subductions of the one from the other: with many other general and special things to be noted, worthy the knowledge.  

¶ The 1. Theorem. The 1. Proposition. If two like plain numbers multiplying the one the other produce any number: the number of them produced shall be a square number.  
SVppose that A and B be two like plain numbers. And let A multiplying B produce the number C. Then I say, that C is a square number. For let A multiplying himself produce D. Wherefore D is a square number. 
Demonstration.  And forasmuch as A multiplying himself produced D, and multiplying B produced C, therefore (by the 17. of the seventh) as A is to B, so is D to C. And forasmuch as A, B, are like plain numbers, therefore (by the 18. of the eight) between A and B there falleth a mean proportional number. But if between two numbers fall numbers in continual proportion, how many numbers fall between them, so many also (by the 8. of the eight) shall there fall between the numbers that have the same proportion with them. Wherefore between C and D also there falleth a mean proportional number. But D is a square number. Wherefore (by the 22. of the eight) C also is a square number: which was required to be proved.   

¶ The 2. Theorem. The 2. Proposition. If two numbers multiplying the one the other produce a square number: those numbers are like plain numbers.  

This proposition is the converse o● t●e form●●.  SVppose that two number● A and B multiplying the one the other do produce C a square number. Then I say, that A and B are like plain numbers. For let A multiplying himself produce D. Wherefore D is a square number. And forasmuch as A multiplying himself produced D, 
Demonstration.  and multiplying B produced C, therefore (by the 17. of the se●enth) as A is to B, so is D to C. And forasmuch as D is a square number, and so likewise is C, therefore D and C are like plain numbers. Wherefore between D and C there is (by the 18. of the eight) one mean proportional number. But as D is to C, so is A to B. Wherefore (by the 8. of the eight) between A and B there is one mean proportional number. But if between two numbers there be one mean proportional number, those numbers are (by the 20. of the eight) like plain numbers. Wherefore A and B are like plain numbers: which was required to be proved.  

A Corollary added by Campane.  
H●●●● it is manifest, th●t two square numbers multiplied the one into thee, other do always produce a squa●● num●●r. 
A Corollary a●ded by Campane.  For they are like superficial numbers, and therefore the number produced of them, is (by the first of this book) a square number. But a square number mul●●plye● into a number not square, produceth a number not square. For if they should produce a square number, they should be like superficial numbers (by this Proposition). But they are not. Wherefore they produce a number not square. But if a square number multiplied into an other number produce a square number, that other number shall be a square number. For by this Proposition that other number is like unto the square number which multiplieth it, and therefore is a square number. But if a square number multiplied into an other number produce a number not square, neither shall that other number also be a square number. For if it should be a square number, then being multiplied into the square number it should produce a square number, by the first part of this Corollary.    

The 3. Theorem. The 3. Proposition. If a cube number multiplying himself produce a number, the number produced shall be a cube number.  
SVppose that A being a cube number multiplying himself, do produce the number B. Then I say that B is a cube number. Take the side of A, and let the same be the number C, and let C multiplying himself produce the number D. Now it is manifest that C multiplying D produceth A (by the 20. definition of the seventh) 
Demonstration.  And forasmuch as C multiplying himself produced D, therefore C measureth D by those unities which are in C. But unity also measureth C by those unities which are in C. Wherefore as unity is to C, so is C to D. Again forasmuch as C multiplying D produceth A: therefore D measureth A by those unities which are in C. But unity measureth C by those unities which are in C: wherefore as unity is to C, so is D to A. But as unity is to C, so is C to D, wherefore as unity is to C, so is C to D & D to A. Wherefore between unity & A there are two mean proportional numbers, namely, C, D. Again forasmuch as A multiplying himself produced B, therefore A measureth B by those unities which are in A. But unity also measureth A by those unities which are in A. Wherefore as unity is to A, so is A to B. But between A and unity, there are two mean proportional numbers. Wherefore between A and B also there are two mean proportional numbers by the 8. of the eight. But if between two numbers, there be two mean proportional numbers, and if the first be a cube number, the fourth also shall be a cube number by the 21. of the eight. But A is a cube number, wherefore B also is a cube number which was required to be proved.   

¶ The 4. Theorem. The 4. Proposition. If a cube number multiplying a cube number, produce any number, the number produced shall be a cube number.  
SVppose that the cube number A multiplying the cube number B, do produce the number C. Then I say that C is a cube number. For let A multiplying himself produce D. Wherefore D is a cube number (by the proposition going before). 
Demonstration.  And forasmuch as A multiplying himself produced D, and multiplying B, it produced C: therefore (by the 17. of the seventh) as A is to B, so is D to C. And forasmuch as A and B be cube numbers, therefore A and B are like solid numbers. Wherefore between A and B (by the 19 of the eight) there are two mean proportional numbers. Wherefore also (by the 8. of the same) between D and C there are two mean proportional numbers. But D is a cube number. Wherefore C also is a cube number (by the 23. of the eight) which was required to be demonstrated.   

¶ The 5. Theorem. The 5. Proposition. If a cube number multiplying any number produce a cube number: the number multiplied is a cube number.  
SVppose that the cube number A, multiplying the number B, do produce a cube number, namely, C. Then I say, that B is a cube number. For let A multiplying himself produce D. Wherefore (by the 3. of the ninth) D is a cube number. And forasmuch as A multiplying himself produced D, 
Demonstration.  and multiplying B, it produced C: therefore (by the 17. of the seventh) as A is to B, so is D to C. And forasmuch as D and C be cube numbers, they are also like solid numbers. Wherefore (by the 19 of the eight) between D and C there are two mean proportional numbers. But as D is to C, so is A to B. Wherefore (by the 8. of the eight) between A and B there are two mean proportional numbers. But A is a cube number. Wherefore B also is a cube number (by the 23. of the eight): which was required to be proved.  

¶ A Corollary added by Campane.  
Hereby it is manifest, that if a cube number multiply a number not cube, it shall produce a number not cube. 
A Corollary added by Campane.  For if it should produce a cube number, than the number multiplied should also be a cube number (by this Proposition) which is contrary to the supposition. For it is supposed to be no cube number. And if a cube number multiplying a number produce a number not cube, the number multiplied shall be no cube number. For if the number multiplied should be a cube number, the number produced should also be a cube number (by the 4. of this book): which is contrary to the supposition, and impossible.    

¶ The 6. Theorem. The 6. Proposition. If a number multiplying himself produce a cube number: then is that number also a cube number.  
SVppose that the number A multiplying himself, do produce B a cube number. Then I say that A also is a cube number. 
Demonstration.  For let A multiplying B produce C. And forasmuch as A multiplying himself produced B, & multiplying B it produced C: therefore C is a cube number. And for that A multiplying himself produced B, and multiplying B it produced C, therefore (by the 17. of the seventh) as A is to B, so is B to C. And for that B and C be cube numbers, they are also like solid numbers. Wherefore (by the 19 of the eight) between C and B there are two mean proportional numbers. But as B is to C, so is A to B: wherefore (by the 8. of the eight) between A and B there are two mean proportional numbers. But B is a cube number. Wherefore A also is a cube number by the 23. of the eight: which was required to be demonstrated.   

¶ The 7. Theorem. The 7. Proposition. If a composed number multiplying any number, produce a number: the number produced shall be a solid number.  
SVppose that the composed number A multiplying the number B, do produce the number C. Then I say that C is a solid number. For forasmuch as A is a composed number, therefore some number measureth it (by the 14. definition). Let D measure it. 
Demonstration.  And how o●ten D measureth A, so many unities let there be in E. Wherefore E multiplying D produceth A. And forasmuch as two numbers D and E, multiplying themselves, produce A, which A again multiplying B produceth C: therefore C produced of three numbers multiplying the one the other, namely, D, E, and B is (by the 18. definition of the seventh) a solid number. And the sides thereof are the numbers D, E, B. If therefore a composed number etc., which was required to be proved.   

¶ The 8. Theorem. The 8. Proposition. If from unity there be numbers in continual proportion how many soever: the third number from unity is a square number, and so are all forward leaving one between. And the fourth number is a cube number, and so are all forward leaving two between. And the seventh is both a cube number and also a square number, and so are all forward leaving five between.  
SVppose that from unity there be these numbers in continual proportion A, B, C, D, E, F. Then I say that the third number from unity, namely, B is a square number, and so are all forward leaving one between, namely, D and F. 
Demonstration of the first part.  And that C the fourth number is a cube number, and so are all forward leaving two between. And that F the seventh number is both a cube number and also a square number, and so are all forward leaving five between. For for that as unity is to A, so is A to B. Therefore how many times unity measureth A, so many times A measureth B. But unity measureth A by those unities which are in A, wherefore A measureth B by those unities which are in A. And forasmuch as A measureth B by those unities which are in A. Therefore A multiplying himself produceth B. Wherefore B is a square number. And forasmuch as these numbers B, C, D, are in continual proportion, and B is a square number, therefore by the 22. of the eight, D also is a square number. And by the same reason also F is a square number. And in like sort may we prove that leaving always one between, all the rest forward are square numbers.  
Now also I say that the fourth number from unity, that is, C, is a cube number, and so are all forward leaving two between. For for that as unity is to the number A, so is B to C, 
The second part demonstrated.  therefore how many times unity measureth the number A, so many times B measureth C. But unity measureth A by those unities which are in A, wherefore B measureth C by those unities which are in A. Wherefore A multiplying B produceth HUNDRED And forasmuch as A multipli●ng himself produced B, and multiplying B it produced C, therefore C is a cube number. And forasmuch as C, D, E, F, are in continual proportion. But G is a cube number, therefore (by the 23. of the eight) F also is a cube number.  
And it is proved, that F being the seventh number from unity is also a square number. 
Demostration of the third part.  Wherefore F is both a cube number, and also a square number. In like sort may we prove, that lea●ing always five between, all the rest forward, are numbers both cube and also square: which was required to be proved.   

¶ The 9 Theorem. The 9 Proposition. If from● unity be numbers in continual proportion how many soever: and if th●● number which followeth next after unity be a square number, than all the rest following also be square numbers. And if that number which followeth next after unity be a cube number, than all the rest following shall be cube numbers.  
SVppose that from unity there be these numbers in continual proportion A, B, C, D, E, F. And let A which followeth next unto unity be a square number. Then I say, that all the rest following also are square numbers. 
Demostration of the first part of this proposition.  That the third number, namely, B, is a square number, & so all forward leaving one between, it is plain by the Proposition next going before. I say also that all the rest are square numbers. For, forasmuch as A, B, C, are in continual proportion, and A is a square number, therefore (by the 22. of the eight) C also is a square number. Again forasmuch as B, C, D, are in continual proportion, and B is a square number, therefore D also (by the 22. of the eight) is a square number. In like sort may we prove, that all the rest are square numbers.  

The second p●rt demonstrated.  But now suppose that A be a cube number. Then I say, that all the rest following are cube numbers. That the fourth from unity, that is, C is a cube number, and so all forward leaving two between, it is plain (by the Proposition going before). Now I say, that all the rest also are cube numbers. For, for that as unity is to A, so is A to B: therefore how many times unity measureth A, so many times A measureth B. But unity measureth A by those unities which are in A. Wherefore A also measureth B by those unities which are in A. Wherefore A multiplying himself produceth B. But A is a cube number. But if a cube number mutiplying himself produce any number, the number produced, is (by the 3. of the ninth) a cube number. Wherefore B is a cube number. And forasmuch as there are four numbers in continual proportion A, B, C, D, and A is a cube number, therefore D also (by the 23. of the eight) is a cube number. And by the same reason E also is a cube number, and in like sort are all the r●st following: which was required to be proved.   

¶ The 10. Theorem. The 10. Proposition. If from unity be numbers in continual proportion how many soever, and if that number which followeth next after unity be not a square number, then is none of the rest following a square number, excepting the third from unity, and so all forward leaving one between. And if that number which followeth next after unity be not a cube number, neither is any of the rest following a cube number, excepting the fourth from unity, and so all forward leaving two between.  
SVppose that from unity be these numbers in continual proportion A, B, C, D, E, F. And let A which followeth next after unity be no square number. Then I say, 
Demonstration of the first part leaving to an absurdity.  that neither is any of the rest a square number, excepting the third from unity, & so all forward leaving one between, namely, B, D, F, which are square numbers (by the 8. of this book). For if it be possible, let C be a square number. But B also is a square number. Wherefore B is unto C in that proportion that a square number is to a square number. But as ● is ●o C, so is A to B. Wherefore A is unto B in th●● proportion that a square number is to ● square number. But B is a square number. Wherefore A also is a square number (by the ●4. of the eight) ● which is contrary to the supposition. Wherefore C is not a square number. And by the same reason none of all the other is a square number, excepting the third from unity, and so all forward leaving one between.  
But now suppose that A be not a cube number● Then I say, 
Demonstration of the ●●cond p●●● leading also to an absurdity.  that none of all the rest is a cube number, excepting the fourth from unity, & so all forward leaving two between, namely, C, and F, which (by the 8. of this book) are cube numbers. For if i● 〈◊〉 possible, l●● D be a cube number. But C also is a cube number (by the 8. of the ninth). For it is the fourth from unity. But as C is to D, so is B to C. Wherefore B is unto C, in that proportion tha● a cube number is to a cube number. But C is a cube number. Wherefore B also is a cube number (by the 25. of the eight). And as unity is to A, so is A to B. But unity measureth A by those unities which are in A. Wherefore A measureth B by those unities which are in A. Wherefore A multiplying him●selfe produceth● B a cube number. But if a number multiplying himself produce a cube number, then is that number also a cube number (by the 6. of the ninth) ● Wherefore A is a cube number: which is contrary to the supposition. Wherefore D is not a cube number. In like sort may we prove, that neither is any of the rest a cube number, excepting the fourth from unity, and so all forward leaving two between: which was required to be proved.   

¶ The 11. Theorem. The 11. Proposition. If from unity be numbers in continual proportion how many soever, the less measureth the greater by some one of them which are before in the said proportional numbers.  
SVppose that from unity A be these numbers in continual proportion B, C, D, E. Then I say that of these numbers B, C, D, ●● E being the less, measureth E the greater by one of these numbers C or D. For for that as unity A is unto the number B, so is D to E, 
Demonstration.  therefore how many times unity A measureth the number B, so many times D measureth E● wherefore alternately (by the 15. of the seventh) how many times unity A measureth the number D, so many times ● measureth E. But unity A measureth D by those unities which are in D. Wherefore B also measureth E by those unities which are in D. Wherefore ● the less, measureth. E the greater by some one of the numbers which went before E in the proportional numbers. And so likewise may we prove that E measureth D by some one of the numbers ●, C, D, namely, by C. And so of the rest. If therefore from unity etc. Which was required to be proved.   

¶ The 12. Theorem. The 12. Proposition. If from unity be numbers in continual proportion how many soever, how many prime numbers measure the least● so many also shall measure● the number which followeth next after unity.  
SVppose that from unity be these numbers in continual proportion A, B, C, D. Th● I say that how many prime numbers measure D, so many also do measure A. Suppose that some prime number namely, E, do measure D. Then I say that E also measureth A, which is next unto unity. For if E do not measure A, and E is a prime number, but every number is to every number which it measureth not a prime number (by the 31. of the seventh). Wherefore A and E are prime numbers the one to the other. And forasmuch as E measureth D, let it measure D by the number F. 
Demonstration leading to an absurdity.  Wherefore E multiplying F produceth D. Again forasmuch as A measureth D by those unities which are in C, therefore A multiplying C produceth D. But E also multiplying F produced D, wherefore that which is produced of the numbers A, C is equal to that which is produced of the numbers E, F. Wherefore as A is to E, so is F to C. But A, E, are prime numbers, yea they are prime and the lest. But the jest numbers measure the numbers that have one and the same proportion with them equally by the 21. of the seventh, namely, the antecedent the antecedent, and the consequent the consequent. Wherefore E measureth C. Let it measure it by G. Wherefore E multiplying G produceth HUNDRED But A also multiplying B produceth C. Wherefore that which is produced of the numbers A, B, is equal to that which is produced of the numbers E, G. Wherefore as A is to E, so is G to B. But A, E are prime numbers, yea they are prime and the lest. But the lest numbers (by the 21. of the seventh) measure the numbers that have one and the same proportion with them equally, namely, the antece●●s the antecedent, & the consequent the conseqent. Wherefore E measureth B. Let it measure it by H. Wherefore B multiplying H produceth B. But A also multiplying himself produceth B, wherefore that which is produced of the numbers E, H, is equal to that which is produced of the number A. Wherefore as E is to A, so is A to H. But AE are prime numbers, yea they are prime & the lest, but the lest numbers (by the 21. of the seventh) measure the numbers that have one and the same proportion with them equally, namely, the antecedent the antecedent, and the consequent the consequent. Wherefore E measureth A and it also doth not measure it by supposition, which is impossible. Wherefore A and E are not prime the one to the other, wherefore they are composed. But all composed numbers are measured of some prime number, wherefore A and E are measured by some prime number. And forasmuch as E is supposed to be a prime number. But a prime number is not (by the definition) measured by any other number but of himself. Wherefore E measureth A and E, wherefore B measureth A, and it also measureth D. Wherefore E measureth these numbers A and D. And in like sort may we prove that how many prime numbers measure D, so many also shall measure A: which was required to be proved.  

An other more brief demonstration after Flussates.  

another demonstration a●ter Flussates.  Suppose that from unity be numbers in continual proportion how many so ever, namely, A, B, C, D. And let some prime number, namely, ● measure the last number which is D. Then I say that th● same E measureth A which is the next number unto unity. For if E do not measure A, then are they prime the one to the other by the 31. of the seventh. And forasmuch as A, B, C, D, are proportional from unity, therefore A multiplying himself produceth B. Wherefore B and E ar● prim●. numbers (by the 27. of the seventh). And forasmuch as A multiplying B produceth C, therefore C is to E also a prime number by the 26. of the seventh. And likewise infinitely A multiplying C produceth D: wherefore D and E are prime numbers the one to the other (b● 〈◊〉 same ●6. of th● s●uenth● Wherefore E measureth not D as it was supposed, which is absurd, wherefore the prime number E measureth A, which is next unto unities which was required to be proved.    

¶ The 13. Theorem. The 13. Proposition. If from unity be numbers in continual proportion how many soever, and if that which followeth next after unity be a prime number: then shall no other number measure the greatest number, but those only which are before in the said proportional numbers.  
SVppose that from unity be these numbers in continual proportion A, B, C, D, and let that which followeth next after unity, that is, A, be a prime number. Then I say, that no other number besides these numbers A, B, C, measureth the greatest number of them which is D. 
Demonstration leading to an absurdity.  For if it be possible, let E measure D. And let E be none of these numbers A, B, C, D. Now it is manifest that E is not a prime number. For if E be a prime number, & do also measure D, it shall likewise measure A being a prime number and not being one and the same with A, by the former Proposition: which is impossible. Wherefore E is not a prime number. Wherefore it is a composed number. But every composed number (by the 33. of the seventh) is measured by some prime number.  
Now I say, that no other prime number besides A shall measure E. For if any other prime number do measure E, & E measureth D, therefore that number also shall measure D (by the 5. common sentence of the seventh). Wherefore it shall also measure A (by the proposition next going before) being a prime number and not being one and the same with A: which is impossible. Wherefore every the prime number A measureth E which measureth the greatest number D.  
And forasmuch as E measureth D, let it measure it by F. Now I say, that F is none of these numbers A, B, C. For if F be one and ●he s●m● with any of these numbers A, B, C, and is measureth D by ●, therefore one of these numbers A, B, C, measureth D by E. But one of these numbers A, B, C, measureth D by some one of these numbers A, B, C, therefore E is one and the same with one of these numbers A, B, C, which is contrary 〈◊〉 th● supposition. Wherefore F●● no● one and the same with any of these ●●●bers A, B, C.  

another demonstration of the same after Campane.  

another demonstration after Campane.  Suppose that E not being one and the same with the numbers A, B, C, D do measure the number D. And let it measure it by the number P. And forasmuch as A being a prime number measureth the number D, which is produced of E into F: therefore by the 3●. of the fourth, A measureth either ● or ●. Let it measure E. Now forasmuch as D is produced of A into G, and also of E into F: therefore by the second part of the 19 of the seventh, A is to E, as F is to C. But A measureth E: wherefore F measureth C. Let it measure it by G. Wherefore by the 32. of the seventh A shall measure either F or G. Let it measure F. Wherefore as before by the second part of the 1●. of the seventh G shall measure B. Let it measure it by H. Now then as before it followeth by the 32. of the seventh that A shall measure either G or H: suppose that it measure G. Wherefore by the second part of the 20. of the seventh H shall measure ● (fo● of A into himself is produced B, and of H into G also it produced B) I● therefore H be not equal unto A, A shall be no prime number. Which is contrary to the supposition. ●●t if it be equal unto A, than ●uery one of these numbers ●, F, E, shall be some one of the numbers A, B, C, D, by th● 11. proposition of the ninth repeated a● of●en as need requireth. Wherefore E is not a number diverse from them, but is one and the same with some one of them: which is contrary to the supposition; wherefore that is manifest which was required to be proved.    

¶ The 14. Theorem. The 14. Proposition. If there be given the lest number, whom certain prime numbers given, do measure: no other prime number shall measure that number, besides those prime numbers given.  
SVppose that the lest number whom these prime numbers B, C, D, do measure, be A. Then I say that no other prime number besides B, C, D, measureth A. For if it be possible, let E being a prime number measure A, and let E be none of these numbers B, C, D. 
Demonstration leading to an absurdity.  And forasmuch as E measureth A, let it measure it by F. Wherefore E multiplying I produceth A: And these prime numbers B, C, D, measure A. ●ut if two numbers multiplying the one the other produce any number. And if some prime number measure that which is produced, it shall also measure one of th●se numbers which were put at the beginning (by the 32. of the seventh) Wherefore those numbers B, C, D, measure one of these numbers E or F. But they measure not E, for E is a prime number, and is not one and the same with any one of these numbers B, C, D. Wherefore they measure F being less than A which is impossible. For A is supposed to be the least whom B, C, D, measure. Wherefore no prime number besides B, C, D, measureth A: which was required to be demonstrated.  

A proposition added by Campane.  
If there be numbers how many soever in continual proportion being the lest in that proportion: a number measuring one of them, shall be a number not prime to one of the two least numbers in that proportion.  
Suppose that there be numbers in continual proportion how many soe●●r namely A, B, C, D, E which let be the lest that have the same proportion with them: 
A proposition added by Campane.  and let the two least numbers in that proportion be F and G. And let some number as H measure some one of the numbers A, B, C, D, E, namely, C. Then I say that H is a number not prime either to F or G. Take (by the 2. of the eight) the three lest numbers in the proportion of A to B: which let be P, Q, R. And afterward four (by the same) which let be K, L, M, N: & so forward till you come to the multitude of the numbers given A, B, C, D, E. Now it is manifest (by the demonstration of the second of the eight) that F multiplied by P, Q, and R produceth K, L, M: and that F multiplied by K, L, M, N produc●th A, B, C, D. And forasmuch as H measureth C: therefore H is either to F or to M not prime (by the corollary of the 32. of the seventh added by Campane) If it be not prime unto F: them i● that manifest which was required to be proved. But if H be not prime unto M. Then shall it not be prime either to F or to R (by the same corollary). If again it be not prime unto F, then is that proved which was required. But if it be not prime unto R, than (by the same corollary) shall it be a number not prime unto G (which produceth R by the 2. of the eight) ● but G is one of the two lest numbers F or G which are in the proportion of the numbers given at the beginning A, B, C, D, E. If therefore there be numbers how many soever. etc. which was required to be proved.    

¶ The 15. Theorem. The 15. Proposition. If three numbers in continual proportion be the lest of all numbers that have one and the same proportion with them: every two of them added together shall be prime to the third.  
SVppose that there be three numbers in continual proportion A, B, C, being the left of all numbers that have one and the same proportion with them. Then I say, that every two of these numbers A, B, C, added together, 
Construction.  are prime to the third: namely, that A, B, is prime to C, and B, C, to A, and A, C, to B. Take (by the 35. of the seventh) two of the numbers that have one and the same proportion with A, B, C, & let the same be the numbers DE, and EF.  
Now it is manifest (by the said 35. Proposition) that DE multiplying himself produced A, 
Demonstration.  and multiplying EF produced B, and moreover EF multiplying himself produced C.  
And forasmuch as DE and EF are the lest in that proportion, they are also prime the one to the other (by the 24. of the seventh). But i● two numbers be prime the one to the other, then both of them added together, shall be prime to either of them (by the 30. of the seventh). Wherefore the whole number DF is prime to either of these numbers DE & EF. But DE also is prime unto EF. Wherefore DF, & DE are prime unto EF. Wherefore that which is produced of DF into DE, is (by the 26. of the seventh) prime unto EF. But if two numbers be prime the one to the other, that which is produced of the one of them into himself, is prime to the other (by the 27. of the seventh). Wherefore that which is produced of DF into DE, is prime to that which is produced of EF into himself. But that which is produced of FD into DE, is the square number which is produced of DE into himself together with that which is produced of DE into EF (by the 3. of the second). Wherefore the square number which is produced of DE together with that which is produced of DE into EF, is prime to that which is produced of EF into himself. But that which is produced of DE into himself, is the number A, & that which is produced of DE into EF, is the number B: and that which is produced of EF into himself, is the number C. Wherefore the numbers A, B, added together are prime unto C.  
By the like demonstration also may we prove, that the numbers B, C, are prime unto the number A.  
Now also I say, that the numbers A, C, are prime unto the number B.  
For forasmuch as DF is prime to either of these DE and EF: therefore that which is produced of DF into himself, is prime to that which is produced of DE into EF. But that which is produced of DF into himself, is equal to the square numbers which are produced of DE and EF together with that number which is produced of DE into EF, twice (by the 4. of the second). Wherefore the square numbers which are produced of DE and EF together with that which is produced of DE into EF twice are prime to that which is produced of DE into EF. And by division also (by the 30. of the seventh) the square numbers produced of DE and EF, together with that which is produced of DE into EF once are prime to that which is produced of DE into EF. Again (by the same 30. of the seventh) the square numbers produced of DE and EF, are prime to that which is produced of DE into EF. But that which is produced of DE into himself is A, and that which is produced of EF into himself is C, and that which is produced of DE into EF, is B. Wherefore the numbers A, C, added together are prime unto the number B: which was required to be demonstrated.  
This latter part of the demonstration, which proveth that the numbers A, & C are prime unto B, is somewhat obscurely put of Theon. And therefore I will here make it plainer.  
Forasmuch as either of the numbers DE, and EF is prime to the whole DF: 
Demonstration to prove that the numbers A and C are prime to B.  (as hath before b●ne proved) therefore that which is produced of DE into EF (which is the number B) is prime unto DF, by the 26. of the seventh. Wherefore by the 27. of the same that which is produced of DF into himself (which is the number composed of A and C and of the double of B by the 4. of the second) shall be prime unto B. Wherefore it followeth that the number composed of A and C is prime unto B. For if a number composed of two numbers, be prime to one of the said two numbers, as here the number composed of A and C taken as one number and of the double of B, is prime unto the double of B: then the two numbers whereof the number is composed, namely, the number composed of A and C, and the double of B shall be prime the one to the other (by the 30 of the seventh). And therefore the number composed of A and C shall be prime to B taken once. For if any number should measure the two numbers, namely the number composed of A and C, and the number B, it should also measure the number composed of A and C, and the double of B (by the 5. common sentence of the seventh): which is not possible, for that they are proved to be prime numbers.  
Here have I added an other demonstration of the former Proposition after Campane, which proveth that in numbers how many soever, which is there proved only touching three numbers: and the demonstration seement somewhat more perspicuous than Theons demonstration. And thus he putteth the proposition.  
If numbers how many soever being in continual proportion be the lest that have one & the same proportion with them: every one of them shallbe to the number composed of the rest prime.  
Secondly I say that this is so in every one of them: namely that C is a prime number to the number composed of A, B, D. For if not, then as before let E measure C, and the number composed of A, B, D: which E shallbe a number not prime either to F or to G (by the former proposition added by Campane) wherefore let H measure them. And forasmuch as H measureth E, it shall also measure the whole A, B, C, D whom E measureth. And forasmuch as H measureth one of these numbers F or G, it shall measure one of the extremes A or D: which are produced of F or G (by the second of the eight) if they be multiplied into the means L or K. And moreover the same H shall measure the meames, BC (by the 5. common sentence of the seventh) when as by supposition it measureth either F or G. which measure B, C (by the second of the eight). But the same H measureth the whole A, B, C, D as we have proved, for that it measureth E. Wherefore it shall also measure the residue, namely, the number composed of the extremes A and D (by the 4. common sentence of the seventh). And it measureth one of these A or D (for it measureth one of these F or G which produce A and D) wherefore the same H shall measure one of these A or D and also the other of them (by the former common sentence) which numbers A and D are by the 3. of the eight prime the one to the other. Which were absurd. This may also be proved in every one of these numbers A, B, C, D. Wherefore no number shall measure one of these numbers A, B, C, D and the numbers composed of the rest. Wherefore they are prime the one to the other: If therefore numbers how many soever. etc.: which was required to be proved.  
Here as I promised, I have added Campanes demonstrations of those Propositions in numbers, which Eucl●de in the second book demonstrated in lines. And that in this place so much the rather, for that Theon as we see in the demonstration of the 15. Proposition seemeth to allege the 3. & 4. Proposition of the second book: which although they concern lines only, yet as we there declared and proved, are they true also in numbers.  

¶ The first Proposition added by Campane.  
That number which is produced of the multiplication of one number into numbers how many soever: is equal to that number which is produced of the multiplication of the same number into the number composed of them.  
This proveth that in numbers which the first of the second proved touching lines. Suppo●● that the number A being multiplied into the number B, and into the number C, 
Demonstratiou.  and into the number D, do produce the numbers E, F and G. Then I say that the number produced of A multiplied into the number composed of B, C, and D is equal to the number composed of E, F, and G. For by the converse of the definition of a number multiplied, what part unity is of A, the self same part is B of E, and C of F, and also D of G. Wherefore by the 5. of the seventh what part unity is of A, the self same part is the number composed of B, C, and D, of the number composed of E, F, and G. Wherefore by the definition that which is produced of A into the number composed of B, C, D, is equal to the number composed of E, F, G: which was required to be proved.   

The second Proposition.  
That number which is produced of the multiplication of numbers how many soever into one number: is equal to that number which is produced of the multiplication of the number composed of them into the same number.  
This is the converse of the former As if the ●●●bers ● and G and D multiplied into the number A do produce the numbers E and F and G. 
This proposition is the converse of the former.  Then the number composed of B, C, D. multiplied into the number A shall produce the number composed of the numbers E, F, G. Which thing is easily proved by the 16. of the seventh and by the former proposition.   

¶ The third Proposition.  
That number which is produced of the multiplication of numbers how many soever into other numbers how many soever, is equal to that number which is produced of the multiplication of the number composed of those first numbers, into the number composed of these latter numbers.  
As if the numbers A, B, C do multiply the numbers D, E, F, each one each other, and if the numbers produced be added together. Then I say that the number composed of the numbers produced is equal to the number produced of the number composed of the numbers A, B, 
Demonstration.  C into the number composed of the numbers D, E, F. For by the former proposition that which is produced of the number composed of A, B, C into D is equal to that which is produced of every one of the said numbers into D: and by the same reason that which is produced of the number composed of A, B, C into E, is equal to that which is produced of every one of the said numbers into E: and so likewise that which is produced of the number composed of A, B, C into F is equal to that which is produced of every one of the said numbers into F. But by the first of these propositions th●● which is produced of the number composed of these numbers A, B, C into every one of these numbers D, E, F is equal to that which is produced of the number composed into the number composed: wherefore that is manifest which was required to be proved.   

¶ The fourth Proposition.  
If a number be deu●●●d into parts how many soever: that number which is produced of the whole into himself, is equal to that number which is produced of the same number into all his parts.  
This proveth in numbers that which the second of the second proved in lines. As if the number A, be divided into the numbers B and C, and D. 
This answereth to the 2. of the second.  Then I say, that that which is produced of A into himself, is equal to that which is produced of A into all the said numbers B, C, and D. For putting the number ● equal to the number A, it is manifest by the first of these propositions that that which is produced of E into A, 
Demonstration.  is equal to that which is produced of E into all the parts of A● But by the common sentence that which is produced of E into A is equal to that which is produced of A into himselfe● and that which is produced of E into the parts of A is equal to that which is produced of A into the self same parts. Wherefore that is manifest which was required to be proved.   

¶ The fift Proposition.  
If a number be divided into two parts that which is produced of the whole in●o one of the parts, as equal ●o that which is produced of the self same par● into himself ● and into the other part.  
This proveth in numbers that which in the 3. of the second was proved in lines. 
This answereth to the 3. of the thirds.  For let the number A be divided into the numbers B and C. Then I say that that which is produced of A into C, is equal to that which is produced of C into himself and into B. For by the 16. of the seventh, that which is produced of A into C is equal to that which is produced of C into A. Now then put the number D equal to the number C. 
Demonstration.  Wherefore that which is produced of A into C is equal to that which is produced of D into A. But by the first of these propositions that which is produced of D into A is equal to that which is produced of D into B and of D into C. Wherefore forasmu●●●● that which is produced of D into A and into B and into C is ●qu●ll to that which is produced of C into A● and into B, and into himself, by reason of the equality of C and D: that is manifest which was required to be proved.   

¶ The sixth Proposition.  
If a number be divided into two parts: that which is produced of the multiplication of the whole ●nto himself, is equal to that which is produced of the multiplication of either of the parts into himself, and of the one into the other awise.  
This proveth in numbers that which the fourth of the second proved touching lines. As if the number A be divided into the numbers B and C. Then I say that that which is produced of A into himself is equal to that which is produced of B into himself, 
This answereth to th● 4. of the second.  and of C into himself, and of B into C twice. For by the 4 of these propositions, that which is produced of A into himself, is equal to that which is produced of A into B, and into C. But that which is produced of A into B, is equal to that which is produced of B into himself and into C (by the former proposition). And by the same that which is produced of A into C is equal to that which is produced of C into himself and into B. 
Demonstration.  And forasmuch as that which is produced of C into B is equal to that which is produced of B into C by the 16. of the seventh, it is manifest that that is true which was required to be proved.   

¶ The seventh Proposition.  
If a number be divided into two equal parts, & into two unequal parts: that which is produced of the greater of the unequal parts into the less, together with the square number of the number set between, is equal to the square number produced of the half of the whole.  
This proveth in numbers that which the 5. of the second proved in lines. As if the number AB be divided into two equal numbers, which let be AC, and CB: and also in two unequal numbers namely, 
This answereth to the 5. of the second.  AD and DB, of which let AD be the greater, and DB the less. Then I say that that which is produced of the whole AD into DB together with the square number of CD, is equal to the square number of CB. For by the former proposition the square of CB is equal to the square of CD and to the square of DB, and to that which is produced of BD into CD twice. But that which is produced of BD into himself and into CD is equal to that which is produced of BD into C B by the first of these propositions, 
Demonstration.  and therefore unto that which is produced of BD into AC. Wherefore that which is produced of BD into himself and into CD twice is equal so that which is produced of BD into AD. Wherefore by the same the square of C B exceedeth that which is produced of BD into AD by the square of CD: wherefore that is manifest which was required to be proved.   

¶ The 8. Proposition.  
If a number be divided into two equal part●s, and if unto it be added an other number: that which is produced of the multiplication of the whole being composed into the number added, together with the square of the half, is equal to the square of the number composed of the half and the number added.  
This proveth in numbers that which the 6. of the second proved touching lines. For suppose that the number AB be divided into equal numbers, which let be AC and CB: and unto it● add the number BD. 
This answereth to the 6. of the second.  Then I say, that that which is produced of the whole AD into DA together with the square of BC, is equal to the square of CD. For by the 6. of these propositions the square of CD is equal to the square of DB, & to the square of BC, and to that which is produced of DB into BC twice. But by the ●. of these propositions, 
Demonstration.  that which is produced of BD into himself and into BC twice is equal to that which is produced of BD into DA (for AC and CB are equal) wherefore the square of CD exceedeth that which is produced of BD into DA by the square of CB. Wherefore that is manifest which was required to be proved.   

¶ The 9 Proposition.  
If a number be divided into two parts: that which is produced of the whole number into himself together with that which is produced of one of the parts into himself, is equal to that which is produced of the whole into the said part twice together with that which is produced of the other part into himself.  
This proveth in numbers that which the 7. of the second proved in lines. For let the number A be divided into the numbers B and D. 
This answereth to the 7. of the second.  Then I say that the square of A together with the square of D is equal to that which is produced of A into D twice together with the square of B. For it is manifest by the 6. of these propositions that the square of A is equal to the squares of B and D together with that which is produced of B into D twice. Wherefore the square of A together with the square of D, is equal to two squares of D ● 
Demonstration.  and to that which is produced of D into B twice together with the square of B. But by the first of these propositions two squares of D, and that which is produced of D into B twice is equal to that which is produced of D into A twice. Wherefore that which is produced of D into A twice together with the square of B, is equal to the square of A together with the square of D: wherefore that is manifest which was required to be proved.   

¶ The 10. proposition.  
If a number be divided into two parts, and unto it be added a number equal to one of the parts: the square of the whole number composed, is equal to the quadruple of that which is produced of the first number into the number added, together with the square of the other part.  
This proveth in numbers, that which the 8. of the second proved in lines. Suppose that the number AB be divided into the numbers AC and CB, unto which add the number BD, which let be equal to the number CB. 
This answereth to the 8. of the second.  Then I say that the square of the whole number composed, namely, AD, is equal to that which is produced of AB into BD four times together with the square of AC. For by the 6. of these propositions the square of AD, is equal to the square AB and to the square of BD together with that which is produced of AB into BD twice. 
Demonstratition.  And forasmuch as the square of BD is equal to the square CB: therefore the square of AD shall be equal to the square of AB and to the square of CB together with that which is produced of AB into BD twice. But by the former proposition the square of AB together with the square of CB, is equal to the square of AC together with that which is produced of AB into BC twice wherefore the square of AD is equal to that which is produced of AB into BD twice, and to that which is produced of AB into BC twice together with the square of AC. And for that that which is produced of AB into BC is equal to that which is produced of AB into BD, therefore is that manifest which was required to be proved.   

The 11. proposition.  
If a number be divided into two equal parts, and into two unequal parts: the squares of the two unequal parts taken together, are double to the square of the half, and to the square of the excess of the greater part above the less, the said two squares being added together.  
This proveth in numbers that which the 9 of the second proved in lines. For suppose that the number AB be divided into two equal parts: which let be AC and CB, and into two unequal parts, which let be AD and D B. 
This answereth to th● 9 of the second.  Then I say that the squares of the two numbers AD & DB, taken together, are double to the two squares of the two numbers AC and CD, taken together. For by the 6. of these propositions the square of AD is equal to the squares of AC and CD, and to that which is produced of AC into CD twice. And forasmuch as the square of AC is equal to the square of CB, the square of AD shall be equal to the square of BC & to the square of CD together with that which is produced of BC into CD twice. 
Demonstration.  Wherefore the square of AD together with the square of BD, is equal to the square of BC, and to the square of CD and to that which is produced of BC into CD twice together with the square of BD. But that which is produced of BC into CD twice together with the square of BD, is equal to the square of BC and to the square of CD by the 9 of these propositions. Wherefore the squares of the two numbers AD and DB are equal to the squares of the two numbers BC and CD, taken twice. And therefore the squares of the two numbers AD and DB are double to the squares of the two numbers BC and CD, that is AC and CD (for the numbers AC and BC are by supposition equal) wherefore that is manifest which was required to be proved.   

The 12. proposition.  
If a number be divided into two equal parts, and unto it be added an other number: the square of the whole number composed together with the square of the number added, is double to the square of the half, together with the square of the number composed of the half and the number added.  
This proveth in numbers that which the 10. of the second proved in lines. Suppose that the number AB be divided into two equal numbers AC and CB: 
This answereth to the 10. o● the second.  and unto it add the number BD. Then I say that the square of AD together with the square of BD, is double to the square of AC together with the square of CD. For forasmuch as the number CD is divided into two parts, and unto it is added the number AC which is equal to one of the parts (namely, to CB) therefore by the 10. of these propositions, 
Demonstration.  the square of AD is equal to that which is produced of CD into CA four times together with the square of BD. And forasmuch as AC is equal to CB, therefore the square of AD, is equal to t●●t which is produced of DC into CB four times together with the square of BD. Wherefore the square of AD together with the square of DB, is equal to that which is produced of DC into CB ●ower times together with two squares of BD. But by the 9 of these propositions, that which is produced of DC into CB four times together with two squares of BD is double to the square of CD together with the square of CB (for the square of CD together with the square of CB is equal to that which is produced of DC into CB twice together with one square of CB). Wherefore forasmuch as the square of CB is equal to the square of AC, that is manifest: which was required to be proved.   

The 13. proposition.  
It is impossible to divide a number in such sort: that that which is contained under the whole and one o● the parts, shall be equal to the square of the other part.  

A negative proposition.  That which the 11. of the second taught to be done in lines is here proved to be impossible to be done in numbers. ●or suppose that there be a number whatsoever namely AB. Then I say, that it is impossible to divide it in such sort as is required in the proposition. For so should it be divided according to a proportion having a mean and two extremes. But i● i● be possible, 
Demonstration lea●ing to an impossibility.  let the number AB be so divided in C. And as AB is to BC, so let BC be to C A. Wherefore AC shall be less than CB. Now then take away from CB a number equal to AC which let be CD. And forasmuch as the proportion of the whole AB to the whole BC, is as the proportion of the part taken away from AB, namely, BC to the part taken away from BC, namely, CD: therefore the proportion of the residue of AB, namely, AC, to the residue of BC, namely, to BD, is as the proportion of the whole AB to the whole BC (by the 11. of the seventh). Wherefore BC is to CD, as CD is to DB. Wherefore CD is greater than DB. Wherefore sub●rahing DE out of CD, so that let DE be equal to DB: the proportion of BC to CD is as the proportion of CD to DE. Wherefore the residue of CB, namely, DB, shall be to the residue of CD, namely, to CE, as the proportion of the whole BC to the whole CD. Wherefore CE may be subtrahed out of ED: wherefore there shallbe no end of this subraction: which is impossible.    

¶ The 16. Theorem. The 16. Proposition. If two numbers be prime the one to the other, the second shall not be to any other number, as the first is to the second.  
SVppose that these two numbers A and B be prime the one to the other. Then I say that B is not to any other number as A is to B. 
Demonstration leading to an absurdity.  For if it be possible, as A is to B, so let B be to C. Now A and B are prime numbers, yea they are prime and the jest by the 23. of the seventh. But (by the 21. of the seventh) the least measure the numbers that have one and the same proportion with them equally, the antecedent the antecedent, and the consequent the consequent. Wherefore the antecedent A, measureth the antecedent B, and it measureth also itself. Wherefore A measureth these numbers A & B being prime the one to the other, which is impossible. Wherefore as A is to B, so is not B to C: which was required to be proved.   

¶ The 17. Theorem. The 17. Proposition. If there be numbers in continual proportion how many soever, and if their extremes be prime the one to the other, the less shall not be to any other number, as the first is to the second.  
SVppose that there be these numbers in continual proportion A, B, C, D, and let their extremes A and D be prime the one to the other. Then I say that D is not to any other number as A is to B. For if it be possible, as A is to B, so let D be to E. Wherefore alternately by the 13. of the seventh, as A is to D, so is B to E. But A and D are prime, 
Demonstration leading to an abjurditie.  yea they are prime and the lest. But the lest numbers (by the 21. of the seventh) measure the numbers that have one and the same proportion with them equally, the antecedent the antecedent, and the consequent. Wherefore the antecedent A measureth the antecedent B: but as A is to B, so is B to C. Wherefore B also measureth C. Wherefore A also measureth C (by the 5. common sentence of the seventh) and forasmuch as B is to C, so is C to D, but B measureth C. Wherefore C measureth D. But A measureth C. Wherefore A also measureth D by the same common sentence, and it also measureth itself. Wherefore A measureth these numbers A and D being prime the one to the other, which is impossible. Wherefore D is not to any other number as A is to B: which was required to be proved.   

¶ The 18. Theorem. The 18. Proposition. Two numbers being given, to search out if it be possible a third number in proportion with them.  
SVppose that the two numbers given be A and B. It is required to search out if it be possible a third number proportional with them. Now A, B are either prime the one to the other or not prime. 
Three cases in this proposition.  If they be prime, then (by the 16. of the ninth) it is maninifest that it is impossible to find out a third number proportional with them. But now suppose that AB be not prime the one to the other. And let B multiplying himself produce C. Now A either measureth C, or measureth it not. First, 
The first case.  let it measure it and that by D. Wherefore A multiplying D produceth HUNDRED 
The second case.  But B also multiplying himself produced C. Wherefore that which is produced of A into D, is equal to that which is produced of B into himself. Wherefore (by the second part of the 19 of the seventh) as A is to B, so is B to D. Wherefore unto these numbers A, B is found out a third number in proportion, namely, D.  
But now suppose that A do not measure C● 
The third case.  Then I say that it is impossible to ●inde out a third number in proportion with these numbers A, B. For if it be possible, let there be found out such a number, and let the same be D. Wherefore that which is produced of A into D, is equal to that which is produced of B into himself, but that which is produced of B into himself is C. Wherefore that which is produced of A into D is equal unto C. Wherefore A multiplying D produced C. Wherefore A measureth G by D. But it is supposed also not to measure it, which is impossible. Wherefore it is not possible to find out a third number in proportion with A & B, whensoever A measureth not C: which was required to be proved.   

¶ The 19 Theorem. The 19 Proposition. Three numbers being given, to search out if it be possible the fourth number proportional with them.  

Divert cases ●n this proposition.  SVppose that the three numbers given be A, B, C. It is required to search out if it be possible a forth number proportional with them. Now A, B, C, are either in continual proportion, and their extremes A, C are prime the one to the other or they are not in continual proportion, and their extremes are yet prime the one to the other: or they are in continual proportion, and their extremes are not prime the one to the other: or they are neither in continual proportion, nor their extremes are prime the one to the other.  

The first case.  If A, B, C, be in continual proportion, and their extremes be prime the one to the other, it is manifest (by the 17. of the ninth) that it is impossible to find out a fourth number proportional with them.  
But now suppose that A do not measure D. Then I say that it is not possible to find out a fourth number proportional with these numbers A, B, C. For if it be possible, let there be found such a number, and let the same be E. Wherefore that which is produced of A into E is equal to that which is produced of B into C. But that which is produced of B into C is D. Wherefore that which is produced of A into E is equal unto D. Wherefore A multiplying E produced D, wherefore A measureth D, but it also measureth it not, which is impossible. Wherefore it is impossible to find out a fourth number proportional, with these numbers A, B, C, whensoever A measureth not D.   

¶ The 20. Theorem. The 20. Proposition. Prime numbers being given how many soever, there may be given more prime numbers.  
SVppose that the prime numbers given be A, B, C. 
Two cases in this Proposition.  Then I say, that there are yet more prime numbers besides A, B, C. Take (by the 38. of the seventh) the jest number whom these numbers A, B, C do measure, and let the same be DE. And unto DE add unity DF. Now EF is either a prime number or not. First let it be a prime number, 
The first case.  then are there found these prime numbers A, B, C, and EF more in multitude then the prime numbers ●irst given A, B, C.  
But now suppose that EF be not prime. 
The second case.  Wherefore some prime number measureth it (by the 24. of the seventh). Let a prime number measure it, namely, G. Then I say, that G is none of these numbers A, B, C. For if G be one and the same with any of these A, B, C. But A, B, C, measure the number DE: wherefore G also measureth DE: and it also measureth the whole EF. Wherefore G being a number shall measure the residue DF being unities which is impossible. Wherefore G is not one and the same with any of these prime numbers A, B, C: and it is also supposed to be a prime number. Wherefore there are ●ound these prime numbers A, B, C, G, being more in multitude then the prime numbers given A, B, C: which was required to be demonstrated.  

A Corollary.  
By this Proposition it is manifest, that the multitude of prime numbers is infinite.    

¶ The 21. Theorem. The 21. Proposition. If even numbers how many soever be added together: the whole shall be even.  
SVppose that these even numbers AB, BC; CD, and DE, be added together. Then I say, that the whole number, namely, AE, is an even number. 
Demonstration.  For forasmuch as every one of these numbers AB, BC, CD, and DE, is an even number, therefore every one of them hath an half. Wherefore the whole AE also hath an half. But an even number (by the definition) is that which may be divided into two equal parts. Wherefore AE is an even number: which was required to be proved.   

¶ The 22. Theorem. The 22. Proposition. If odd numbers how many soever be added together, & if their multitude be even, the whole also shall be even.  
SVppose that these odd numbers AB, BC, CD, and DE, being even in multitude, be added together. Then I say, that the whole AE is an even number. For forasmuch as every one of these numbers AB, BC, CD, and DE, is an odd number, is ye take away unity from every one of them, 
Demonstration.  that which remaineth o● every one of them is an even number. Wherefore they all added together, are (by the 21. of the ninth) an even number: and the multitude of the unities taken away is even. Wherefore the whole AE is an even number: which was required to be proved.   

¶ The 23. Theorem. The 23. Proposition. If odd numbers how many soever be added together, and if the multitude of them be odd, the whole also shall be odd.  
SVppose that these odd numbers, AB, BC, and CD being odd in multitude be added together. Then I say that the whole AD is an odd number. Take away from CD, unity DE, wherefore that which remaineth CE is an even number. But AC also (by the 22. of the ninth) is an even number. 
Demonstration  Wherefore the whole AE is an even number. But DE which is unity being added to the even number AE, maketh the whole AD a● odd number: which was required to be proued●   

¶ The 24. Theorem. The 24. Proposition. If from an even number be taken away an even number, that which remaineth shall be an even number.  
SVppose that AB be an even number, and from it take away an even number CB. 
Demonstration.  Then I say that that which remaineth, namely, AC is an even number. For forasmuch as AB is an even even number, it hath an half, and by the same reason also BC hath an half. Wherefore the residue CA hath an half. Wherefore AC is an even number: which was required to be demonstrated.   

¶ The 25. Theorem. The 25. Proposition. If from an even number be taken away an odd number, that which remaineth shall be an odd number.  
SVppose that AB be an even number, and take away from it BC an odd number. Then I say that the residue CA is an odd number. 
Demonstration.  Take away from BC unity CD. Wherefore DB is an even number. And AB also is an even number, wherefore the residue AD is an even number (by the ●ormer proposition) But CD which is unity, being taken away from the even number AD maketh the residue AC an odd number: which was required to be proved.   

¶ The 26. Theorem. The 26. Proposition. If from an odd number be taken away an odd number, that which remaineth shall be an even number.  
SVppose that AB be an odd number, and from it take away an odd number BC. Then I say that the residue CA is an even number. 
Demonstration.  For forasmuch as AB is an odd number, take away from it unity BD. Wherefore the residue AD is even. And by the same reason CD is an even number: wherefore the residue CA is an even number (by the 24. of this book) ● which was required to be proved.   

¶ The 27. Theorem. The 27. Proposition. If from an odd number be taken a way an even number, the residue shall be an odd number.  
SVppose that AB be an odd number, and from it take away an even number BC. 
Demonstration.  Then I say that the residue CA is an odd number. Take away from AB unity AD. Wherefore the residue DB is an even number, & BC is (by supposition) even. Wherefore the residue CD is an even number. Wherefore DA which is unity, being added unto CD which is an even number maketh the whole AC an ●dde number: which was required to be proved.   

¶ The 28. Theorem. The 28. Proposition. If an odd number multiplying an even number produce any number, the number produced shall be an even number.  
SVppose that A being an odd number multiplying B being an even number, do produce the number C. Then I say that C is an even number. 
Demonstration.  For forasmuch as A multiplying B produced C, therefore C is composed of so many numbers equal unto B as there be in unities in A. But B is an even number: wherefore C is composed of so many even numbers, as there are unities in A. But if even numbers how many soever be added together, the whole (by the 21. of the ninth) is an even number: wherefore C is an even number: which was required to be demonstrated.   

¶ The 29. Theorem. The 29. Proposition. I● an odd number multiplying an odd number produce any number, the number produced shallbe an odd number  
SVppose that A being an odd number multiplying B being also an odd number, do produce the number C. Then I say that C is an odd number. For forasmuch as A multiplying B produced C, therefore C is composed of so many numbers equal unto B as there be unities in A. 
Demonstration.  But either of these numbers A and B is an odd number. Wherefore C is composed of odd numbers, whose multitude also is odd. Wherefore (by the 23. of the ninth) C is an odd number: which was required to be demonstrated.  

A proposition added by Campane.  

A proposition added by Campane.  If an odd number measure an even number, it shall measure it by an even number.  
For if it should measure it by an odd number, then of an odd number multiplied into an odd number should be produced an odd number, which by the former proposition is impossible.   

another proposition added by him.  

An other added by him.  If an odd number measure an odd number, it shall measure it by an odd number.  
For if it should measure it by an even number, then of an odd number multiplied into an even number should be produced an odd number which by the 28. of this book is impossible.    

¶ The 30. Theorem. The 30. Proposition. If an odd number measure an even number, it shall also measure the half thereof.  
SVppose that A being an odd number do measure B being an even number. Th●● I say that it shall measure the half thereof. For forasmuch as A measureth B let i● measure it by C. 
Demonstration leading to an absurdity.  Then I say that C is an even number. For if not then, if it be possible le● i● be odd. And forasmuch as A measureth B by C: therefore A multiplying C produceth B. Wherefore B is composed of odd numbers whose multitude also is odd. Wherefore B is an odd number (by the 29. of this book) which is absurd for it is supposed to be even: wherefore C is an even num●er. Wherefore A measureth B by an even number: and C measureth B by A. But either of these numbers C and B hath an half part wherefore as C is to B, so is the half to the half. But C measureth B by A. Wherefore the half of C measureth the half of B by A: wherefore A multiplying the half of C produceth the half of B. Wherefore A measureth the half of B: and it measureth it by the half of C. Wherefore A measureth the half of the number B: which was required to be demonstrated.   

¶ The 31. Theorem. The 31. Proposition. If an odd number be prime to any number, it shall also be prime to the double thereof.  
SVppose that A being an odd number be prime unto the number B: and let the double of B be C. 
Demonstration.  Then I say, that A is prime unto C. For if A and C be not prime the one to the other, some one number measureth them both. Let there be such a number which measureth them both, and let the same be D. But A is an odd number. Wherefore D also is an odd number. (For if D which measureth A should be an even number, than should A also be an even number (by the 21. of this book): which is contrary to the supposition. For A is supposed to be an odd number: & therefore D also is an odd number). And forasmuch as D being an odd number measureth C, but C is an even number (for that it hath an half, namely, B). Wherefore (by the Proposition next going before) D measureth the half of C. But the half of C is B. Wherefore D measureth B: and it also measureth A. Wherefore D measureth A and B being prime the one to the other: which is absurd. Wherefore no number measureth the numbers A & C. Wherefore A is a prime number unto C. Wherefore these numbers A and C are prime the one to the other: which was required to be proved.   

¶ The 32. Theorem. The 32. Proposition. Every number produced by the doubling of two upward, is evenly even only.  
SVppose that A be the number two: and from A upward double numbers how many soever; as B, C, D. Then I say, that B, C, D, are numbers evenly even only. That every one of them is evenly even, it is manifest: 
Demonstration.  for every one of them is produced by the doubling of two. I say also, that every one of them is evenly even only. Take unity E. And forasmuch as from unity are certain numbers in continual proportion, & A which followeth next after unity is a prime number, therefore (by the 13. of the third) no number measureth D being the greatest number of these numbers A, B, C, D, besides the self same numbers in proportion. But every one of these numbers A, B, C, is evenly even. Wherefore D is evenly even only. In like sort may we prove, that every one of these numbers A, B, C, is evenly even only: which was required to be proved.   

¶ The 33. Theorem. The 33. Proposition. A number whose half part is odd, is evenly odd only.  
SVppose that A be a number whose half part is odd. Then I say that A is evenly odd only. That it is evenly odd it is manifest: for his half being odd measureth him by an even number, namely, by 2. 
Demonstration leading to an absurdity.  (by the definition). I say also that it is evenly odd only. For if A be evenly even, his half also is even. For (by the definition) an even number measureth him by an even number. Wherefore that even number which measureth him by an even number shall also measure the half thereof being an odd number by the 4. common sentence of the seventh which is absurd. Wherefore A is a number evenly odd only: which was required to be proved.  

another demonstration to prove the same.  
Suppose that the number A have to his half an odd number, namely, B. Then I say that A is evenly odd only. That it is evenly odd needeth no proof: forasmuch as the number 2. an even number measureth it by the half thereof which is an odd number. 
another demonstration.  Let C be the number 2. by which B measureth A (for that A is supposed to be double unto B). And let an even number, namely, D measure A (which is possible for that A is an even number by the definition) by F. And forasmuch as that which is produced of C into B is equal to that which is produced of D into F, therefore by the 19 of the seventh, as C is to D, so is B to F. But C the number two measureth D being an even number: wherefore F also measureth B which is the half of A. Wherefore F is an odd number. For if F were an even number than should it in the B whom it measureth an odd number also by the 21. of this book, which is contrary to the supposition. And in like manner may we prove that all the even numbers which measure the number A● do measure it by odd numbers. Wherefore A is a number evenly odd only: which was required to be proved.    

¶ The 34. Theorem. The 34. Proposition. If a number be neither doubled from two, nor hath to his half part an odd number, it shall be a number both evenly even, and evenly odd.  
SVppose that the number A be a number neither doubled from the number two, neither also let it have to his half part an odd number. Then I say that A is a number both evenly even, and evenly odd. 
Demonstration.  That A is evenly even it is manifest, for the half thereof is not odd, and is measured by the number 2. which is an even number. Now I say that it is evenly odd also. For if we divide A into two equal parts, and so continuing still, we shall at the length light upon a certain odd number which shall measure A by an even number. For if we should not light upon such an odd number, which measureth A by an even number, we should at the length come unto the number two, and so should A be one of those numbers which are doubled from two upward, which is contrary to the supposition. Wherefore A is evenly odd. And it is proved that it is evenly even: wherefore A is a number both evenly even and evenly odd: which was required to be demonstrated.  
This proposition and the two former manifestly declare that which we noted upon the tenth definition of the seventh book namely, that Campane and Flussates and divers other interpreters of Euclid (only Theon except) did not rightly understand the 8. and 9 definitions of the same book concerning a number evenly even, and a number evenly odd. For in the one definition they add unto Euclides words extant in the Greek this word only (as we there noted) and in the other this word all. So that after their definitions a number can not be evenly even unless it be measured only by even numbers: likewise a number can not be evenly odd unless all the even numbers which do measure it, do measure it by an odd number. The contrary whereof in this proposition we manifestly see. For here Euclid proveth that one number may be both evenly even and evenly odd. And in the two former propositions he proved that some numbers are evenly even only, and some evenly odd only: which word only had been in vain of him added, if no number evenly even could be measured by an odd number, or if all the numbers that measure a number evenly odd must needs measure it by an odd number. Although Campane and Flussates to avoid this absurdity have wrested the 32. proposition of this book from the true sense of the Greek and as it is interpreted of Theon. So also hath Flussates wrested the 33. proposition. For whereas Euclid saith, Every number produced by the doubling of two upward, is evenly even only: they say, only the numbers produced by the doubling of two, are evenly even. Likewise whereas Euclid saith, A number whose hafle part is odd, is evenly odd only, Flussates saith, only a number whose half part is odd, Is evenly odd. Which their interpretation is not true, neither can be applied to the propositions as they are extant in the Greek. In deed the said 32. and 33. propositions as they put them are true touching those numbers which are evenly even only, or evenly odd only. For no number is evenly even only, but those only which are doubled from two upward. Likewise no numbers are evenly odd only, but those only whose half is an odd number. But this letteth not, but that a number may be evenly even although it be not doubled from two upward & also that a number may be evenly odd although it have not to his half an odd number. As in this 34. proposition Euclid hath plainly proved. Which thing could by no means be true, if the foresaid 32. & 33. propositons of this book should have that sense and meaning wherein they take it.   

¶ The 35. Theorem. The 35. Proposition. If there be numbers in continual proportion how many soever, and if from the second and last be taken away numbers equal unto the first, as the excess of the second is to the first, so is the excess of the last to all the numbers going before the last.  
SVppose that these numbers A, BC, D, and EF, be in continual proportion beginning at A the lest. And from BC, which is the second, take away CG equal unto the first, namely, to A, and likewise from EF the last take away FH equal also unto the first, namely, to A. Then I say, that as the excess BG is to A the first, so is HE the excess, to all the numbers D, BC, and A, which go before the last number, namely, EF. 
Demonstration.  Forasmuch as EF is the greater (for the second is supposed greater than the first) put the number FL equal to the number D, and likewise the number FK equal to the number BC. And forasmuch as FK is equal unto CB, of which FH is equal unto GC, therefore the residue HK is equal unto the residue GB. And for that as the whole F●, is to the whole FL, so is the part taken away FL, to the part taken away FK, therefore the residue LE is to the residue KL, as the whole ●E is to the whole FL (by the 11. of the seventh). So likewise for that FL is to FK, as FK is to FH, KL shall be to HK, as the whole FL is to the whole FK (by the same Proposition). But as FE is to FL, and as FL is to FK, and FK to FH, so were FE to D, and D to BC, and BC 〈◊〉 A. Wherefore as LE is to KL, and as KL is to HK, so is D to BC. Wherefore alternately (by the 23. of the seventh) as LE is to D, so is KL to be BC, and as KL is to BC, so is HK to A. Wherefore also as one of the antecedentes is to one of the consequentes; so are all the antecedentes to all the consequentes. Wherefore as KH is to A, so are HK, KL, and LE, to D, BC, and A (by the 12. of the seventh). But it is proved, that KH is equal unto BG. Wherefore as BG, which is the excess of the second, is to A, so is EH the excess of the last unto the numbers going before D, BC, and A. Wherefore as the excess of the second is unto the first, so is the excess of the last to all the numbers going before the last: which was required to be proved.   

¶ The 36. Theorem. The 36. Proposition. If from unity be taken numbers how many soever in double proportion continually, until the whole added together be a prime number, and if the whole multiplying the last produce any number, that which is produced is a perfect number.  

This proposition teach●th how to find out a perfect number.  SVppose that from unity be taken these numbers A, B, C, D, in double proportion continually, so that all those numbers A, B, C, D, & unity added together, make a prime number: and let E be the number composed of all those numbers A, B, C, D, & unity added together: and let E multiplying D, which is the last number, produce the number FG. Then I say, that FG is a perfect number.  
 

Construction.  How many in multitude A, B, C, D, are, so many in continual double proportion take beginning at E, which let be the numbers E, HK, L, and M. Wherefore of equality (by the 13. of the seventh) as A is to D, so is E to M. Wherefore that which is produced of E into D, is equal to that which is produced of A into M. But that which is produced of E into D, is the number FG. Wherefore that which is produced of A into M, is equal unto FG. Wherefore A multiplying M produceth FG. Wherefore M measureth FG by those unities which are in A. But A is the number two. Wherefore FG is double to M. And the numbers M, L, HK, and E, are also in continual double proportion. Wherefore all the numbers E, HK, L, M, and FG, are continually proportional in double proportion. Take from the second number KH, and from the last FG a number equal unto the first, namely, to E: and let those numbers taken be HN, & FX. 
Demonstration.  Wherefore (by the Proposition going before) as the excess of the second number is to the first number, so is the excess of the last to all the numbers going before it. Wherefore as NK is to E, so is XG to these numbers M, L, KH, and E. But NK is equal unto E (for it is the half of HK, which is supposed to be double to E). Wherefore XG is equal unto these numbers M, L, HK, and E. But XF is equal unto E, and E is equal unto these numbers A, B, C, D, and unto unity. Wherefore the whole number FG is equal unto these numbers E, HK, L, M, 
Demonstration leading to an absurdity.  and also unto these numbers A, B, C, D, and unto unity. Moreover I say, that unity and all the numbers A, B, C, D, E, HK, L, and M, do measure the number FG. That unity measureth it, it needeth no proof. And forasmuch as FG is produced of D into E, therefore D and E do measure it. And forasmuch as the double from unity, namely, the numbers A, B, C, do measure the number D (by the 13. of this book) therefore they shall also measure the number FG (whom D measureth) by the ●. common sentence. By the same reason forasmuch as the numbers E, HK, L, and M, are unto FG, as unity and the numbers A, ●, C, are unto D (namely, in subduple proportion) and unity and the numbers A, B, C, do 〈◊〉 D, therefore also the numbers E, HK, L, and M, shall measure the number FG: No● I say also, that no other number measureth FG besides these numbers A, B, C, D, E, HK, L, M, and unity. For if it be possible, let O measure FG. And let O not be any of these numbers A, B, C, D, E, HK, L, and M. And how often O measureth FG, 〈◊〉 unities let there be in P. Wherefore O multiplying P produceth FG. But E also multiplying D produced FG. Wherefore (by the 19 of the seventh) as E is to O, so is P to D. Wherefore alternately (by the 9 of the seventh) as E is to P, so is O to D. And forasmuch as from unity are these numbers in continual proportion A, B, C, D, and the number A which is next after unity is a prime number, therefore (by the 13. of the ninth) no other number measureth D besides the numbers A, B, C. And it is supposed that O is not one and the same with any of these numbers A, B, C. Wherefore O measureth not D. But as O is to D, so is E to P. Wherefore neither doth E measure P. And E is a prime number. But (by the 31. of the seventh) every prime number, is to every number that it measureth not, a prime number. Wherefore E and P are prime the one to the other: yea they are prime and the lest. But (by the 21. of the seventh) the least measure the numbers that have one and the same proportion with them equally, the antecedent the antecedent, and the consequent the consequent. And as E is to P, so is O to D. Wherefore how many times E measureth O, so many times P measureth D. But no other number measureth D besides the numbers A, B, C (by the 13. of this book). Wherefore P is one and the same with one of these numbers A, B, C. Suppose that P be one and the same with B, & how many B, C, D, are in multitude, so many take from E upward, namely, E, HK, and L. But E, HK, and L, are in the same proportion that B, C, D, are. Wherefore of equality, as B is to D, so is E to L. Wherefore that which is produced of B into L, is equal to that which is produced of D into E. But that which is produced of D into E, is equal to that which is produced of P into O. Wherefore that which is produced of P into O, is equal to that which is produced of B into L. Wherefore as P is to B, so is L to O: and P is one & the same with B: wherefore L also is one and the same with O: which is impossible. For O is supposed not to be one and the same with any of the numbers given. Wherefore no number measureth FG besides these numbers A, B, C, D, E, HK, L, M, and unity. And it is proved, that FG is equal unto these numbers A, B, C, D, E, HK, L, M, and unity, which are the parts thereof (by the 39 of the seventh). But a perfect number (by the definition) is that which is equal unto all his parts. Wherefore FG is a perfect number● which was required to be proved.   
The end of the ninth book of Euclides Elements.   

¶ The tenth book of Euclides Elements.  
IN THIS TENTH BOOK doth Euclid entreat of lines and other magnitudes rational & irrational, 
The Argument of the tenth book.  but chief of irrational magnitudes, commensurable and incommensurable: of which hitherto, in all his former 9 books he hath made no mention at all. And herein differeth number from magnitude, 
Difference between number and magnitude.  or Arithmetic from Geometry: for that although in Arithmetic, certain numbers be called prime numbers in consideration of themselves, or in respect of an other, and so are called incommensurable, for that no one number measureth them, but only unity. Yet in deed and to speak absolutely and truly, there are no two numbers incommensurable, but have one common measure which measureth them both, if none other, yet have they unity, which is a common part and measure to all numbers, and all numbers are made of unities, as of their parts. 
A line is not made of points as number is made of unities.  As hath before been showed in the declaration of the definitions of the seventh book. But in magnitude it is far otherwise, for although many lines, plain figures, and bodies, are commensurable, and may have one measure to measure them, yet all have not so, nor can have. For that a line is not made of points, as number is made of unities, and therefore cannot a point be a common part of all lines, and measure them, as unity is a common part of all numbers and measureth them. Unity taken certain times maketh any number. For there are not in any number infinite unities: but a point taken certain times, yea as often as ye list, never maketh any line, for that in every line there are infinite points. Wherefore lines, figures, and bodies in Geometry, are oftentimes incommensurable and irrational. Now which are rational, and which irrational, which commensurable, and which incommensurable, how many and how sundry sorts and kinds there are of them, what are their natures, passions, and properties, doth Euclid most manifestly show in this book, and demonstrate them most exactly.  
This tenth book hath ever hitherto of all men, and is yet thought & accounted, to be the hardest book to understand of all the books of Euclid. 
This book the hardest to understand of all the books of Euclid.  Which common received opinion, hath caused many to shrink, and hath (as it were) deterred them from the handling and treaty thereof. There have been in deed in times past, and are presently in these our days, many which have dealt, and have taken great and good diligence in commenting, amending, and restoring of the six first books of Euclid, and there have stayed themselves and gone no farther, being deterred and made afraid (as it seemeth, by the opinion of the hardness of this book) to pass forth to the books following. Truth it is that this book hath in it somewhat an other & stranger manner of matter entreated of, 
In this book is entreated of a stranger manner of matter then in the former.  then the other books before had, and the demonstrations also thereof, & the order seem likewise at the first somewhat strange and unaccustomed, which things may seem also to 'cause the obscurity thereof, and to fear away many from the reading and diligent study of the same, so much that many of the well learned have much complained of the darkness and difficulty thereof, and have thought it a very hard thing, and in manner impossible to attain to the right and full understanding of this book, 
Many even of the well learned have thought that this book can not well be understanded without Algebra.  without the aid and help of some other knowledge and learning, and chief without the knowledge of that more secret and subtle part of Arithmetic, commonly called Algebra, which undoubtedly first well had and known, would give great light thereunto: yet certainly may this book very well be entered into, and fully understand without any strange help or succour, only by diligent observation of the order, and course of Euclides writings. So that he which diligently hath perused and fully understandeth the 9 books going before, and marketh also earnestly the principles and definitions of this ●enth book, he shall well perceive that Euclid is of himself a sufficient teacher and instructor, 
The nine former books & the principles of this ●ooke well understood, this book will not be hard to understand.  and needeth not the help of any other, and shall soon see that this tenth book is not of such hardness and obscurity, as it hath been hitherto thought. Yea, I doubt not, but that by the travel and industry taken in this translation, and by additions and emendations gotten of others, there shall appear in it no hardness at all, but shall be as easy as the rest of his books are.  

Definitions.  


The f●rst definition.  1 Magnitudes commensurable are suchwhich one and the self same measure doth measure.  
First he showeth what magnitudes are commensurable one to an other. To the better and more clear understanding of this definition, note that that measure whereby any magnitude is measured, is less than the magnitude which it measureth, or at lest equal unto it. For the greater can by no means measure the less. Farther it behoveth, that that measure if it be equal to that which is measured, taken once make the magnitude which is measured: if it be less, than oftentimes taken and repeated, it must precisely tender and make the magnitude which it measureth. Which thing in numbers is easily seen, for that (as was before said) all numbers are commensurable one to an other. And although Euclid in this definition comprehendeth purposedly, only magnitudes which are continual quantities, as are lines, superficieces, and bodies, yet undoubtedly the explication of this and such like places, is aptly to be sought of numbers as well rational as irrational. For that all quantities commensurable have that proportion the one to the other, which number hath to numbers. In numbers therefore, 9 and 12 are commensurable, because there is one common measure which measureth them both, namely, the number 3 First it measureth 12, for it is less than 12. and being taken certain times, namely, 4 times, it maketh exactly 12: 3 times 4 is 12, it also measureth 9, for it is less than 9 and also taken certain times, namely, 3 times, it maketh precisely 9: 3 times 3 is 9 Likewise is it in magnitudes, if one magnitude measure two other magnitudes, those two magnitudes so measured, are said to be commensurable. As for example, if the line C being doubled, do make the line B, and the same line C tripled, do make the line A, then are the two lines A and B, lines or magnitudes commensurable. For that one measure, namely, the line C measureth them both. First, the line C is less than the line A, and alsolesse than the line B, also the line C taken or repeated certain times, namely, 3 times, maketh precisely the line A, and the same line C taken also certain times, namely, two times, maketh precisely the line B. So that the line C is a common measure to them both, and doth measure them both. And therefore are the two lines A and B lines commensurable. And so imagine ye of magnitudes of other kinds, as of superficial figures, and also of bodies.   

2 Incommensurable magnitudes are such, which no one common measure doth measure. 
The second definition.   
This definition needeth no explanation at all, 
Contraries made manifest by the comparing of the one to the other.  it is easily understanded by the definition going before of lines commensurable. For contraries are made manifest by comparing of the one to the other: as if the line C, or any other line oftentimes iterated, do not tender precisely the line A, nor the line B, then are the lines A and B incommensurable. Also if the line C, or any other line certain times repeated, do exactly tender the line A, and do not measure the line B: or if it measure the line B, and measureth not also the line A, the lines A and B, are yet lines incommensurable: & so of other magnitudes as of superficieces, and bodies.   

3 Right lines commensurable in power are such, whose squares one and the self same superficies, area, or plat doth measure. 
The third definition.   
To the declaration of this definition we must first call to mind what is understanded & meant by the power of a line: which as we have before in the former books noted is nothing else but the square thereof, or any other plain figure equal to the square thereof. And so great power & ability ●s a line said to have, as is the quantity of the square, which it is able to describe, or a figure superficial equal to the square thereof. 
What the power of a line is.  〈◊〉 This i● also to be noted that of lines, some are commensurable in length, the one to the other, and some are commensurable the one to the other in power. Of lines commensurable in length the one to the other, was given an example in the declaration of the first definition, namely, the lines A and B, which were commensurable in length, one and the self measure, namely, the line C measured the length of either of them. Of the other kind is given this definition here set: for the opening of which take this example. Let there be a certain line, namely, the line BC, and let the square of that line be the square BCDE. Suppose also an other line, namely, the line FH, & let the square thereof be the square FHIK, and let a certain superficies, namely, the superficies A, measure the square BCDE, taken 16. times: which is the number of the little areas, squares, plaits, or superficieces contained and described within the said squares each of which is equal to the superficie: A. Again let the same superficies A measure the square FHIK 9 times taken, according to the number of the field●s or superficieces contained and described in the same. You see then that one and the self same superficies, namely, the superficies A, is a common measure to both these squares, and by certain repetitions thereof, measureth them both. Wherefore the two lines BC and FH, which are the sides or lines producing these squares, and whose powers these squares are, are by this definition lines commensurable in power.   

4 Lines incommensurable are such, whose squares no one plat or superficies doth measure. 
The fourth definition.   
This definition is easy to be understanded by that which was said in the definition last set before this, and needeth no farther declaration. And thereof take this example. If neither the superficies A, nor any other superficies do measure the two squares B CDE, and FHIK: or if it measure the one, ●●rely BCDE, and not the other FHIK, or if it measure the square FHIK, and not the square BCDE: the two lines BC and FH, are in power incommensurable, and therefore also incommensurable in length. For whatsoever lines are incommensurable in power, the same are also incommensurable in length as shall afterward in the 9 proposition of this book be proved. And therefore such lines are here defined to be absolutely incommensurable. These things thus standing it may easily appear, that if a line be assigned and laid before us, there may be innumerable other lines commensurable unto it, and other incommensurable unto it: of commensurable lines some are commensurable in length and power, and some in power only.   

5 And that right line so set forth is called a rational line.  
Thus may ye see, how to the supposed line first set may be compared infinite lines, 
Unto the supposed line first set may be compared infinite lines.  some commensurable both in length & power, and some commensurable in power only, and incommensurable in length, and some incommensurable both in power & in length. And this first line so set, whereunto, and to whose squares the other lines and their squares are compared, is called a rational line, commonly of the most part of writers. 
Why some mislike that the line first set should be called a rational line.  But some there are which mislike that it should be called a rational line, & that not without just cause. In the Greek copy it is called 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉, rete, which signifieth a thing that may be spoken, & expressed by word, a thing certain, granted and appointed. Wherefore Flussates, a man which bestowed great travel and diligence in restoring of these elements of Euclid, leaving this word rational, calleth this line supposed and first set, a line certain, because the parts thereof into which it is divided are certain, 
Flussates calleth this line, a line certain.  and known, and may be expressed by voice, and also be counted by number: other lines, being to this line incommensurable, whose parts are not distinctly known, but are uncertain, nor can be expressed by name nor assigned by number, which are of other men called irrational, he calleth uncertain and furred lines. Petrus Montaureus although he doth not very well like of the name, yet he altereth it not, but useth it in all his book. Likewise will we do here, for that the word hath been and is so universally received. And therefore will we use the same name, and call it a rational line. For it is not so great a matter what names we give to things, so that we fully understand the things which the names signify.  
This rational line thus here defined, is the ground and foundation of all the propositions almost of this whole tenth book. 
This rational line the ground in a manner of all the propositions in this tenth book.  And chief from the tenth proposition forwards. So that unless ye first place this rational line, and have a special and continual regard unto it before ye begin any demonstration, ye shall not easily understand it. For it is as it were the touch and trial of all other lines, by which it is known whether any of them be rational or not. 
Note.  And this may be called the first rational line, 
The line Rational of purpose.  the line rational of purpose or a rational line set in the first place, and so made distinct and severed from other rational lines, of which shall be spoken afterward. And this must ye well commit to memory.   


The sixth definition.  6 Lines which are commensurable to this line, whether in length and power, or in power only, are also called rational.  
This definition needeth no declaration at all, but is easily perceived, if the first definition be remembered, which showeth what magnitudes are commensurable, and the third, which showeth what lines are commensurable in power. Here not●, how aptly & naturally, Euclid in this place useth these words commensurable either in length and power, or in power only. Because that all lines which are commensurable in length, are also commensurable in pours when he speaketh of lines commensurable in length, he ever addeth and in power, but when he speaketh of lines commensurable in power, he addeth this word Only, and addeth not this word in length, as he in the other added this word in power. For not all lines which are commensurable in power, are strait way commensurable also in length. Of this definition, take this example. Let the first line rational of purpose, which is supposed and laid forth, whose parts are certain & known, and may be expressed, named, and numbered be AB, the quadrate whereof let be ABCD: then suppose again an other line, namely, the line EF, which let be commensurable both in length and in power to the first rational line, that is, (as before was taught) let one line measure the length of each line, and also l●t one superficies measure the two squares of the said two lines, as here in the example is supposed and also appeareth to the eye, then is the line E F also a rational line.  
Moreover if the line EF be commensurable in power only to the rational line AB first set and supposed, so that no one line do measure the two lines AB and EF: As in example y● see to be (for that the line EF, is made equal to the line AD, which is the diameter of the square ABCD, of which square the line AB is a side, it is certain that the ●ide of a square is incommensurable in length to the diameter of the same square: if there be yet found any one superficies, which measureth the two squares ABCD, and EFGH (as here doth the triangle ABD, or the triangle ACD noted in the square ABCD, or any of the four triangles noted in the square EFGH, as appeareth somewhat more manifestly in the second example, in the declaration of the last definition going before) the line EF is also a rational line. Note that these lines which here are called rational lines, are not rational lines of purpose, or by supposition, as was the first rational line, but are rational only by reason of relation and comparison which they have unto it, because they are commensurable unto it either in length and power, or in power only. Farther here is to be noted, that these words length, and power, and power only, are joined only with these worde● commensurable or incommensurable, and are never joined with these words rational or irrational. So that no lines can be called rational in length, or in power, nor like wise can they be called irrational in length, or in power. Wherein undoubtedly Campanus was deceived, 
Camp●nus ●ath caused much obscurity in this tenth book.  who using those words & speeches indifferently, caused & brought in great obscurity to the propositions and demonstrations of this book, which he shall easily see which marketh with diligence the demonstrations of Campanus in this book.   

7 Lines which are incommensurable to the rational line, are called irrational. 
The seventh definition.   
By lines incommensurable to the rational line supposed in this place, he understandeth such as be incommensurable unto it both in length and in power. For there are no lines incommensurable in power only: for it cannot be that any lines should so be incommensurable in power only, that they be not also incommensurable in length. What so ever lines be incommensurable in power, the same be also incommensurable in length. Neither can Euclid here in this place mean lines incommensurable in length only, for in the definition before, he called them rational lines, neither may they be placed amongst irrational lines. Wherefore it remaineth that in this diffintion he speaketh only of those lines which are incommensurable to the rational line first given and supposed, both in length, and in power. Which by all means are incommensurable to the rational line, & therefore most aptly are they called irrational lines. This definition is easy to be understanded by that which hath been said before. Yet for the more plainness see this example. Let the ●●rst rational line supposed, be the line AB, whose square or quadrate, let be ABCD. And let there be given an other line EF which l●t be to the rational line incommensurable in length and power, so that let no one line measure the length of the two lines, AB and EF: and let the square of the line EF be EFGH. Now if also there be no one superficies which measureth the two squares ABCD, and EFGH, as is supposed to be in this example, them is the line EF an irrational line, which word irrational (As before did this word rational) misliketh many learned in this knowledge of Geometry. Flussates, 
Flussates in steed of this word irrational useth this word uncertain.  as he left the word rational, and in stead thereof used this word certain, so here he leaveth the word irrational, and useth in place thereof this word uncertain, and ever nameth these lines uncertain lines. Petrus Montaureus also misliking the word irrational, would rather have them to be called furred lines, yet because this word irrational hath ever by custom and long use, so generally been receive he useth continually the same. In Greek such lines are called 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉, alogois, which signifieth nameless, unspeakable, uncertain, in determinate, 
Why they are called irrational lines.  and with out proportion: not that these irrational lines have no proportion at all, either to the first rational line, or between themselves: but are so named, for that their proportions to the rational line cannot be expressed in number. That is undoubtedly very untrue, which many writ, that their proportions are unknown both to us and to nature. Is it not think you a thing very absurd to say that there is any thing in nature, and produced by nature, to be hid from nature, and not to be known of nature? it can not be said that their proportions are utterly hid and unknown to us (much less unto nature) although we cannot give them their names, and distinctly express them by numbers: otherwise should Euclid have taken all this travel and wonderful diligence bestowed in this booke● in vain and to no vse● in which he doth nothing ell● but teach the proprieties and passions of these irrational lines● and showeth the proportions which they have the one to the other.  
Here is also to be noted, which thing also Tartalea hath before diligently noted● that Campanus and many other writers of Geometry● over much ●●●ed and were deceived in that they wrote and taught, that all these lines whose squares were not signified and might be expressed by a square number (although they might by any other number: as by 11. 12. 14. and such others not square numbers) are irrational lines. Which is manifestly repugnant to the grounds and principles of Euclid, who will, that all lines which are commensurable to the rational line, whether it be in length and power, or in power only, should be rational. Undoubtedly this hath been one of the chiefest and greatest causes of the wonderful confusion and darkness of this book, 
The cause of the obscurity and confusedness in this book.  which so hath tossed, and turmoiled the wits of all both writers and readers, masters and scholars, and so overwhelmed them, that they could not with out infinite travel and sweat, attain to the truth and perfect understanding thereof.   


The eighth definition.  8 The square which is described of the rational right line supposed, is rational.  
Until this definition hath Euclid set forth the nature and propriety of the first kind of magnitude, namely, of lines how they are rational or irrational, now he begins to ●hew how the second kind of magnitudes, namely superficies are one to the other rational or irrational. This definition is very plain. Suppose the line AB, to be the rational line, having his parts and divisions certainly known: the square of which line let be the square ABCD. Now because it is the square of the rational line AB, it is also called rational: and as the line AB, is the first rational line, unto which other lines compared are counted rational, or irrational so is the quadrat or square thereof, the ●irst rational superficies unto which all other squares or figures compared, are counted and named rational or irrational.   

9 Such which are commensurable unto it, are rational.  

The ninth definition.  In this definition, where it is said, such as are commensurable to the square of the rational line, are not understand only other squares or quadrates, but all other kinds of rectiline figures plain plaits & superficieses. What so ever so that if any such figure be commensurable unto that rational square● it is also rational. As suppose that the square of the rational line, which is also rational, be ABCD: suppose 〈◊〉 so some other square as the square EFGH, to be commensurable to the same: them is the square EFGH also rational. So also if the rectiline figure KLMN, which is a figure on the one side longer, be commensurable unto the said square as is supposed in this example● it is also a rational superficies and so of all other superficieses.   

10 Such which are incommensurable unto it, are irrational.  

The tenth definition.  Where it is said in this definition such which are incommensurable, it is generally to be taken: as was this word commensurable in the definition before. For all superficieses, whether they be squares or figures on the one side longer, or otherwise what manner of right lined figure so ever it be, if they be incommensurable unto the rational square supposed, them are they irrational. As let th● square ABCD be the square of the supposed rational line which square therefore is also rational: suppose also also an other square, namely the square E, suppose also any other figure, as for example sake a figure of one side longer, which let be F: Now if the square E and the figure F, be both incommensurable to the rational square ABCD, then is 〈◊〉 of these figures E & F irrational. And so of other.   

11 And these lines whose poweres they are, are irrational. If they be squares, then are their sides irrational. If they be not squares, 
The eleventh definition.  but some other rectiline figures, then shall the lines, whose squares are equal to these rectiline figures, be irrational.  
Suppose that the rational square be ABCD. Suppose also an other square, namely the square E, which let be incommensurable to the rational square, & therefore is it irrational: and let the side or line which produceth this square be the line FG: then shall the line FG by this definition be an irrational line: because it is the side of an irrational square. Let also the figure H being a figure on the one side longer (which may be any other rectiline figure rectangled or not rectangled, triangle, pentagon, trapezite, or what so ever else) be incommensurable to the rational square ABCD, then because the figure H is not a square, it hath no side or root to produce it yet may there be a square made equal unto it: for that all such figures may be reduced into triangles, and so into squares, by the 14. of the second. Suppose that the square Q be equal to the irrational figure H. The side of which figure Q let be the line KL: then shall the line KL be also an irrational line, because the power or square thereof, is equal to the irrational figure H: and thus conceive of others the like.  
These irrational lines and figures are the chiefest matter and subject, which is entreated of in all this tenth book: the knowledge, of which is deep, and secret, and pertaineth to the highest and most worthy part of Geometry, wherein standeth the pith and marry of the hole science: the knowledge hereof bringeth light to all the books following, with out which they are hard and cannot be at all understood. And for the more plainness, ye shall note, that of irrational lines there be di●ers sorts and kinds. But they, whose names are set in a table here following, and are in number 13. are the chief, and in this tenth book sufficiently for Euclides principal purpose, discoursed on.  

A medial line.  
A binomial line.  
A first bimedial line.  
A second bimedial line.  
A greater line.  
A line containing in power a rational superficies and a medial superficies.  
A line containing in power two medial superficieces.  
A residual line.  
A first medial residual line.  
A second medial residual line.  
A less line.  
A line making with a rational superficies the whole superficies medial.  
A line making with a medial superficies the whole superficies medial.  
Of all which kinds the definitions together with there declarations shallbe set here after in their due places.     

¶ The 1. Theorem. The 1. Proposition. Two unequal magnitudes being given, if from the greater be taken away more than the half, and from the residue be again taken away more than the half, and so be done still continually, there shall at length be left a certain magnitude lesser than the less of the magnitudes first given.  
SVppose that there be two unequal magnitudes AB, and C, of which let AB be the greater. Then I say, that if from AB, be taken away more than the half, 
Construction.  and from the residue be taken again more than the half, and so still continually, there shall at the length be left a certain magnitude, lesser than the less magnitude given, namely, then C. For forasmuch as C is the less magnitude, therefore C may be so multiplied, that at the length it will be greater than the magnitude AB (by the 5. definition of the fift book). 
Demonstration.  Let it be so multiplied, and let the multiplex of C greater than AB, be DE. And divide DE into the parts equal unto C, which let be DF, FG, and GOE And from the magnitudes AB take away more than the half, which let be BH: and again from AH, take away more than the half, which let be HK. And so do continually until the divisions which are in the magnitude AB, be equal in multitude unto the divisions which are in the magnitude DE. So that let the divisions AK, KH, and HB, be equal in multitude unto the divisions DF, FG, and GOE And forasmuch as the magnitude DE is greater than the magnitude AB, and from DE is taken away less than the half, that is, EG (which detraction or taking away is understand to be done by the former division of the magnitude DE into the parts equal unto C: for as a magnitude is by multiplication increased, so is it by division diminished) and from AB is taken away more than the half, that is, BH: therefore the residue GD is greater than the residue HA (which thing is most true and most easy to conceive, if we remember this principle, that the residue of a greater magnitude, after the taking away of the half or less than the half, is ever greater than the residue of a less magnitude, after the taking away of more than the half). And forasmuch as the magnitude GD is greater than the magnitude HA, and from GD is taken away the half, that is, GF: and from AH is taken away more than the half, that is, HK: therefore the residue DF is greater than the residue AK (by the foresaid principle). But the magnitude DF is equal unto the magnitude C (by supposition). Wherefore also the magnitude C is greater than the magnitude AK. Wherefore the magnitude AK is less than the magnitude C. Wherefore of the magnitude AB is left a magnitude AK less than the less magnitude given, namely, then C which was required to be proved. In like sort also may it be proved if the halves be taken away.  

A Corollary.  

A Corollary.  Of this Proposition it followeth, that any magnitude being given how little soever it be, there may be given a magnitude less than it: so that it is impossible that any magnitude should be given then which can be given no loss.  

¶ An other demonstration of the same.  
Suppose that the two unequal magnitudes given be AB and C. And let C be the less. And forasmuch as C is the less, therefore C may so be multiplied, that it shall at the length be greater than A●. 
Construction.  Let it be so multiplied, and let the multiplex of C exceeding AB be the magnitude FM. And divide FM into his parts equal unto C, that is, into the magnitudes MH, HG, and GF. And from AB take away more than the half, which let be the magnitude BE: and likewise from EA take away again more than the half, namely, the magnitude ED. And thus do continually until the divisions which are in the magnitude FM, be equal in multitude to the divisions which are in the magnitude AB: and let those divisions be the magnitudes BE, ED, and DA. And how multiplex the magnitude FM is to the magnitude C, so multiplex let the magnitude KX be to the magnitude DA. 
Demonstration.  And divide the magnitude KX into the magnitudes equal to the magnitude DA: which let be KL, LN, and NX. Now than the divisions which are in the magnitude KX, are equal unto the divisions which are in the magnitude MF. And forasmuch as BE is greater than the half of AB, therefore BE is greater than the residue EA Wherefore BE is much more greater then DA. But DA is equal unto XN. Wherefore BE is greater than XN. Again forasmuch as DE is greater than the half of EA, therefore DE is greater than the residue DA: but DA is equal unto LN: wherefore DE is greater than LN. Wherefore the whole magnitude DB is greater than the whole magnitude XL. But DA is equal unto LK. Wherefore the whole magnitude AB is greater than the whole magnitude KX. And the magnitude MF is greater than the magnitude BA: wherefore MF is much greater than KX. And forasmuch as those magnitudes XN, NL, & LK, are equal the one to the other, & likewise these magnitudes MH, HG, and GF, are equal the one to the other, & the multitude of those magnitudes which are in MF, is equal to the multitude of those magnitudes which are in KX: therefore as KL is to FG, so is LN to GH, and NX to HM. Wherefore (by the 12. of the fift) as one of the antecedentes, namely, KL, is to one of the consequences, namely, to FG, so are all the antecedentes, namely, the whole KX to all the consequentes, namely, to the whole FM. But FM is greater than KX. Wherefore FG is greater than LK. But FG is equal unto C: and KL unto DA (by supposition). Wherefore the magnitude C is greater than the magnitude AD: which was required to be proved.     

¶ The 2. Theorem. The 2. Proposition. Two unequal magnitudes being given, if the less be continually taken from the greater, & that which remaineth measureth at no time the magnitude going before: then are the magnitudes given incommensurable.  

This proposition teacheth that incontinuall quantity which the first of the seventh taught in discrete quantity.  SVppose that there be two unequal magnitudes AB, and CD, and let AB be the less: and taking away continually by a certain alternate detraction the less from the greater, let not the residue measure the magnitude going before. Then I say, that those two magnitudes AB and CD, are incommensurable. For if they be commensurable, then (by the first definition of the tenth) some one magnitude shall measure them both: Let there be such a magnitude, if it be possible, and let the same be E. And let AB measuring DF, 
Construction.  leave a less than itself, namely, CF (that is, from the greater magnitude CD, take away a certain part as DF, which let be equal to AB, or if it be not equal unto it, yet let it be such, that that less magnitude AB being more than once repeated may make the magnitude DF● For this is the meaning of this let AB measuring DF. etc. And this detraction made, of the less I say out of the greater, let there be left of the greater a certain portion CF, less than the magnitude AB. And this is the meaning of that which in the Theorem was said, And that which remaineth measureth at no time the magnitude going before). Likewise let CF measuring BG leave a less than itself, namely, AG: and do this continually as often as need requireth, until there be found such a magnitude that is less than E: which must needs at the length happen (by the Proposition going before) ● Let there be found such a magnitude less than E, which let be AG. And forasmuch as the magnitude E measureth the magnitude AB, but AB measureth DF: therefore E measureth the magnitude DF (by this common sentence, If a magnitude measure an other magnitude it shall also measure every magnitude whom it measureth). 
Demonstration leading to an absurdity.  And it measureth also the whole CD (for it is supposed to be a common measure to the magnitudes AB and CD). Wherefore also it measureth the residue CF (by this common sentence, If a magnitude measure an whole and a part taken away, it shall also measure the residue). And forasmuch as E measureth CF, but CF measureth BG, therefore E also measureth BG (by the first foresaid common sentence): and it measureth the whole AB. Wherefore it shall also measure the residue AG (by the other foresaid common sentence) namely, the greater magnitude shall measure the less: which is impossible. Wherefore no magnitude measureth these magnitudes AB and CD. Wherefore the magnitudes AB and CD are incommensurable. If therefore two unequal magnitudes being given, the less be continually taken from the greater, and that which remayneth● measureth at no time the magnitude going before, then are the magnitudes given incommensurable: which was required to be demonstrated.  

¶ A Corollary added by Montaureus.  
By this proposition it is manifest, that if the two unequal magnitudes given be not incommensurable, but commensurable, than the less being continually subtrahed out of the greater, the residue shall of necessity measure that which went before.    

¶ The 1. Problem. The 3. Proposition. Two magnitudes commensurable being given, to find out, their greatest common measure.  
SVppose that the two commensurable magnitudes given be AB & CD, of which let AB be the less. It is required to find out the greatest common measure of the magnitudes AB and CD. 
Two cases in this proposition.  Now AB either measureth CD, or not. If it measure it, and seeing it also measureth itself, 
The first case.  wherefore AB is a common measure unto the magnitudes AB and CD. 
This proposition teacheth, that in continual quantity, which the 2. of the s●●ith taught in numbers.  And it is manifest that it is the greatest common measure to them. For no magnitude greater than AB can measure AB. But now suppose that AB do not measure CD. And taking continually the less from the greater that which remaineth shall at length measure that which goeth before (by the corollary before added), for that AB and CD are commensurable. Now then let AB measuring ED which is a part of the magnitude CD, leave a magnitude less than itself, namely, EC. 
The second case.  And let EC measuring the magnitude FB, which is a part of the magnitude AB, leave a less than itself, namely, FA, and let FA precisely measure the magnitude CE (And this is the meaning of this, That which remaineth shall at the length measure that which goeth before, when there is nothing left after the measuring made) And forasmuch as AF measureth the magnitude CE, but CE measureth FB: wherefore AF also measureth FB, & it measureth itself: wherefore AF measureth also the whole AB. But AB measureth DE, wherefore AF also measureth DE. And AF also measureth CE, wherefore it measureth the whole magnitude CD. Wherefore the magnitude AF measureth both the magnitudes AB and CD. Wherefore AF is a common measure unto AB and CD. I say also that it is the greatest common measure unto them. For if not, then is there some magnitude greater than the magnitude AF which measureth both the magnitudes AB and CD. Let there be such a one if it be possible, and let the same be the magnitude G. 
Demonstration leading to an absurdity.  And forasmuch as G measureth AB, and AB measureth ED: therefore G also measureth ED, and by supposition, it measureth the whole CD, wherefore G measureth also the residue CE. But CE measureth FB, wherefore G also measureth FB. And by supposition it measureth the whole AB. Wherefore it measureth the residue AF, namely, the greater magnitude the less, which is impossible. Wherefore no magnitude greater than AF, measureth these magnitudes AB and CD. Wherefore AF is the greatest common measure unto the magnitudes AB and CD. Wherefore unto two commensurable magnitudes given, namely, AB and CD is found out their greatest common measure, namely, the magnitude AF● which was required to be done.  

¶ Corollary.  
Hereby it is manifest that if a magnitude measure two magnitudes, 
A Corollary.  it shall also measure their greatest common measure. For if it measure the wholes and the parts taken away, it shall also measure the residues, of which one is the greatest common measure, as we may see by the latter end of the former demonstration.  

Monta●reus reduceth this Problem into a Theorem after this manner.  
Two unequal and commensurable magnitudes being given, if the less do measure the greater, 
This Problem reduced to a Theorem.  than is it the greatest common measure to them both. But if not, than the less being continually by a mutual detraction (as before hath been taught) taken out of the greater, whensoever the residue precisely measureth that which w●nt before leaving nothing, the said residue shall be the greatest common measure to both the magnitudes given.     

¶ The 2. Problem. The 4. Proposition. Three magnitudes commensurable being given, to find out their greatest common measure.  

This proposition teacheth, that in continual quantity, which the 3. of the second taught in numbers.  SVppose that the three commensurable magnitudes given be A, B, C. It is required of these three magnitudes to find out the greatest common measure. Take (by the former proposition) the greatest common measure of the two magnitudes A, B and let the same be D. Now then this magnitude D either measureth the third magnitude, which is C, or not. First let it measure C. And forasmuch as D measureth C, and it measureth also the magnitudes A, B● therefore D measureth the three magnitudes A, B, C. Wherefore D is a common measure unto the magnitudes A, B, C. And it is manifest, that it is the greatest common measure. 
Construction.  For no magnitude greater than D can measure the magnitudes A, B, C. 
Two cases in this Proposition.  For if it be possible let the magnitude E being greater than the magnitude D, measure the magnitude A, B, C. 
The first case.  And forasmuch as E measureth the magnitudes A, B, C, it measureth the two first magnitudes AB. 
Demonstration leading to an absurdity.  Wherefore it shall also (by the Corollary of the former proposition) measure the greatest common measure of the magnitudes A, B which is D, namely, the greater shall measure the less, which is impossible.  

The second case.  But now let D not measure the magnitude C. First I say that the magnitudes C, D, are commensurable. For forasmuch as the magnitudes A, B, C are commensurable, some magnitude shall measure them, which shall also measure the magnitudes AB taken a part. 
A Le●ma necessary to be pr●●●d be●o●e 〈◊〉 ●all to the demonstration.  Wherefore it shall also (by the corollary going before) measure the greatest common" measure of A, B, that is, D: and (by supposition) it measureth the magnitude C. Wherefore that said magnitude shall measure the magnitudes C and D. 
Construction.  Wherefore the magnitudes C, D are commensurable. Take by the third of this tenth, their greatest common measure, & let the same be the magnitude E. And forasmuch as E measureth D, but D measureth the magnitudes A, B. Wherefore E also measureth the magnitudes A, B, and it also measureth the magnitude C. Wherefore E is a common measure to ●he three magnitudes A, B, C. 
Demonstration leading to an absurdity.  I say also that it is the greatest common measure. For if it be possible, let there be a magnitude, namely, F, greater than the magnitude E. And let F measure the three magnitudes A, B, C. A●d forasmuch as F measureth the magnitudes A, B, C, it also measureth the two first magnitudes A, B. Wherefore (by the corollary going before) it shall also measure the greatest common measure of the magnitudes A, B. But the greatest common measure of the magnitudes A, B, is D. Wherefore F measureth D, and it also measureth C. Wherefore F measureth the magnitudes CD. Wherefore F shall also measure the greatest common measure of the magnitudes C, D. But the greatest common measure of the magnitudes C, D, is E. Wherefore F measureth E, namely, the greater, the less, which is impossible. Wherefore no magnitude greater than E measureth the magnitudes A, B, C. Wherefore E is the greatest common measure of the magnitudes ABC. If D do not measure the magnitude C. But if D do measure C, then is D the greatest common measure. Wherefore three magnitudes commensurable being given, there is found their greatest common measure: which was required to be done.  

¶ Corollary.  
Hereby it is manifest, that if a magnitude measure three magnitudes, 
A Corollary.  it shall also measure their greatest common measure. In like sort also in magnitudes commensurable how many soever being given, may be found out their greatest common measure, and the corollary will ever be true.  

This Problem also Montaureus reduceth into a Theorem after this manner. 
This Problem reduced to a Theorem.   
Three magnitude● being commensurable, if the greatest common measure to two of them, do measure the third it shall be the greatest common measure to all the three magnitudes given. But if it do not measure it, the greatest common measure of the third and of the greatest common measure of the two first, is the greatest common measure of all the three magnitudes.     

¶ The 3. Theorem. The 5. Proposition. Magnitudes commensurable, have such proportion the one to the other, as number hath to number.  
SVppose that A and B be magnitudes commensurable. Then I say that A hath unto B such proportion as number hath to number. For forasmuch as A and B are commensurable, therefore some magnitude measureth them, let there be a magnitude that measureth them, and let the same be C. 
Construction.  And how often C measureth A, so many unities let there be in the number D. And how often C measureth B, so many unities let there be in the number E, and let F be unity. 
Demonstration.  And forasmuch as the magnitude C measureth the magnitude A by those unities which are in the number D, and unity F measureth the number D by those unities which are in the number D: therefore how many times unity measureth the number D, so many times doth the magnitude C measure the magnitude A: wherefore as the magnitude C is to the magnitude A, so is unity F to the number D. Wherefore contrary wise (by the corollary of the fourth proposition of the fift book) as the magnitude A is to the magnitude C, so is the number D to unity F. Again forasmuch as the magnitude C measureth the magnitude B by those unities which are in the number E, and unity F measureth the number E by those unities which are in the number E: therefore how many times unity F measureth the number E, so many times doth the magnitude C measure the magnitude B. Wherefore as the magnitude C is to the magnitude B, so is unity F to the number E, and it is proved that as the magnitude A is to the magnitude C, so is the number D to unity F. Wherefore of equality (by the 22. of the fift) as the magnitude A is to the magnitude B, so is the number D to the number E. Wherefore the commensurable magnitudes A and B have that proportion the one to the other that the number D hath to the number E. Magnitudes therefore commensurable, have such proportion the one to the other, as number hath to number: which was required to be proved.  
Magnitudes are said to have such proportion the one to the other, as number hath to number, 
How magnitudes are said to be in proportion the on● to the other, as number is to number.  when as what soever proportion is between those magnitudes, the same is found between some certain numbers: as if a magnitude be unto a magnitude either equal, as the number 2. is to the number 2, or double, as the number 4. to the number 2, or triple, as 6. to 2, or in any other multiplex proportion. And so also touching the other kinds o● proportion, either superparticular, or superpartient.   

¶ The 4. Theorem. The 6. Proposition. I● two magnitudes have such proportion the one to the other, as number hath to number: those magnitudes are commensurable.  
SVppose that these two magnitudes A and B have that proportion the one to the other, 
This proposition is the converse of former.  that the number D hath to the number E. Then I say, that the magnitudes A, B, are commensurable. How many unities there are in the number D, into so many equal parts divide the magnitude A (by the 9 of the sixth) and let the magnitude C be equal to one o● the parts thereof. And how many unities there are in the number E, o● so many magnitudes equal unto the magnitude C let the magnitude F be composed. And let G be unity. 
Construction.  Now forasmuch as how many unities there are in the number D, so many magnitudes also are there in the magnitude A equal unto the magnitude C: 
Demonstration.  therefore what part unite G is of the number D, the same part is the magnitude C of the magnitude A. Wherefore as the magnitude C is to the magnitude A, so is unity G to the number D. But unity G measureth the number D. Wherefore the magnitude C also measureth the magnitude A. And for that as the magnitude C is to the magnitude A, so is unity G to the number D: therefore contrariwise (by the Corollary of the fourth of the fift) as the magnitude A is to the magnitude C, so is the number D to unity G. Again forasmuch as how many unities there are in the number E, so many magnitudes also are there in the magnitude F equal unto the magnitude C: therefore as the magnitude C is to the magnitude F, so is unity G to the number E. And it is proved that as the magnitude A is to the magnitude C, so is the number D to unity G. Wherefore of equality (by the 22. of the fift) as the magnitude A is to the magnitude F, so is the number D to the number E. But as the number D is to the number E, so is the magnitude A to the magnitude B. Wherefore (by the 11. of the fift) as the magnitude A is to the magnitude B, so is the same magnitude A to the magnitude F: wherefore A hath unto either of these magnitudes B and F one and the same proportion. Wherefore (by the 9 of the fift) the magnitude B is equal unto the magnitude F. But the magnitude C measureth the magnitude F: wherefore it also measureth the magnitude B: and it likewise measureth the magnitude A. Wherefore the magnitude C measureth the magnitudes A and B. Wherefore the magnitudes A & B are commensurable. If therefore two magnitudes have such proportion the one to the other, as number hath to number, those magnitudes are commensurable: which was required to be proved.  

Corollary.  
Hereby it is manifest, that if there be two numbers, 
A Corollary.  as D and E, and a right line, as A, it is possible to give an other line, unto which the line A shall have the same proportion, that the number D hath to the number E. For divide the line A into so many equal parts as there are unities in the number D (by the 9 of the sixth). And take an other line, as F, which let be composed of so many parts equal to the parts of the line A, as there be unities in the number E. Wherefore the line A shall be to the line F, as the number D is to the number E. And by this means you may unto any line given give an other line commensurable in length. For if two lines be in proportion the one to the other, as number is to number, they shall also be commensurable in length, by this 6. Theorem.   

¶ An Assumpt.  
Two numbers being given, and also a right line: as the one number is to the other, so to make the square of the line given to be to the square of an other line.  
Suppose that the numbers given be D and E: and let the right line given be A. It is required, as the number D is to the number E, so to make the square of the line A to be to the square of an other line. As the number D is to the number E, so let the line A be to the line F (by the former Corollary). 
Construction.  And take between those two lines A and F the mean proportional (by the 13. of the sixth) which ●et be the line B: 
Demonstration.  Now for that as the number D is to the number E, so is the line A to the line F: and as the line A is to the line F, so is the square of the line A to the square of the line B (by the second Corollary of the 20. of the sixth). Wherefore as the number D is to the number E, so is the square of the line A to the square of the line B: which was required to be done.   

¶ An other demonstration of the 6. Proposition.  
Suppose that these two magnitudes given A and B, have that proportion the one to the other, that the number C hath to the number D. Then I say, that th●se magnitudes are commensurable. 
Construction.  How many unities there are in the number C, into so many equal parts let the magnitude A be divided, & let the magnitude E be equal unto one of those parts. 
Demonstration.  Wherefore as unity is to the number C, so is the magnitude E to the magnitude A. And as the number C is to the number D, so is the magnitude A to the magnitude B. Wherefore of equality (by the 22. of the fift) as unity is to the number D, so is the magnitude E to the magnitude B. But unity measureth the number D. Wherefore the magnitude E measureth the magnitude B. And it also measureth the magnitude A (●or that unity measureth the number C). Wherefore the magnitude E measureth either of these magnitudes A and B. Wherefore ●he magnitu●es A and B are commensurable, and the magnitude E is their common measure.    

¶ The 5. Theorem. The 7. Proposition. Magnitudes incommensurable, have not that proportion the one to the other, that number hath to number.  
SVppose that the magnitudes A and B be incommensurable. Then I say, that A hath not to B, 
Demonstration leading to an absurdity.  that proportion that number hath to number. For if A have unto B that proportion that number hath to number, then is A comensurable unto B (by the 6. of this tenth). But (by supposition) it is not. Wherefore A hath not unto B that proportion that number hath to number. Magnitudes incommensurable therefore have not that proportion the one to the other that number hath to number: which was required to be demonstrated.   

¶ The 6. Theorem. The 8. Proposition. If two magnitudes have not that proportion the one to the other that number hath to number, those magnitudes are incommensurable.  
SVppose that these two magnitudes A and B, have not that proportion the one to the other that number hath to number. Then I say, that A and B are magnitudes incommensurable. 
This is the 〈…〉 demonstration.  For if A and B be commensurable, then shall A have unto B, that proportion that number hath to number (by the 5. of this tenth). But (by supposition) it hath not that proportion that number hath to number. Wherefore A and B are incommensurable magnitudes. If therefore two magnitudes have not that proportion the one to the other that number hath to number, those magnitudes are incommensurable: which was required to be proved.   

¶ The 7. Theorem. The 9 Proposition. Squares described of right lines commensurable in length, have that proportion the one to the other, that a square number hath to a square number. And squares which have that proportion the one to the other that a square number hath to a square number, shall also have their sides commensurable in length. But squares described of right lines incommensurable in length, have not that proportion the one to the other, that a square number hath to a square number. And squares which have not that proportion the one to the other that a square number hath to a square number, have not their sides commensurable in length.  
SVppose that A and B be lines commensurable in length. 
The first part demonstrated.  Then I say that the square of the line A hath unto the square of the line B, that proportion that a square number hath to a square number. For forasmuch as the lines A and B are commensurable in length: therefore the line A hath unto the line B that proportion that number hath to number (by the 5. of this tenth). Let the line A have unto the line B that proportion, that the number C hath to the number D, Now for that as the line A is to the line B, so is the number C to the number D: but the square of the line A is unto the square of the line B in double proportion of that which the line A is unto the line B (for like rectiline figures (by the first corollary of the 20. of the sixth) are in double proportion of that which the sides of like proportion are) and likewise the square number produced of C is to the square number produced of D, in double proportion of that which the number C is to the number D (for by the 11. of the eight between two square numbers there is one mean proportional number & a square number is to a square number in double proportion of that which side is unto side). Wherefore as the square of the line A is to the square of the line B, so is the square number produced of the number C, to the square number produced of the number D.  

another demonstration to prove the same.  
Forasmuch as the lines A and B are commensurable, therefore (by the 5 of this tenth) A hath unto B the same proportion that number hath to number. 
another demonstration of the first part.  Let them have that proportion that the number C hath to the number D. And let the number C multiplying himself produce the number E, and multiplying the number D, let it produce the number F: and let the number D multiplying himself produce the number G. And forasmuch as the number C multiplying himself produced the number E, and multiplying the number D it produced the number F: therefore (by the 17. of the seventh) as the number C is to the number D, that is, the line A to the line B so is the number E to the number F. But as the line A is to the line B, so is the square of the line A to the parallelogram contained under the lines A and B (by the first of the sixth) Wherefore as the square of the line A is to that which is contained under the lines A and B, so is the number E to the number F. Again for as much as the number C multiplying the number D produced the number F, & the number D multiplying himself produced the number G: therefore (by the 17. of the seventh) as the number C is to the number D, that is, as the line A is to the line B, so is the number F to the number G. But as the line A is to the line B, so is parallellograme contained under the lines A and B to the square of the line B (by the first of the sixth). Wherefore as that which is contained under the lines A and B is to the square of the line B, so is the number F to the number G. But as the square of the line A is to that which is contained under the lines A and B, so is the number E to the number F. Wherefore of equality (by the 22. of the fift) as the square of the line A is to the square of the line B, so is the number E to the number G. But either of these numbers E and G is a square number. For E is produced of the number C multiplied into himself, and G is produced of the number D multiplied into himself. Wherefore the square of the line A hath unto the square of the line B that proportion that a square number hath to a square number: which was required to be demonstrated.   

another demonstration of the same first part after Montaureus.  
Suppose that there be two lines commensurable in length A and B. Then I say that the squares described of those lin●s shallbe in proportion the one to the other as a square number is to a square number. 
another demonstration o● the same first part after Montaureus.  For forasmuch as the lines A and B are commensurable in length, they shallbe in proportion the one to the other as number is to number (by the 5. of this book). Let A be to B in duple proportion, which is in such proportion as number is to number, namely, as 4. is to 2, and 6. to 3, and so of many other. And (by the s●cond o● the eight) take three lest numbers in continual proportion, and in duple proportion, & let the same be the numbers 4.2.1: wherefore by the corollary of the second of the eight, the numbers 4. and 1. shallbe square numbers. (For as 4. is a square number produced of 2. multiplied into himself, so is 1. also a square number, for it is produced of unity multiplied into himself.) I say moreover that those are the square numbers, whose proportion the squares of the lines A and B have the one to the other. For as the number 4. is to the number 2. so is the line A to the line B (●or either proportion is double by supposition): but as the line A is to the line B, so is the square of the line A to the parallelogram contained under the lines A and B (by the first of the sixth). Wherefore as the number 4. is to the number 2: so shall the square of the line A be to the parallelogram contained under the lines A and B. Likewise as the number 2. is to the number 1. so is the line A to the line B (For either proportion is duple by supposition): but as the line A is to the line B, so is the parallelogram contained under the lines A and B to the square of the line B (by the self same first of the sixth). Wherefore as the number 2. is to 1. so is the parallelogram contained under the lines A and B to the square of the line B: wherefore of equality (by the 22. of the fifth) as the square of the line A is to the square of the line B, so is the number 4. to 1. which are proved to be square numbers.  

Demonstration of the second part which is the converse of the former.  But now suppose that the square of the line A, be unto the square of the line B, as the square number produced of the number C is to the square number produced of the number D. Then I say that the lines A & B are commensurable in length. For for that as the square of the line A is to the square of the line B, so is the square number produced of the number C to the square number produced of the number D: but the proportion of the square of the line A, is unto the square of the line B double to that proportion which the line A hath unto the line B (by the corollary of the 20. of the sixth). And the proportion of the square number which is produced of the number C to the square number produced of the number D is (by the 11. of the eight) double to that proportion which the number C hath unto the number D. Wherefore as the line A is to the line B, so is the number G to the number D. Wherefore the line A hath unto the line B the same proportion that the number C hath to the number D. Wherefore (by the 6. of this book) the lines A and B are commensurable in length: which was required to be proved.   

another demonstration to prove the same.  
●Vppose again that the square of the line A have unto the square of the line B, 
another demonstration of the second part.  the same proportion, that the square number E hath to the square number G. Then I say that the lines A and B are commensurable in length. For suppose that the side of the square number E be the number C, & let the side of the square number G be the number D: and let the number C multiplying the number D produce the number F. Wherefore these numbers E, F, G, are in continual proportion, and in the same proportion that the number C is to the number D (by the 17. and 18. of the seventh). 
This Assumpt followeth as a Corollary of the 25: but (so as it might also be here in Method, placed) you shall ●inde it after the 53. of this book, absolutely demonstrated: for there it serveth to the 54. his demonstration.  And forasmuch as the mean proportional between the squares of the lines A and B is that which is contained under the lines A and B. (Which though it might briefly be proved: yet we take it as now). And likewise the mean proportional between the numbers E and G is the number F (by the 20. of the seventh): therefore as the square of the line A is to that which is contained under the lines A and B, so is the number E to the number F, and as that which is contained under the lines A and B is to the square of the line B, so is the number F to the number G. But as the square of the line A is to that which is contained under the lines A and B, so is the line A to the line B (by the first of the sixth). Wherefore the lines A and B are in the same proportion that the number E is to the number F, that is, that the number C is to the number D. Wherefore the lines A and B are commensurable in length (by the 6. of this tenth) which was required to be proved.  
But now suppose that the lines A and B be incommensurable in length. Then I say that the square of the line A hath not unto the square of the line B that proportion that a square number hath to a square number. 
Demonstration of the third part.  For if the square of the line A have unto the square of the line B, the same proportion that a square number hath to a square number, them shall the lines A and B be commensurable in length (by the second part of this proportion). But by supposition they are not. Wherefore the square of the line A, hath not unto the square of the line B that proportion that a square number hath to a square number: which was required to be proved. 
Demonstration of the fourth part which is the converse of the ●.   
Again suppose that the square of the line A have not unto the square of the line B, the same proportion that a square number hath to a square number. Then I say that the lines A and B are incommensurable in length. For if the lines A and B be commensurable in length, than the square of the line A should have unto the square of the line B, the same proportion that a square number hath to a square number, by the first part of this proposition, but by supposition it hath not, wherefore the lines A and B are not commensurable in length. 
Conclusion of the whole proposition.  Wherefore they are incomensurable in length. Wherefore squares made of right lines commensura- in length, have that proportion the one to the other, that a square number hath to a square number. And squares which have that proportion the one to the other, that a square number hath to a squa●e number, shall also have the sides commensurable in length. But squares described of right lines incommensurable in length, have not that proportion the one to the other that a square number hath to a square number. And squares which have not that proportion the one to the other, that a square number hath to a square number, have not also their sides commensurable in length: which was all that was required to be proved.   

¶ corollary.  
Hereby it is manifest, that right lines commensurable in length, are also ever commensurable in power. But right lines commensurable in power, are not always commensurable in length. 
A Corollary.  And right lines incommensurable in length are not always incommensurable in power. But right lines incommensurable in power, are ever also incommensurable in length.  
For forasmuch as squares made of right lines commensurable in length, have that proportion the one to the other, 
Pro●e of the first part of the Corollary.  that a square number hath to a square number (by the first part of this proposition), but magnitudes which have that proportion the one to the other, that number simply hath to number, are (by the sixth of the tenth) commensurable. Wherefore right lines commensurable in length, are commensurable not only in length, but also in power.  

Proof of the second part.  Again forasmuch as there are certain squares which have not that proportion the one to the other that a square number hath to a square number, but yet have that proportion the one to the other which number simply hath to number: their sides in deed are in power commensurable, for that they describe squares which have that proportion which number simply hath to number, which squares are therefore commensurable (by the 6. of this book): but the said sides are incommensurable in length by the latter part of this proposition. Wher●fore it is true that lines commensurable in power, are not strait way commensurable in length also. 
Proof of the third p●rt.   
And by the sel●e same reason is proved also that third part of the corollary, that lines incommensurable in length, are not always incommensurable in power. For they may be incommensurable in length, but yet commensurable in power. As in those squares which are in proportion the one to the other, as number is to number, but not as a square number is to a square number.  

Pro●e o● the fourth part.  But right lines incommensurable in power, are always also incommensurable in length. For i● they be commensurable in length, they shall also be commensurable in power by the first part of this Corollary. But they are supposed to be incommensurable in length, which is absurd. Wherefore right lines incommensurable in power, are ever incommensurable in lengthy  
For the better understanding of this proposition and the other following, I have here added certain annotations taken out of Montaureus. 
Certain annotations ●ut of Montau●●us.  And first as touching the signification o● words and terms herein used, wh●ch ar● such, that unless they be well marked and poised, the matter will be obscure and hard, and in a manner inexplicable.  
First, this ye must note, that lines to be commensurable in length, and lines to be in proportion the one to the other, as number is to number is all one. So that whatsoever lines are commensurable in length, are also in proportion the one to the other, as number is to number. And conversedly what so ever lines are in proportion the one to the other, as number is to number, are also commensurable in length, as it is manifest by the 5 and 6 of this book. Likewise lines to be incommensurable in length, and not to be in proportion the one to the other, as number is to number is all one, as it is manifest by the 7. and 8. of this book. Wherefore that which is said in this Theorem, aught to be understand of lines commensurable in length, and incommensurable in length.  
This moreover is to be noted, that it is not all one, numbers to be square numbers, and to be in proportion the one to the other, as a square number is to a square number. For although square numbers be in proportion the one to the other, as a square number is to a square number, yet are not all those numbers which are in proportion the one to the other, as a square number is to a square number, square numbers. For they may be like superficial numbers, and yet not square numbers, which yet are in proportion the one to the other, as a square number is to a square number. Although all square numbers are like superficial numbers. For between two square numbers there falleth one mean proportional number (by the 11. of the eight). But if between two numbers, there fall one mean proportional number, those two numbers are like superficial numbers (by the 20. of the eight). So also if two numbers be in proportion the one to the other, as a square number is to a square number, they shall be like superficial numbers by the first corollary added after the last proposition of the eight book.  
And now to know whether two superficial numbers given, be like superficial numbers or not, 
Rules to know whether two superficial numbers be like or no.  it is thus found out. First if between the two numbers given, there fall no mean proportional, then are not these two numbers like superficial numbers (by the 18. of the eight. But if there do fall between them a mean proportional, then are they like super●iciall numbers (by the 20. of the eight) Moreover two like superficial numbers multiplied the one into the other, do produce a square number (by the firs● of the ninth). Wherefore if they do not produce a square number, then are they not like superficial numbers. And if the one being multiplied into the other, they produce a square number, then are they like superficial (by the 2. of the ninth). Moreover if the said two superficial numbers be in superperticular, or superbipartient proportion, then are they not like superficial numbers. For if they should be like, then should there be a mean proportional between them (by the 20. of the eight). But that is contrary to the Corollary of the 20. of the eight.  
And the easilier to conceive the demonstrations following, take this example of that which we have said.   

¶ An Assumpt.  
Forasmuch as in the eight book in the 26. proposition it was proved, that like plain numbers have that proportion the one to the other, that a square number hath to a square number: and likewise in the 24. of the same book it was proved, that if two numbers have that proportion the one to the other, 
This assumpt is the converse of the 26. of the eight.  that a square number hath to a square number, those numbers are like plain numbers. Hereby it is manifest, that unlike plain numbers, that is, whose sides are not proportional, have not that proportion the one to the other, that a square number hath to a square number. For if they have, than should they be like plain numbers, which is contrary to the supposition. Wherefore unlike plain numbers have not that proportion the one to the other, that a square number hath to a square number. And therefore squares which have that proportion the one to the other, that unlike plain numbers have, shall have their sides incommensurable in length (by the last part of the former proposition) for that those squares have not that proportion the one to the other that a square number hath to a square number.    

¶ The 8. Theorem. The 10. Proposition. If four magnitudes be proportional, and if the first be commensurable unto the second, the third also shall be commensurable unto the fourth. And if the first be incommensurable unto the second, the third shall also be incommensurable unto the fourth.  
SVppose that these four magnitudes A, B, C, D, be proportional. As A is to B, so let C be to D, and let A be commensurable unto B. Then I say that C is also commensurable unto D. 
Demonstration o● the first part.  For forasmuch as A is commensurable unto B, it hath (by the fift of the tenth) that proportion that number hath to number. But as A is to B, so is C to D. Wherefore C also hath unto D that proportion that number hath to number. Wherefore C is commensurable unto D (by the 6. of the tenth). But now suppose that the magnitude A be incommensurable unto the magnitude B. 
Demonstration of the second part●  Then I say that the magnitude C also is incommensurable unto the magnitude D. For forasmuch as A is incommensurable unto B, therefore (by the 7. of this book) A hath not unto B such proportion as number hath to number. But as A is to B, so is C to D. Wherefore C hath not unto D such proportion as number hath to number. Wherefore (by the 8. of the tenth) C is incommensurable unto D. If therefore there be four magnitudes proportional, and if the first be commensurable unto the second, the third also shall be commensurable unto the fourth. And if the first be incommensurable unto the second, the third shall also be incommensurable unto the fourth: which was required to be proved.  

¶ A Corollary added by Montaureus.  
If there be four lines proportional, and if the two first, or the two last be commensurable in power only, the other two also shall be commensurable in power only. 
A Corollary.  This is proved by the 22. of the sixth, and by this tenth proposition. And this Corollary Euclid useth in the 27. and 28. propositions of this book, and in other propositions also.    

¶ The 3. Problem. The 11. Proposition. Unto a right line first set and given (which is called a rational line) to find out two right lines incommensurable, the one in length only, and the other in length and also in power.  
SVppose that the right line first set and given, which is called a rational line of purpose be A. It is required unto the said line A, to find out two right lines incommensurable, the one in length only, the other both in length and in power. 
To find out the first line incommensurable in length only to the line given.  Take (by that which was added after the 9 proposition of this book) two numbers B and C, not having that proportion the one to the other, that a square number hath to a square number, that is, let them not be like plain numbers (for like plain numbers by the 26. of the eight have that proportion the one to the other that a square number hath to a square number). And as the number B is to the number C, so let the square of the line A be unto the square of an other line, namely, of D (how to do this was taught in the assumpt put before the 6. proposition of this book.) Wherefore the square of the line A, is unto the square of the line D commensurable (by the sixth of the tenth.) And forasmuch as the number B hath not unto the number C, that proportion that a square number hath to a square number, therefore the square of the line A hath not unto the square of the line D, that proportion that a square number hath to a number. Wherefore by the 9 of the tenth, the line A is unto the line D incommensurable in length only. And so is found out the first line, namely, D incommensurable in length only to the line given A. 
To find out the second line incommensurable both in length and in power to the line given.  Again take (by the 13. of the sixth) the mean proportional between the lines A and D, and let the same be E. Wherefore as the line A is to the line D, so is the square of the line A to the square of the line E (by the Corollary of the 20. of the sixth). But the line A is unto the line D incommensurable in length. Wherefore also the square of the line A is unto the square of the line E incommensurable by the second part of the former proposition. Now forasmuch as the square of the line A is incommensurable to the square of the line E, it followeth (by the definition of incommensurable lines) that the line A is incommensurable in power to the line E. Wherefore unto the right line given, and first set, A, which is a rational line, and which is supposed to have such divisions and so many parts as ye list to conceive in mind, as in this example 11, whereunto, as was declared in the 5. definition of this book, may be compared infinite other lines, either commensurable or incommensurable, is found out the line D incommensurable in length only. Wherefore the line D is rational (by the sixth definition of this book) for that it is incommensurable in length only to the line A, which is the first line set, and is by supposition rational. There is also found out the line E, which is unto the same line A incommensurable, not only in length but also in power, which line E compared to the rational line A, is by the definition irrational. For Euclid always calleth those lines irrational, which are incommensurable both in length and in power to the line first set, and by supposition rational.   

¶ The 9 Theorem. The 12. Proposition. Magnitudes commensurable to one and the self same magnitude: are also commensurable the one to the other.  
SVppose that either of these magnitudes A and B, be commensurable unto the magnitude C: Then I say that the magnitude A is commensurable unto the magnitude B. 
Construction.  For forasmuch as the magnitude A is commensurable unto the magnitude C, therefore (by the 5. of the tenth) A hath unto C such proportion as number hath to number. Let A have unto C that proportion that the number D hath to the number E. Again forasmuch as B is commensurable unto C, therefore (by the self same) C hath unto B that proportion that number hath to number. Let C have unto B that proportion that the number F hath unto the number G. Now then take the lest numbers in continual proportion and in these proportions given, namely, that the number D hath to the number E, and that the number F hath to the number G (by the 4. of the eight): which let be the numbers ●, K, L. 
Demonstration.  So that as the number D is to the number E, so let the number H be to the number K, and as the number F is to the number G, so let the number K be to the number L. Now for that as A is to C, so is D to E, but as D is to E so is H to K, therefore as A is to C, so is H to K. Again for that as C is to B, so is F to G, but as F is to G so is K to L: therefore as C is to B, so is K to L. But it is now proved that as A is to C, so is H to K. Wherefore of equality (by the 22. of the fift) as A is to B, so is the number H to the number L. Wherefore A hath unto B such proportion as number hath to number. Wherefore (by the sixth of the tenth) the magnitude A is commensurable unto the magnitude B. Magnitudes therefore commensurable to one and the self same magnitude, 
T●is is wi●h Zambert an A●●●mpt, but v●●e●ly improperly: ●l●ssate● ma●eth i● a Corollary, but the Gree●e and Montaureus ma●e it a proposition: but every way an ●nfallible truth 〈…〉.  are also commensurable the one to the other: which was required to be proved.  

¶ An Assumpt.  
If there be two magnitudes compared to one and the self same magnitude, and if the one of them be commensurable unto it, and the other incommensurable: those magnitudes are incommensurable the one to the other.  
SVppose that there be two magnitudes, namely, 
Demonstration leading to an absurdity.  A and B and let C be a certain other magnitude. And let A ●e commensurable unto C, and let B be commensurable unto the self same C. Then I say that the magnitude A is incommensurable unto B. For if A be commensurable unto B, forasmuch as A is also commensurable unto C therefore (by the 12. of the tenth) B is also commensurable unto C: which is contrary to the supposition.    

¶ The 10. Theorem. The 13. Proposition. If there be two magnitudes commensurable, and if the one of them be incommensurable to any other magnitude: the other also shall be incommensurable unto the same.  
SVppose that these two magnitudes A, B be commensurable the one to the other, and let the one of them, namely A, be incommensurable unto an other magnitude, 
Demonstration leading to an absurdity.  namely, unto C. Then I say that the other magnitude also, namely B, is incommensurable unto C. For if B be commensurable unto C, then forasmuch as A is commensurable unto B, therefore (by the 12. of the tenth) the magnitude A also is commensurable unto the magnitude C. But it is supposed to be incommensurable unto it, which is impossible. Wherefore the magnitudes B and C are not commensurable. Wherefore they are incommensurable. If therefore there be two magnitudes commensurable, and if the one of them be incommensurable to any other magnitude, the other also shallbe incommensurable unto the same: which was required to be proved.  

¶ A Corollary added by Montaureus.  
Magnitudes commensurable to magnitudes incommensurable, are also incommensurable the one to the other. 
A Corollary.   
Suppose that the magnitudes A and B be incommensurable the one to the other, and let the magnitude C be commensurable to A, and let the magnitude D be commensurable unto B. Then I say that the magnitu●●s C and D are incommensurable the one to the other. For A and C are commensurable, of which the magnitude A is incommensurable unto B, wherefore by this 13. proposition the magnitudes C and B are also incommensurable: but the magnitude● B and D are commensurable wherefore by the same, or by the former assumpt, the magnitudes C and D are incommensurable the one to the other. This corollary, Theon useth often times as in the 22. 26. and 36 propositions of this book, and in other propositions also.   

¶ An Assumpt.  
Two unequal right lines being given, to fi●de out how much the greater is in power more than the less.  

A Corollary.  And like in sort, two right lines being given, by this means may be found out a right line which containeth them both in power. Suppose that the two right lines given be AD and DB. It is required to ●inde out a right line that containeth them both in power. Let the lines AB and DB be so put, that they comprehend a right angle ADB, and draw a right line from A to B. Now again it is manifest (by the 47. of the ●irst) that the line AB containeth in power the lines AD and DB.    

¶ The 11. Theorem. The 14. Proposition. If there be sour right lines proportional, and if the first be in power more than the second by the square of a right line commensurable in length unto the first, the third also shallbe in power more than the fourth, by the square of a right line commensurable unto the third. And if the first be in power more than the second by the square of a right line incommensurable in length unto the first, the third also shall be in power more than the fourth by the square of a right line incommensurable in length to the third.  
SVppose that these four right lines A, B, C, D, be proportional. As A is to B, so let C be to D. And let A be in power more than B, by the square of the line E. And likewise let C be in power more than D, by the square of the line F. 
Demonstration.  Then I say that if A be commensurable in length unto the line E, C also shall be commensurable in length unto the line F. And if A be incommensurable in length to the line E, C also shall be incommensurable in length to the line F. For for that as A is to B, so is C to D, therefore as the square of the line A is to the square of the line B, so is the square of the line C to the square of the line D (by the 22. of the sixth). But by supposition unto the square of the line A are equal, the squares o● the lines E and B, and unto the square of the line C are equal the squares of the of the lines D and F: Wherefore as the squares of the lines E and B (which are equal to the square of the line A) are to the square of the line B, so are the squares of the lines D and F (which are equal to the square of the line C) to the square of the line D (by the seventh of the fift). Wherefore (by the 17. of the fift) as the square of the line ● i● to the square of the line B, so is the square of the line F to the square of the line D. Wherefore also as the line E is to the line ●, so is the line F to the line D (by the second part of the 22. of the sixth) wherefore contrariwise (by the Corollary of the fourth of the fift) as B is to E so is D to F. But (by supposition) ●s A is to B, so is C to D, Wherefore of equality (by the 22. of the fift) a● A is to ●, so is C is F. If therefore A be commensurable in length unto E, C also shall be commensurable in length unto F: and if it ba incommensurable in length unto E, C also shallbe incommensurablel in length unto F, by the 10. of this book. If therefore there be four right tlines proportional, and if the first be in power more than the secondby the square of a right line commensurable in length unto the first, the third also shall be in power more than the fourth, by the square of a right line commensurable in length unto the third and if the first be in power more than the second, by the square of a right line incommensurable in length unto the first, the third also shall be in power more the the fourth, by the square of a right line incommensurable in length to the third: which was required to be proved.  
Note that the line A may be proved to be in proportion to the line E, as the line C is to the line F, by an other way, namely, by conversion of proportion (of some● as we have before noted, 
another way to prove that the lines A, E, C, F, are proportional.  called inuerse proportion) by the 19 of the fift. For, forasmuch as the four lines A, B, C, D, are proportional: therefore (by the 22. of the sixth) their squares also are proportional. And forasmuch as the antecedent, namely, the square of the line A exceedeth the consequent, namely, the square of the line B, by the square of the line E: and the other antecedent, namely, the square of the line C, exceedeth the other consequent, namely, the square of the line D, by the square of the line F, therefore as the square of the line A is to the excess, namely, to the square of the line E, so is ●he square of the line C to the excess; namely, to the square of the line F. Wherefore (by the second part of the 22. of the sixth) as the line A is to the line E, so is the line C to the line F.   

¶ The 12. Theorem. The 15. Proposition. If two magnitudes commensurable be composed, the whole magnitude composed also shall be commensurable to either of the two parts. And if the whole magnitude composed be commensurable to any one of the two parts, those two parts shall also be commensurable.  
LEt these two commensurable magnitudes AB and BC, be composed or added together. Then I say, that the whole magnitude AC is commensurable to either of these parts AB and BC: For forasmuch as AB and BC are commensurable, therefore (by the first definition of the tenth) some one magnitude measureth them both. 
Demonstration of the first part.  Let there be a magnitude that measureth them, and let the same be D. Now forasmuch as D measureth AB and BC, it shall also measure the whole magnitude composed AC, by this common sentence, what soever magnitude measureth two other magnitudes, shall also measure the magnitude composed of them. But the same D measureth AB and BC (by supposition). Wherefore D measureth AB, BC, and AC. Wherefore AC is commensurable to either of these magnitudes AB and BC.  
But now suppose that the whole composed magnitude AC be commensurable to any one of these two magnitudes AB or BC, let it be commensurable I say unto AB. 
Demonstration of the second pa●t which is the converse of the first.  Then I say, that the two magnitudes AB and BC are commensurable. For forasmuch as AB and AC are commonsurable, some one magnitude measureth them (by the first definition of the tenth). Let some magnitude measure them, and let the same be D. Now forasmuch as D measureth AB and AC, it also measureth the residue BC, by this common sentence, what soever measureth the whole and the part taken away, shall also measure the residue. But the same D measureth the magnitude AB (by supposition). Wherefore D measureth either of these magnitudes AB and BC. Wherefore the magnitudes AB and BC are commensurable. If therefore two magnitudes commensurable be composed, the whole magnitude composed also shall be commensurable to either of the two parts. And if the whole magnitude composed be commensurable to any one of the two parts, those two parts shall also be commensurable: which was required to be demonstrated.  

¶ A Corollary added by Montaureus.  

A Corollary.  If an whole magnitude be commensurable to one of the two magnitudes which make the whole magnitude, it shall also be commensurable to the other of the two magnitudes. For if the whole magnitude AC be commensurable unto the magnitude BC, then by the 2. part of this 15 Proposition: the magnitudes AB and BC are commensurable. Wherefore (by the first part of the same) the magnitude AC shall be commensurable to either of these magnitudes AB and BC. This Corollary Theon useth in the demonstration of the 17. Proposition and also of other Propositions. Howbeit Euclid left it out, for that it seemed easy as in a manner do all other Corollaryes.    

¶ The 13. Theorem. The 16. Proposition. If two magnitudes incommensurable be composed, the whole magnitude also shall be incommensurable unto either of the two parts componentes. And if the whole be incommensurable to one of the parts componentes, those first magnitudes also shall be incommensurable.  

Demonstration of the first part by an argument leadindg to an absurdity.  LEt these two incommensurable magnitudes AB & BC, be composed, or added together. Then I say, that the whole magnitude AC, is incommensurable to either of these magnitudes AB and BC. For if AC and AB be not incommensurable, than some one magnitude measureth them (by the first definition of the tenth). Let there be such a magnitude, if it be possible, and let the same be D. Now forasmuch as D measureth CA and AB, it also measureth the residue BC, & it likewise measureth AB. Wherefore D measureth AB and BC. Wherefore (by the first definition of the tenth) the magnitudes AB and BC are commensurable. But it is supposed that they are incommensurable: which is impossible. Wherefore no magnitude doth measure the magnitudes AB and AC. Wherefore the magnitudes CA and AB are incommensurable. In like sort also may we prove, that the magnitudes AC and CB are incommensurable.  

Demonstration of the second pa●t leading also to an impossibility. And this second part is the converse of the first.  But now suppose that the magnitude AC be incommensurable to one of these magnitudes AB or BC, and first let it be incommensurable unto AB. Then I say, that the magnitudes AB and BC are incommensurable. For if they be commensurable some one magnitude measureth them. Let some one magnitude measure them, & let the same be D. Now forasmuch as D measureth AB and BC, it also measureth the whole magnitude AC. And it measureth AB: Wherefore D measureth these magnitudes CA and AB. Wherefore CA & AB. are commensurable. And they are supposed to be incōmensurable● which is impossible. Wherefore no magnitude measureth AB and BC. Wherefore the magnitudes AB and BC are incommensurable. And in like sort may they be proved to be incommensurable, if the magnitude AC be supposed to be incommensurable unto BC. If therefore there be two magnitudes incommensurable composed, the whole also shall be incommensurable unto either of the two parts component, and if the whole be incommensurable to one of the parts component, those first magnitudes shall be incommensurable: which was required to be proved.  

¶ A Corollary added by Montaureus.  
If an whole magnitude be incommensurable to one of the two magnitudes which make the whole magnitude, it shall also be incommensurable to the other of the two magnitudes. For if the whole magnitude AC be incommensurable unto the magnitude BC, then by the 2 part of this 16. Theorem, the magnitudes AB and BC shall be incommensurable. Wherefore by the first part of the same Theorem, the magnitude AC shall be incommensurable to either of these magnitudes AB and BC. This Corollary 〈◊〉 useth in the demonstration of the ●3. Theorem, & also of other Propositions.   

¶ An Assumpt.  
If upon a right line be applied a parallelogram wanting in figure by a square: the parallelogram so applied, is equal to that parallelogram which is contained under the segments of the right line, which segments are made by reason of that application.  
This Assumpt I before added as a Corollary out of Flussates after the 28. Proposition of the sixth book.    

¶ The 14. Theorem. The 17. Proposition. If there be two right lines unequal, and if upon the greater be applied a parallelogram equal unto the fourth part of the square of the less line, and wanting in figure by a square, if also the parallelogram thus applied divide the line where upon it is applied into parts commensurable in length: then shall the greater line be in power more than the less, by the square of a line commensurable in length unto the greater. And if the greater be in power more than the less by the square of a right line commensurable in length unto the greater, and if also upon the greater be applied a parallelogram equal unto the fourth part of the square of the less line, and wanting in figure by a square: then shall it divide the greater line into parts commensurable.  
But now suppose that the line BC be in power more than the line A, by the square of a line commensurable in length unto the line BC. 
Demonstration of the second part which is the converse of the first.  And upon the line BC let there be applied a rectangle parallelogram equal unto the fourth part of the square of the line A, and wanting in figure by a square, and let the said parallelogram be that which is contained under the lines BD and DC. Then must we prove that the line BD is unto the line DC commensurable in length. The same constructions and suppositions, that were before, remaining, we may in like sort prove that the line BC is in power more than the line A, by the square of the line FD. But by supposition the line BC is in power more than the line A by the square of a line commensurable unto it in length. Wherefore the line BC is unto the line FD commensurable in length. Wherefore the line composed of the two lines BF and DC is commensurable in length unto the line FD (by the second part of the 15. of the tenth). Wherefore (by the 12. of the tenth or by the first part of the 15. of the tenth) the line BC is commensurable in length to the line composed of BF and DC. But the whole line conposed BF and DC is commensurable in length unto DC. For BF (as before hath been proved) is equal to DC. Wherefore the line BC is commensurable in length unto the line DC (by the 12. of the tenth). Wh●●●fore also the line BD is commensurable in length unto the line DC (by the second part of th● 15. of the te●th). If therefore there be two right lines unequal, and if upon the greater be applied a parallelogram equal unto the fourth part of the square of the less and wanting in figure by a square, if also the parallelogram thus applied divide the line whereupon it is applied into parts commensurable in length: then shall the greater line be in power more than the less by the square of a line commensurable in length unto the greater. And if the greater be in power more than the less by the square of a line commensurable in length unto the greater, and if also upon the greater be applied a parallelogram equal unto the fourth part of the square made of the less and wanting in figure by a square: then shall it divide the greater line into parts commensurable in length: which was required to be proved.  
Campan● after this proposition reacheth how we may readily apply upon the line BC a parallelogram equal to the fourth part of the square of half of the line A, and wanting in figure by a square after this manner.  
Divide the line BC into two lines in such sort that half of the line A shallbe the mean proportional between those two lines, which is possible, when as the line BC is supposed to be greater than the line A, and may thus be done. 
How to divide the line BC readily in such sort as i● required in the proposition.  Divide the line BC into two equal parts in the point E and describe upon the line BC a semicircle BHC. And unto the line BC, and from the point C erect a perp●dicular line CK and put the line CK equal to half of the line A● And by the point K draw unto the line EC a parallel line KH cutting the semicircle in the point H, (which it must needs cut, forasmuch as the line BC is greater than the line A). And from the point H draw unto the line BC a perpendicular li●e HD: which line HD● forasmuch as by the 34: of the first it is equal unto the line KC, shall also be equal to half of the line A: draw the lines BH and HC. Now then by the ●●. of the third the angle BHC is a right a●gle. Wherefore by the corollary of the eight of the sixth book the line HD is the mean proportional between the lines BD and DC. Wherefore the half of the line A which is equal unto the line HD is the mean proportional between the lines BD and DC. Wherefore that which is contained under the lines BD and DC is equal to the fourth part of the square of the line A. And so if upon the line BD be described a rectangle parallelogram having his other side equal to the line DC, there shallbe applied upon the line BC a rectangle parallelogram equal unto the square of half of the line A, and wanting in figure by a square: which was required to be done.   

¶ The 15. Theorem. The 18. Proposition. If there be two right lines unequal, and if upon the greater be applied a parallelogram equal unto the fourth part of the square of less, and wanting in figure by a square, if also the parallelogram thus applied divide the line whereupon it is applied into parts incommensurable in length: the greater line shallbe in power more than the less line by the square of a line incommensurable in length unto the greater line. And if the greater line be in power more than the less line, by the square of a line incommensurable in length unto the greater, and if also upon the greater be applied a parallelogram equal unto the fourth part of the square of the less and wanting in figure by a square: then shall it divide the greater line into parts incommensurable in length.  
But now suppose that the line BC be in power more than the line A by the square of a line incommensurable in length unto BC. 
Demonstration of the second part which is the converse of t●e former.  And upon the line BC let there be applied a parallelogram equal unto the fourth part of the square of the line A, and wanting in figure by a square, and let the said parallelogram be that which is contained under the lines BD & DC. Then must we prove that the line BD is unto the line DC incommensurable in length. The same order of construction and demonstration being kept, we may in like sort prove that the line BC is in power more than the line A by the square of the line FD. But now (by supposition) the line BC is in power more than the line A by the square of a line incommensurable in length unto BC. Wherefore the line BC is unto the line FD incommensurable in length. Wherefore the line composed of BF and DC taken as one line, shall be incommensurable in length to the line FD (by the second part of the 16. of the tenth): wherefore also by the first part of the same, the line BC shall be incommensurable in length to the line composed of the lines BF and DC. But the line composed of the lines BF and DC is commensurable in length to the line DC (for that BF (as before hath been proved) is equal to DC). Wherefore the line BC is incommensurable in length to the line DC (by the 13. of the tenth). Wherefore by the second part of the 16. of the tenth, the line BD is incommensurable in length unto the line DC. If therefore there be two right lines unequal, and if upon the greater be applied a parallelogram equal unto the fourth part of the square of the less line, & wanting in ●igure by a square, if also the parallelogram thus applied divide the line whereupon it is applied into parts incommensurable in length: the greater line shall be in power more than the less line by the square of a line incommensurable in length unto the greater. And if the greater line be in power more than the less, by the square of a line incommensurable in length unto the greater, and if also upon the greater be applied a parallelogram equal unto the fourth part of the square of the less line, and wanting in figure by a square, then shall it divide the greater line into parts incommensurable in length: which was required to be demonstrated.  
This Proposition may also be demonstrated by the former proposition, namely, the first part of this by the second part of the former, and the second part of this by the first part of the former, by an argument leading to an absurdity. 
An other demonstration●y an argument leading to an absurdity.  For as touching the first part of this proposition, the line BC containing in power more than the line A by the square of the line FD, if the line BG be not incommensurable unto the line FD, then is it commensurable unto it. Wherefore (by the second part of the 17. proposition) the lines BD and DC also are commensurable, which is impossible, for they are supposed to be incommensurable. So likewise as touching the second part of the same, the line BC containing in power more than the line A by the square of the line FD, if the line DB be not incommensurable to the line DC, then is it commensurable unto it: wherefore (by the first part of the ●●, proposition) the lines BC and FD are also commensurable, which were absurd. For the lines BC and FD are supposed to be incommensurable: which was required to be proved.  

¶ An assumpt.  
Forasmuch as it hath been proved that lines commensurable in length, 
An Assumpt.  are always also commensurable in power, but lines commensurable in power are not always commensurable in length, but may be in length both commensurable and also incommensurable: it is manifest, that if unto the line propounded, which is called rational of purpose, a certain line be commensurable in length, it aught to be called rational and commensurable unto it, not only in length, but also in power: for lines commensurable in length are also always commensurable in power. But if unto the line propounded which is called rational of purpose, a certain line be commensurable in power, then if it be also commensurable unto it in length, it is called rational and commensurable unto it both in length and in power. But again if unto the said line given which is called rational, a certain line be commensurable in power, and incommensurable in length, that also is called rational, commensurable in power only.   

An annotation of Proclus.  
He calleth those lines rational, which are unto the rational line first set commensurable in length & in power, or in power only. And there are also other right lines, which are unto the rational line first set, incommensurable in length, and are unto it commensurable in power only, and therefore they are called rational, & commensurable the one to the other● for which cause they are rational. But even these lines may be commensurable the one to the other, either in length, and therefore in power, or else in power only. Now if they be commensurable in length, then are those lines called rational, commensurable in length, but yet so that they be understand to be in power commensurable: but if they be commensurable the one to the other in power only, they also are called rational commensurable in power only.   

¶ A Corollary.  
And that two lines or more being rational and commensurable in length to the rational line first set, are also commensurable the one to the other in length, hereby it is manifest: 
A Corollary added by Montaureu●.  for forasmuch as they are rational and commensurable in length to the rational line first set, but those magnitudes which are commensurable to one and the self same magnitude, are also commensurable the one to the other (by the 12. of the tenth) wherefore the rational lines, commensurable in length to the rational line first set, are also commensurable in length the one to the other. And as touching those which are rational commensurable in power only to the rational line first set, they also must needs be at the lest commensurable in power the one to the other. For forasmuch as their squares are rational they shall be commensurable to the square of the rational line first set. Wherefore by the 12. of this book, they are also commensurable the one to the other. Wherefore their lines are at the lest commensurable in power the one to the other. And it is possible also that they may be commensurable in length the one to the other. For suppose that A be a rational li●e first set, and let the line B be unto the same rational line A commensurable in power only, that is, incommensurable in length unto it. Let there be also an other line C commensurable in length to the line B (which is possible by the principles of this book.) Now by the 13. of the tenth, it is manifest that the line C is incommensurable in length unto the line A. But the square of the line A is commensurable to the square of the line B by supposition, and the square of the line C is also commensurable to the square of the line B by supposition. Wherefore by the 12. of this book, the square of the line C is commensurable to the square of the line A. Wherefore by the definition, the line C shall be rational commensurable in power only to the line A, as also is the line B. Wherefore there are given two rational lines commensurable in power only to the rational line first set, and commensurable in length the one to the other.  
Here is to be noted, which thing also we before noted in the definitions, that Campane and others which followed him, brought in these phrases of speeches, to call some lines rational in power only, 
Cause.  and other some rational in length and in power, which we cannot find that Euclid ever used. For these words in length and in power are never referred to rationality or irrationalitie, but always to the commensurabilitie or incommensurablitie of lines. Which perverting of words (as was there declared) hath much increased the difficulty and obscureness of this book. 
cause of increasing the difficulty of this book.  And now I think it good again to put you in mind, that in these propositions which follow, we must ever have before our eyes the rational line first set, 
Note.  unto which other lines compared are either rational or irrational, according to their commensurability or incommensurabilitie.    

¶ The 16. Theorem. The 19 Proposition. A rectangle figure comprehended under right lines commensurable in length, being rational according to one of the foresaid ways: is rational.  
SVppose that this rectangle figure AC be comprehended under these right lines AB and BC being commensurable in length, and rational according to one of the foresaid ways. Then I say that the superficies AC is rational, describe (by the 46. o● the first) upon the line AB a square AD. 
Construction.  Wherefore that square AD is rational by the definition. 
Demonstration.  And forasmuch as the line AB is commensurable in length unto the line BC, and the line AB is equal unto the line BD, therefore the line BD is commensurable in length unto the line BC. And as the line BD is to the line BC, so is the square DA to the superficies AC (by the first of the sixth): but it is proved that the line BD is commensurable unto the line BC, wherefore (by the 10. of the tenth) the square DA is commensurable unto the rectangle superficies AC. But the square DA is rational, wherefore the rectangle superficies AC also is rational by the definition. A rectangle figure therefore comprehended under right lines commensurable in length, being rational according to one of the foresaid ways is rational: which was required to be proved.  

divers ca●es in this proposition.  Where as in the former demonstration the square was described upon the less line, we may also demonstrate the Proposition, if we describe the square upon the greater line, and that after this manner. Suppose that the rectangle superficies BC be contained of these unequal lines AB and AC, which let be rational commensurable the one to the other in length. And let the line AC be the greater. 
The second case.  And upon the line AC describe the square DC. Then I say, that the parallelogram BC is rational. 
The first kind of rational lines commensurable in length.  For the line AC is commensurable in length unto the line AB by supposition, and the line DA is equal to the line AC. Wherefore the line DA is commensurable in length to the line AB. But what proportion the line DA hath to the line AB, the same hath the square DC to the para●lelogramme C● (by the first of the sixth). Wherefore (by the 10. of this book) the square DC is commensurable to the parallelogram CB. But it is manifest, that the square DC is rational, for that it is the square of a rational line, namely, AC. Wherefore (by the definition) the parallelogram also CB is rational.  
Moreover, forasmuch as those two former demonstrations seem to speak of that parallelogram which is made of two lines, of which any one may be the li●e first set, which is called the first rational line, from which (we said) aught to be taken the measures of the other lines compared unto it, and the other is commensurable in length to the same first rational line, 

This particle in the proposition (according to any of the foresaid ways) was not in vain put.  
The second kind of rational lines commensurable in length.   which is the first kind of rational lines commensurable in length: I think it good here to set an other case of the other kind of rational lines, of lines I say rational commensurable in length compared to an other rational line first set, to declare the general truth of this Theorem, and that we might see that this particle according to any of the foresaid ways was not here in vain put. Now then suppose first a rational line AB. Let there be also a parallelogram CD contained under the lines CE and ED, which lines let be rational, that is commensurable in length to the ●irst rational line propounded AB. Howbeit, let those two lines CE and ED be divers and unequal lines unto the first rational line AB. Then I say, that the parallelogram CD is rational. 
The third case.  Describe the square of the line DE, which let be DF. First it is manifest (by the 12. of this book) that the lines CE & ED, are commensurable in length the one to the other. For either of them is supposed to be commensurable in length unto the line AB. But the line ED is equal to the line EF. Wherefore the line CE is commensurable in length to the line BF. But 〈◊〉 the line CE is ●o the line ● F, ●o is the parallelogram CD to the square DF (by the first of the sixth). Wherefore (by the 10. of this book) the parallelogram CD shall be commensurable to the square DF. But the square DF is commensurable to the square of the line AB which is the first rational line propounded. Wherefore (by the 12. of this book) the parallelogram CD is commensurable to the square of the line AB. But the square of the line AB is rational (by the definition). Wherefore by the definition also of rational figures, the parallelogram CD shall be rational.  
Now resteth an other ca●e of the third kind of rational lines commensurable in length the one to the other, which are to the rational line AB first set commensurable in power only, and yet are therefore rational lines. And let the lines CE and ED be commensurable in length the one to the other. 
The third kind of rational lines commensurable in length.  Now then let the self same construction remain that was in the former: so that let the lines CE and ED be rational commensurable in power only unto the line AB. But let them be commensurable in length the one to the other. Then I say, that in this case also the parallelogram CD is rational. First it may be proved as before, that the parallelogram CD is commensurable to the square DF. Wherefore (by the 12. of this book) the parallelogram CD shall be commensurable to the square of the line AB● But the square of the line AB is rational. 
The fourth case.  Wherefore (by the definition) the parallelogram CD shall be also rational. This case is well to be noted. For it serveth to the demonstration and understanding of the 25. Proposition of this book.   

¶ The 17. Theorem. The 20. Proposition. If upon a rational line be applied a rational rectangle parallelogram: the other side that maketh the breadth thereof shall be a rational line and commensurable in length unto that line whereupon the rational parallelogram is applied.  
SVppose that this rational rectangle parallelogram AC, be applied upon the line AB, which let be rational according to any one of the foresaid ways (whether it be the first rational line set, 
This proposition is the converse of the former proposition.  or any other line commensurable to the rational line first set, and that in length and in power, or in power only: for one of these three ways, as was declared in the Assumpt put before the 19 Proposition of this book, is a line called rational) and making in breadth the line BC. Then I say, that the line BC is rational and commensurable in length unto the line BA. 
Construction.  Desrcribe (by the 46. of the first) upon the line BA a square AD. Wherefore (by the 9 definition of the tenth) the square AD is rational. But the parallelogram AC also is rational (by supposition). 
Demonstration.  Wherefore (by the conversion of the definition of rational figures, or by the 12. of this book) the square DA is commensurable unto the parallelogram AC. But as the square DA is to the parallelogram AC, so is the line DB to the line BC (by the first of the sixth). Wherefore (by the 10. of the tenth) the line DB is commensurable unto the line BC. But the line DB is equal unto the line BA. Wherefore the line AB is commensurable unto the line BC. But the line AB is rational. Wherefore the line BC also is rational and commensurable in length unto the line BA. If therefore upon a rational line be applied a rational rectangle parallelogram, the other side that maketh the breadth thereof shall be a rational line commensurable in length unto that line whereupon the rational parallelogram is applied: which was required to be demonstrated.  

¶ An Assumpt.  
A line containing in power an irrational superficies, is irrational.  

An Assumpt.  Suppose that the line AB contain in power an irrational superficies, that is, let the square described upon the line AB, be equal unto an irrational superficies. Then I say, that the line AB is irrational. For if the line AB be rational, them shall the square of the line AB be also rational. For so was it put in the definitions. But (by supposition) it is not. Wherefore the line AB is irrational. A line therefore containing in power an irrational superficies, is irrational.    

¶ The 18. Theorem. The 21. Proposition. A rectangle figure comprehended under two rational right lines commensurable in power only, is irrational. And the line which in power containeth that rectangle figure is irrational, & is called a medial line.  
SVppose that this rectangle figure AC be comprehended under these rational right lines AB and BC commensurable in power only. Then I say, that the superficies AC is irrational: and the line which containeth it in power is irrational, and is called a medial line. 
Construction.  Describe (by the 46. of the first) upon the line AB a square AD. 
Demonstration.  Wherefore the square AD is rational. And forasmuch as the line AB is unto the line BC incommensurable in length, for they are supposed to be commensurable in power only, and the line AB is equal unto the line BD, therefore also the line● BD is unto the line BC incommensurable in length. And 〈◊〉 ●h● lin● 〈…〉 is to the line ● C, so 〈◊〉 the square AD to the parallelogram AC (by the first of the fiu●). Wherefore (by the 10. of the tenth) the square DA is unto the parallelogram AC incommensurable. But the square DA is rational. Wherefore the parallelogram AC is irrational. Wherefore also the line that containeth the superficies AC in power, that is whose square is equal unto the parallelogram AC, is (by the Assumpt going before) irrational. And it is called a medial line, for that the square which is made of it, is equal to that which is contained under the lines AB and BC, and therefore it is (by the second part of the 17. of the sixth) a mean proportional line between the lines AB and BC. A rectangle figure therefore comprehended under rational right lines which are commensurable in power only, is irrational. And the line which in power containeth that rectangle figure is irrational, and is called a medial line.  
At this Proposition doth Euclid first entreat of the generation and production of irrational lines. And here he searcheth out the first kind of them, which he calleth a medial line. And the definition thereof is fully gathered and taken out of this 21. Proposition, which is this. A medial line is an irrational line whose square is equal to a rectangled figure contained of two rational lines commensurable in power only. 
Definition of a medial line.  It is called a medial line, as Theon rightly saith, for two causes, first for that the power or square which it produceth● is equal to a medial superficies or parallelogram. For as that line which produceth a rational square, is called a rational line, and that line which produceth an irrational square, or a square equal to an irrational figure generally is called an irrational line: so i● tha● line which produceth a medial square, or a square equal to a medial superficies, called by special name a medial line. Secondly it is called a medial line, because it is a mean proportional between the two lines commensurable in power only which comprehend the medial superficies.  

¶ A Corollary added by Flussates.  
A rectangle parallelogram contained under a rational line and an irrational line, is irrational. 
A Corollary.  For if the line AB be rational, and if the line CB be irrational, they shall be incommensurable. But as the line BD (which is equal to the line BA) is to the line BC, so is the square AD to the parallelogram AC. Wherefore the parallelogram AC shall be incommensurable to the square AD which is rational (for that the line AB whereupon it is described is supposed to be rational). Wherefore the parallelogram AC which is contained under the rational line AB, and the irrational line BC, is irrational.   

¶ An Assumpt.  
If there be two right lines, as the first is to the second, so is the square which is described upon the first to the parallelogram which is contained under the two right lines.  
Suppose that there be two right lines AB and BC. 
This assumpt is nothing else but a part of the first proposition of the sixth book.  Then I say that as the line AB is to the line BC, so is the square of the line AB, ●● that which is contained under the lines AB and BC. Describe (by the 46. of the first) upon the line AB a square AD. And make perfect the parallelogram AC. Now for that as the line AB is to the line BC (for the line AB, is equal to the line BD); so is the square AD to the parallelogram CA by the first of the six● and AD is the square which is made of the line AB, and AC is that which is contained under the lines BD and BC, that is, under the lines AB & BC: therefore as the line AB is to the line BC, so is the square described upon the the line AB to the rectangle figure contained under the lines AB & BC. And conversedly as the parallelogram which is contained under the lines AB and BC is to the square of the line AB, so is the line CB to the line BA.    

¶ The 19 Theorem. The 22. Proposition. If upon a rational line be applied the square of a medial line: the other side that maketh the breadth thereof shallbe rational, and incommensurable in length to the line whereupon the parallelogram is applied.  
SVppose that A be a medial line, and let BC be a line rational, and upon the line BC describe a rectangle parallelogram equal unto the square of the line A, and let the same be BD making in breadth the line CD. Then I say that the line CD is rational and incommensurable in length unto the line CB. 
〈◊〉  For forasmuch as A is a medial line, it containeth in power (by the 21. of the tenth) a rectangle parallelogram comprehended under rational right lines commensurable in power only. Suppose that is contain in power the parallelogram GF: and by supposition it also containeth in power the parallelogram BD. Wherefore the parallelogram BD is equal unto the parallelogram GF: and it is also equiangle unto it, for that they are each rectangle. But in parallelograms equal and equiangle the sides which contain the equal angles, are reciprocal (by the 14. of the sixth): Wherefore what proportion the line BC hath to the line EG, the same hath the line EF to the line CD. Therefore (by the 22. of the sixth) as the square of the line BC is to the square of the line EG, so is the square of the line EF to the square of the line CD. But the square of the line BC is commensurable unto the square of the line EG (by supposition). For either of them is rational. Wherefore (by the the 10. of the tenth) the square of the line EF is commensurable unto the square of the lin● CD. But the square of the line EF is rational. Wherefore the square of the line CD is likewise rational. Wherefore the line CD is rational. And forasmuch as the line EF is incommensurable in length unto the line EG (for they are supposed to be commensurable in power only). But as the line EF is to the line EG, so (by the assumpt going before) is the square of the line EF to the parallelogram which is contained under the lines EF and EG. Wherefore (by the 10. of the tenth) the square of the line EF is incommensurable unto the parallelogram which is contained under the lines FE and EG. But unto the square of the line EF the square of the line CD is commensurable, for it is proved that ●ither of them is a rational lin●. And that which is contained under the lines DC and CB is commensurable unto that which is contained under the lines FE and EG. For they are both equal to the square of the line A. Wherefore (by the 13. of the tenth) the square of the line CD is incommensurable to that which is contained under the lines DC and CB. But as the square of the line CD is to that which is contained under the lines DC and CB, so (by the assumpt going before) is the line DC to the line CB. Wherefore the line DC is incommensurable in length unto the line CB. Wherefore the line CD is rational and incommensurable in length unto the line CB. If therefore upon a rational line be applied the square of a medial line, the other side that maketh the breadth thereof shallbe rational, and incommensurable in length to the line whereupon the parallelogram is applied: which was required to be proved.  
A square is said to be applied upon a line, when it, or a parallelogram equal unto it, is applied upon the said line. 
How a square is said to be applied upon a line.  If upon a rational line given we will apply a rectangle parallelogram equal to the square of a medial line given, and so of any line given, we must, by the 11. of the sixth, find out the third line proportional with the rational line and the medial line given: so yet that the rational line be the first, and the medial line given, (which containeth in power the square to be applied) be the second. For then the superficies contained under the first and the third, shallbe equal to the square of the middle line, by the 17. of the sixth.   

¶ The 20. Theorem. The 23. Proposition. A right line commensurable to a medial line, is also a medial line.  
SVppose that A be a medial line. And unto the line A let the line B be commensurable, either in length, & in power, or in power only. 
Construction.  Then I say that B also is a medial line. Let there be put a rational line CD. And upon the line CD, apply a rectangle parallelogram CE, equal unto the square of the line A, and making in breadth the line ED. Wherefore (by the proposition going before) the line ED is rational and incommensurable in length unto the line CD. 
Demonstration.  And again upon the line CD apply a rectangle parallelogram CF equal unto the square of the line B, and making in breadth the line DF. And forasmuch as the line A is commensurable unto the line B, therefore the square of the line A is commensurable to the square of the line B. But the parallelogram EC is equal to the square of the lin● A, and the parallelogram CF is equal to the square of the line B: wherefore the parallelogram EC is commensurable unto the parallelogram CF. But as the parallelogram EC, is to the parallelogram CF, so is the line ED to the line DF (by the first of the sixth). Wherefore (by the 10. of the tenth) the line ED is commensurable in length unto the line DF. But the line ED is rational and incommensurable in length unto the line DC, wherefore the line DF is rational and incommensurable in length unto the line DC (by the 13. of the tenth). Wherefore the lines CD and DF are rational commensurable in power only. But a rectangle figure comprehended under rational right lines commensurable in power only, is (by the ●1. of the tenth) irrational, and the line that containeth it in power is irrational, and is called a medial line. Wherefore the line that containeth in power that which is comprehended under the lines CD and DF is a medial line. But the line B containeth in power the parallelogram which is comprehended under the lines CD and DF Wherefore the line B is a medial line. A right line therefore commensurable to a medial line, is also a medial line: which was required to be proved.  

¶ Corollary.  
Hereby it is manifest that a superficies commensurable unto a medial superficies, is also a medial superficies.  
For the lines which contain● in power those superficieces are commensurable in power, of which the one is a medial line (by the definition of a medial line in the 21. of this tenth): wherefore the other also is a medial line by this 23. proposition. And as it was said of rational lines so also is it to be said o● medial lines, namely, that a li●e commensurable to a medial line, is also a medial line, a line I say which is commensurable unto a medial line, whether it be commensurable in length, and also in power, or else in power only. For universally it is true, that lines commensurable in length, are also commensurable in power. Now if unto a medial line there be a line commensurable in power, if it be commensurable in length, them are those lines called medial lines commensurable in length & in power. But if they be commensurable in power only, th●y are called medial lines commensurable in power only.  
There are also other right lines incommensurable in length to the medial line, and commensurable in power only to the same: and these lines are also called medial, for that they are commensurable in power to the medial line. And in a● mu●h as they are medial lines, they are commensurable in power the one to the other. But being compared the one to the other, they may be commensurable either in length, and therefore in power, or else in power only. And then if they be commensurable in length, they are called also medial lines commensurable in length, and so consequently they are understanded to be commensurable in power. But i● they be commensurable in power only, yet notwithstanding they also are called medial lines commensurable in power only.  
Flussates after this proposition teacheth how to come to the understanding of medial superficieces and lines, by furred numbers, after this manner. Namely to express the medial superficieces by the roots of numbers which are not square numbers: and the lines containing in power such medial superficieces, by the roots of roots of numbers not square. medial lines also commensurable, are expressed by the roots of roots of like superficial numbers, but yet not square, but such as have that proportion that the squares of square numbers have. For the roots of those numbers and the roots of roots are in proportion as numbers are, namely, if the squares be proportional the sides also shallbe proportional (by the 22. of the sixth). But medial lines incommensurable in power, are the roots of roots of numbers, which have not that proportion, that square numbers have. For their roots are the powers of medial lines, which are incommensurable (by the 9 of the tenth). But medial lines commensurable in power only, are the roots of roots of numbers, which have that proportion that simple square numbers have, and not which the squares of squares have. For the roots (which are the powers of the medial lines) are commensurable, but the roots of roots (which express the said medial lines) are incommensurable.  
Wherefore there may be found out infinite medial lines incommensurable in pow●r, by comparing infinite unlike plain numbers the one to the other. For unlike plain numbers, which have not the proportion of square numbers, do make the roots which express the superficieces of medial lines incommensurable (by the 9 of the tenth). And therefore the medial lines containing in power those superficieces are incommensurable in length. For lines incommensurable in power, are always incommensurable in length (by the corollary of the 9 of the tenth).    

¶ The 21. Theorem. The 24. Proposition. A rectangle parallelogram comprehended under medial lines commensurable in length, is a medial rectangle parallelogram.  
SVppose that the rectangle parallelogram AG, be comprehended under these medial right lines AB and BC, which let be commensurable in length. 
Construction.  Then I say, that AC is a medial rectangle parallelogram. Describe (by the 46. of the first) upon the line AB a square AD. 
Demonstration.  Wherefore the square AD is a medial superficies. And forasmuch as the line AB is commensurabl● in length unto the line BC, and the line AB is equal unto the line BD, therefore the line BD is commensurable in length unto the line BC. But 〈◊〉 the line DB is to the line BC, so is the square DA to the parallelogram AC (by the first of the sixth). Wherefore (by the 10. of the tenth) the square DA is commensurable unto the parallelogram AC. But the square DA is medial, for that it is described upon a medial line. Wherefore AC also is a medial parallelogram (by the former Corollary). A rectangled etc.: which was required to be proved.   

¶ The 22● Theorem. The 25. Proposition. A rectangle parallelogram comprehended under medial right lines commensurable in power only, is either rational, or medial.  
And now if the line HK be commensurable in length unto the line HM, that is, unto the line FG, 
Note.  which is equal to the line HM, than (by the 19 of the tenth) the parallelogram NH is rational. But if it be incommensurable in length unto the line FG, than the lines HK and HM are rational commensurable in power only. And so shall the parallelogram HN be medial. Wherefore the parallelogram HN is either rational, or medial. But the parallelogram HN is equal to the parallelogram AG. Wherefore the parallelogram AC is either rational, or medial. A rectangle parallelogram therefore comprehended under medial right lines commensurable in power only, is either rational, or medial: which was required to be demonstrated.  
How to find medial lines commensurable in power only containing a rational parallelogram, and also other medial lines commensurable in power containing a medial parallelogram, shall afterward be taught in the 27. and 28. Propositions of this book.  

¶ A Corollary.  
Hereby it is manifest, that a rectangle parallelogram contained under two right lines, is the mean proportional between the squares of the said lines. 
A Corollary.  As it was manifest (by the first of the sixth) that that which is contained under the lines AB and BC, is the mean proportional between the squares AD and CX. This Corollary is put after the 53. Proposition of this book as an Assumpt, and there demonstrated, which there in his place you shall find. But because it followeth of this Proposition so evidently and briefly without farther demonstration, I thought it not amiss here by the way to note it.    

¶ The 23. Theorem. The 26. Proposition. A medial superficies exceedeth not a medial superficies, by a rational superficies.  
FOr if it be possible, let AB being a medial superficies, exceed AC being also a medial superficies, by DB being a rational superficies. And let there be put a rational right line EF. 
Construction.  And upon the line EF apply a rectangle parallelogram FH, equal unto the medial superficies AB, whose other side let be EH: and from the parallelogram FH take away the parallelogram FG, equal unto the medial superficies AC. 
Demonstration leading to an absurdity.  Wherefore (by the third common sentence) the residue BD is equal to the residue KH. But (by supposition) the superficies DB is rational. Wherefore the superficies KH is also rational. And forasmuch as either of these superficieces AB and AC is medial, and AB is equal unto FH, & AC unto FG: therefore either of these superficieces FH and FG is medial: and they are applied upon the rational line EF. Wherefore (by the 22. of the tenth) either of these lines HE and EG is rational & incommensurable in length unto the line EF. And forasmuch as the superficies DB is rational, and the superficies KH is equal unto it: therefore KH is also rational: and it is applied upon the rational line EF (for it is applied upon the line GK, which is equal to the line EF). Wherefore (by the 20. of the tenth) the line GH is rational and commensurable in length unto the line GK. But the line GK is equal to the line EF. Wherefore the line GH is rational and commensurable in length unto the line EF. But the line EG is rational and incommensurable in length to the line EF. Wherefore (by the 13. of the tenth) the line EG is incommensurable in length unto the line GH. And as the line EG is to the line GH, so is the square of the line EG to the parallelogram contained under the lines EG and GH (by the Assumpt put before the 21. of the tenth). Wherefore (by the 10. of the tenth) the square of the line EG is incommensurable unto the parallelogram contained under the lines EG and GH. But unto the square of the line EG are commensurable the squares of the lines EG and GH, for either of them is rational, as hath before been proved. Wherefore the squares of the lines EG and GH are incommensurable unto the parallelogram contained under the lines EG and GH. But unto the parallelogram contained under the lines EG and GH, is commensurable that which is contained under the lines FG and GH twice (for they are in proportion the one to the other as number is to number, namely, as unity is to the number 2, or as 2. is to 4: and therefore (by the 6. of this book) they are commensurable). Wherefore (by the 13. of the tenth) the squares of the lines EG and GH are incommensurable unto that which is contained under the lines EG and GH twice. (This is more brie●ly concluded by the corollary of the 13. of the tenth). But the squares of the lines EG and GH together with that which is contained under the lines EG and GH twice are equal to the square of the line EH (by the 4. of the second). Wherefore the square of the line EH is incommensurable to the squares of the lines EG and GH (by the 16. of the tenth). But the squares of the lines FG & GH are rational. Wherefore the square of the line EH is irrational. Wherefore the line also EH is irrational. But it hath before been proved to be rational: which is impossible. Wherefore a medial superficies exceedeth not a medial superficies by a rational superficies: which was required to be proved.   

¶ The 4. Problem. The 27. Proposition. To find out medial lines commensurable in power only, containing a rational parallelogram.  
LEt there be put two rational lines commensurable in power only, namely, A and B. 
Construction.  And (by the 13. of the six) take the mean proportional between the lines A and B, and let the same line be C. And as the line A is to the line B, so (by the 12. of the sixth) let the line C be to the line D. 
Demonstration.  And forasmuch as A and B are rational lines commensurable in power only, therefore (by the 21. of the tenth) that which is contained under the lines A and B, that is, the square of the line C. For the square of the line C is equal to the parallelogram contained under the lines A an● B (by the 17. of the sixth) is medial, ●herfore C also is a medial line. And for that as the line A is to the line B, so is the line C to the line D, therefore as the square of the line A is to the square of the line B, so is the square of the line C to the square of the line D (by the 22. of the sixth). But the squares of the lines A and B are commensurable, for the li●●s A and B a●e supposed to be rational commensurable in power only. Wherefore also the squares of the lines C and D are commensurable (by the 10. of the tenth) wherefore the lines C and D are commensurable in power only. And C is a medial line. Wherefore (by the 23. of the tenth) D also is a medial line. Wherefore C and D are medial lines commensurable in power only. Now also I say that they contain a rational parallelogram. For for that as the line A is to the line B, so is the line C to the line D: therefore alternately also (by the 16. of the fift) as the line A is to the line C, so is the line B to the line D. But as the line A is to the line C, so is the line C to the line B: wherefore as the line C is to the line B, so is the line B to the line D. Wherefore the parallelogram contained under the lines C and D is equal to the square of the line B. But the square of the line B is rational. Wherefore the parallelogram which is contained under the lines C and D is also rational. Wherefore there are found out medial lines commensurabl● in pow●r only containing a rational parallelogramme● which 〈◊〉 required to be done.   

The 5. Problem. The 28. Proposition. To find out medial right lines commensurable in power only, containing a medial parallelogram.  
LEt there be put three rational right lines commensurable in power only, namely, A, B, and C, 
Construction.  and (by the 13. of the sixth) take the mean proportional between the lines A and B, & let th● same be D. And as the line B is to the line C, so (by the 12. of the sixth) let the line D be to the line E. And forasmuch as the lines A and B are rational commensurable in power only, 
Demonstration.  therefore (by the 21. of the tenth) that which is contained under the lines A and B, that is the square of the line D, is medial. Wherefore D is a medial line. And forasmuch as the lines B and C are commensurable in power only; and as the line B is to the line C, so is the line D to the line E: wherefore the lines D and E are commensurable in power only (by the corollary of the tenth of this book) but D is a medial line. Wherefore E also is a medial line (by the 23. of this book.) Wherefore D & E are medial lines commensurable in power only. I say also that they contain a medial parallelogram. For for that as the line B is to the line C, so is the line D to the line E: therefore alternately (by the 16 of the fift) as the line B is to the line D, so is the line C to the line E. But as the line B is to the line D, so is the line D to the line A● by converse proportion (which is proved by the corollary of the fourth of the fifth) Wherefore as the line D is to the line A, so is the line C to the line E. Wherefore that which is contained under the lines A & C, is (by the 16. of the six●) equal to that which is contained under the lines D & E. But that which is contained under the lines A and C is medial (by the 21. of the tenth.) Wherefore that which is contained under the lines D and E is medial. Wherefore there are found out medial lines commensurable in power only, containing a medial superficies: which was required to be done.  

An Assumpt.  
To find out two square numbers, which added together make a square number.  
Let there be put two like superficial numbers AB and BC (which how to find out, hath been taught after the 9 proposition of this book) And let them both be either even numbers or odd. And let the greater number be AB. And forasmuch as if from any even number be taken away an even number, or from an odd number be taken away an odd number, the residue shall be even (by the 24. and 26 of the ninth). If therefore from AB being an even number be taken away BC an even number, or from AB being an odd number be taken away BC being also odd: the residue AC shall be even. Divide the number AC into two equal parts in D: wherefore the number which is produced of AB into BC together with the square number of CD, is (by the sixth of the second, as Barlaam demonstrateth it in numbers) equal to the square number of BD. But that which is produced of AB into BC is a square number. For it was proved (by the first of the ninth) that if two like plain numbers multiplying the one the other, produce any number, the number produced shall be a square number. Wherefore there are found out two square numbers, the one being the square number which is produced of AB into BC, and the other the square number produced of CD, which added together make a square number, namely, the square number produced of BD multiplied into himself, forasmuch as they were demonstrated equal to it.   


* A Corollary.  A Corollary.  

To find out two square numbers exceeding the one the other by a square ●umber.  And hereby it is manifest, that there are found out two square numbers, namely, the 〈◊〉 the square number of BD, and the other the square number of CD, so that that number wherein th'one exceedeth the other, the number (I say) which is produced of AB into BC, is also a square number: namely, when A● & BC are like plain numbers. But when they are not like plain numbers, then are there found out two square numbers, the square number of BD, and the square number of DC, whose excess, that is, the number whereby the greater exceedeth the less, namely, that which is produced of AB into BC, is not a square number.   

¶ An Assumpt.  
To find out two square numbers which added together make not a square number.  

An Assumpt.  Let AB and BC be like plain numbers, so that (by the first of the ninth) that which is produced of AB into BC is a square number, and let AC be an even number. And divide C● into two equal par●es in D. Now by that which hath before been said in the former assumpt, it is manifest that the square number produced of AB into BC, together with the square number of CD, is equal to the square number of BD. Take away from CD unity DE. Wherefore that which is produced of AB into BC together with the square of CE is less than the square number of BD. Now then I say that the square num●er produced of AB into BC added to the square number of CE, make not a square number. For if they do make a square number, than that square number which they make, is either greater than the square number of BE, or equal unto it, or less than it. First, greater it cannot be, for it is already proved that the square number produced of AB into BC, together with the square number of CE, is less than the square number of BD. But between the square number of BD, and the square number of BE, there is no mean square number. For the number BD exceedeth the number BE only by unity: which unity can by no means be divided into numbers. Or if the number produced of AB into BC together with the square of the number CE, should be greater than the square of the number BE, then should the self same number produced of AB into BC together with the square of the number CE, be equal to the square of the number BD, the contrary whereof is already proved. Wherefore if it be possible, let that which is produced of AB into BC together with the square number of the number CE be equal to the square number of BE. And let GA be double to unity DE, that is, let it be the number two. Now forasmuch as the whole number AC is by supposition double to the whole number CD, of which the number AG is double to unity DE, therefore (by the 7. of the seventh) the residue, namely, the number GC is double to the residue, namely, to the number EC. Wherefore the number GC is divided into two equal parts in E. Wherefore that which is produced of GB into BC together with the square number of CE is equal to the square number of BE. But that which is produced of AB into BC, together with the square number of CE, is supposed to be equal to the square number of BE wherefore that which is produced of GB into BC together with the square number of CE is equal to that which is produced of AB into BC, together with the square number of CE. Wherefore taking away the square number of CE, which is common to them both, the number AB shall be equal to the number GB, namely, the greater to the less, which is impossible. Wherefore that which is produced of AB into BC together with the square number of CE, is not equal to the square number of BE, I say also that that which is produced of AB into BC together with the square number of CE is not less than the square number of BE. For if it be possible, them shall it be equal to some square number less than the square number of BE. Wherefore let the number produced of AB into BC together with the square of the number CE be equal to the square number of BF. And let the number HA be double to the number DF. Then also it followeth that the number HC is double to the number CF, so that HC also is divided into two equal parts in F, and therefore also the number which is produced of HD into BC, together with the square number of FC, is equal to the square number of the number BF. But by supposition, the number which is produced of AB into BC together with the square number of CE is equal to the square number of BF. Wherefore it followeth that the number produced of AB into BC together with the square number of CE, is equal to that which is produced of HB into BC together with the square number CF, which is impossible. For if it should be equal, then forasmuch as the square of CF is less than the square of CE, the number produced of HB into BC should be greater than th● number produced of AB into BC. And so also should the number HB be greater than the number AB, when yet it is less than it. Wherefore the number produced of AB into BC together with the square number of CE, is not less than the square number of ● E. And it is also proved that it cannot be equal to the square number of BE, neither greater than it. Wherefore that which is produced of AB into BC added to the square number of CE, maketh not a square number. And although it be possible to demonstrate this many other ways, yet this seemeth to us sufficient, lest the matter being over long, should seem to much tedious.    

¶ The 6. Problem. The 29. Proposition. To find out two such rational right lines commensurable in power only, that the greater shall be in power more than the less, by the square of a right line commensurable in length unto the greater.  
LEt there be put a rational line AB, and take also two such square numbers CD and DE, 
Construction.  that their excess CE be not a square number (by the corollary of the first assumpt of the 28. of the tenth) And upon the line AB describe a semicircle AFB. And by the corollary of the 6. of the tenth, as the number DC is to the number CE, so let the square of the line BA be to the square of the line AF. 
Demonstration.  And draw a line from F to B. Now for that as the square of the line BA is to the square of the line AF, so is the number CD to the number CE, therefore the square of the line BA hath to the square of the line AF, that proportion that the number CD hath to the number CE. Wherefore the square of the line BA is commensurable to the square of the line AF (by the 6. of the tenth). But the square of the line AB is rational. Wherefore also the square of the line AF is rational. Wherefore also the line AF is rational. And forasmuch as the number CD hath not unto the number CE that proportion that a square number hath to a square number, therefore neither also hath the square of the line AB to the square of the line AF that proportion that a square number hath to a square number. Wherefore (by the 9 of the tenth) the line AB is unto the line AF incommensurable in length. Wherefore the lines AF and AB are rational commensurable in power only. And for that as the number DC is to the number CE, so is the square of the line AB to the square of the line AF: therefore by conversion or everse proportion which is demonstrated (by the corollary of the 19 of the fifth) as the number CD is to the number DE, so is the square of the line AB to the square of the line BF, which is the excess of the square of the line AB above the square of the line AF (by the assumpt put before the 14. of this book). But the number CD hath to the number DE that proportion that a square number hath to a square number: wherefore the square of the line AB hath to the square of the line BF, that proportion that a square num●er hath to a square number. Wherefore (by the 9 of the tenth) the line AB is commensurable in length unto the line BF. And (by the 47. of the first) the square of the line AB is equal to the squares of the lines AF and FB. Wherefore the line AB is in power more than the line AF by the square of the line BF which is commensurable in length unto the line AB. Wherefore there are found out two such rational lines commensurable in power only, namely, AB and AF, so that the greater line AB is in power more than the less line AF, by the square of the line FB, which is commensurable in length unto the line AB: which was required to be done.   

¶ The 7. Theorem. The 30. Proposition. To find out two such rational lines commensurable in power only, 
Montaureus maketh this an Assumpt: as the Grecke text seemeth to do likewise but without a cause.  that the greater shallbe in power more than the less by the square of a right line incommensurable in length to the greater.  
LEt there be put a rational line AB, and take also (by the 2. assumpt of the 28. of the tenth) two square numbers CE and ED, which being added together make not a square number, and let the numbers CE and ED added together make the number CD. 
Construction.  And upon the line AB describe a sencircle AFB. And (by the corollary of the 6. of the tenth) as the number DC is to the number CE, so let the square of the line AB be to the square of the line AF, and draw a line from F to B. And we may in like sort, 
Demonstration.  as we did in the former proposition, prove that the lines BA and AF are rational commensurable in power only. And for that as the number DC is to the number CE, so is the square of the line AB to the square of the line AF: therefore by conversion (by the corollary of the 19 of the fift) as the number CD is to the number DE, so is the square of the line AB to the square to the line FB. But the number C D hath not to the number DE that proportion that a square n●mbe● h●th to a square number. Wherefore neither also the square of the line AB hath to the square of the line BF that proportion that a square number hath to a square number. Wherefore the line AB is (by the 9 of the tenth) incommensurable in length to the line BF. And the line AB is in power more than the line AF by the square of the right line BF, which is incommensurable in length unto the line AB. Wherefore the lines AB and AF are rational commensurable in power only. And the line AB is in power more than the line AF by the square of the line FB which is commensurable in length unto the line AB ● which was required to be done.  

¶ An Assumpt.  
If there be two right lines having between themselves any proportion: as the one right line is to the other, so is the parallelogram contained under both the right lines to the square of the less of those two lines.  
Suppose that these two right AB and BC be in some certain proportion. 
This Assumpt setteth fo●th nothing ●ls but that which the first o● the s●●t jetteth ●orth, and therefore in s●me examplars it is not found.  Then I say that as the line AB is to the line BC, so is the parallelogram contained under AB and BC to the square of BC. Describe the square of the line BC and let the same be CD, and make perfect the parallelogram AD now it is manifest that as the line AB is to the line BC, so is the parallelogram AD to the parallelogram or square BE (by the first of the sixth). But the parallelogram AD is that which is bontained under the lines AB and BC, for the line BC is equal to the line BD and the parallelogram BE is the square of the line BC. Wherefore as the line AB is to the line BC so is the parallelogram coutained under the lines AB and BC to the square of the line BC, which was required to be proved.    

¶ The 8. Problem. The 31. Proposition. To find out two medial lines commensurable in power only, comprehending a rational superficies, so that the greater shall be in power more than the less by the square of a line commensurable in length unto the greater.  
LEt there be taken (by the 29. of the tenth) two rational lines commensurable in power only A and B, 
Construction.  so that let the line A being the greater be in power more than the line B, being the less by the square of a line commensurable in length unto the line A ● And let the square of the line C be equal to the parallelogram contained under the lines A and B which is done by finding out the mean proportional line, namely the line C between the lines A and B (by the 13. of the sixth). Now the parallelogram contained under the lines A and B is medial (by the 21. of this book). Wherefore (by the corollary of the 23. of the tenth) the square also of the line C is medial. Wherefore the line C also is medial: Unto the square of the line B let the parallelogram contained under the lines C and D be equal (by finding out a third line proportional) namely the line D to the two lines C and B (by the 11. of the sixth). But the square of the line B is rational. 
Demonstration.  Wherefore the parallelogram contained under the line C and D is rational. And for that as the line A is to the line B, so is the parallelogram contained under the lines A and B to the square of the line B (by the assumpt going before). But unto the parallelogram contained under the lines A and B is equal the square of the line C, and unto the square of the line B is equal the parallelogram contained under the lines C and D, as it hath now been proved: therefore as the line A is to the line B, so is the square of the line C to the parallelogram contained under the lines C, D. But as the square of the line C is to that which is contained under the lines C and D, so is the line C to the line D. Wherefore as the line A is to the line B, so is the line C to the line D. But (by supposition) the line A is commensurable unto the line B in power only. Wherefore (by the 11. of the tenth) the line C also is unto the line D commensurable in power only. But the line C is medial. Wherefore by the 23● of the tenth) the line D also is medial. And for that as the line A is to the line B, so is the line C to the line D: but the line A is in power more than the line B, by the square of a line commensurable in length unto the line A (by supposition). Wherefore the line C also is in power more than the line D by the square of a line commensurable in length unto the line C. Wherefore there are found out two medial lines C and D commensurable in power only comprehending a rational superficies, and the line C is in power more than the line D, by the square of a line commensurable in length unto the line C. And in like sort may be found out two medial lines commensurable in power only containing a rational superficies, so that the greater shallbe in power more than the less by the square of a line incommensurable in length to the greater, namely, when the line A is in power more than the line B by the square of a line incommensurable in length unto the line A, which to do is taught by the 30. of this book.  
The self same construction remaining, that part of this proposition from these words. And for that as the line A is to the line B, to these words, But (by supposition) the line A is commensurable unto the line B, may more easily be demonstrated after this manner. The lines C, B, D, are in continual proportion by the second part of the 17. of the sixth. But the lines A, C, D are also in continual proportion by the same. Wherefore by the 11. of the fifth, as the line A is to the line C, so is the line B to the line D. Wherefore alternately as the line A is to the line B so is the line C to the line D. etc. which was required to be done.  

¶ An assumpt.  
If there be three right lines having between themselves any proportion: as the first is to the third, so is the parallelogram contained under the first and the second, to the parallelogram contained under the second and the third.  
Suppose that these three lines AB, B C, and CD be in some certain proportion. Then I say that as the line AB is to the line CD, so is the parallelogram contained under the lines AB and BC to the parallelogram contained under the lines BC and CD. 
Construction.  From the point A raise up unto the line AB a perpendicular line AE, and let AE be equal to the line BC: and by the point E draw unto the line AD a parallel line EK: and by every one of the points B, C, and D draw unto the line AE parallel lines BF, CH, and DK. And for that as the line AB is to the line BC so is the parallelogram AF to the parallelogram BH (by the first of the sixth): 
Demonstration.  and as the line BC is to the lin● CD, so is the parallelogram BH to the parallelogram CK. Wherefore of equality as the line AB is to the line CD, so is the parallelogram AF to the parallelogram CK. But the parallelogram AF is that which is contained under the lines AB and BC, for the line AE is put equal to the line BC. And the parallelogram CK is that which is contained under the lines BC and CD, for the line BC is equal to the line CH, for that the line CH is equal to the line AE (by the 34. of the first). If therefore there be three right lines having between themselves any proportion: as the first is to the third, so is the parallelogram contained under the first and the second, to the parallelogram contained under the second and the third: which was required to be demonstrated.    

¶ The 9 Problem. The 32. Proposition. To find out two medial lines commensurable in power only, comprehending a medial superficies, so that the greater shall be in power more than the less, by the square of a line commensurable in length unto the greater.  
LEt there be taken three rational lines commensurable in power only, A, B, C, so, that (by the 29. of the tenth) let the line A be in power more than the line C, by the square of a line commensurable in length unto the line A. 
Construction.  And unto the parallelogram contained under the lines A & B, let the square of the line D be equal. But that which is contained under the lines A and B is medial. Wherefore (by the Corollary of the 23. of the tenth) the square of the line D also is medial. Wherefore the line D also is medial. And unto that which is contained under the lines B and C, let be equal that which is contained under the lines D and E (which is done by ●inding out a fourth line proportional unto the lines D, B, C, which let be the line E). 
Demonstration.  And for that (by the Assumpt going before) as that which is contained under the lines A and B is to that which is contained under the lines B and C, so is the line A to the line C. But unto that which is contained under the lines A & B, is equal the square of the line D, and unto that which is contained under the lines B & C, is equal that which is contained under the lines D and E. Wherefore as the line A is to the line C, so i● the square of the line D, to that which is contained under the lines D and E. But as the square of the line D is to that which is contained under the lines D and E, so is the line D to the line E (by the Assumpt put before the 22. of the tenth). Wherefore as the line A is to the line C, so is the line D to the line E. But the line A is unto the line C commensurable in power only. Wherefore the line D is unto the line E commensurable in power only. But D is a medial line. Wherefore (by the 23. of the tenth) E also is a medial line. And for that as the line A is to the line C, so is the line D to the line E, and the line A is in power more than the line C, by the square of a line commensurable in length unto the line A. Wherefore (by the 14. of the tenth) D is in power more than E, by the square of a line commensurable in length unto the line D. I say moreover that that which is contained under the lines D and E is medial. For forasmuch as that which is contained under the lines B & C, is equal to that which is contained under the lines D and E: but that which is contained under the lines B and C is medial. Wherefore that which is contained under the lines D and E is also medial. Wherefore there are found out two medial lines D and E, commensurable in power only, comprehending a medial superficies, so that the greater is in power more than the less, by the square of a line commensurable in length to the greater: which was required to be done. And thus i● i● eu●de●t, how in like sort may be found out two medial lines comm●●surable in power. only, 
A Corollary.  containing a medial superficies, so that the greater shall be in power more the● the less, by the square of a line incommensurable in length unto the greater. When the line A is in power more than the line C, by the square of a line incommensurable in length unto the line A: as the thirteth teacheth us.  

1. ¶ An Assumpt.  
Suppose that there be a rectangle triangle ABC, having the angle BAC a right angle. And (●y the 12. of the first) from the point A to the right line BC, a perpendicular line being drawn AD: then I say first, that the parallelogram contained under the lines C● and BD, is equal to the square of the line BA. Secondly I say, that the parallelogram contained under the lines BC and CD, is equal to the square of the line CA Thirdly I say, that the parallelogram contained under the lines BD and DC, is equal to the square of the line AD. And fourthly I say, that the parallelogram contained under the lines BC & AD, is equal to the parallelogram contained under the lines BA & AC.  
As touching the second, that the parellelogramme contained under the lines B● and CD, is equal to the square of the line AC is by the self same reason proved. For the triangle ABC is like to the triangle ADC. Wherefore as the line BC is to the line AC, 
I Dee  so is the line, AC to the line DC. 
* The second Corollary.  Wherefore the parallelogram contained under the lines BC and CD, is equal to the square of the line AC. 
* Therefore if you divide the square of the side AC, by the side BC, the portion DC, will be the product. etc. as in the former Corollary.  As touching the third, that the parallelogram contained under the lines BD and DC, is equal to the square of the line DA, is thus proved. For, forasmuch as if in a rectangle triangle be drawn from the right angle to the base a perpendicular line, the perpendicular so drawn is the mean proportional between the segments of the base (by the corollary of the 8. of the sixth): therefore as the line BD is to the line DA, so is th● line AD to the line DC. Wherefore (by the 1●. of the sixth) the parallelogram contained under the lines BD and DC, is equal to the square of the line DA. As touching the fourth, that the parallelogram contained under the lines BC and AD, is equal to the parallelogram contained under the lines BA and AC, is thus proved. For forasmuch as (as we have already declared) the triangle ABC is like, and therefore equiangle, to the triangle ABD, therefore as the line BC is to the line AC, 
I D●e  so is the line BA to the line AD (by the 4. of the sixth). 
* The third Corollary.  But if there be four right lines proportional, that which is contained under the first and the last, is equal to that which is contained under the two means (by the 16. of the sixth). Wherefore that which is contained under the lines BC and AD, is equal to that which is contained under the lines BA and AC. 
* Therefore if the parallelogram of BA, and AC, be divided by BC, the product will give the perpendicular D A. These three Corollaryes in practice logistical and Geometrical are profitable.   
I say moreover, that if there be made a parallelogram complete, 
another demonstration of this fourth part of the determination.  contained under the lines BC and AD, which let be EC: and if likewise be made complete the parallelogram contained under the lines BA and AC, which let be AF, it may by an other way be proved that the parallelogram EC is equal to the parallelogram AF. For, forasmuch as either of them is double to the triangle ACB (by the 41. of the first): and things which are double to one and the self same thing, are equal the one to the other. Wherefore that which is contained under the lines BC and AD, is equal to that which is contained under the lines BA and AC.   

2. ¶ An Assumpt.  
If a right line be divided into two unequal parts: as the greater part is to the less, 
An Assumpt.  so is the parallelogram contained under the whole line and the greater part, to the parallelogram contained under the whole line and the less part.  
This Assumpt differeth little from the first Proposition of the sixth book.   

3. ¶ An Assumpt.  
If there be two unequal right lines, and if the less be divided into two equal parts: the parallelogram contained under the two unequal lines, is double to the parallelogram contained under the greater line & half of the less line.  
Suppose that there be two unequal right lines AB and BC, of which le● AB be the greater, and divide the line BC into two equal parts in the point D. Th●n I say, that the parallelogram contained under the lines AB & BC, is double to the parallelogram contained under the lines AB and BD. From the point B raise up upon the right line BC, a perpendicular line BE, and let BE be equal to the line BA. And drawing from the point C and D, the lines CG and DF parallels and equal to BE: and then drawing the right line GFE, the figure is complete. N●● for that a●●he line DB is to the line DC, so is the parallelogram BF to the parallelogram DG (by the 1. of the sixth): therefore by composition of proportion, as the whole line BC is to the line DC, so is the parallelogram BG to the parallelogram DG (by the 18. of the fift). But the line BC is double to the line DC. Wherefore the parallelogram BG is double to the parallelogram DG. But the parall●logramme BG is contained under the lines AB and BC) for the line AB is equal to the line BE (and the parallelogram DG is contained under the lines AB and BD) for the line BD is equal to the line DC, and the line AB to the line DF: which was required to be demonstrated.    

¶ The 10. Problem. The 33. Proposition. To ●inde out two right lines incommensurable in power, whose squares added together make a rational superficies, and the parallelogram contained under them make a medial superficies.  
TAke by the 30. of the tenth, two rational right lines commensurable in power only, namely, AB and BC, so that let the line AB, being the greater, be in power more than the line BC being the less, 
Construction.  by the square of a line incommensurable in length unto the line AB. And by the 10. of the first, divide the line BC into two equal parts in the point D. And upon the line AB apply a parallelogram equal to the square either of the line BD or of the line DC, and wanting in figure by a square, by the 28. of the sixth, and let that parallelogram be that which is contained under the lines AE and EB. And upon the line AB describe a semicircle AFB. And by the 11. of the first, from the point E, raise up unto the line AB, a perpendicular line EF, cutting the circumference in the point F. And draw lines from A to F, 
Demonstration.  and from F to B. And forasmuch as there are two unequal right lines AB and BC, and the line AB is in power more than the line BC, by the square of a line incommensurable in length unto AB, and upon the line AB is applied a parallelogram equal to the fourth part o● the square of the line BC, that is, to the square of the half of the line BC, and wanting in ●igure by a square, and the said parallelogram is that which is contained under the lines AE and EB. wherefore by the 2. part of the 18. of the tenth, the line AE is incommensurable in length unto the line EB. But as the line AE is to the line EB, so is the parallelogram contained under the lines BA and AE, to the parallelogram contained under the lines AB and BE by the second assumpt before put. And that which is contained under the line BA and AE is equal to the square of the line AF, by the second part of the first assumpt before put. And that which is contained under the lines AB and BE is by the first part of the same assumpt, equal to the square of the line BF. Wherefore the square of the line AF is incommensurable to the square of the line BF. Wherefore the lines AF and BF are incommensurable in power. And forasmuch as AB is a rational line (by supposition) therefore (by the 7 definition of the tenth) the square of the line AB is rational. 
The first part of the determination concluded.  Wherefore also the squares of the lines AF and FB added together make a rational superficies. For (by the 47. of the first) they are equal to the square of the line AB. Again forasmuch as (by the third part of the first assumpt going before) that which is contained under the lines AE and EB, is equal to the square of the line EF. But by supposition that which is contained under the lines AE and EB is equal to the square of the line BD. Wherefore the line FE is equal to the line BD. Wherefore the lin● BG is double to the line ● E. Wherefore (by the third assumpt going before) that which is contained under the lines AB and BC, is double to that which is contained under the lines AB and EF. But that which is contained under the lines AB and BC, is by supposition medial. Wherefore (by the corollary of the 23. of the tenth) that which is contained under the lines AB and EF is also medial, but that which is contained under the lines AB and EF, 
The second part concluded.  is (by the last part of the first assumpt going before) equal to that which is contained under the lines AF and FB. Wherefore that which is contained under the lines AF & FB is a medial superficies. And it is proved, that that which is composed of the squares of the lines AF and FB added together is rational. 
The total conclusion.  Wherefore there are found out two right lines AF and FB incommensurable in power, whose squares added together, make a rational superficies, and the parallelogram contained under them, is a medial superfici●s● which was required to be done.   

¶ The 11. Problem. The 34. Proposition. To find out two right lines incommensurable in power, whose squares added together make a medial superficies, and the parallelogram contained under them, make a rational superficies.  
TAke (by the 31. of the tenth) two medial lines AB and BC, commensurable in power only, comprehending a rational superficies, 
Construction.  so that let the line AB be in power more than the line BC by the square of a line incommensurable in length unto the line AB. And describe upon the line AB a semicircle ADB. And by the 10. of the first, divide the line BC unto two equal parts in the point E. And (by the 28. of the sixth) upon the line AB apply a parallelogram equal to the square of the line BE, and wanting in figure by a square, and let that parallelogram be that which is contained under the lines AF and FB. Wherefore the line AF is incommensurable in length unto the line FB (by the 2. part of the 18. of the tenth.) And from the point F unto the right line AB, raise up (by the 11. of the first) a perpendicular line FD, and draw lines from A to D, and from D to B. 
Demonstration.  And forasmuch as the line AF is incommensurable unto the line FB: but (by the second assumpt going before the 33. of the tenth) as the line AF is to the line FB, so is the parallelogram contained under the lines BA and AF, to the parallelogram contained under the lines BA and BF, wherefore (by the tenth of the tenth) that which is contained under the lines BA and AF is incommensurable to that which is contained under the lines AB and BF: but that which is contained under the lines BA and AF is equal to the square of the line AD, and that which is contained under the lines AB and BF is also equal to the square of the line DB (by the second part of the first assumpt going before the 33. of the tenth) wherefore the square of the line AD is incommensurable to the square of the line DB. Wherefore the lines AD and DB are incommensurable in power. And forasmuch as the square of the line AB is medial, therefore also the superficies made of the squares of the lines AD and DB added together is medial. 
The first part of the determination concluded.  For the squares of the lines AD and DB are (by the 47. of the first) equal to the square of the line AB. And forasmuch as the line BC is double to the line FD (as it was proved in the proposition going before) therefore the parallelogram contained under the lines AB and and BC is double to the parallelogram contained under the lines AB and FD (by the third assumpt going before the 33. proposition) wherefore it is also commensurable unto it (by the sixth of the tenth) But that which is contained under the lines AB and BC is supposed to be rational. Wherefore that which is contained under the lines AB and FD is also rational. But that which is contained under the lines AB and FD, is equal to that which is contained under the lines AD and DB (by the last part of the first assumpt going before the 33. of the tenth) Wherefore that which is contained under the lines AD and DB is also rational. 

The second part concluded.  
The total conclusion.   Wherefore there are ●ound out two right lines AD and DB incommensurable in power, whose squares added together, make a medial superficies, and the parallelogram contained under them, make a rational superficies: which was required to be done.   

¶ The 12. Problem. The 35. Proposition. To find out two right lines incommensurable in power, whose squares added together, make a medial superficies, and the parallelogram contained under them, make also a medial superficies, which parallelogram moreover, shall be incommensurable to the superficies made of the squares of those lines added together.  
TAke (by the 32. of the tenth) two medial lines AB and BC commensurable in power only, 
Construction.  comprehending a medial superficies, so that let the line AB be in power more than the line BC by the square of a line incommensurable in length unto the line AB. And upon the line AB describe a semicircle ADB, and let the rest of the construction be as it was in the two former propositions. And forasmuch as (by the 2 part of the 18. of the tenth) the line AF is incommensurable in length unto the line FB, 
Demonstration.  therefore the line AD is incommensurable in power unto the line DB (by that which was demonstrated in the proposition going before). And forasmuch as the square of the line AB is medial, therefore that also which is composed of the squares of the lines AD and DB (which squares are equal to the square of the line AB by the 47. of the first) is medial. 
The first part concluded.  And forasmuch as that which is contained under the lines AF and FB, is equal to either of the squares of the lines EB and FD, for by supposition the parallelogram contained under the lines AF and FB is equal to the square of the line EB, and the same parallelogram is equal to the square of the line DF (by the third part of the first assumpt going before the 33. of the tenth). Wherefore the line BE is equal to the line DF. Wherefore the line BC is double to the line FD. Wherefore that which is contained under the lines AB and BC is double to that which is contained under the lines AB and FD. Wherefore they are commensurable by the sixth of this book: but that which is contained under the lines AB and BC is medial by supposition. Wherefore also that which is contained under the lines AB and FD is medial (by the corollary of the 23, of the tenth) but that which is contained under the lines AB and FD, is (by the fourth part of the first assumpt going before the 33. of the tenth) equal to that which is contained under the lines AD and DB: 
The second part concluded.  wherefore that which is contained under the lines AD and DB is also medial. And forasmuch as the line AB is incommensurable in length unto the line BC. But the line BC is commensurable in length unto the line BE. Wherefore (by the 13● of the tenth) the line AB is incommensurable in length unto the line BE. Wherefore the square of the line AB is incommensurable to that which is contained under the lines AB and BE (by the first of the sixth and 10. of this book) But unto the square of the line AB are equal the squares of the lines AD and DB added together (by the 47. of the first): and unto that which is contained under the lines AB and BE, is equal that which is contained under the lines AB and FD, that is, which is contained under AD and DB. For the parallelogram contained under the lines AB and FD is equal to the parallelogram contained under the lines AD and DB (by the last part of the first assumpt going before the 33. of this tenth book). 
The third part concluded.  Wherefore that which is composed of the squares of the lines AD and DB is incommensurable to that which is contained under the lines AB and DB. Wherefore there are found out two right lines AD and DB incommensurable in power, whose squares added together, make a medial superficies, 
The total conclusion.  and the parallelogram contained under them, make also a medial superficies, which parallelogram moreover is incommensurable to the superficies composed of the squares of those lines added together, which was required to be done.   

The beginning of the Senaries by Composition. ¶ The 2●. Theorem. The 36. Proposition. If two rational lines commensurable in power only be added together: 
The first Senary by composition.  the whole line is irrational, and is called a binomium, or a binomial line.  
〈…〉 B and BC is incommensurable to the square of the line BC. But unto the parallelogram contained under the lines AB and BC is commensurable the parallelogram contained under AB and BC twice (by the 6. of the tenth): wherefore that which is contained under AB and BC twice is incommensurable to the square of the line BC (by the 13 of the tenth). But unto the square of the line BC is commensurable that which is composed of the squares of the lines AB and BC (by the 15. of the tenth), for by supposition the lines AB and BC are commensurable in power only. Wherefore (by the 13. of the tenth) that which is composed of the squares of the lines AB and BC added together is incommensurable to that which is contained under the lines AB and BC twice. Wherefore (by the 16. of the tenth) that which is contained under AB and BC twice together with the squares of the lines AB and BC, which (by the 4. of the second) is equal to the square of the whole line AC, is incommensurable to that which is composed of the squares of AB and BC added together. But that which is composed of the squares of AB and BC added together is rational, for it is commensurable to either of the squares of the lines AB and BG of which either of them is rational by supposition: wherefore the square of the line AC is (by the 10. definition of the tenth) irrational. Wherefore the line AC also is irrational, and is called a binomial line.  

Definition of a binomial line.  This proposition showeth the generation and production of the second kind of irrational lines which is called a binomium, or a binomial line. The definition whereof is fully gathered out of this proposition, and that thus.  
A binomium or a binomial line, is an irrational line composed of two rational lines commensurable the one to the other in power only. And it is called a binomium, that is, having two names, because it is made of two such lines as of his parts which are only commensurable in power and not in length: and therefore each part or line, or at the lest the one of them, as touching length, is uncertain and unknown. Wherefore being joined together their quantity cannot be expressed by any one number or name, but each part remaineth to be severally named in such sort as it may. And of these binomial lines there are six several kinds, 
Six kinds of binomial lines.  the first binomiall, the second, the third, the fourth, the fifth, and the sixth, of what nature and condition each of these is shallbe known by their definitious which are afterward set in their due place.   

¶ The 25. Theorem. The 37. Proposition. If two medial lines commensurable in power only containing a rational superficies, be added together: the whole line is irrational, and is called a first bimedial line.  
LEt these two medial lines AB and BC being commensurable in power only, and containing a rational superficies (the 27. of the tenth teacheth to find out two such lines) be composed. Then I say that the whole line AC is irrational. For as 〈◊〉 said in the proposition next going before that which is composed of the squares of the 〈◊〉 AB and BC is incommensurable to that which is contained under the lines AB and BC twis● wherefore (by the 16. of the tenth) that which is composed of the squares of the lines AB and BC together with that which is contained under the lines AB and BC twice, 
Demonstration.  that is, the square of the line AC is incommensurable to that which is contained under the lines AB and BC twice. But that which is contained under the lines AB and BC twice i● commensurable to that which is contained under the lines AB and BC once (by the 6. of the tenth) wherefore the square of the whole line AC, is (by the 13● of the tenth) incommensurable ●o that which is contained under the lines AB and BC once. But by supposition the lines AB and BC comprehend a rational superficies. Wherefore the square of the whole line AC is irrational: wherefore also the line AC is irrational. And it is called a first bimedial line.  
The third irrational line which is called a first bimedial line, is showed by this proposition, and the definition thereof is by it made manifest, 
Definition of a first bimedial line.  which is this. A first bimedial line, is an irrational line, which is composed of two medial lines commensurable in power only containing a rational parallelogram. It is called a first bimedial line, by cause the two medial lines or parts whereof it is composed contain a rational superficies, which is preferred before an irrational.   

¶ The 26. Theorem. The 38. Proposition. If two medial lines commensurable in power only containing a medial superficies, be added together: the whole line is irrational, and is called a second bimedial line.  
LEt these two medial lines AB and BC being commensurable in power only, and containing a medial superficies (the 28. of the tenth teacheth to find● out two such lines) be added together. 
Construction.  Then I say that the whole line AC is irrational. Take a rational line DE. And (by the 44. of the first) upon the line DE apply the parallelogram DF, equal to the square of the line AC. whose other side let be the line DG. And forasmuch as the square of the line AC is (by the 4. of the second) equal to that which is composed of the squares of the lines AB and BC, 
Demonstration.  together with that which is contained under the lines AB and BC twice: but the square of the line AC is equal to the parallelogram DF. Wherefore the parallelogram DF is equal to that which is composed of the squares of the lines AB and BC together with that which is contained under the lines AB and BC twice. Now then again (by the 44, of the first) upon the line DE apply the parallelogram EH equal to the squares of the lines AB and BC. Wherefore the parallelogram remaining, namely, HF, is equal to that which is contained under the lines AB and BC twice. And forasmuch as either of these lines AB and BC is medial, therefore the squares of the lines AB and BC are also medial. And that which is contained under the lines AB and BC twice is (by the corollary of the ●4. of the tenth) medial. For by the 6. of this book it is commensurable ●● that 〈◊〉 is contained under the lines AB and BC once, which is by supposition medial. 〈…〉 squares of the lines. AB and BC is equal the parallelogram EH, and unto that 〈◊〉 contained under the lines AB and BC twice is equal the parallelogram HE 〈…〉 either of these parallelograms HE and AF is medial: and they are applied upon the rational line ED. Wherefore (by the 22. of the tenth) either of these lines DH and HG is a rational line, and incommensurable in length unto the line DE. And forasmuch as (by supposition) the line AB is incommensurable in length unto the line BC. But as the line AB is to the line BC ● so is the square of the line AB to the parallelogram which is contained under the lines AB and BC (by the first of the sixth). Wherefore (by the 10 of this book●) the square of the line AB is incommensurable to the parallelogram contained under the lines AB and BC. But to the square of the line AB is commens●able that which is composed of the squares of the lines AB and BC (by the 15. of the tenth). For the squares of the lines AB and BC are commensurable (when as the lines AB and BC are put to be commensurable in power only). And to that which is contained under the lines AB and BC is commensurable that which is contained under the lines AB and BC twice (by the 6 of the tenth) wherefore that which is composed of the squares of the lines AB and BC is incommensurable to that which is contained under the lines AB and BC twice. But to the squares of the lines AB and BC is equal the parallelogram EH. And to that which is contained under the lines AB and BC twice is equal the parallelogram FH. Wherefore the parallelogram FH is incommensurable to the parallelogram HERALD Wherefore the line DH is incommensurable in length to the line HG (by the 1 of the sixth and 10 of this book). And it is proved that they are rational lines. Wherefore the lines DH & HG are rational commensurable in power only. Wherefore (by the 36. of the tenth) the whole line DG is irrationally And the line DE is rational. But a rectangle superficies comprehended under a rational line and an irrational line is (by the corollary added after the 21 of the tenth) irrational. Wherefore the superficies DF is irrational. And the line also which containeth it in power is irrational. But the line AC containeth in power the superficies DF. Wherefore the line AC is irrational. And it is called a second bimedial line.  
This Proposition showeth the generation of the fourth irrational line, called a second bimedial line. 
Definition of a second bimedial line.  The definition whereof is evident by this Proposition, which is thus. A second bimedial line is an irrational line, which is made of two medial lines commensurable in power only joined together, which comprehend a medial superficies. And it is called a second bimediall, because the two medial lines of which it is composed, contain a medial superficies, and not a rational. Now a medial is by nature & in knowledge after a rational.   

¶ The 27. Theorem. The 39 Proposition. If two right lines incommensurable in power be added together, having that which is composed of the squares of them rational, and the parallelogram contained under them medial: the whole right line is irrational, and is called a greater line.  
LEt t●ese two right lines AB and BC being incommensurable in power only, and making that which is required in the Proposition (The 33. of the tenth teacheth to find out two such lines) be added together. Then I say, that the whole line AC is irrational. 
Demonstration.  For forasmuch as (by supposition) the parallelogram contained under the lines AB and BC is medial, therefore the parallelogram contained twice under the lines AB and BC is medial. (For that which is contained under AB and BC twice, is commensurable to that which is contained under AB and BC once (by the 6. of the tenth). Wherefore (by the Corollary of the 23. of the tenth) that which is contained under AB & BC twice, is medial). But by supposition that which i● composed of the squares of the lines AB and BC, is rational. Wherefore that which is contained under the lines AB and BC twice is incommensurable to that which is composed of the squares of the lines AB and BC. Wherefore (by the 16. of the tenth) that which is composed of the squares of the lines AB and BC together with that which is contained under the lines AB & BC twice, which is (by the 4. of the second) equal to the square of the line AC, is incommensurable to that which is composed of the squares of the lines AB and BC. But that which is composed of the squares of the lines AB and BC, is rational. Wherefore the square of the whole line AC, is irrational. Wherefore the line AC also is irrational. And is called a greater line. And it is called a greater line for that that which is composed of the squares of the lines AB & BC which are rational, is greater than that which is contained under the lines AB and BC twice, which are medial. Now it is meet that the name should be given according to the property of the rational.  

An Assumpt.  
This Proposition teacheth the production of the fift irrational line, which is called a greater line: which is by the sense of this Proposition thus defined.  
A greater line is an irrational line, which is composed of two right lines which are incommensurable in power, 
Definition of a greater line.  the squares of which added together, make a rational superficies, and the parallelogram which they contain, is medial. It is therefore called a greater line, as Theon saith, because the squares of the two lines of which it is composed, added together being rational, are greater than the medial superficies contained under them twice. And it is convenient that the denomination be taken of the propriety of the rational part, rather than of the medial part.    

¶ The 28. Theorem. The 40. Proposition. If two right lines incommensurable in power be added together, having that which is made of the squares of them added together medial, and the parallelogram contained under them rational: the whole right line is irrational, and is called a line containing in power a rational and a medial superficies.  
In this Proposition is taught the generation of the sixth irrational line, which is called a line whose power is rational and medial. The definition of which is gathered of this Proposition after this manner.  
A line whose power is rational and medial, is an irrational line which is made of two right lines incommensurable in power added together, 
Definition of a line whose power is rational and medial.  whose squares added together make a medial superficies, but that supersicies which they contain is rational. The reason of the name is before set forth in the Proposition.   

¶ The 29. Theorem. The 41. Proposition. If two right lines incommensurable in power be added together, having that which is composed of the squares of them added together medial, and the parallelogram contained under them medial, and also incommensurable to that which is composed of the squares of them added togethers the whole right line is irrational, and is called a line containing in power two medials.  
In this proposition is taught the nature of the 7. kind of irrational lines which is called a line whose power is two medials. The definition whereof is taken of this proposition after this manner.  
A line whose power is two medials, is an irrational line which is composed of two right lines incommensurable in power, 
Definition of a li●e containing in power two medials.  the squares of which added together, make a medial superficies, and that which is contained under them is also medial, and moreover it is incommensurable to that which is composed of the two squares added together.  
The reason why this line is called a line whose power is two medials, was before in the end of the demonstration declared.  
And that the said irrational lines are divided one way only, that is, in one point only, into the right lines of which they are composed, and which make every one of the kinds of those irrational lines, shall strait way be demonstrated: but first will we demonstrate two assumptes here following.  

¶ An Assumpt.  
Take a right line and let the same be AB, and divide it into two unequal parts in the point C, 
An Assumpt.  and again divide the same line AB into two other unequal parts, in an other point namely, in D, and let the line AC (by supposition) be greater than the line DB. Then I say that the squares of the lines AC and BC added together, are greater than the squares of the lines AD and DB added together. Divide the line AB (by the 10. of the first) into two equal parts in the point E. And forasmuch as the line AC is greater than the line DB, take away the line DC which is common to them both: wherefore the residue AD is greater than the residue CB, but the line AE is equal to the line EB. Wherefore the line DE is less than the line EC. Wherefore the points C and D are not equally distant from the point E, which is the point of the section into two equal parts. And forasmuch as (by the 5. of the second) that which is contained under the lines AC and CB together with the square of the line EC is equal to the square of the line EB. And by the same reason that which is contained under the lines AD and DB together with the square of the line DE, is also equal to the self same square of the line EB: wherefore that which is contained under the lines AC and CB together with the square of the line EC is equal to that which is contained under the lines AD and DB together with the square of the line DE: of which the square of the line DE is less than the square of the line EC (for it was proved that the line DE is less than the line EC). Wherefore the parallelogram remaining, contained under the lines AC and CB is less than the parallelogram remaining contained under the lines AD and DB. Wherefore also that which is contained under the lines AC and CB twice is less than that which is contained under the lines AD and DB twice. But (by the fourth of the second) the square of the whole line AB is equal to that which is composed of the squares of the lines AC and CB together with that which is contained under the lines AC and CB twice, and by the same reason the square of the whole line AB is equal to that which is composed of the squares of the lines AD and DB together with that which is contained under the lines AD and DB twice: wherefore that which is composed of the squares of the lines AC and CB together with that which is contained under the lines AC and CB twice, is equal to that which is composed of the squares of the lines AD and DB, together with that which is contained under the lines AD and DB twice. But it is already proved that that which is contained under the lines AC and CB twice, is less than that which is contained under the lines AD & DB twice. Wherefore the residue, namely, that which is composed of the squares of the lines AC and CB is greater than the residue, namely, then that which is composed of the squares of the lines AD and DB: which was required to be demonstrated.   

¶ An Assumpt.  
A rational superficies exceedeth a rational superficies, by a rational superficies.  
Let AD be a rational superficies, and let it exceed AF being also a rational superficies by the superficies ED. Then I say that the superficies ED is also rational. For the parallelogram AD is commensurable to the parallelogram AF, for that either of them is rational. Wherefore (by the second part of the 15. of the tenth) the parallelogram AF is commensurable to the parallelogram ED. But the the parallelogram AF is rational. Wherefore also the parallelogram ED is rational.    

¶ The 30. Theorem. The 42. Proposition. A binomial line is in one point only divided into his names. 
The second Senary by composition.   
SVppose that AB be a binomial line, and in the point G let it be divided into his names, that is, into the lines whereof the whole line AB is composed. Wherefore these lines AC and CB are rational commensurable in power only. Now I say that the line AB cannot in any other point besides C be divided into two rational lines commensurable in power only. 
Demonstration leading to an impossibility.  For if it be possible, let it be divided in the point D, so that let the lines AD and DE b● rational commensurable in power only. First this manifest, that neither of these points C and D divideth the right line AB into two equal parts. Otherwise the lines AC and CB should be rational commensurable in length and so likewise should the lines AD and DB be. For every line measureth itself, and any other line equal to itself. Moreover the line DB is either one and the same with the line AC● that is, is equal to the line AC, o● else it is greater-then the line AC, either else it is less than it. If DB be equal to the line AC, then putting the line DB upon the line AC each ends of the one, shall agreed with each ends of the other. Wherefore putting the point B upon the point A, the point D also shall fall upon the point C, and the line AD which is the rest of the line AC, shall also be equal to the line CB, which is the rest of the line DB. Wherefore the line AB is divided into his names in the point C. And so also shall the line AB being divided in the point D be divided in the self ●ame point that the self same line AB was before divided in the point C, which is contrary to the supposition. For by supposition it was divided in sundry points, namely, in C & D. But if the line DB be greater● the the line AC, let the line AB be de●ided into two equal parts in the point E. Wherefore the points C & D shall not equally be distant from the point E (Now (by the first assupt going before this proposition) that which is composed of the squares of the lines AD & DB is greater than that which is composed of the squares of the lines AC & CB● But that which is composed of the squares of the lines AD & DB together with that which is contained under the lines AD & DB twice, is equal to that which is composed of the squares of the lines AC & CB together with that which is contained under the lines AC and CB twice, for either of them is equal to the square of the whole line AB (by the 4. of the second) wherefore how much that which is composed of the squares of the lines AD and DB added together is greater than that which is composed of the squares of the lines AC and CB added together, so much is that which is contained under the lines AC and CB twice greater than that which is contained under the lines AD and DB twice. But that which is composed of the squares of the lines AD and DB exceedeth that which is composed of the squares of the lines AC and CB by a rational superficies (by the 2. assumpt going before this propositions) For that which is composed of the squares of the lines AD and DB is rational, and so also is that which is composed of the squares of the lines AC and CB: for the lines AD and DB are put to be rational commensurable in power only, and so likewise are the lines AC and CB. Wherefore also that which is contained under the lines AC and CB twice, exceedeth that which is contained under the lines AD & DB twice by a rational superficies, when yet notwithstanding they are both medial superficieces (by the 21. of the tenth) which (by the 26. of the same) is impossible. And if the line DB be less than the line AC, we may by the like demonstration prove the self same impossibility. Wherefore a binomial line is in one point only divided into his names: Which was required to be demonstrated.  

〈…〉 ollary added by Flussates.  
Two ration 〈…〉 surable in power only being added together cannot be equal to two other rational line 〈…〉 in power only added together. 
A Corollary.  For either of them should make a binomia 〈…〉 so should a binomial line be divided into his names in more points than on●●●ch by this proposition is proved to be impossible. The like shall follow in the five 〈◊〉 irrational lines as touching their two names.    

¶ The 31. Problem. The 43. Proposition. A first bimedial line is in one point only divided into his names.  
SVppose that AB be a first bimedial line, and let it be divided into his parts in the point C, so that let the lines AC and CB be medial commensurable in power only, and containing a rational superficies. Then I say that the line AB can not be divided into his names in any other point then in C. For if it be possible let it be divided into his names in the point D, 
Demonstration leading to an impossibil●●e.  so that let AD & DB be medial lines commensurable in power only, comprehending a rational superficies. Now forasmuch as how much that which is contained under the lines AD and DB twice di●ferreth from that which is contained under the lines AC and CB twice, so much differreth that which is composed of the squares of the lines AD and DB from that which is composed of the squares of the lines AC and CB: but that which is contained under the lines AD and DB twice differreth from that which is contained under the lines AC and CB twice, by a rational superficies (by the second assumpt going before the 41. of the tenth). For either of those superficieces is rational. Wherefore that which is composed of the squares of the lines AC and CB differeth from that which is composed of the squares of the lines AD and DB by a rational superficies, when yet they are both medial superficieces: which is impossible. Wherefore a first bimedial line is in one point only divided into his names: which was required to be proved.   

¶ The 32. Theorem. The 44. Proposition. A second bimedial line is in one point only divided into his names.  
SVppose that the line AB being a second bimedial line, be divided into his names in the point C: so that let the lines AC and CB be medial lines commensurable in power only, comprehending a medial superficies. It is manifest that the point C divideth not the whole line AB into two equal parts. 
Demonstration leading to an impossibility.  For the lines AC and CB are not commensurable in length the one to the other. Now I say that the line AB cannot be divided into his names in any other point but only in C. For if it be possible, let it be divided into his names in the point D, so that let not the line AC be one and the same, that is, let it not be equal, with the line DB. But let it be greater than it. Now it is manifest (by the first assumpt going before the 42. proposition of this book) that the squares of the lines AC and CB are greater than the squares of the lines AD and DB. And also that the lines AD and DB are medial lines commensurable in power only, comprehending a medial supersicies. Take a rational line EF. And (by the 44. of the first) upon the line EF apply a rectangle parallelogram EK equal to the square of the line AB. From which parallelogram take away the parallelogram EG equal to the squares of the lines AC and CB Wherefore the residue, namely, the parallelogram HK is equal to that which is contained under the lines AC and CB twice. Again from the the parallelogram EK take away the parallelogram EL equal to the squares of the lines AD and DB which are less than the squares of the lines AC and CB. Wherefore the residue, namely, the parallelogram MK is equal to that which is contained under the lines AD and DB twice. And forasmuch as the squares of the lines AC and CB are medial, therefore the parallelogram EG also is medial. 
Demonstration leading to an impossibility.  And it is applied upon the rational line EF ● wherefore the line EH is rational and incommensurable in length to the line EF. And by the same reason, the parallelogram HK is medial (for that which is equal unto it, namely, that which is contained under the lines AC and CB twice is medial) therefore the line HN is also rational and incommensurable in length unto the line EF. And forasmuch as the lines AC and CB are medial lines commensurable in power only, therefore the line AC is incommensurable in length unto the line CB. But as the line AC is to the line CB, so is the square of the line AC to that which is contained under the lines AC and CB (by the 1. of the sixth). Wherefore the square of the line AC is incommensurable to that which is contained under the lines AC and CB. Butt (by the 16. of the tenth) unto the square of the line AC are commensurable the squares of the lines AC and CB added together, for the lines AC and CB are commensurable in power only. And unto that which is contained under the lines AC and CB is commensurable that which is contained under the lines AC and CB twice. Wherefore that which is composed of the squares of the lines AC and CB, is incommensurable to that which is contained under the lines AC and CB twice. But to the squares of the lines AC and CB is equal the parallelogram EG, and to that which is contained under the lines AC and CB twice is equal the paralelograme HK. Wherefore the parallelogram EG is incommensurable to the parallelogram HK. Wherefore also the line EH is incommensurable in length to the line HN. And the lines EH and HN are rational. Wherefore they are rational commensurable in power only: but if two rational lines commensurable in power only be added together, the whole line is irrational, and is called a binomial line (by the 36. of the tenth). Wherefore the binomial line EN is in the point H divided into his names. And by the same reason also may it be proved that the lines EM and MN are rational lines commensurable in power only. Wherefore EN being a binomial line is divided into his names in sundry points, namely, in H and M, neither is the line EH one and the same, that is, equal with MN. For the squares of the lines AC and CB are greater than the squares of the lines BD and AD (by the 1. assumpt put after the 41. of the tenth). But the squares of the lines AD and DB are greater than that which is contained under the lines AD and DB twice (by the assumpt put after the 39 of the tenth). Wherefore the squares of the lines AC and CB, that is, the parallelogram EG is much greater than that which is contained under the lines AD and DB twice, that is then the parallelogram MK. Wherefore (by the first of the sixth) the line EH is greater than the line MN. Wherefore EH is not one and the same with MN. Wherefore a binomial line is in two sundry points divided into his names. Which is impossible. The self same absurdity also will follow if the line AC be supposed to be less than the line DB. A second binomial line therefore is not divided into his names in sundry points. Wherefore it is divided in one only: which was required to be demonstrated.   

¶ The 33. Theorem. The 45. Proposition. A greater line is in one point only divided into his names.  
LEt AB being a greater line be divided into his names in the point C, so that let the lines AC and CB be rational incommensurable in power, having that which is composed of the squares of the lines AC and CB rational, and that which is contained under the lines AC and CB medial. 
Demonstration leading to an impossibi●●●e.  Then I say that the line AB can not in any other point then in C be divided into his names. For if it be possible, let it be divided into his names in the point D, so that let AD and DB be lines incommensurable in power, having that which is composed of the squares of the lines AD and DB rational, and that which is contained under the lines AD and DB medial. Now forasmuch as how much the squares of the lines AC and CB di●●er from the squares of the lines AD and DB, so much differeth that which is contained under the lines AD and DB twice from that which is contained under the lines AC and CB twice, by those things which have be●e said in the demonstration of the 42. proposition. But the squares of the lines AC and CB exceed the squares of the lines AD and DB by a rational superficies (for they are either of them rational). Wherefore that which is containe●●●der the lines AD and DB twice exceedeth that which is contained under the lines AC and CB twice by a rational superficies: when as either of them is a medial supers●●●es. Which is impossible (by the 26. of the tenth). Wherefore a greater line is in one point only divided into his names: which was required to be proved.   

¶ The 34. Theorem. The 46. Proposition. A line containing in power a rational and a medial, is in one point only divided into his names.  
LEt AB being a line containing in power a rational and a medial, be divided into his names in the point C, so that let the lines AC & CB be incommensurable in power, having that which is composed of the squares of the lines AC and CB medial, and that which is contained under the lines AC and CB rational. 
Demonstration leading to an impossibility.  Then I say, that the line AB can not in any other point be divided into his names but only in the point C. For if it be possible, let it be divided into his names in the point D, so that let the lines AD and DB be incommensurable in power, having that which is composed of the squares of the lines AD and DB medial, and that which is contained under the lines AD and DB rational. Now forasmuch as how much that which is contained under the lines AD & DB twice differeth from that which is contained under the lines AC and CB twice, so much differ the squares of the lines AC & CB added together from the squares of the lines AD and DB added together. But that which is contained under the lines AC and CB twice exceedeth that which is contained under the lines AD and DB twice by a rational superficies (for either of them is rational). Wherefore also the squares of the lines AC and CB added together, exceed the squares of the lines AD and DB added together by a rational superficies, when yet each of them is a medial superficies: which is impossible. Wherefore a line containing in power a rational and a medial, is in one point only divided into his names: which was required to be demonstrated.   

¶ The 35. Theorem. The 47. Proposition. A line containing in power two medials, is in one point only divided into his names.  
SVppose that AB being a line containing in power two medials, be divided into his names in the point C, so that let the lines AC and CB be incommensurable in power, having that which is composed of the squares of the lines AC & CB medial, and that also which is contained under the lines AC and CB medial, and moreover incommensurable ●o that which is composed of the squares of the lines AC and CB. Then I say, 
Construction.  that the line AB can in no other point be divided into his names but only in the point C. For if it be possible, let it be divided into his names in the point D, so that let not the line AC be one and the same, that is, equal with the line DB: but by supposition let the line AC be the greater. And take a rational line EF. And (by the 43. of the first) upon the line EF apply a rectangle parallelogram EG equal to that which is composed of the squares of the lines AC and CB: and likewise upon the line HC, which is equal to the line EF, apply the parallelogram HK equal to that which is contained under the lines AC and CB twice. Wherefore the whole parallelogram EK is equal to the square of the line AB. 
Demonstration leading to an absurdity.  Again upon the same line EF describe the parallelogram EL equal to the squares of the lines AD and DB. Wherefore the residue, namely, that which is contained under the lines AD and DB twice, is equal to the parallelogram remaining, namely, to MK. And forasmuch as that which is composed of the squares of the lines AC and CB, is (by supposition) medial, therefore the parallelogram EG which is equal unto it, is also medial: and it is applied upon the rational line EF. Wherefore (by the 22. of the tenth) the line HE is rational and incommensurable in length unto the line EF. And by the same reason also the line HN is rational and incommensurable in length to the same line EF. And forasmuch as that which is composed of the squares of the lines AC and CB is incommensurable to that which is contained under the lines AC and CB twice (for it is supposed to be incommensurable to that which is contained under the lines AC and CB once): therefore the parallelogram EG is incommensurable to the parallelogram H ●. Wherefore the line EH also is incommensurable in length to the line HN, and they are rational lines: wherefore the lines EH and HN are rational commensurable in power only. Wherefore the whole line EN is a binomial line, and is divided into his names in the point H. And in like sort may we prove, that the same binomial line EN is divided into his names in the point M, and that the line EH is not one and the same that is equal with the line MN, as it was proved in the end of the demonstration of the 44. of this book. Wherefore a binomial line is divided into his names in two sundry points: which is impossible (by the 42. of the tenth). Wherefore a line containing in power two medials, is not in sundry points divided into his names. Wherefore it is divided in one point only: which was required to be demonstrated.  

¶ Second Definitions.  
IT was showed before that of binomial lines there were six kinds, 
Six kinds of binomial lines.  the definitions of all which are here now set, and are called second definitions. All binomial lines, as all other kinds of irrational lines, are conceived, considered, and perfectly understanded only in respect of a rational line (whose parts as before is taught, are certain, and known, and may be distinctly expressed by number) unto which line they are compared. This rational● line must ye ever have before your eyes, in all these definitions, so shall they all be ●asie enough.  

A binomial line co●●●ste●h of two pa●t●s.  A binomial line (ye know) is made of two parts or names, whereof the one is greater than the other. Wherefore the power or square also of the one is greater than the power or square o● the other. The three first kinds of binomial lines, namely, the first, the secon●, & the third, are produced, when the square of the greater name or part of a binom●all e●cedeth the square of the less name or part, by the square of a line which is commensurable in length to it, namely, to the greater. The three last kinds, namely, the fourth, the ●i●t, and the sixth, are produced, when the square of the greater name or part ●●●●edeth the square of the less name or part, by the square of a line incommensurable in length unto it, that is, to the greater part.  


Firs● definition.  A first binomial line is, whose square of the greater part exceedeth the square of t●e less part ●y the square of a line commensurable in length to the greater part, and the greater part is also commensurable in length to t●e rational line first set.  
As l●t the rational line first set be AB: whose parts are distinctly known: suppose also that the line CE be a binomial line, whose names or parts let be CD and DE. And let the square of the line CD the greater part exceed the square of the line DE the less part by the square of the line FG: which line FG, let b●e commensurable in length to the line CD, which is the greater part of the binomial line. And moreover let the line CD the greater pa●t be commensurble in length to the rational line first set, namely, to AB. So by this d●●inition the binomial line CE is a first binomial line.   


Secon● definition.  A second binomial line is, when the square of the greater part exceedeth the square of the less part by the square of a line commensurable in length unto it, and the less part is commensurable in length to the rational line first set.  
As (supposing ever the rational line) let CE be a binomial line divided in the point D. The square of whose greater part CD let exceed the square of the less part DE by the square of the line FG, which line ●G let be commensurable in length unto the line CD t●e greater p●●● o● the binomial line. And let also the line DE the less part of the binomial line be commensurable in length to the rational line first set AB. So by this definition the binomial line CE is a second binomial line.   


Third ●●●●●●ition.  A third binomial line is, when the square of the greater part exceedeth the square of the less part, by the square of a line commensurable in length unto it. And neither part is commensurable in length to the rational line given.  
As suppose the line CE to be a binomial line: whose parts are joined together in the point D: and let the square of the line CD the greater part exceed the square of the less part DE by the square of the line FG, and let the line FG be commensurable in length to the line CD the greater part of the binomial. Moreover, let neither the greater part CD, nor the less part DE, be commensurable in length to the rational line AB, then is the line CE by this definition a third binomial line.   

A fourth binomial line is, 
Fourth definition.  when the square of the greater part exceedeth the square of the less by the square of a line incommensurable in length unto the greater part. And the greater is also commensurable in length to the rational line.  
As let the line CE be a binomial line, whose parts let be CD & DE, & let the square of the line CD the greater part exceed the square of the line DE the less, by the square of the line FG. And let the line FG be incommensurable in length to the line CD the greater. Let also the line CD the greater part be commensurable in length unto the rational line AB. Then by this definition the line CE is a fourth binomial line.   

A fift binomial line is, 
Fifth definition.  when the square of the greater part exceedeth the square of the less part, by the square of a line incommensurable unto it in length. And the less part also is commensurable in length to the rational line given.  
As suppose that CE be a binomial line, whose greater part let be CD, and let the less part be DE. And let the square of the line CD exceed the square of the line DE by the square of the line FG, which let be incommensurable in length unto the line CD the greater part of the binomial line. And let the line DE the second part of the binomial line be commensurable in length unto the rational line AB. So is the line CE by this definition a fift binomial line.   

A sixth binomial line is, 
Sixth definition.  when the square of the greater part exceedeth the square of the less, by the square of a line incommensurable in length unto it. And neither part is commensurable in length to the rational line given.  
As let the line CE be a binomial line, divided into his names in the point D. The square of whose greater part CD let exceed the square of the less part DE by the square of the line FG, and let the line FG be incommensurable in length to the line CD the greater part of the binomial line. Let also neither CD the greater part, nor DE the less part be commensurable in length to the rational line AB. And so by this definition the line CE is a sixth binomial line. So ye see that by these definitions, & their examples, and declarations, all the kinds of binomial lines are made very plain.  
This is to be noted that here is nothing spoken of those lines, both whose portions a●e commensurable in length unto the rational line first set, for that such lines cannot be binomial lines. ●or binomial lines are composed of two rational lines commensurable in Power only (by the 36. of this book). But lines both whose portions are commensurable in length to the rational line first set are not binomial lines. For that the parts of such lines should by the 12. of this book be commensurable in length the one to the other. And so should they not be such lines as are required to the composition of a binomial line. Moreover such lines should not be irrational but rational, for that they are commensurable t●●ch of the parts whereof they are composed (by the 15. o● this book). And therefore they should be rational for that the lines which compos● them are rational.     

¶ The 13. Problem. The 48. proposition. To find out a first binomial line.  

The third Senary by composition.  TAke two numbers AC and CB, & let them be such, that the number which is made of them both added together, namely, AB, have unto one of them, 〈◊〉, unto BC that proportion that a square number hath to a square number. ●ut unto the other, namely, unto CA let it not have that proportion that a square number hath to a square number (such as is every square number which may be divided into a square number and into a number not square). 
Construction.  Take also a certain rational line, and let the same be D. And unto the line D let the line EF be commensurable in length. Wherefore the line EF is rational. And as the number AB is to the number AC, so let the square of the line EF be to the square of an other ●i●e, namely, of FG (by the corollary of the sixth of the tenth). Wherefore the square of the line EF hath to the square of the line FG that proportion that number hath to number. Wherefore the square of the line EF is commensurable to the square of the line FG (by the 6. of this book) And the line EF is rational. 
Demonstration.  Wherefore the line FG also is rational. And forasmuch as the number AB hath not to the number AC, that proportion that a square number hath to a square number, neither shall the square of the line EF have to the square of the line FG that proportion that a square number hath to a square number. Wherefore the line EF is incommensurable in length to the line FG (by the 9 of this book). Wherefore the lines EF and FG are rational commensurable in power only. Wherefore the whole line EG is a binomial line (by the 36. of the tenth). I say also that it is a ●irst binomial line. For for that as the 〈◊〉 BA is to the number AC, so is the square of the line EF to the square of the line ●G: but the number BA is greater than the number AC: wherefore the square of the line ●F is also greater than the square o● the line FG. Unto the square of the line EF let the squares of the lines FG and H be equal (which how to find out is taught in the assumpt put ●ft●r the 13. of the t●nth). And f●r that as th● number BA is to the number AC, so is the square of the line EF to the square of the line FG: therefore (by co●uersion or e●ersion of proportion (by the corollary of the 19 of the fift) as the number AB is to the number BC, so is the square of the line EF to the square of the line H. But the number AB hath to the number BC that proportion that a square number hath to a square number. Wherefore also the square of the line EF hath to the square of the line H that proportion that a square number hath to a square number. Wherefore the line EF is commensurable in length to the line H (by the 9 of this book. Wherefore the line EF is in power more than the line FG by the square of a line commensurable in length to the line EF. And the lines EF and FG are rational commensurable in power only. And the line EF is commensurable in length to the rational line D. Wherefore the line EG is a first binomial line: which was required to be done.   

The 14. Problem. The 49. Proposition. To find out a second binomial line.  
TAke two numbers AC and CB, and let them be such that the number made of them both added together, namely, 
Construction.  AB have unto BC that proportion that a square number hath to a square number, and unto the number CA let it not have that proportion that a square number hath to a square number, as it was declared in the former proposition. Take also a rational line, and let the same be D, and unto the line D let the line FG be commensurable in length Wherefore FG is a rational line. And as the number CA is to the number AB so let the square of the line GF be to the square of the line FE (by the 6. of the tenth), Wherefore the square of the lin● GF is commensurable to the square of the line FE. Wherefore also FE is a rational line. 
Demonstration.  And forasmuch as the number CA hath not unto the number AB that proportion that a square number hath to a square number, therefore neither also the square of the line GF hath to the square of the line FE that proportion that a square number hath to a square number. Wherefore the line GF is incommensurable in length unto the line FE (by the 9 of the tenth): wherefore the lines FG and FE are rational commensurable in power only. Wherefore the whole line EG is a binomial line. I say moreover that the lin● EG is a second binomial line. For for that by contrary proportion as the number BA is to the number AC, so is the square of the line EF to the square of the line FG. But the number BA is greater than the number AC, wherefore also the square of the line EF is greater than the square of the line FG. Unto the square of the line EF, let the squares of the lin●s GF and H be equal. Now by conversion (by the corollary of the 19 of of the fift) as the number AB is to the number BC, so is the square of the line EF to the square of the line H. But the number AB hath to the number BC that proportion that a square number hath to a square number. Wherefore the square of the line EF hath to the square of the line H, that proportion that a square number hath to a square number. Wherefore (by the 9 of the tenth) the line EF is commensurable in length unto the line H. Wherefore the line EF is in power more than the line FG by the square of a line commensurable in length unto the line EF: and the lines EF and FG are rational commensurable in power only, and FG being the less name ●s commensurable in length unto the rational line given, namely, to D. Wherefore EG is a second binomial line: which was required to be done.   

¶ The 15. Problem. The 50. Proposition. To find out a third binomial line.  

Construction.  TAke two numbers AC and CB, and let them be such that the number made of them both added together, namely, AB, have to the number BC that proportion that a square number hath to a square number. But to the number AC let it not have that proportion that a square number hath to a square number, as it was declared in the two former. And take also some other number that is not a square number, and let the same be D, and let not the number D have either to the number BA, or to the number AC that proportion that a square number hath to a square number. And take a rational line and let the the same be E. And as the number D is to the number AB, so let the square of the line E be to the square of the line FG. Wherefore the square of the line E is commensurable to the square of the line FG, but the line E is rational, wherefore the line FG also is rational. And for that the number D hath not to the number AB that proportion that a square number hath to a square number, neither also shall the square of the line E have to the square of the line FG that proportion that a square number hath to a square number. Wherefore the line E is incommensurable in length to the line FG (by the 9 of the tenth.) Now again as the number AB is to the number AC, so let the square of the line FG be to the square of the line GH. Wherefore the square of the line FG is commensurable to the square of the line GH. 
Demonstration.  And the line FG is rational. Wherefore also the line GH is rational. And for that the number BA hath not to the number AC, that proportion that a square number hath to a square number, therefore neither also hath the square of the line FG to the square of the line GH that proportion that a square number hath to a square number. Wherefore the line FG is incommensurable in length to the line GH. Wherefore the lines FG & GH are rational commensurable in power only. Wherefore the whole line FH is a binomial line. I say moreover that it is a third binomial line. For for that as the number D is to the number AB, so is the square of the line E to the square of the line FG: but as the number AB is to the number AC, so is the square of the line FG to the square of the line GH● therefore o● equality (by the 22. of the fift) as the number D is to the number AC, so is the square of the line E to the square of the line GH. But the number D hath not to the number A. C that proportion that a square number hath to a square number. Wherefore neither also hath the square of the line E to the square of the line GH that proportion tha● a square number hath so a square number. Wherefore the line E is incommensurable in length to the line GH. And for that as the number AB is to the number AC, so is the square of the line FG to the square o● the line GH, therefore the square of the line FG is greater than the square of the line GH. Unto the square of the line FG, let the squares of the lines GH and K be equal. Wherefore (by averse proportion by the corollary of the 19 of the ●ift) as the number AB is to the number BC, so is the square of the line FG to the square of the line K. But the number AB hath to the number BC that proportion that a square number hath to a square number. Wherefore also the square of the line FG hath to the square of the line K that proportion that a square number hath to a square number. Wherefore the line FG is commensurable in length to the line K. Wherefore the line FG is in power more than the line GH by the square of a line commensurable in length unto it. And the lines FG and GH are rational commensurable in power only, and neither of them is commensurable in length unto the rational line E: wherefore the line FH is a third binomial line: which was required to be done.   

¶ The 16. Problem. The 51. Proposition. To find out a fourth binomial line.  
TAke two numbers AC and CB, 
Construction.  & let them be such that the number made of them both added together, namely, AB, have to neither of the numbers AC and CB that proportion that a square number hath to a square number (such as is every square number to two numbers not square, which are less than it & make the said square number). And take a rational line, and let the same be D. And unto the line D let the line EF be commensurable in length. 
Demonstration.  Wherefore EF is a rational line, and as the number BA is to the number AC, so let the square of the line EF be to the square of the line FG. Wherefore the square of the line EF is commensurable to the square of the line FG, and the line BF is a rational line. Wherefore also the line FG is a rational line. And for that the number BA hath not to the number AC that proportion that a square number hath to a square number, neither also shall the square of the line EF have to the square of the line FG that proportion that a square number hath to a square number. Wherefore the line EF is incommensurable in length to the line FG. Wherefore the lines EF and FG are rational commensurable in power only. Wherefore the whole line EG is a binomial line. I say moreover that it is a fourth binomial line. For for that as the number BA is to the number AC, so is the square of the line EF to the square of the line FG. But the number BA is greater than the number AC. Wherefore also the square of the line EF is greater than the square of the line FG. Unto the square of the line EF let the squares of the lines FG and H be equal. Wherefore by conversion (by the corollary of the 19 of the fift) as the number AB is to the number BC, so is the square of the line EF to the square of the line H. But the number AB hath not to the number BC that proportion that a square number hath to a square number: therefore neither also hath the square of the line EF to the square of the line H that proportion that a square number hath to a square number. Wherefore (by the 9 of the tenth) the line EF is incommensurable in length unto the line H. Wherefore the line EF is in power more than the line FG by the square of a line incommensurable in length unto it. And the lines EF and FG are rational commensurable in power only, and the line EF is commensurable in length to the rational line D. Wherefore the line EG is a fourth binomial line: which was required to be found out.   

¶ The 17. Problem. The 52. Proposition. To find out a fift binomial line.  
TAke two numbers AC and CB, and let them be such, that the number AB have to neither of the numbers AC or CB that proportion that a square number hath to a square number, 
Construction.  as in the former proposition. And take a rational line and let the same be D. And unto the line D let the line FG be commensurable in length. Wherefore the line FG is rational. And as the number CA is to the number AB, so let the square of the line GF be to the square of the line EF. 
Demonstration.  Wherefore the square of the line GF is commensurable to the square of the line FE. Wherefore also the line FE is rational. An● for that the number CA hath not to the number AB that proportion that a square number hath to a square number, therefore neither also hath the square of the line GF to the square of the line FE that proportion that a square number hath to a square number. Wherefore (by the 9 of the tenth) the line GF is incommensurable in length to the line FE. Wherefore the lines EF and FG are rational commensurable in power only. Wherefore the whole line EG is a binomial line. I say moreover that it is a fift binomial line. For for that as the number CA is to the number AB, so is the square of the line GF to the square of the line FE, therefore contrariwise, as the number BA is to the number AC, so is the square of the line EF to the square of the line FG. but the number BA is greater than the number AC. Wherefore also the square of the line EF is greater than the square of the line FG. Unto the square of the line EF, let the squares of the lines FG and H be equal. Wherefore by conversion (by the corollary of the 19 of the fift) as the number AB is to the number BE so is the square of the line EF to the square of the line H. But the number AB hath not to the number BC that proportion that a square number hath to a square number. Wherefore neither also hath the square of the line EF to the square of the line H that proportion that a square number hath to a square number. Wherefore (by the 9 of the tenth) the line EF is incommensurable in length to the line H. Wherefore the line EF is in power more than the line FG by the square of a line incommensurable in length unto it. And the lines EF and FG are rational commensurable in power only. And the line FG being the less name, is commensurable in length to the rational line given, namely, to D. Wherefore the whole line EG is a fift binomial line: which was required to be found out.   

¶ The 18. Problem. The 53. Proposition. To find out a sixth binomial line.  
TAke two numbers AC & CB, and let them be such that the number which is made of them both added together, 
Construction.  namely, AB, have to neither of the numbers AC or CB that proportion that a square number hath to a square number. Take also any other number which is not a square number, and let the same be D. And let not the number D have to any one of these numbers AB and AC that proportion that a square number hath to a square number. Let there be put moreover a rational line, and let the same be E. And as the number D is to the number AB, so let the square of the line E be to the square of FG. Wherefore (by the 6. of the tenth) the line E is commensurable in power to the line FG, & the line E is rational. Wherefore also the line FG is rational. And for that the number D hath not ●o the number AB that proportion that a square number hath to a square number, therefore neither also shall the square of the line E have to the square of the line FG that proportion that a square number hath to a square number. Wherefore the line FG is incommensurable in length to the line E. Again, as the number BA is to the number AC, so let the square of the line FG be to the square of the line GH. 
Demonstration.  Wherefore (by the 6. of the tenth) the square of the line FG is commensurable to the square of the line GH. And the square of the line FG is rational. Wherefore the square of the line GH is also rational. Wherefore also the line GH is rational. And for that the number AB hath not to the number AC, that proportion that a square number hath to a square number: therefore neither also hath the square of the line FG to the square of the line GH, that proportion that a square number hath to a square number. Wherefore the line FG is incommensurable in length to the line GH. Wherefore the lines FG and GH are rational commensurable in power only. Wherefore the whole line FH is a binomial line. I say moreover, that it is a sixth binomial line. For for that as the number D is to the number AB, so is the square of the line E to the square of the line FG. And as the number BA is to the number AC, so is the square of the line FG to the square of the line GH. Wherefore of equality (by the 22. of the fift) as the number D is to the number AC, so is the square of the line E to the square of the line GH. But the number D hath not to the number AC that proportion that a square number hath to a square number. Wherefore neither also hath the square of the line E to the square of the line GH that proportion that a square number hath to a square number. Wherefore the line E is incommensurable in length to the line GH. And it is already proved, that the line FG is also incommensurable in length to the line E. Wherefore either of these lines FG and GH is incommensurable in length to the line E. And for that as the number ●A is to the number AC; so is the square of the line FG to the square of the line GH: therefore the square of the line FG is greater than the square of the line GH. Unto the square of the line FG, let the squares of the lines GH and K be equal. Wherefore by eversion of proportion, as the number AB is to the number BC, so is the square of the line FG to the square of the line K. But the number AB hath not to the number BC that proportion that a square number hath to a square number. Wherefore neither also hath the square of the line FG to the square of the line K that proportion that a square number hath to a square number. Wherefore the line FG is incommensurable in length unto the line K. Wherefore the line FG is in power more than the line GH, by the square of a line incommensurable in length to it. And the lines FG and GH are rational commensurable in power only. And neither of the lines FG & GH is commensurable in length to the rational line given, namely, to E. Wherefore the line FH is a sixth binomial line: which was required to be found out.  

¶ A Corollary added out of Flussates.  
By the 6. former Propositions it i● manifest, ho● 〈◊〉 divide any right line given into the names of every one of the six● foresaid binomiall lines. 
A Corollary added by Flussates.  For if it be required to divide a right line given into a first binomial line, then by the 48● of this book find out a first binomial line. And this right line being so found out divided into his names, you may by the 10. of the sixth, divide the right line given in like sort. And so in the other five following.  
Although I here note unto you this Corollary out of 〈…〉, in very conscience and of grateful ●inde● I am enforced to certify you, that, i● any yeare●, before the travails of Flussas (upon Eu●li●●● Geometrical Elements) were published, the order how to divide, not only the 6. binomial lines into their names, but also to add to the 6. Resid●●ls their due parts: ●nd f●rthermore to divide all the other irrational lines (of this tenth book) into the parts distinct, of which they are composed: with many other strange conclusions Mathematical, to the better understanding of this tenth book and other Mathematical books, most necessary, were by M. john Dee invented and demonstrated: 
M. Dee his book called Ty●●c●ni●m Mathematicum.  as in his book, whose title is Tyrocinium Mathematicum (dedicated to Petru● Nonnius, An. 1559.) may at large appear. Where also is one new art, with sundry particular points, whereby the Mathematical Sciences, greatly may be enriched. Which his book, I hope, God will one day allow him opportunity to publish: with divers other his Mathematical and Metaphysical labours and inventions.   

¶ An Assumpt.  
Is a right line be divided into two parts how soever: the rectangle parallelogram contained under both the parts, is the mean proportional between the squares of the same parts. And the rectangle parallelogram contained under the whole line and one of the parts, is the mean proportional between the square of the whole line and the square of the said part.  

This Assumpt as was before noted, followeth most ●ri●fly without farther demonstration of the 25. of this book.  Suppose that there be two squares AB and BC, and let the lines DB and BE so be put that they both make one right line. Wherefore (by the 14. of the first) the lines FB and BG make also both one right line. And make perfect the parallelogram AC. Then I say, that the rectangle parallelogram DG is the mean proportional between the squares AB and BC: and moreover, that the parallelogram DC is the mean proportional between the squares AC and CB. First the parallelogram AG is a square. For forasmuch as the line DB is equal to the line BF, and the line BE unto the line BG, therefore the whole line DE is equal to the whole line FG. But the line DE is equal to either of these lines AH & KC, and the line FG is equal to either of these lines AK and HC (by the 34. of the first). 
Demonstration.  Wherefore the parallelogram AC is equilater, it is also rectangle (by the 29. of the first). Wherefore (by the 46. of the first) the parallelogram AC is a square. Now for that as the line FB is to the line BG, so is the line DB to the line BE. But as the line FB is to the line BG, so (by the 1. of the sixth) is the parallelogram AB, which is the square of the line DB, to the parallelogram DG, and as the line DB is to the line BE, so is the same parallelogram DG to the parallelogram BC, which is the square of the line BE. Wherefore as the square AB is to the parallelogram DG, so is the same parallelogram DG to the square BC. Wherefore the parallelogram DG is the mean proportional between the squares AB and BC. I say moreover, that the parallelogram DC is the mean proportional between the squares AC and CB. For for that as the line AD is to the line DK, so is the line KG to the line GC (for they are each equal to each). Wherefore by composition (by the 18. of the fift) as the line AK is to the line KD, so is the line KC to the line CG. But as the line AK is to the line KD, so is the square of the line AK, which is the square AC, to the parallelogram contained under the lines AK and KD, which is the parallelogram CD: and as the line KC is to the line CG, so also is the parallelogram DC to the square of the line GC, which is the square BC. Wherefore as the square AC is to the parallelogram DC, so is the parallelogram DC to the square BC. Wherefore the parallelogram DC is the mean proportional between the squares AC and BC: which was required to be demonstrated.   

¶ An Assumpt.  
Magnitudes that are mean proportionalls between the self same or equal magnitudes, are also equal the one to the other.  
Suppose that there be three magnitudes A, B, C. 
An Assumpt.  And as A is to B, so let B be to C. And likewise as the same magnitude A is to D, so let D be to the same magnitude C. Then I say that B and D are equal the one the other. For the proportion of A unto C is double to that proportion which A hath to B (by the 10. definition of the fift) and likewise the self same proportion of A to C is (by the same definition) double to that proportion which A hath to D. But magnitudes whose equemultiplices are either equal or the self same, are also equal. Wherefore as A is to B, so is A to D. Wherefore (by the 9 of the fift) B and D are equal the one to the other. So shall if also be if there be other magnitudes equal to A and C, namely, E and F, between which let the magnitude D be the mean proportional.    

¶ The 36. Theorem. The 54. Proposition. If a superficies be contained under a rational line & a first binomial line: the line which containeth in power that superficies is an irrational line, & a binomial line. 
The fourth Senary by composition.   
SVppose that the superficies ABCD, be contained under the rational line AB, and under a first binomial line AD. Then I say that the line which containeth in power the superficies AC, is an irrational line, and a binomial line. 
Construction.  For forasmuch as the line AD is a first binomial line, it is in one only point divided into his names (by the 42. of this tenth): let it be divided into his names in the point E. And let AE be the greater name. Now it is manifest that the lines AE and ED are rational commensurable in power only, and that the line AE is in power more than the line ED, by the square of a line commensurable in length to the line AE, and moreover that the line AE is commensurable in length to the rational line given AB by the definition of a first binomial line set before the 48. proposition of this tenth. Divide (by the 10. of the first) the line ED into two equal parts in the point F. And forasmuch as the line AE is in power more than the line ED by the square of a line commensurable in length unto the line AE, therefore if upon the greater line, namely, upon the line AE be applied a parallelogram equal to the fourth part of the square of the less line, that is, to the square of the line EF, & wanting in form by a square, it shall divide the greater line, namely, AE into two parts còmmensurable in length the one to the other (by the second part of the 17. of the tenth.) Apply therefore upon the line AE a parallelogram equal to the square of the line EF, and wanting in form by a square by the 28. of the sixth, and let the same be that which is contained under the lines AG and GOE Wherefore the line AG is commensurable in length to the line GOE Draw by the points G, E, and F, to either of these lines AB and DG these parallel lines GH, EK, and FL (by the 31. of the first) And (by the 14. of the second) unto the parallelogram AH describe an equal square SN. And unto the parallelogram GK describe (by the same) an equal square NP. And let these lines MN & NX be so put, that they both make one right line. Wherefore (by the 14. of the first) the lines RN and NO make also both one right line. 
Demonstration.  Make perfect the parallelogram SP. Wherefore the parallelogram SP is a square by those things which were demonstrated after the determination in the first assumpt going before. And forasmuch as that which is contained under the lines AG and GE is equal to the square of the line EF (by construction): therefore as the line AG is to the EF, so is the line EF to the line EG (by the 14. or 17. of the sixth) Wherefore also (by the 1. of the sixth) as the parallelogram AH is to the parallelogram EL, so is the parallelogram EL to the parallelogram GK. Wherefore the parallelogram EL is the mean proportional between the parallelograms AH and GK. But the parallelogram AH is equal to the square SN, and the parallelogram GK is equal to the square NP by construction. Wherefore the parallelogram EL is the mean proportional between the squares SN and NP (by the 7. of the fifth) But (by the first assumpt going before) the parallelogram MR is the mean proportional between the squares SN and NP. Wherefore the parallelogram MR is equal to the parallelogram EL (by the last assumpt going before) But the parallelogram MR is equal to the parallelogram OX (by the 43. of the first) and the parallelogram EL is equal to the parallelogram FC by construction, and by the first of the sixth. Wherefore the whole parallelogram EC is equal to the two parallelograms MR & OX. And the parallelograms AH and GK are equal to the squares SN and NP by construction. Wherefore the whole parallelogram AC is equal to the whole square SP, that is, to the square of the line MX. 
The first part of this demonstration concluded.  Wherefore the line MX containeth in power the parallelogram AC. I say moreover that the line MX is a binomial line. For forasmuch as (by the 17. of the tenth) the line AG is commensurable in length to the line EG. Therefore (by the 15. of the tenth) the whole line AE is commensurable in length to either of th●se lines AG and GOE But by supposition the line AE is commensurable in length to the line AB. Wherefore (by the 12. of the tenth) either of the lines AG & GE are commensurable in length to the line AB. But the line AB is rational. Wherefore either of these lines AG and GE is rational. Wherefore (by the 19 of the tenth) either of these parallelograms AH and GK is rational. Wherefore (by the first of the sixth, and 10. of the tenth) the parallelogram AH is commensurable to the parallelogram GK. But the parallelogram AH is equal to the square SN, 
The second part of the demonstration concluded.  and the parallelogram GK is equal to the square NP Wherefore the squares SN and NP, which are the squares of the lines MN and NX are rational and commensurable. And forasmuch as (by supposition) the line AE is incommensurable in length to the line ED. But the line AE is commensurable in length to the line AG. And the line DE is commensurable in length to the line EF, for it is double to it by construction. Wherefore (by the 13. of the tenth) the line AG is incommensurable in length to the line EF. Wherefore the parallelogram AH is incommensurable to the parallelogram EL. But the parallelogram AH is equal to the square SN, and the parallelogram EL is equal to the parallelogram MR. Wherefore the square SN is incommensurable to the parallelogram MR. But as the square SN is to the parallelogram MR, so is the line ON to the line NR by the 1. of the sixth. 
The third part concluded.  Wherefore the line ON is incommensurable to the line NR. But the line ON is equal to the line MN, and the line NR to the line NX. Wherefore the line MN is incommensurable to the line NX. And it is already proved that the squares of the lines MN and NX are rational and commensurable. Wherefore the lines MN and NX are rational commensurable in power only. 
The total conclusion.  Wherefore the whole line MX is a binomial line, and it containeth in power the parallelogram AC: which was required to be proved.   

¶ The 37. Theorem. The 55. Proposition. If a superficies be comprehended under a rational line and a second binomial line: the line that containeth in power that superficies is irrational, and is a first bimedial line.  
SVppose that the superficies ABCD be contained under a rational line AB, and under a second binomial line AD. Then I say that the line that containeth in power the superficies AC is a first bimedial line. For forasmuch as AD is a second binomial line, it can in one only point be divided into his names, by the 43. of this tenth: let it therefore by supposition be divided into his names in the point E, so that let AE be the greater name. Wherefore the lines AE and ED are rational commensurable in power only, and the line AE is in power more than the line ED by the square of a line commensurable in length to AE, and the less name, namely, ED is commensurable in length to the line AB by the definition of a second binomial line, set before the 48. proposition of this tenth. Divide the line ED (by the tenth of the first) into two equal parts in the point F. And (by the 28. of the sixth) upon the line AE apply a parallelogram equal to the square of the line EF, and wanting in figure by a square. And let the same parallelogram be that which is contained under the lines AG and GOE Wherefore (by the second part of the 17. of this tenth) the line AG is commensurable in length to the line GOE And (by the 31. of the first) by the points G, E, F, draw unto the lines AB and CD these parallel lines, GH, EK, & FL. And (by the 14. of the second) unto the parallelogram AH describe an equal square SN. And to the parallelogram GK describe an equal square NP, and let the lines MN & NX be so put that they both make one right line: wherefore (by the 14. of the first) the lines also RN, and NO make both one right line. 
Demonstration.  Make perfect the parallelogram SP. Now it is manifest (by that which hath been demonstrated in the proposition next going before) that the parallelog●ame MR is the mean proportional between the squares SN and NP, and is equal to the parallelogram EL, and that the line MX containeth in power the superficies AC. Now resteth to prove that the line MX is a first bimedial line. Forasmuch as the line AE is incommensurable in length to the line ED, and the line ED is commensurable in length to the line AB, therefore (by the 13. of the tenth) the line AE is incommensurable in length to the line AB. And forasmuch as the line AG is commensurable in length to the line GE therefore the whole line AE is (by the 15. of the tenth) commensurable in length to either of these lines AG and GOE But the line AE is rational: wherefore either of these lines AG and GE is rational. And forasmuch as the line AE is incommensurable in length to the line AB, but the line AE is commensurable in length to either of these lines AG and GE: wherefore either of the lines AG and GE are incommensurable in length to the line AB (by the 13. of the tenth). Wherefore the lines AB, AG, and GE are rational commensurable in power only. Wherefore (by the 21. of the tenth) either of these parallelograms AH and GK is a medial superficies. Wherefore also either of these squares SN and NP is a medial superficies by the corollary of the 23. of the tenth. Wherefore the lines MN & NX are medial lines by the 21. of this tenth. 
The first part of this demonstration concluded.  And forasmuch as the line AG is commensurable in length to the line GE, therefore (by the 1. of the sixth and 11. of the tenth) the parallelogram AH is commensurable to the parallelogram GK that is, the square SN to the square NP, that is, the square of the line MN to the square of the line NX. Wherefore the lines MN and NX are medials commensurable in power. 
The third part concluded.  And forasmuch as the line AE is incommensurable in length to the line ED, but the line AE is commensurable in length to the line AG, and the line ED is commensurable in length to the line E●, therefore (by the 13. of the tenth) the line AG is incommensurable in length to the line E●. Wherefore (by the ●. of the sixth and 11. of the tenth) the parallelogram AH is incommensurable to the parallelogram EL, that is, th● square SN to the parallelogram MR, that is, the line ON is incommensurable to the line NR, that is the line MN to the line NX. And it is proved that the lines MN and NX are medial lines commensurabl● in power. 
The fourth part conclude●.  Wherefore the lines MN and NX are medial lines commensurable in power only. Now I say moreover that they comprehend a rational superficies. For forasmuch as by supposition the line DE is commensurable in length to either of these lines AB and EF, therefore the line FE is commensurable in length to the line EK which is equal to the line AB (by th● 12. of the tenth). And either of these lines EF and EK is a rational line. Wherefore the parallelogram EL, that is the parallelogram MR, is a rational superficies (by the 19 of the tenth). 
The fift part concluded.  But the parallelogram MR is that which is contained under the lines MN and NX. But if two medial lines commensurable in power only and comprehending a rational superficies be added together, the whole line is irrational and is called a first bimediall (by the 37. of the tenth). 
The total conclusion.  Wherefore the line MX is a first bimedial line: which was required to be demonstrated.   

¶ The 38. Theorem. The 56. Proposition. If a superficies be contained under a rational line and a third binomial line: the line that containeth in power that superficies is irrational, and is a second bimedial line.  
SVppose that the superficies ABCD be comprehended under the rational line AB and a third binomial line AD, and let the line AD be supposed to be divided into his names in th● point E, of which let AE be the greater name. Then I say, that the line that containeth in power the superficies AC is irrational, and is a second bimedial line. Let the same construction of the figures be in this, that was in the two Propositions nex● going before. And now forasmuch as the line AD is a third binomial line, therefore these lines AE and ED are rational commensurable in power only. 
Demonstration.  And the line AE is in power more than the line ED by the square of a line commensurable in length to the line AE, and neither of the lines AE nor ED is commensurable in length to the line AB by the definition of a third binomial line set before the 48. Proposition. As in the former Propositions it was demonstrated, so also may it in this Proposition be proved, that the line MX containeth in power the superficies AC, and that the lines MN and NX are medial lines commensurable in power only. Wherefore the line MX is a bimedial line. Now resteth to prove that it is a second bimedial line. Forasmuch as the line DE is (by supposition) incommensurable in length to the line AB, that is, to the line EK. But the line ED is commensurable in length to the line EF. Wherefore (by the 13. of the tenth) the line EF is incommensurable in length to the line EK. And the lines FE and EK are rational. For by supposition the line ED is rational, unto which the line FE is commensurable. Wherefore the lines FE and EK are rational lines commensurable in power only. Wherefore (by the 21. of the tenth) the parallelogram EL, that is, the parallelogram MR which is contained under the lines MN and NX is a medial superficies. Wherefore that which is contained under the lines MN and NX is a medial superficies. Wherefore the line MX is a second bimedial line (by the 38. Proposition and definition annexed thereto): which was required to be proved.   

¶ The 39 Theorem. The 57 Proposition. If a superficies be contained under a rational line, and a fourth binomial line: the line which containeth in power that superficies is irrational, and is a greater line.  
SVppose that the superficies AC be comprehended under a rational line AB and a fourth binomial line AD, & let the binomial line AD be supposed to be divided into his names in the point E, so that let the line AE be the greater name. Then I say, that the line which containeth in power the superficies AC is irrational, and is a greater line. 
Construction.  For, forasmuch as the line AD is a fourth binomial line, therefore the lines AE and ED are rational commensurable in power only. And the line AE is in power more than the line ED by the square of a line incommensurable in length to AE. And the line AE is commensurable in length to the line AB. Divide (by the 10. of the first) the line DE into two equal parts in the point F. And upon the line AE apply a parallelogram equal to the square of EF and wanting in figure by a square: and let the same parallelogram be that which is contained under the lines AG & GOE Wherefore (by the second part of the 18. of the tenth) the line AG is incommensurable in length to the line EG. Draw unto the line AB, by the points G, E, F, parallel lines GH, EK, and PL, and let the rest of the construction be as it was in the three former Propositions. 
Demonstration.  Now it is manifest, that the line MX containeth in power the superficies AC. Now resteth to prove that the line MX is an irrational line, and a greater line. Forasmuch as the line AG is incommensurable in length to the line EG, therefore (by the 1. of the sixth, and 11. of the tenth) the parallelogram AH is incommensurable to the parallelogram GK, that is, the square SN to the square NP. Wherefore the lines MN and NK are incommensurable in power. And forasmuch 〈◊〉 the line AE is commensurable in length to the rational line A●, therefore the parallelogram AK is rational. And it is equal to the squares of the lines MN and NX. Wherefore that which is composed of the squares of the lines MN and NX added together is rational. And forasmuch as the line ED is incommensurable in length to the line AB, that is, to the line EK, but the line ED is commensurable in length to the line EF, therefore the line EF is incommensurable in length to the line EK. Wherefore the lines EK and EF are rational commensurable in power only. Wherefore (by the 21. of the tenth) the parallelogram LE, that is, the parallelogram MR is medial. And the parallelogram MR is that which is contained under the lines MN, and NX. Wherefore that which is contained under the lines MN and NX is medial. And that which is composed of the squares of the lines MN & NX is proved to be rational, & the line MN is demonstrated to be incommensurable in power to the line NX. But if two lines incommensurable in power be added together, having that which is made of the squares of them added together rational, & that which is under them medial, the whole line is irrational, and is called a greater line (by the 39 of the tenth). Wherefore the line MX is irrational, and is a greater line, and it containeth in power the superficies AC: which was required to be demonstrated.   

¶ The 40. Theorem. The 58. Proposition. If a superficies be contained under a rational line and a fift binomial line: the line which containeth in power that superficies is irrational, and is a line containing in power a rational and a medial superficies.  
SVppose that the superficies AG be contained under the rational line A●, and under a fift binomial line Addend let the same lin● AD be supposed to be divided into his names in the point E, so that let the line AE be the greater name. Then I say, that the line which containeth in power the superficies AC is irrational, and is a line containing in power a rational and a medial superficies. 
Demonstration.  Let the self same constructions be in this, that were in the four Proposition next going before. And it is manifest, that the line MX containeth in power the superfici●● AG. Now testeth to prove that the line MX is a line containing in power a rational & a medial superficies. Forasmuch as the line AG is incommensurable in length to the line GE, therefore (by the 1. of the sixth, and 10. of the tenth) the parallelogram AH is incommensurable to the parallelogram HE, that is, the square of the line MN to the square of the line NX. Wherefore the lines MN and NX are incommensurable in power. And forasmuch as the line AD i● a fif● binomial line, and his less name or part is the line ED, therefore the line ED is commensurable in length to the line AB. But the line AE is incommensurable in length to the line ED. Wherefore (by the 13. of the tenth) the line AB is incommensurable in length to the line AE. Wherefore the lines AB and AE are rational commensurable in power only. Wherefore (by the 21. of the tenth) the parallelogram AK is medial, that is, that which is composed of the squares of the lines MN & NX added together. And forasmuch as the line DE is commensurable in length to the line AB, that is, to the line EK, but the line DE is commensurable in length to the line EF, wherefore (by the 12. of the tenth) the line EF is also commensurable in length to the line EK. And the line EK is rational. Wherefore (by the 19 of the tenth) the parallelogram EL, that is, the parallelogram MR, which is contained under the lines MN and NX is rational. Wherefore the lines MN and NX are incommensurable in power, having that which is composed of the squares of them added together, medial, and that which is contained under them, Rational. Wherefore (by the 40. of the tenth) the whole line MX is a line containing in power a rational and a medial superficies, and it containeth in power the superficies AC: which was required to be proved.   

¶ The 41. Theorem. The 59 Proposition. If a superficies be contained under a rational line, and a sixth binomial line, the line which containeth in power that superficies, is irrational, & is called a line containing in power two medials.  
SVppose that the superficies ABCD be contained under the rational line AB, and under a sixth binomial line AD, and let the line AD be supposed to be divided ●●to his names in the point E, so that let the line AE be the greater name. Then I say that the line that containeth in power the superficies AC is irrational, and is a line containing in power two medials. 
Demonstration.  Let the self same constructions be in this, that were in the former propositions. Now it is manifest that the line MX containeth in power the superficies AC, and that the line MN is incommensurable in power to the line NX. And forasmuch as the line AE is incommensurable in length to the line AB, therefore the lines AE and AB are rational commensurable in power only. Wherefore (by the 〈◊〉 of the tenth) the parallelogrām● AK that is, that which is composed of the squares of the lines MN and NX added together is medial. Again forasmuch as the line ED is incommensurable in length to the line AB, therefore also the line EF is incommensurable in length to the line EK. Wherefore the lines EF and EK are rational commensurable in power only. Wherefore the parallelogram EL, that is, the parallelogram MR which is contained under the lines MN and NX is medial. And forasmuch as the line AE is incommensurable in length to the line EF, therefore the parallelogram AK is also incommensurable to the parallelogram EL (by the first of the sixth, and 100LS of the tenth.) But the parallelogram AK is equal to that which is composed of the squares of the lines MN and NX added together. And the parallelogram EL is equal to that which is contained under the lines MN and NX. Wherefore that which is composed of the squares of the lines MN and NY added together, is incommensurable to that which is contained under the l●nes MN and NX● and e●●her of them, namely, that which is composed of the squares of the lines MN and NX added together, and that which is contained v●der the lines MN and N●, is proved medial; and the lines MN and NX are proved incommensurable in power. Wherefore (by the 41. of the tenth) the whole line MX is a line containing in power two medials, and it containeth in power the superfices AC: which was required to be demonstrated.  

An A●●umpt.  
If a right line be divided into two unequal parts, the squares which are made of the unequal parts, are greater than the rectangle parallelogram contained under the unequal parts, twice.  
Suppose that AB be a right line, and let it be 〈…〉 point C. And let the line AC be the greater part. 〈…〉 and ●B, are greater than that which is contained under the lines A● and CB twice. D●●id● (by the 10. of the first) the line AB into two equal parts, in the point D. Now forasmuch as the right line AB is divided into two equal parte● in the point D, and into two unequal parte● in the point C, therefore (by the 5. of the second) that which is contained under the lines 〈…〉 line CD, is equal to the square of the line AD. 〈…〉 the lines AC and CB, (omitting the square of the line CD) is less than the square of the AD (by the 9 common sentence, and the seventh of the fifth:) Wherefore that which is contained under the lines AC and CB, twice, is less than the double of the square of the line AD (that is, then twice the square of the line AD) by 
* Look after the Assumpt concluded at this mark: for plainer opening of this place.  alternate proportion, and the 14. of the fift. But the squares of the lines AC and CB are double to the squares of the lines AD and DC (by the 9 of the second). Therefore the squares of AC and CB are more than double to the square of AD alone, (leaving out the square of DC) by the 8. of the fift. But the parallelogram contained under the lines AC and ●B twice, is proved less than the double of the square of the line AD. Therefore the same parallelogram contained under the lines AC and CB twice, is much less than the squares of the lines AC and CB. If a right line therefore be divided into two unequal parts, the squares which are made of the unequal parts, are greater than the rectangle parallelogram contained under the unequal parts, twice: which was required to be demonstrated.  
In numbers I need not to have so alleged, for the 17. of the seventh had confirmed the doubles to be one to the other, as their singles were, but in our magnitudes, it likewise is true and evident by alternate proportion, thus. As the parallelogram of the lines AC and CB is to his double, so is the square of the line AD to his double (each being half). Wherefore, alternately, as the parallelogram is to the square, so is the parallelogram his double to the double of the square. But the parallelogram was proved less than th● square: wherefore his double is less than the square his double, by the 14. of the fifth.  
This Assumpt is in some books not read, for that in manner it seemeth to be all one with that which was put after the 39 of this book: but for the divers manner of demonstrating, it is necessary. 
The use of this Assumpt is in the next proposition, & other following.  For the fear of invention is thereby furthered. And though Zambert did in the demonstration hereof, omit that which P. Montaureus could not supply● but plainly doubted of the sufficiency of this proof, yet M. Dee, by only allegation of the due places of credit, whose pith & force● Theon his words do contain, hath restored to the demonstration sufficiently, both light and authority, as you may perceive, and chief such may judge, who can compare this demonstration here (thus furnished) with the Greek of Theon, or latin translation of Zambert.    

¶ The 42. Theorem. The 60. Proposition. 
The fift Senary by composition.  The square of a binomial line applied unto a rational line, maketh the breadth or other side a first binomial line.  
SVppose that the line AB be a binomial line, and let it be supposed to be divided into his names in the point C, so that let AC be the greater name. 
Construction.  And take a rational line DE ● And (by the ●4. of the first) unto the line DE apply a rectangle parallelogram DEFG equal to the square of the●line AB and making in breadth the line DG. Then I say that the line DG is a first binomial line. Unto the line DE apply the parallelogram DH equal to the square of the line AC, and unto the line KH which is equal to the line DE apply the parallelogram KL equal to the square of the line BC. Wherefore the residue namely, that which is contained under the lines AC & CB twice is equal to the residue, namely, to the parallelogram MF, by 〈◊〉 4● of the second Divide (by the 1●. of the first) the line MG into two equal parts in the point N. 
Demonstration.  And (by the 31. of the first) draw the line NX parallel to either of these lines ML and GF. Wherefore either of these parallelograms MX and NF is equal to that which is contained under the lines AC and CB once, by the 15. of the fifth. And forasmuch as the line AB is a binomial line, and is divided into his names in the point C therefore the ●ines AC and CB are rational commensurable in power only. Wherefore the squares of the lines AC and CB are rational, and therefore commensurable the one to the other. Wherefore (by the 15. of the tenth) that which is made of the squares of the lines AC and CB added together is commensurable to either of the squares of the lines AC or CB: wherefore that which is made of the squares of the line● AC and CB added together is rational. And it is equal to the parallelogram DL by construction. Wherefore the parallelogram DL is rational. And it is applied unto the rational line DE: wherefore (by the 20. of the tenth) the line DM is rational, and commensurable in length to the line DE. Again forasmuch as the lines AC and CB are rational commensurable in power only, therefore that which is contained under the lines AC and CB twice, that is, the parallelogram MF is medial by the 21 of the tenth, and it is applied unto the rational line ML. Wherefore the line MG is rational and incommensurable in length to the line ML (by the 22. of this tenth) that is, to the line DE. But the line MD is proved rational and commensurable in length to the line DE. Wherefore (by the 13. of the tenth) the line DM is incommensurable in length to the line MG. Wherefore the lines DM & M● are rational commensurable in power only. 
Concluded that DG is a binomial line.  Wherefore (by the 36. of the tenth) the whole line DG is a binomial line. Now resteth to prove that it is a first binomial line. Forasmuch as (by the the assumpt going before the 54. of the tenth) that which is contained under the lines AC and CB is the mean proportional between the squares of the lines AC and CB, therefore the parallelogram MX is the mean proportional between the parallelograms DH and KL. Wherefore as the parallelogame DH ● is to the parallelogram MX, so is the parallelogram MX to the parallelogram KL, that is, as the line DK is to the line MN, so is the line MN to the line MK. Wherefore that which is contained under the lines DK and KM is equal to the square of the line MN. And forasmuch as the square of the line AC is commensurable to the square of the line CB, the parallelogram DH is commensurable to the prarallelograme KL. Wherefore (by the 1. of the sixth and 10. of the tenth) the line DE is commensurable in length to the line KM. And forasmuch as the squares of the lines AC and CB are greater than that which is contained under the lines AC and CB twice by the assumpt going before this proposition, or by the assumpt after this 39 of the tenth, therefore the parallelogram DL is greater than the parallelogram MF. Wherefore (by the first of the sixth) the line DM is greater than the line MG. And that which is contained under the lines DK and KM is equal to the square of the line MN, that is to the fourth part of the square of the line MG. But (by the 17. of the tenth), if there be two unequal right lines, and if upon the greater be applied a parallelogram equal to the fourth part of the square made of the less line and wanting in figure by a square, if also the parallelogram thus applied de●ide the line whereupon it is applied into parts commensurable in length, then shall the greater line be in power more than the less by the square of a line commensurable in length to the greater. Wherefore the line DM is in power more than the line MG by the square of a line commensurable in length unto the line DM. And the lines DM and MG are proved rational commensurable in power only. And the line DM is proved the greater name and commensurable in length to the rational line given DE. Wherefore by the definition of a first binomial line se● before the 48. proposition of this book, the line DG is a first binomial line: which was required to be proved.  
This proposition and the five following are the converses of the six former propositions.   

¶ The 43. Theorem. The 61. Proposition. The square of a first bimedial line applied to a rational line, maketh the breadth or other side a second binomial line.  
SVppose that the line AB be a first bimedial line, and let it be supposed to be divided into his parts in the point C, of which let AC be the greater part. Take also a rational line DE, and (by the 44. of the first) apply to the line DE the parallelogram DF equal to the square of the line AB, & making in breadth the line DG. 
Construction.  Then I say, that the line DG is a second binomial line. Let the same constructions be in this, that were in the Proposition going before. And forasmuch as the line AB is a first bimedial line, 
Demonstration.  and is divided into his parts in the point C, therefore (by the 37. of the tenth) the lines AC and CB are medial commensurable in power only, comprehending a rational superficies. Wherefore also the squares of the lines AC and CB are medial. Wherefore the parallelogram DL is medial (by the Corollary of the 23. of the tenth) and it is applied upon the rational line DE. Wherefore (by the 22. of the tenth) the line MD is rational and incommensurable in length to the line DE. Again forasmuch as that which is contained under the lines AC and CB twice is rational, therefore also the parallelogram MF is rational, and it is applied unto the rational line ML. Wherefore the line MG is rational and commensurable in length to the line ML, that is, to the line DE (by the 20. of the tenth). Wherefore the line DM is incommensurable in length to the line MG, and they are both rational. Wherefore the lines DM and MG are rational commensurable in power only. Wherefore the whole line DG is a binomial line. 
Concluded that DG is a binomial line.  Now resteth to prove that it is a second binomial line. Forasmuch as the squares of the lines AC and CB are greater than that which is contained under the lines AC and CB twice (by the Assumpt before the 60. of this book): therefore the parallelogram DL is greater than the parallelogrrmme MF. Wherefore also (by the first of the sixth) the line DM is greater than the line MG. And forasmuch as the square of the line AC is commensurable to the square of the line CB, therefore the parallelogram DH is commensurable to the parallelogram KL. Wherefore also the line DK is commensurable in length to the line KM. And that which is contained under the lines DK and KM is equal to the square of the line MN, that is, to the fourth part of the square of the line MG. Wherefore (by the 17. of the tenth) the line DM is in power more than the line MG, by the square of a line commensurable in length unto the line DM: and the line MG is commensurable in length to the rational line put, namely, to DE. Wherefore the line DG is a second binomial line: which was required to be proved.   

¶ The 44. Theorem. The 62. Proposition. The square of a second bimedial line, applied unto a rational line: maketh the breadth or other side thereof, a third binomial line.  
SVppose that AB be a second bimedial line, and let AB be supposed to be divided into his parts in the point C, so that let AC be the greater part. And take a rational line DE. And (by the 44. of the first) unto the line DE apply the parallelogram DF equal to the square of the line AB, and making in breadth the line DG. Then I say that the line DG is a third binomial line. Let the self same constructions be in this that were in the propositions next going before. And forasmuch as the line AB is a second bimedial line, 
Construction.  and is divided into his parts in the point C, 
Demonstration.  therefore (by the 38. of the tenth) the lines AC and CB are medials commensurable in power only, comprehending a medial superficies. Wherefore 
†  that which is made of the squares of the lines AC and CB added together, is medial, and it is equal to the parallelogram DL by construction. Wherefore the parallelogram DL is medial, and is applied unto the rational line DE, wherefore (by the 22. of the tenth) the line MD is rational and incommensurable in length to the line DE. And by the like reason also 
*  the line MG is rational and incommensurable in length to the line ML, that is, to the line DE. Wherefore either of these lines DM and MG is rational, and incommensurable in length to the line DE. And forasmuch as the line AC is incommensurable in length to the line CB, but as the line AC is to the line CB, so (by the assumpt going before the 22. of the tenth) is the square of the line AC to that which is contained under the lines AC and CB. Wherefore the square of the line AC is inc●mmmensurable to that which is contained under the lines AC and CB. Wherefore that 
‡  that which is made of the squares of the lines AC and CB added together, is incommensurable to that which is contained under the lines AC and CB twice, that is, the parallelogram DL to the parallelogram MF. Wherefore (by the first of the sixth, and 10. of the tenth) the line DM is incommensurable in length to the line MG. And they are proved both rational, 
DG, concluded a binomial line.  wherefore the whole line DG is a binomial line by the definition in the 36. of the tenth. Now resteth to prove that it is a third binomial line. As in the former propositions, so also in this may we conclude that the line DM is greater than the line MG, and that the line DK is commensurable in length to the line KM. And that that which is contained under the lines DK and KM is equal to the square of the line MN. Wherefore the line DM is in power more than the line MG by the square of a line commensurable in length unto the line DM, and neither of the lines DM nor MG is commensurable in length to the rational line DE. Wherefore (by the definition of a third binomi●ll line) the line DG is a third binomial line: which was required to be proved.  

¶ Here follow certain annotations by M. Dee, made upon three places in the demonstration, which were not very evident to young beginners.  
† (The squares of the lines AC and C● are medials (〈◊〉 i● taught after the 21● of this tenth) and therefore forasmuch as they are (by supposition) commensurable th'one to the other: (by the 15. of the tenth) the compound of them both is commensurable to each part. But the parts are medials, therefore (by the corollary of the 23. of the tenth) the compound shall be medial.  
● For that MX is equal (by construction) to that which is contained under the lines AC and CB, which is proved medial: therefore (by the corollary of the 23. of this tenth) MX is medial, and therefore (by the same corollary) his double MF is medial. And it is applied to a rational line, ML (being equal to D●) therefore by the 22. of the tenth, the line MG is rational and incommensurable in length to ML, that is, to DE.  
‡ Because the compound of the two squares (of the lines AC and C●) being commensurable one to the other, is also to either square (by the 15.) commensurable, therefore to the square of AC: But the square of AC is proved incommensurable to that which is contained under AC & CB once. Wherefore (by the 13. of the tenth) the compound of the two squares (of the lines AC and CB) is incommensurable, to that which is contained under the lines AC and C● once. But to that which is twice contained under the same lines AC and CB, the paralleloganrme once contained, is commensurable (for it is as 1. is to 2.) therefore that which is made of the squares of the lines AC and CB is incommensurable to the parallelogram contained under AC and CB twice, by the said 13. of this tenth.   

¶ A Corollary.  

A Corollary added by M. Dee.  Hereby it is evident, that the squares made of the two parts of a second bimedial line, composed, is a compound medial, and that the same compound is incommensurable to the parallelogram contained under the two parts of the second bimedial line.  
The proof hereof, is in the first and third annotations here before annexed.    

¶ The 45. Theorem. The 63. Proposition. The square of a greater line applied unto a rational line, maketh the breadth or other side a fourth binomial line.  
SVppose that the line AB be a greater line, and let it be supposed to be divided into his parts in the point C, so that let AC be the greater part. And take a rational line DE. And (by the 44. of the first) unto the line DE, apply the parallelogram DF equal to the square of the line AB, and making in breadth the line DG. Then I say, that the line DG is a fourth binomial line. Let the self same construction be in this, that was in the former Propositions. 
Construction.  And forasmuch as the line AB is a greater line, & is divided into his parts in the point C: 
Demonstration.  therefore the lines AC and CB are incommensurable in power, having that which is made of the squares of them added together rational, and the parallelogram which is contained under them, medial. Now forasmuch as that which is made of the squares of the lines AC and CB added together is rational, therefore the parallelogram DL is rational. Wherefore also the line MD is rational and commensurable in length to the line DE (by the 20. of this tenth). Again forasmuch as that which is contained under the lines AC and CB twice is medial, that is, the parallelogram MF, and it is applied unto the rational line ML, therefore (by the ●2. of the tenth) the line MG is rational and incommensurable in length to the line DE, Therefore (by the 13. of the tenth) the line DM is incommensurable in length to the line MG. Wherefore the lines DM and MG are rational commensurable in power only. Wherefore the whole line DG is a binomial line. Now resteth to prove, that it is also a fourth binomial line. Even as in the former Propositions, so also in this may we conclude, that the line DM is greater than the line MC. And that that which is contained under the lines DK and KM is equal to the square of the line MN. Now forasmuch as the square of the line AC is incommensurable to the square of the line CB, therefore the parallelogram DH is incommensurable to the parallelogram KL. Wherefore (by the 1. of the sixth, and 10. of the tenth) the line DK is incommensurable in length to the line KM. But if there be two unequal right lines, and if upon the greater be applied a parallelogram equal to the fourth part of the square made of the less, and wanting in figure by a square, and if also the parallelogram thus applied divide the line whereupon it is applied into parts incommensurable in length, the greater line s●all be in power more than the less, by the square of a line incommensurable in length to the greater (by the 18. of the tenth). Wherefore the line DM is in power more than the line MG, by the square of a line incommensurable in length to DM. And the lines DM and MG are proved to be rational commensurable in power only. And the line DM is commensurable in length to the rational line given DE. Wherefore the line DG is a fourth binomial line: which was required to be proved.   

¶ The 46. Theorem. The 64. Proposition. The square of a line containing in power a rational and a medial superficies applied to a rational line, maketh the breadth or other side a fift binomial line.  
SVppose that the line AB be a line containing in power a rational and a medial superficies, and let it be supposed to be divided into his parts in the point C, so that let AC be the greater part, and take a rational line DE. And (by the 44. of the first) unto the line DE apply the parallelogram DF equal to the square of the line AB, and making in breadth the line DG. Then I say, that the line DG is a fift binomial line. 
Construction.  Let the self same construction be in this, that was in the former. And forasmuch as AB is a line containing in power a rational and a medial superficies, 
Demonstration.  and is divided into his parts in the point C, therefore the lines AC & CB are incommensurable in power, having that which is made of the squares of them added together medial, and that which is contained under then rational. Now forasmuch as that which is made of the squares of the lines AC and CB added together is medial, therefore also the parallelogram DL is medial. Wherefore (by the 22. of the tenth) the line DM is rational and incommensurable in length to the line DE. Again forasmuch as that which is contained under the lines AC and CB twice, that is, the parallelogram MF, is rational, therefore by the 20. the line MG is rational & commensurable in length to the line DE. Wherefore (by the 13. of the tenth) the line DM is incommensurable in length to the line MG. Wherefore the lines DM and MG are rational commensurable in power only. Wherefore the whole line DG is a binomial line. I say moreover, that it is a fift binomiall. For, as in the former, so also in this may it be proved, that that which is contained under the lines DK and KM is equal to the square of MN the half of the less: and that the line DK is incommensurable in length to the line KM. Wherefore (by the 18. of the tenth) the line DM is in power more than the line MG by the square of a line incommensurable in length to the line DM. And the lines DM and MG are rational commensurable in power only, and the less line, namely, MG is commensurable in length to the rational line given DE. Wherefore the line DG is a fift binomial line: which was required to be demonstrated.   

¶ The 47. Theorem. The 65. Proposition. The square of a line containing in power two medials applied unto a rational line, maketh the breadth or other side a sixth binomial line.  
SVppose that the line AB be a line containing in power two medials, and let it be supposed to be divided into his parts in the point C. And take a rational line DE. And (by the 44. of the first) unto the rational line DE apply the parallelogram DF equal to the square of the line AB and making in breadth the line DG. Then I say that the line DG is a sixth binomial line. 
Construction.  Let the self same construction be in this that was in the former. And forasmuch as the line AB is a line containing in power two medials, 
Demonstration.  and is divided into his parts in the point C, therefore the lines AC & CB are incommensurable in power having that which is made of the squares of them added together medial, and that which is contained under them, medial, and moreover incommensurable to that which is made of the squares of them added together. Wherefore by those things which have been before proved, either of these parallelograms DL and MF is medial, and either of them is applied upon the rational line DE. Wherefore (by the 22. of the tenth) either of these lines DM and MG is rational and incommensurable in length to the line DE. And forasmuch as ●hat which is mad● of the squares of the lines AC and CB, added together is incommensurable to that which is contained under the lines AC and CB twice, therefore the parallelogram DL is incommensurable to the parallelogram MF. Wherefore (by the 1. of the sixth and 10. of the tenth) the line DM is incommensurable in length to the line MG. Wherefore the lines DM and MG are rational commensurable in power only. Wherefore the whole line DG is a binomial line. I say also that is a sixth binomial line. For even as in the other propositions it hath been proved so also in this may it be proved, that that which is contained under the lines DK and KM, is equal to the square of the line MN, and that the line DK is incommensurable in length to the line KM, and therefore (by the 18. of the tenth) the line DM is in power more than the line MG by the square of a line incommensurable in length to the line DM. And neither of these lines DM nor MG is commensurable in length to there an●nall line given DE. Wherefore the line DG is a sixth ●●nomiall line: which was required to be demonstrated.   

¶ The 48. Theorem. The 66. Proposition. A line commensurable in length to a binomial line, is also a binomial line of the self same order.  
SVppose that the line AB be a binomial line, 
The sixth Senary.  and unto the line AB let the line CD be commensurable in length. Then I say that the line CD is a binomial line and of the self same order that the line AB is. For forasmuch as AB is ● binomial line let it be divided into his names in the point E, 
Construction.  and let AE be the greater name. Wherefore the lines AE and EB are rational commensurable in power only. And as the line AB is to the line CD, (so by the 12. of the sixth) let the line AE be to the line CF. 
Demonstration.  Wherefore (by the 19 of the fift) the residue, namely, the line EB is to the residue, namely, to the line FD, as the line AB is to the line CD. But (by supposition) the line AB is commensurable in length to the line CD. Wherefore (by the 10. of the tenth) the line AE is commensurable in length to the line CF, and the line EB to the line FD. And the lines AE and EB are rational. Wherefore the lines CF and FD are also rational. And for that as the line AE is to the line CF, so is the line EB. to the line FD, therefore alternately (by the 16. of the fift) as the line A●E is to the line EB, so is the line CF to the line FD. But the lines AE and EB are commensurable in power only, wherefore the lines CF and FD are also commensurable in power only, and they are rational. Wherefore the whole line CD is a binomial line. I say also that it is of the self same order of binomial lines that the line AB is. For the line AE is in power more than the line EB either by the square of a line commensurable in length to the line AE, or by the square of a line incommensurable in length to the line AE. If the line AE be in power more than the line EB by the square of a line commensurable in length to the line AE, the line also CF, by the 14. of the tenth) shallbe in power more than the line FD by the square of a line commensurable in length to CF. And if the line● AE be commensurable in length to a rational line given, the line CF also shallbe commensurable in length to the same (by the 12. of the tenth). And so either of these lines AB and CD is a first binomial line, that is, they are both of one and the self same order. But if the line EB be commensurable in length to the rational line put, the line FD also shallbe commensurable in length to the same. And by that means again the lines AB and CD are both of one and the self same order● for either of them is a second binomial line. But if neither of the lines AE nor E● be commensurable in length to the rational line put, neither also of these lines CF nor FD shallbe commensurable in length to the same. And so neither of the lines AB and CD is a third binomial line. But if the line AE be in power more than the line EB by the square of a line incommensurable in length to the line AE, the line also CF shallbe in power more than the line FD by the square of a line incommensurable in length to the line CF, (by the 14. of the tenth). And then if the line AE be commensurable in length to the rational line put the line CF also shallbe commensurable in length to the same, and so either of the lines AB, and CD shallbe a fourth binomial line. And if the line EB be commensurable in length to the rational line given, the line FD also shallbe commensurable in length to the same. And so either of the lines AB and CD shallbe a fift binomial line. But if neither of the lines AE nor EB be commensurable in length to the rational line given, neither also of the lines CF nor FD shallbe commensurable in length to the same, and so either of the lines AB and CD shallbe a sixth binomial line. A line, therefore commensurable in length to a binomial line, is also a binomial line of the self same order: which was required to be proved.   

¶ The 49. Theorem. The 67. Proposition. A line commensurable in length to a bimedial line, is also a bimedial line and of the self same order.  
SVppose that the line AB be a bimedial line, And unto the line AB, let the line CD be commensurable in length. Then I say that the line CD is a bimedial line, and of the self order that the line AB is. Divide the line AB into his parts in the point E. 
Construction.  And forasmuch as the line AB is a bimedial line, and is divided into his parts in the point E, therefore (by the 37. and 38. of the tenth) the lines AE and EB are medials commensurable in power only, And (by the 12. of the sixth) as the line AB is to the line CD, so let the line AE be to the line CF. Wherefore (by the 19 of the fift) the residue, 
Demonstration.  namely, the line EB is to the residue, namely, to the line FD, as the line AB is to the line CD. But the line AB is commensurable in length to the line CD. Wherefore the line AE is commensurable in length to the line CF, and the line EB to the line FD. Now the lines AE and EB are medial, wherefore (by the 23. of the tenth) the lines CF and FD are also medial. And for that as the line AE is to the line EB, so is the line CF to the line FD. But the lines AE and EB are commensurable in power only, wherefore the lines CF and FD are also commensurable in power only. And it is proved that they are medial. Wherefore the line CD is a bimedial line. I say also that it is of the self same order that the line AB is. For, for that as the line AE is to the line EB, so is the line CF to the line FD, but as the line CF is to FD, so is the square of the line CF to the parallelogram contained under the lines CF and FD, by the first of the sixth. Therefore as the line AE is to the line EB, so (by the 11. of the fift) is the square of the line CF to the parallelogram contained under the lines CF and FD: but as AE is to EB, so by the 1. of the sixth, is the square of the line AE, to the parallelogram contained under the lines AE and EB, therefore (by the 11. of the fift) as the square of the line AE is to that which is contained under the lines AE and EB, so is the square of the line CF to that which is contained under the lines CF and FD. Wherefore alternately (by the 16. of the fift) as the square of the line AE is to the square of the line CF, so is that which is contained under the lines AE and EB to that which is contained under the lines CF & FD. But the square of the line AE is commensurable to the square of the line CF, because AE and CF are commensurable in length. Wherefore that which is contained under the lines AE and EB in commensurable to that which is contained under the lines CF and FD. If therefore that which is contained under the lines AE and EB be rational, that is, if the line AB be a first bimedial line, that also which is contained under the lines CF and FD is rational. Wherefore also the line CD is a first bimedial line. But if that which is contained under the lines AE and EB be medial, that is, if the line AB be a second bimedial line, that also which is contained under the lines CF and FD is medial: wherefore also the line CD is a second bimedial line. Wherefore the lines AB and CD are both of one and the self same order: which was required to be proved.  

¶ A Corollary added by Flussates: but first noted by P. Monta●reus.  
A line commensurable in power only to a bimedial line, is also a bimedial line, and of the self same order.  
Suppose that AB be a bimedial line, either a first or a second, whereunto let the line GD be commensurable in power only. 
A Corollary added by Flussetes.  Take also a rational line EZ, upon which (by the 45. of the first) apply a rectangle parallelogram equal to the square of the line AB, which let be EZFC, and let the rectangle parallelogram CFIH be equal to the square of the line GD. And forasmuch as upon the rational line EZ is applied a rectangle parallelogram EF equal to the square of a first bimedial line, therefore the other side thereof, namely, EC, is a second binomial line, by the 61. of this book. And forasmuch as by supposition the squares of the lines AB & GD are commensurable, therefore the parallelograms EF and CI (which are equal unto them) are also commensurable. And therefore by the 1. of the sixth, the lines EC and CH are commensurable in length. But the line EC is a second binomial line. Wherefore the line CH is also a second binomial line, by the 66. of this book. And forasmuch as the superficies CI is contained under a rational line EZ or CF, and a second binomial line CH, therefore the line which containeth it in power, namely, the line GD is a first bimedial line, by the 55. of this book. And so is the line GD in the self same order of bimedial lines that the line AB is. The like demonstration also will serve if the line AB be supposed to b● a second bimedial line. For so shall it make the breadth EC a third binomial line whereunto the line CH shall be commensurable in length, and therefore CH also shall be a third binomial line, by means whereof the line which containeth in power the superficies CI, namely, the line GD shall also be a second bimedial line. Wherefore a line commensurable either in length, or in power only to a bimedial line, is also a bimedial line of the self same order.  
But so is it not of necessity in binomial lines, for if their powers only be commensurable, it followeth not of necessity that they are binomialls of one and the self same order, 
Note.  but they are each binomialls either of the three first kinds, or of the three last. As for example. Suppose that AB be a first binomial line, whose greater name let be AG, and unto AB let the DZ be commensurable in power only. Then I say, that the line DZ is not of the self same order that the line AB is. For if it be possible, let the line DZ be of the self same order that the line AB is. Wherefore the line DZ may in like sort be divided as the line AB is, by that which hath been demonstrated in the 66. Proposition of this booke● let it be so divided in the point E. Wherefore it can not be so divided in any other point, by the 42● of this book. And for that the line AB ●● to the line DZ, as the line AG is to the line DE, but the lines AG & DE, namely, the greater names, are commensurable in length the one to the other (by the 10. of this book) for that they are commensurable in length to 〈◊〉 and the self same rational line, by the first definition of binomial lines. Wherefore the lines AB and DZ are commensurable in length, by the 13. of this book. But by supposition they are commensurable in power only: which is impossible.  
The self same demonstration also will serve, if we suppose the line AB to be a second binomial line: for the less names GB and EZ being commensurable in length to one and the self same rational line, shall also be commensurable in length the one to the other. And therefore the lines AB and DZ which are in the self same proportion with them, shall also be commensurable in length the one to the other: which is contrary to the supposition. Farther, if the squares of the lines AB and DZ be applied unto the rational line CF, namely, the parallelograms CT and HL, they shall make the breadthes CH and HK first binomial lines, of what order soever the lines AB & DZ (whose squares were applied unto the rational line) are, (by the 60. of this book). Wherefore it is manifest, that under a rational line and a first binomial line, are confusedly contained all the powers of binomial lines (by the 54. of this book). Wherefore the only commensuration of the powers doth not of necessity bring forth one and the self same order of binomial lines. The self same thing also may be proved, if the lines AB and DZ be supposed to be a fourth or fifth binomial line, whose powers only are conmmensurable, namely, that they shall as the first bring forth binomial lines of divers orders. Now forasmuch as the powers of the lines AG and GB, and DE and EZ are commensurable & proportional, it is manifest, that if the line AG be in power more than the line GB by the square of a line commensurable in length unto AG, the line DE also shall be in power more than the line EZ by the square of a line commensurable in length unto the line DE (by the 16. of this book). And so shall the two lines AB and DZ be each, of the three first binomial lines. But if the line AG be in power more than the line GB by the square of a line incommensurable in length unto the line AG, the line DE shall also be in pow●r 〈◊〉 than the line EZ by the square of a line incomensurable in length unto the line DE, by the self same Proposition. And so shall each of the lines AB and DZ be of the three last binomial lines. But why it is not so in the third and sixth binomial lines, the reason is: For that in them neither of the name● is commensurable in length to the rational line put FC.    

¶ The 50. Theorem. The 68 Proposition. A line commensurable to a greater line, is also a greater line.  
SVppose that the line AB be a greater line. And unto the line AB let the line CD be commensurable. 
Construction.  Then I say that the line CD also is a greater line. Divide the line AB into his parts in the point E. Wherefore (by the 39 of the tenth) the lines AE and EB are incommensurable in power, having that which is made of the squares of them added together rational, and that which is contained under them medial. And let the rest of the construction be in this, as it was in the former. And for that as the line AB is to the line CD, 
Demonstration.  so is the line AE to the line CF, & th● line EB to the line FD, but the line AB is commensurable to the line CD by supposition. Wherefore the line AE is commensurable to the line CF, and the line EB to the line FD. And for that as the line AE is to the line CF, so is the line EB to the line FD. Therefore alternately (by the 16. of the fift) as the line AE is to the line EB, so is the line CF to the line FD. Wherefore by composition also (by the 18. of the fift) as the line AB is to the line EB, so is the line CD to the line FD. Wherefore (by the 22. of the sixth) as the square of the line AB is to the square of the line EB, so is the square of the line CD to the square of the line FD. And in like sort may we prove that as the square of the line AB is to the square of the line AE, so is the square of the line CD, to the square of the line CF. Wherefore (by the 11. of the fift) as the square of the line AB is to the squares of the lines AE and EB, so is the square of the line CD to the squares of the lines CF and FD. Wherefore alternately (by the 16. of the fift) as the square of the line AB is to the square of the line CD, so are the squares of the lines AE and EB to the squares of the lines CF and FD. But the square of the line AB is commensurable to the square of the line CD (for the line AB is commensurable to the line CD by supposition). Wherefore also the squares of the lines AE and EB are commensurable to the squares of the lines CF and FD. But the squares of the lines AE and EB are incommensurable, and being added together are rational. Wherefore the squares of the lines CF and FD are incommensurable, & being added together, are also rational. And in like sort may we prove that that which is contained under the lines AE and EB twice is commensurable to that which is contained under the lines CF and FD twice. But that which is contained under the lines AE and EB twice, is medial, wherefore also that which is contained under the lines CF and FD twice is medial. Wherefore the lines CF and FD are incommensurable in power, having that which is made of the squares of them added together rational, and that which is contained under them medial. Wherefore (by the 39 of the tenth) the whole line CD is irrational, & is called a greater line. A line therefore commensurable to a greater line, is also a greater line.  

another demonstration of Peter Montaureus to prove the same.  
Suppose that the line AB be a greater line, and unto it let the line CD be commensurable any way, that is, either both in length and in power, or else in power only. 
another demonstration after P. Montaureus.  Then I say that the line CD also is a greater line. Divide the line AB into his parts in the point E. and let the rest of the construction be in this as it was in the former. And for that as the line AB is to the line CD, so is the line AE to the line CF, and the line EB to the line FD, therefore as the line AE is to the line CF, so is the line EB to the line FD, but the line AB is commensurable to the line CD. Wherefore also the line AE is commensurable to the line CF, and likewise the line EB to the line FD. And for th●● as the line AE is to the line CF, so is the line EB to the line FD, therefore alternately as the line AE is to the line EB, so is the line CF to the line FD. Wherefore (by the 22. of the sixth) as the square of the line AE is to the square of the line EB, so is the square of the line CF to the square of the line FD. Wherefore by composition (by the 18. of the fift) as that which is made of the squares of the lines A● and E● added together is to the square of the line EB, so is that which is made of the square● of the lyne● C● and FD added together to the square of the line FD. Wherefore by contrary proportion as the square of the line EB is to that which is made of the squares of the lines AE and E● added together, so is the square of the line FD to that which is made of the squares of the lines CF and FD added together. Wherefore alternately as the square of the line EB is to the square of the line FD, so is that which is made of the squares 〈◊〉 the l●nes AE and EB added together to that which is made of the squares of the lines CF and FD added together. But the square of the line EB is commensurable to the square of the line FD, for it hath already been proved that the lines EB and FD are commensurable. Wherefore that which is made of the squares of the lines AE & EB added together is commensurable to that which is made of the squares of C● & FD added together. But that which is made of the squares of the lines AE and EB added together is rational by supposition. Wherefore that which is made of the squares of the lines CF and FD added together is also rational. And as the line AE is to the line EB, so is the line CF to the line FD: But as the line AE is to the line EB, so is the square of the line A 〈…〉 contained under the lines AE and EB: therefore at the line CF is to the line FD, so is the square of the line AE to the parallelogram contained under the lines AE and EB & as the line CF is to the line FD, so is the square of the line CF to the parallelogram contained under the lines ●F & FD. Wherefore as the square of the line AE is to the parallelogram con●●●●ed under the lines AE and EB, so is the square of the line CF to the parallelogram contained under the lines CF and FD. Wherefore 〈◊〉 ●s the square of the line AE is to the square of the line CF, so is the parallelogram contained under the lines AE and EB to the parallelogram 〈◊〉 vnde● the lines ●● and ●●. But the square of the line AE is commensurable to the square of the line CF, for it is already pr●●●d that the lines AE and CF are commensurable. Wherefore the parallelogram contained under the lines AE and EB is commensurable to the parallelogram contained under the lines CF and FD. But the parallelogram contained under the lines AE and EB is medial by supposition. Wherefore the parallelogram contained under the lines CF and ●D also is medial. And (as it hath already been proved) as the line AE is to the line EB, so is the line CF to the line FD. But the line AE was by supposition incommensurable in power to the line EB. Wherefore (by the 10. of the tenth) the line CF is incommensurable in power to the line FD. Wherefore the lines CF and FD are incommensurable in power, having that which is made of the squares of them added together rational, and that which is contained under them medial. Wherefore the whole line CD is (by the 39 of the tenth) a greater line. Wherefore a line commensurable to a greater line is also a greater line: which was required to be demonstrated.   

An other more brief demonstration of the same after Campane.  
Suppose that A be a greater line, unto which let the line B be commensurable, either in length and power, or in power only. 
another demonstration after Campane.  And take a rational line CD. And upon it apply the superficies C● equal to the square of the line A: and also upon the line FE (which is equal to the rational line CD) apply the parallelogram FG equal to the square of the line B. And forasmuch as the squares of the two lines A and ● are commensurable by supposition, the superficies C●, shallbe commensurable unto the superficies FG: and therefore by the first of the sixth and tenth of this book, the line DE is commensurable in length to the line GB. And forasmuch as (by the ●3. of this book) the line DE is a fourth binomial line, therefore by the ●6. of this book the line GE is also a fourth binomial line: wherefore by the 57 of this book the line B which containeth in power the superficies FG is a greater line.    

¶ The 51. Theorem. The 69. Proposition. A line commensurable to a line containing in power a rational and a medial: is also a line containing in power a rational and a medial.  
SVppose that AB be a line containing in power a rational and a medial. And unto the line AB let the line CD be commensurable, whether in length and power, or in power only. Then I say that the line CD is a line containing in power a rational & a medial. Duide the line AB into his parts in the point E. 
Construction.  Wherefore (by the 40. of the tenth) the lines AE and EB are incommensurable in power, having that which is made of the squares of them added together medial, and that which is contained under them national. Let the same construction be in this that was in the former. 
Demonstration.  And in like sort we may prove that the lines CF and FD are incommensurable in power, and that that which is made of the square of the lines AE and EB is commensurable to that which is made of the squares of the lines CF and FD, and that that also which is contained under the lines AE and EB is commensurable to that which is contained under the lines CF and FD. Wherefore that which is made of the squares of the lines CF and FD is medial, and that which is contained under the lines CF and FD is rational. Wherefore the whole line CD is a line containing in power a rational and a medial: which was required to be demonstrated.  

another demonstration of the same after Campane.  
Suppose that AB be a line containing in power a rational and a medial: 
another demonstration af●●r Campane.  whereunto let the line GD be commensurable either in length and power, or in power only. Then I say that the line GD is a line containing in power a rational and a medial. Take a rational line EZ, upon which by the 45. of the first apply a rectangle parallelogram EZFC equal to the square of the line AB: and upon the line CF (which is equal to the line EZ) apply the parallelogram FCHI equal to the square of the line GD● and let the breadths of the said parallelograms be the lines EG and CH. And forasmuch as the line AB is commensurable to the line GD at the lest in power only, therefore the parallelograms EF and FH (which are equal to their squares) shallbe commensurable. Wherefore by the 1. of the sixth the right lines EC and CH are commensurable in length. And forasmuch as the parallelogram EF (which is equal to the square of the line A● which containeth in power ● rational and a medial) is applied upon the rational EZ, making in breadth the line EC, therefore the line EC is a fifth binomial line (by the 64. of this book) unto which line EC the line CH is commensurable in length, wherefore by the 66. of this book the line CH is also a fifth binomial line. And forasmuch as the superficies CI is contained under the rational line EZ (that is CF) and a fifth binomall line CH, therefore the line which containeth in power the superficies CI, which by supposition is the line GD is a line containing in power a rational and a medial by the 58. of this book. A line therefore commensurable to a line containing in power a rational and a medial. etc.    

¶ The 52. Theorem. The 70. Proposition. A line commensurable to a line containing in power two medials, is also a line containing in power two medials.  
SVppose that AB be a line containing in power two medials. And unto the line AB let the line CD be commensurable, whether in length & power, or in power only. Then I say, that the line CD is a line containing in power two medials. Forasmuch as the line AB is a line containing in power two medials, 
Construction.  let it be divided into his parts in the point E. Wherefore (by the 41. of the tenth) the lines AE and EB are incommensurable in power, having that which is made of the squares of them added together medial, and that also which is contained under them medial, and that which is made of the squares of the lines AE & EB is incommensurable to that which is contained under the lines AE and EB. Let the self same construction be in this, that was in the former. 
Demonstration.  And in like sort may we prove, that the lines CF & FD are incommensurable in power, and that that which is made of the squares of the lines AE and EB added together, is commensurable to that which is made of the squares of the lines CF and FD added together, and that that also which is contained under the lines AE and EB is commensurable to that which is contained under the lines CF and FD. Wherefore that which is made of the squares of the lines CF and FD is medial (by the Corollary of the 23. of the tenth): and that which is contained under the lines CF and FD is medial (by the same Corollary) ● and moreover, that which is made of the squares of the lines CF & FD is incommensurable to that which is contained under the lines CF and FD. Wherefore the line CD is a line containing in power two medials: which was required to be proved.  

¶ An Assumpt added by Montaureus.  

An Assumpt.  That that which is made of the squares of the lines CF and FD added together, is incommensurable to that which is contained under the lines CF and FD is thus proved. For, because as that which is made of the squares of the lines AE and EB added together is to the square of the line AE, so is that which is made of the squares of the lines CF and FD added together, to the square of the line CF, as it was proved in the Propositions going before: therefore alternately, as that which is made of the squares of AE and EB added together is to that which is made of the squares of CF and FD added together, so is the square of the line AE to the square of the line CF. But before, namely, in the 68 Proposition, it was proved, that as the square of the line AE is to the square of the line CF, so is the parallelogram contained under the lines AE and EB to the parallelogram contained under the lines CF and FD. Wherefore as that which is made of the squares of the lines AE and EB is to that which is made of the squares of the lines CF and FD, so is the parallelogram contained under the lines AE and EB to the parallelogram contained under the lines CF and FD. Wherefore alternately, as that which is made of the squares of the lines AE and EB is to the parallelogram contained under the lines AE and EB, so is that which is made of the squares of the lines CF and FD to the parallelogram contained under the lines CF and FD. But by supposition that which is made of the squares of the lines AE and EB, is incommensurable to the parallelogram contained under the lines AE & EB. Wherefore that which is made of the squares of the lines CF and FD added together, is incommensurable to the parallelogram contained under the lines CF and FD: which was required to be proved.   

another demonstration after Campane.  
Suppose that AB be a line containing in power two medials: whereunto let the line GD be commensurable either in length, and in power, or in power only. 
another demonstration after Campan●.  Then I say, that the line GD is a line containing in power two medials. Let the same construction be in this, that was in the former. And forasmuch as the parallelogram EF is equal to the square of the line AB, and is applied upon a rational line EZ, it maketh the breadth EC a sixth binomial line, by the 65. of this book. And forasmuch as the parallelograms EF & CI (which are equal unto the squares of the lines AB and GD, which are supposed to be commensurable) are commensurable, therefore the lines EC and CH are commensurable in length, by the first of the sixth. But EC is a sixth binomial line: Wherefore CH also is a sixth binomial line, by the 66. of this book. And forasmuch as the superficies CI is contained under the rational line CF and a sixth binomial line CH, therefore the line which containeth in power the superficies CI, namely, the line GD is a line containing in power two medials, by the 59 of this book. Wherefore a line commensurable to a line containing in power two medials. etc.   

An Annotation.  
If other to hath been spoken of six Senaries, of which the first Senary containeth the production of irrational lines by composition: the second, the division of them, namely, that those lines are in one poin● only deuide● the third, the finding out of binomial lines, of the first, I say, the second, the third, the fourth, the fift, and the sixth: after that beginneth the forth Senary, containing the difference of irrational lines between themselves. For by the nature of every one of the binomial lines are demonstrated the differences of irrational lines. The fifth entre●teth of the applications of the squares of every irrational line, namely, what irrational lines are the breadthes of every superficies so applied. In the sixth Senary is proved, that any line commensurable to any irrational line, is also an irrational line of the same nature. And now shall be spoken of the seventh Senary, wherein again are plainly set forth the rest of the differences of the said lines between themselves.  
And the●e is even in those irrational lines an arithmetical proportionality. 
Note.  And that line which is the arithmetical mean proportional between the parts of any irrational line, is also an irrational line of the self same kind. First it is certain that there is an arithmetical proportion between those parts. For suppose that the line AB be any of the foresaid irrational lines, as for example, let it be a binomial line, & let it be divided into his names in the point C. And let AC be the greater name, from which take away the line AD equal to the less name, namely, to CB. And divide the line CD into two equal parts in the point E. It is manifest that the line AE is equal to the line EB. Let the line FG be equal to either of them. It is plain that how much the line AC differeth from the line FG, so much the same line FG differeth from the line CB: for in each is the difference of the line DE or EC, which is the property of arithmetical proportionality. And it is manifest, that the line FG is commensurable in length to the line AB, for it is the half thereof. Wherefore (by the 66. of the tenth) the line FG is a binomial line. And after the self same manner may it be proved touching the rest of the irrational lines.    

¶ The 53. Theorem. The 71. Proposition. If two superficieces, namely, a rational and a medial superficies be composed together, the line which containeth in power the whole superficies, is one of these four irrational lines, either a binomial line, or a first bimedial line, or a greater line, or a line containing in power a rational and a medial superficies.  
But now let the line EH be in power more than the line HK by the square of a line incommensurable in length to the line EH: 
Second part of the first case.  now the greater name that is, EH is commensurable in length to the rational line given EF. Wherefore the line EK is afourth binomial line. And the line EF is rational. But if a superficies be contained under a rational line and afourth binomial line, the line that containeth in power the same superficies is (by the 57 of the tenth) irrational, and is a greater line. Wherefore the line which containeth in power the parallelogram EI is a greater line. Wherefore also the line containing in power the superficies AD is a greater line.  

The second case.  But now suppose that the superficies AB which is rational, be less than the superficies CD which is medial. Wherefore also the parallelogram EG is less than the parallelogram high. Wherefore also the line EH is less than the line HK. Now the line HK is in power more than the line EH either by the square of a line commensurable in length to the line HK, or by the square of a line incommensurable in length unto the line HK. First let it be in power more by the square of a line commensurable in length unto HK: 
First part of the second case.  now the less name, that is EH is commensurable in length to the rational line given EF, as it was before proved. Wherefore the whole line EK is a second binomial line. And the line EF is a rational line. But if a superficies be contained under a rational line and a second binomial line, the line that containeth in power the same superficies, is (by the 55. of the tenth) a first bimedial line. Wherefore the line which containeth in power the parallelogram EI is a first bimedial line. Wherefore also the line that containeth in power the superficies AD is a first bimedial line.  

Second part of the second case.  But now let the line HK be in power more than the line EH, by the square of a line incommensurable in length to the line HK, now the less name, that is, EH is commensurable in length to the rational line given EF. Wherefore the whole line EK is a fift binomial line. And the line EF is rational. But if a superficies be contained under a rational line, and a fift binomial line, the line that containeth in power the same superficies, is (by the 58. of the tenth) a line containing in power a rational and a medial. Wherefore the line that containeth in power the parallelogram EI is a line containing in power a rational and a medial. Wherefore also the line that containeth in power the superficies AD is a line containing in power a rational and a medial. If therefore a rational and a medial superficies be added together, the line which containeth in power the whole superficies, is one of these four irrational lines, namely, either a binomial line, or a first bimedial line, or a greater line, or a line containing in power a rational and a medial: which was required to be demonstrated.   

¶ The 54. Theorem. The 72. Proposition. If two medial superficieces incommensurable the one to the other be composed together: the line containing in power the whole superficies is one of the two irrational lines remaining, namely, either a second bimedial line, or a line containing in power two medials.  
LEt these two medial superficieces AB and CD being incommensurable the one to the other be added together. Then I say, that the line which containeth in power the superficies AD is either a second bimedial line, or a line containing in power two medials. 
Construction.  For the superficies AB is either greater or less than the superficies CD (for they can by no means be equal, when as they are incommensurable). 

Two cases in this Proposition.  
The first case.   First let the superficies AB be greater than the superficies CD. And take a rational line EF. And (by the 44. of the first) unto the line EF apply the parallelogram EG equal to the superficies AB, and making in breadth the line EH: and unto the same line EF, that is, to the line HG, apply the parallelogram high equal to the superficies CD, & making in breadth the line HK. And forasmuch as either of these superficieces AB & CD is medial, therefore also either of these parallelograms EG and high is medial. And they are each applied to the rational line EF, making in breadth the lines EH and HK. Wherefore (by the 22. of the tenth) either of these lines EH and HK is rational and incommensurable in length to the line EF. And forasmuch as the superficies AB is incommensurable to the superficies CD, and the superficies AB is equal to the parallelogram E●, and the superficies CD to the parallelogram high: therefore the parallelogram EG is incommensurable to the parallelogram high. But (by the 1. of the sixth) as the parallelogram EG is to the parallelogram high, so is the line EH● to the line HK. Wherefore (by the 10. of the tenth) the line EH i●●spans● HK● Wherefore the ●●nes EH ●nd HK are rational commensurable in power only. Wherefore the whole line EK is a binomial line. And as in the former Proposition so als● in this may it be proved, that the line EH is greater than the line HK. Wherefore the line EH is in power more than the line HK, either by the square of a line commensurable in length to the line EH, or by the square of a line incommensurable in length to the line EH. First let it be greater by the square of a line commensurable in length unto the line EH. 
The first part of the first case.  Now neither of these lines EH and HK is commensurable in length to the rational line given EF. Wherefore the whole line EK is a third binomial line. And the line EF is a rational line. But if a superficies be contained under a rational line & a third binomial line, the line that containeth in power the same superficies, is (by the 56. of the tenth) a second bimedial line. Wherefore the line that containeth in power the superficies EI, that is, the superficies AD, is a second bimedial line.  

〈◊〉 second 〈◊〉 of the 〈◊〉 case.  But now suppose that the line EH be in power more than the line HK by the square of a line incommensurable in length to the line EH. And forasmuch as either of these lines EH and HK is incommensurable in length to the rational line given EF, therefore the line EK is a sixth binomial line. But if a superficies be contained under a rational line and a sixth binomial line, the line that containeth in power the same superficies, is (by the 59 of the tenth) a line containing in power two medials. Wherefore the line that containeth in power the superficies AD, is a line containing in power two medials. 
The second case.  And after the self same manner, if the superficies AB be less than the superficies CD, may we prove, that the line that containeth in power the superficies AD, is either a second bimedial line, or a line containing in power two medials. If therefore two medial superficieces incommensurable the one to the other be added together, the line containing in power the whole superficies is one of the two irrational lines remaining, namely, either a second bimedial line, or a line containing in power two medials: which was required to be proved.  

¶ A Corollary following of the former Propositions.  
A binomial line and the other irrational lines following it, are neither medial lines, nor one and the same between themselves. For the square of a medial line applied to a rational line, 
A Corollary  maketh the breadth rational and incommensurale in length to the rational line, whereunto it is applied (by the 22. of the tenth). The square of a binomial line applied to ● rational line, maketh the breadth a first binomial line (by the 60. of the tenth). The square of a first bimedial line applied unto a rational line, maketh the breadth a second binomial line (by the 61. of the tenth). The square of a second bimedial line applied unto a rational line, maketh the breadth a third binomial line (by the 62. of the tenth). The square of a greater line applied to a rational line, maketh the breadth a fourth binomial line (by the 63. of the tenth). The square of a line containing in power a rational & a medial superficies, maketh the breadth a fift binomial line (by the 64. of the tenth). And the square of a line containing in power two medials, applied unto a rational line, maketh the breadth a sixth binomial line (by the 65. of the tenth). Seeing therefore that these foresaid breadthes differ both from the first breadth, for that it is rational, and differ also the one from the other, for that they are binomials of divers orders: it is manifest that those irrational lines differ also the one from the other.    

Here beginneth the Senaries by substraction. ¶ The 55. Theorem. The 73. Proposition. If from a rational line be taken away a rational line commensurable in power only to the whole line: the residue is an irrational line, and is called a residual line.  
SVppose that AB be a rational line, and from AB take away a rational line BC commensurable in power only to the whole line AB. 
The first Senary by substraction.  Then I say that the line remaining, namely AC is irrational and is called a residual line. For forasmuch as the line AB is incommensurable in length unto the line BC, and (by the assumpt going before the 22. of the tenth) as the line AB is to the line BC, 
Demonstration.  so i● the square of the line AB to that which is contained under the lines AB and BC: wherefore (by the 10. of the tenth) the square of the line AB is incommensurable to that which is contained under the lines AB and BC. But unto the square of the line AB are commensurable the squares of the lines AB and BC (by the 15. of the tenth). Wherefore the squares of the lines AB and BC are incommensurable to that which is contained under the lines AB and BC. But unto that which is contained under the lines AB and BC is commensurable that which is contained under the lines AB and BC twice. Wherefore the squares of the lines AB and BC are incommensurable to that which is contained under the lines AB and BC twice. But the squares of the lines AB and BC are equal to that which is contained under the lines AB and BC twice, and to the square of the line AC (by the 7. of the second). Wherefore that which is contained under the lines AB and BC twice together with the square of the line AC is incommensurable to that which is contained under the lines AB and BC twice. Wherefore (by the 2 part of the 16. of the tenth) that which is contained under the lines AB and BC twice, is incommensurable to the square of the line AC. Wherefore (by the first part of the same) that which is contained under the lines AB and BC twice together with the square of the line AC, that is, the squares of the lines AB and BC are incommensurable to the square of the line AC. But the squares of the lines AB and BC are rational, for the lines AB and BC are put to be rational: wherefore the line AC is irrational and is called a residual line: which was required to be proved.  

another demonstration after Campane.  
Campane demonstrateth this Proposition by a figure more briefly after this m●ner. 
another demonstration after Campane.  Let the superficies EG be equal to the squares of the lines AB and BC added together: which shall be rational (for that the lines AB and BC are supposed to be rational commensurable in power only). Fron which superficies take away the superficies DF equal to that which is con●●ya●d under the lines AB & DC twice, which shall be medial (by the 21. of this book). Now by the 7. of the second, the superficies FG is equal to the square of the line AC. And forasmuch as the superficies EG is incommensurable to the superficies DF (for that the one is rational and the other medial): therefore (by the 16. of this book) the 〈◊〉 superficies EG is incommensurable to the superficies FG. Wherefore the superficies FG is irrational. And therefore the line AC which containeth it in power is irrational: which was required to be proved.   

An annotation of P. Monta●re●s.  
This Theorem teacheth nothing else but that that portion of the greater name of a binomial line which remaineth after the taking away of the less name from the greater name is irrational, which is called a residual line, that is to say, if from the greater name of a binomial line, which greater name is a rational line commensurable in power only to the less name, be taken away the less name, which self less name is also commensurable in power only to the greater name (which greater name this Theorem calleth the whole line) the rest of the line which remaineth is irrational, which he calleth a residual line. Wherefore all the lines which are entreated in this Theorem, and in the five other which follow are the portions remaining of the greater parts of the whole lines which were entreated of in the 36.37.38.39.40.41. propositions, after the taking away the less part from the greater.  
In this proposition is set forth the nature of the eight kind of irrational lines which is called a residual line, the definition whereof by this proposition is thus.  

Definition of the eight irrational line.  A residual line is an irrational line which remaineth, when from a rational line given, is taken away a rational line commensurable to the whole line in power only.    

¶ The 56. Theorem. The 74. Proposition. If from a medial line be taken away a medial line commensurable in power only to the whole line, and comprehending together with the whole line a rational superficies: the residue is an irrational line, and is called a first medial residual line.  
Out of this proposition is taken the definition of the ninth kind of irrational lines, which is called a first residual medial line the definition whereof is thus.  

Definition of 〈◊〉 ●inth irrational line.  A first residual medial line is an irrational line which remaineth, when from a medial line is taken away a medial line commensurable to the whole in power only, and the part taken away and the whole line contain a medial superficies.  

another demonstration after Campane.  
Let the line DE be rational, upon which apply the superficies DF equal to that which is contained under the lines AB and BC twice, and let the superficies GE be equal to that which is composed of the squares of the lines AB and BC: 
another demonstration after Campane.  wherefore by the 7. of the second, the superficies FG is equal to the square of the line AC. And forasmuch as (by supposition) the superficies EG is medial, therefore (by the 22. of the tenth) the line DG is rational commensurable in power only to the rational line DE. And forasmuch as by supposition the superficies EH is rational, therefore by the 20. of the tenth, the line DH is rational commensurable in length unto the rational line DE. Wherefore the lines DG and DH are rational commensurable in power only (by the assumpt put before the 13. of this book). Wherefore by the 73 of this book, the line GH is a residual line, and is therefore irrational. Wherefore (by the corollary of the 21. of this book) the superficies FG is irrational. And therefore the line AC which containeth it in power is irrational, and is called a first medial residual line.    

¶ The 57 Theorem. The 75. Proposition. If from a medial line be taken away a medial line commensurable in power only to the whole line, and comprehending together with the whole line a medial superficies, the residue is an irrational line, and is called a second medial residual line.  
SVppose that AB be a medial line, and from AB take away a medial line CB commensurable in power only to the whole line AB, and comprehending together with the whole line AB a medial superficies, namely, the parallelogram contained under the lines AB and BC. 
Construction.  Then I say that the residue, namely; the line AC is irrational, and is called a second medial residual line. Take a rational line DIEGO, and (by the 44. of the first) unto the line DIEGO apply the parallelogram DE equal to the squares of the lines AB & BC, and making in breadth the line DG. 
Demonstration.  And unto the same line DIEGO apply the parallelogram DH equal to that which is contained under the lines AB & BC twice, and making in breadth the line DF. Now the parallelogram DH is less than the parallelogram DE, for that also the square of the lines AB and BC are greater than that which is contained under the lines AB and BC twice, by the square of the line AC by the 7. of the second. Wherefore the parallelogram remaining, namely, FE, is equal to the square of the line AC. And forasmuch as the squares of the lines AB and BC are medial, therefore also the parallelogram DE is medial, and is applied to the rational line DIEGO, making in breadth the line DG. Wherefore (by the 22. of the tenth) the line DG is rational and incommensurable in length to the line DI. Again forasmuch as that which is contained under the lines AB and BC is medial, therefore also that which is contained under the lines AB and BC twice is medial, but that which is contained under the lines AB and BC twice is equal to the parallelogram DH. Wherefore the parallelogram DH is medial and is applied to the rational line DIEGO making in breadth the line DF. Wherefore the line DF is rational and incommensurable in length to the line DI. And forasmuch as the lines AB and BC are commensurable in power only, therefore the line AB is incommensurable in length to the line BC. Wherefore (by the assumpt going before the 22. of the tenth, and by the 10. of the tenth) the square of the line AB is incommensurable to that which is contained under the lines AB and BC. But unto the square of the line AB are commensurable the squares of AB and BC (by the 15. of the tenth) And unto that which is contained under the lines AB and BC is commensurable to that which is contained under the lines AB and BC twice. Wherefore the squares of the lines AB and BC are incommensurable to that which is contained under the lines AB and BC twice. But unto the squares of the lines AB and BC is equal the parallelogram DE, and to that which is contained under the lines AB and BC twice, is equal the parallelogram DH. Wherefore the parallelogram DE is incommensurable to the parallelogram DH. But as the parallelogram DE is to the parallelogram DH, so i● the line GD to the line DF. Wherefore the line GD is incommensurable in length to the line DF. And either of them is rational. Wherefore the lines GD and DF are rational commensurable in power only. Wherefore the line FG is a residual line (by the 73. proposition of the tenth) And the line DE is a rational line. but a superficies comprehended under a rational line, and an irrational line is irrational (by the 21 of the te●●●● and the line which containeth in power the same superficies is irrational (by the assumpt going before the same) Wherefore the parallelogram FE is irrational. But the line AC containeth in power the parallelogram FE. Wherefore the line AC is an irrational line and is called a second medial residual line. And this second medial residual line is that part of the greater part of a bimedial line which remaineth after the taking away of the less part from the greater: which was required to be proved.  

another demonstrtion more brief after Campane.  
This proposition setteth forth the nature of the tenth kind of irrational lines, which is called a second residual medial line, which is thus defined.  

Definition of the tenth irrational line.  A second residual● medial line is an irrational line which remaineth, when from a medial line is taken away a medial line commensurable to the whole in power only, and the part taken away & the whole line contain a medial superficies.    

¶ The 58. Theorem. The 76. Proposition. I●●rom a right line be taken away a right line incommensurable in power to the whole, and if that which is made of the squares of the whole line and of the line taken away added together be rational, and the parallelogram contained under the same lines medial: the line remaining is irrational, and is called a less line.  
In this Proposition is contained the definition of the eleventh kind of irrational lines, which is called a less line, whose definition is thus.  
A less line is an irrational line which remaineth, 
Definition of the eleventh irrational line.  when from a right line is taken away a right line incommensurable in power to the whole, and the square of the whole line, & the square of the part taken away added together, make a rational superficies, and the parallelogram contained of them is medial.  
This Proposition may after Campanes way be demonstrated, if you remember well the order & positions which he in the three former Propositions used.   

¶ The 19 Theorem. The 77. Proposition. If from a right line be taken away a right line incommensurable in power to the whole line, and if that which is made of the squares of the whole line and of the line taken away added together be medial, and the parallelogram contained under the same lines rational: the line remaining is irrational, and is called a line making with a rational superficies the whole superficies medial.  
In this Proposition is declared the nature of the twelfth kind of irrational lines, which is called a line making with a rational superficies the whole superficies medial, whose definition is thus.  

●●●●i●ition of the twelfth irra●ionall line.  A line making with a rational superficies the whole superficies medial, is an irrational line which remaineth, when from a right line is taken away a right line incommensurable in power to the whole line, and the square of the whole line & the square of the part taken away added together make a medial superficies, and the parallelogram contained of them is rational.  
This Proposition also may after Campanes way be demonstrated, observing the former caution.   

¶ The 60. Theorem. The 78. Proposition. If from a right line be taken away a right line incommensurable in power to the whole line, and if that which is made of the squares of the whole line and of the line taken away added together be medial, and the parallelogram contained under the same lines be also medial, and incommensurable to that which is made of the squares of the said lines added together: the line remaining is irrational, and is called a line making with a medial superficies the whole superficies medial.  
This proposition may thus more briefly be demonstrated: forasmuch as that which is composed of the squares of the lines AB and BC is medial, and that also which is contained under them is medial, therefore the parallelogramm●s DE and DH which are equal unto them are medial: but a medial superficies exceedeth not a medial superficies by a rational superficies. Wherefore the superficies FE which is the excess of the medial superficies DE above the medial superficies DH is irrational. And therefore the line AC which containeth it in power is irrational. etc.  
In this proposition is showed the condition and nature of the thirteenth and last kind of irrational lines, which is called a line making with a medial superficies the whole superficies medial, whose definition is thus.  
A line making with a medial superficies the whole superficies medial is an irrational line which remaineth, 
Definition of the thirteenth and last irrational line.  when from a right li●e is taken away a right line incommensurable in power to the whole line, and the squares of the whole line and of the line taken away added together make a medial superficies, and the parallelogram contained of them is also a medial superficies, moreover the squares of them are incommensurable to the parallelogram contained of them.  

An assumpt of Campane.  
If there be four quantities, & if the difference of the first to the second, be as the difference of the third to the fourth, then alternately, as the difference of the first is to the third, so is the difference of the second to the fourth.  
This is to be understand of quantities in like sort referred the one to the other, that is if the first be greater than the second, 
An Assumpt of Campane.  the third aught to be greater than the fourth and if the first be less than the second, the third aught to be less than the fourth: and is also to be understand in arithmeticiall proportionality. As for example let the difference of A be unto B as the difference of C is to D. 

I Dee.  
Though Campanes lemma be true, ye● the manner of demonstrating it, (narrowly considered) is not artificial.   Then I say that as the difference of A is to C, so is the difference of B to D. For (by this common sentence, the difference of the extremes is composed of the differences of the extremes to the means), the difference of A to C is composed of the difference of A to B and of the difference of B to C. And (by the same common sentence) the difference of B to D is composed of the difference of B to C, and of ●he difference of C to D. And forasmuch as (by supposition) the difference of A to B is as the difference of C to D, and the difference of B to C is common to them both. Wherefore it followeth, that as the difference of A is to C, so is the difference of B to D: which was required to be proved.    

¶ The 61. Theorem. The 79. Proposition. Unto a residual line can be joined one only right line rational, and commensurable in power only to the whole line.  

Second Senary.  LEt AB be a residual line, and unto it let the line BC be supposed to be joined, so that let the lines AC and BC be rational commensurable in power only. Then I say that unto the line AB cannot be joined any other rational line commensurable in power only to the whole line. 
Demonstration leading to an impossibility.  For if it be possible, let BD be such a line added unto it. Wherefore the lines AD and DB are rational commensurable in power only. And forasmuch as how much the squares of the lines AD and DB do exceed that which is contained under the lines AD and DB twice, so much also do the squares of the lines AC and CB exceed that which is contained under the lines AC and CB twice, for the excess of each is one and the same, namely, the square of the line AB (by the 7. of the second.) Wherefore alternately (by the ●ormer assumpt of Campanus) how much the squares of the lines AD and DB do exceed the squares of the lines AC & CB, so much also exceedeth that which is contained under the lines AD and DB twice, that which is contained under the lines AC and CB twice. But that which is made of the squares of the lines AD and DB added together, exceedeth that which is made of the squares of the lines AC and CB added together by a rational superficies: (for they are either of them rational). Wherefore that which is contained under the lines AD and DB twice, exceedeth that which is contained under the lines AC and CB twice by a rational superficies. But that which is contained under the lines AD and DB twice, is medial, for it is commensurable to that which is contained under the lines AD and DB once, which superficies is medial (by the 21. of the tenth) and by the same reason also that which is contained under the lines AC and CB twice is medial. Wherefore a medial superficies differeth from a medial superficies by a rational superficies, which (by the 26. of the tenth) is impossible. Wherefore unto the line AB cannot be joined any other rational line besides BC commensurable in power only, to the whole line. Wherefore unto a residual line can be joined one only right line rational and commensurable in power only to the whole line: which was required to be demonstrated.   

¶ The 62. Theorem. The 80. Proposition. Unto a first medial residual line can be joined one only medial right line, commensurable in power only to the whole line, and comprehending with the whole line a rational superficies.  
SVppose that AB be a first medial residual line, & unto AB join the line BC, so that let the lines AC and BC be medial commensurable in power only, & let that which is contained under the lines AC and BC be rational. Then I say that unto the line AB cannot be joined any other medial line commensurable in power only to the whole line, and comprehending together with the whole line a rational superficies. 
Demonstration leading to an absurdity.  For if it be possible let the line BD be such a line. Wherefore the lines AD and DB are medial commensurable in power only, and that which is contained under the lines AD and DB is rational. And forasmuch as how much the squares of the lines AD and DB exceed that which is contained under the lines AD and DB twice, so much also exceed the squares of the lines AC & BC, that which is contained under the lines AC and CB twice (for the excess of each is one and the same, namely, the square of the line AB). Wherefore alternately (as it was said in the former proposition) how much the squares of the lines AD and DB exceed the squares of the lines AC and CB, so much also that which is contained under the lines AD and DB twice, exceedeth that which is contained under the lines AC and CB twice. But that which is contained under the lines AD and DB twice, exceedeth that which is contained under the lines AC and CB twice by a rational superficies, for they are either of them a rational supersicies. Wherefore that which is made of the squares of the lines AD & DB exceedeth that which is made of the squares of the lines AC & CB by a rational superficies, which (by the 26. of the tenth) is impossible. For they are either of them medial (for those four lines were put to be medial.) Wherefore unto a first medial residual line can be joined only one right medial line commensurable in power only to the whole line, and comprehending with the whole line a rational superficies: which was required to be proved.   

¶ The 63. Theorem. The 81. Proposition. Unto a second medial residual line can be joined only one medial right line, commensurable in power only to the whole line, and comprehending with the whole line a medial superficies.  
SVppose that AB be a second medial residual line, & unto the line AB join the line BC, so that let the lines AC and CB be medial commensurable in power only, and let that which is comprehended under the lines AC and CB be medial. Then I say, that unto the line AB can not be joined any other medial right line commensurable in power only to the whole line, and comprehending together with the whole line a medial superficies. For if it be possible, let the line BD be such a line. Wherefore the lines AD & DB are medial commensurable in power only, and that which is contained under the lines AD and DB is also medial. Take a rational line EF. 
Construction.  And (by the 44. of the first) unto the line EF apply the parallelogram EG equal to the squares of the lines AC and CB, and making in breadth the line EM: and from that parallelogram EG take away the parallelogram HG equal to that which is contained under AC and CB twice, and making in breadth the line HM. Wherefore the parallelogram remaining, namely, EL, is (by the 7. of the second) equal to the square of the line AB. Wherefore the line AB containeth in power the parallelogram EL. Again, unto the line EF apply (by the 44. of the first) the parallelogram EI equal to the squares of the lines AD and DB, and making in breadth the line EN. But the squares of the lines AD and DB are equal to that which is contained under the lines AD and DB twice, and to the square of the line AB. 
Demonstration leading to an absurdity.  Wherefore the parallelogram EI is equal to that which is contained under the lines AD & DB twice, and to the square of the line AB. But the parallelogram EL is equal to the square of the line AB. Wherefore the parallelogram remaining, namely, high, is equal to that which is contained under the lines AD and DB twice. And forasmuch as the lines AC and CB are ●●●iall, therefore the squares also of the lines AC and CB are medial: and they are equal to the parallelogram EG: wherefore the parallelogram EG is (by that which was spoken in the 75. Proposition) medial: and it is applied unto the rational line EF, making in breadth the line EM. Wherefore (by the 22. of the tenth) the line EM is rational, and incommensurable in length to the line EF. Again, forasmuch as that which is contained under the lines AC and CB is medial, therefore (by the Corollary of the 23. of the tenth) that which is contained under the lines AC and CB twice is also medial: and it is equal to the parallelogram HG: wherefore also the parallelogram HG is medial, and is applied to the rational line EF, making in breadth the line HM. Wherefore (by the 22 of the tenth) the line HM is rational, and incommensurable in length to the line EF. And forasmuch as the lines AC and CB are commensurable in power only, therefore the line AC is incommensurable in length to the line CB. But as the line AC is to the line CB, so (by the Assumpt going before the 22. of the tenth) is the square of the line AC to that which is contained under the lines AC & CB. Wherefore (by the 10. of the tenth) the square of the line AC is incommensurable to that which is contained under the lines AC and CB. But unto the square of the line AC are commensurable the squares of AC & CB, and unto that which is contained under the lines AC and CB, is commensurable that which is contained under the lines AC and CB twice. Wherefore the squares of the lines AC & CB are incommensurable to that which is contained under the lines AC and CB twice. But unto the squares of the lines AC and CB is equal the parallelogram EG, and unto that which is contained under the lines AC & CB twice, is equal the parallelogram GH. Wherefore the parallelogram EG is incommensurable to the parallelogram HG. But as the parallelogram EG is to the parallelogrmme HG, so is the line EM to the line HM. Wherefore the line EM is incommensurable in length to the line HM. And they are both rational lines. Wherefore the lines EM and MH are rational commensurable in power only. Wherefore the line EH is a residual line, and unto it is joined a rational line HM commensurable in power only to the whole line EM. In like sort also may it be proved, that unto the line EH is joined the line HN, being also rational, and commensurable in power only to the whole line EN. Wherefore unto a residual line is joined ●●re than one only line commensurable in power only to the whole line: which (by the 79. of the tenth) is impossible. Wherefore unto a second medial residual line can be joined only one medial right line commensurable in power only to the whole line, and comprehending with the whole line a medial superficies: which was required to be demonstrated.   

¶ The 64. Theorem. The 82. Proposition. Unto a less line can be joined only one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squares added together rational, and that which is contained under them medial.  
SVppose that AB be a less line, and to AB join the line BC, so that let BC be such a line as is required in the Theorem. Wherefore the lines AC and CB are incommensurable in power, having that which is made of the squares of them added together rational, and that which is contained under them medial. Then I say that unto AB cannot be joined any other such right line. 
Demonstration leading to an absurdity.  For if it be possible, l●t the line BD be such a line. Wherefore the lines AD & DB are incommensurable in power, having that which is made of the squares of them added together, rational, and that which is contained under them medial. And for that how much the squares of the lines AD and DB exceed the squares of the lines AC and CB, so much that which is contained under the lines AD and DB twice, exceedeth that which is contained under the lines AC and CB twice (by those things which were spoken in the 79. proposition) But that which is made of the squares of the lines AD and DB added together exceedeth that which is made of the squares of the lines AC and CB added together by a rational superficies, for they are either of them rational by supposition. Wherefore that which is contained under the lines AD and DB twice, exceedeth that which is contained under the lines AC and CB twice by a rational superficies: which (by the 26. of the tenth) is impossible, for either of them is medial by supposition. Wherefore unto a less line can be joined only one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squares added together rational, and that which is contained under them medial: which was required to be demonstrated.   

¶ The 65. Theorem. The 83. Proposition. Unto a line making with a rational superficies the whole superficies medial, can be joined only one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squares added together medial, and that which is contained under them rational.  
SVppose that AB be a line making with a rational superficies the whole superficies medial, and unto it let the line BC be joined, so that let BC be such a line as is required in the Theorem. Wherefore the lines AC and CB are incommensurable in power, ha●ing that which is made of the squares of the lines AC and CB added together medial, and that which is contained under the lines AC and CB rational. 
Demonstration leading to an impossibility.  Then I say that unto the line AB cannot be joined any other such line. For if it be possible, let the line BD be such a line. Wherefore the lines AD and DB are incommensurable in power, having that which is made of the squares of the lines AD and DB added together medial, and that which is contained under the lines AD and DB rational. Now for that how much the squares of the lines AD and DB exceed the squares of the lines AC and CB, so much that which is contained under the lines AD and DB twice exceedeth that which is contained under the lines AC and CB twice, by that which was spoken in the 79. proposition. But that which is contained under the lines AD and DB twice, exceedeth that which is contained under the lines AC and CB twice by a rational superficies, for they are either of them rational by supposition. Wherefore that which is made of the squares of the lines AD and DB added together, exceedeth that which is made of the squares of the lines AC and CB added together by a rational superficies, which by the 26. of the tenth, is impossible, for they are either of them medial by supposition. Wherefore unto the line AB cannot be joined any other line besides BC, making that which is required in the proposition. Wherefore unto a line making with a rational superficies the whole superficies medial can be joined only one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squa●es added together medial, and that which is contained under them rational: which was required to be proved.   

¶ The 66. Theorem. The 84. Proposition. Unto a line making with a medial superficies the whole superficies medial, can be joined only one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squares added together medial, and that which is contained under them medial, and moreover making that which is made of the squares of them added together incommensurable to that which is contained under them.  
SVppose that AB be a line making with a medial superficies the whole superficies medial, and unto it let the line BC be joined, so that let BC be such a line as is required in the Theorem. Wherefore the lines AC and CB are incommensurable in power, having that which is made of the squares of the lines AC and CB added together medial, & that which is contained under the lines AC and CB medial, and moreover that which is made of the squares of the lines AC and CB is incommensurable so that which is contained under the lines AC and CB. Then I say, that unto the line AB can be joined no other such line. For if it be possible, let BD be such a line. Wherefore the lines AD and DB are incommensurable in power, having that which is made of the squares of the lines AD and DB added together medial, and that which is contained under the lines AD and DB medial, and moreover that which is made of the squares of the lines AD and and DB added together, is incommensurable to that which is contained under the lines AD and DB. Take a rational line EF. 
Construction.  And (by the 44. of the first) unto the line EF apply the parallelogram EG equal to the squares of the lines AC and CB, and making in breadth the line EM● and from the parallelogram EG take away the parallelogram HG equal to that which is contained under the lines AC & CB twice, and making in breadth the line HM. Wherefore the residue, namely, the square of the line AB is equal to the parallelogram EL (by the 7. of the second). Wherefore the line AB containeth in power the parallelogram EL. Again (by the 44. of the first) unto the line EF apply the parallelogram E● equal to the squares of the lines AD and DB, and making in breadth the line EN. But the square of the line AB is equal to the parallelogram EL. Wherefore the residue, namely, the parallelogram high is equal to that which is contained under the lines AD and DB twice. 
Demonstration 〈◊〉 an abjurd●t●●●  And forasmuch as that which is made of the squares of the lines AC and CB is medial, and is equal to the parallelogram EG, therefore also the parallelogram EG is medial. And it is applied unto the rational line EF, making in breadth the line EM. Wherefore (by the 22. of the tenth) the line EM is rational and incommensurable in length to the line EF. Again, forasmuch as that which is contained under the lines AC and CB twice is medial, and is equal to the parallelogram HG. Wherefore the parallelogram H●G is medial, which parallelogram HG is applied to the rational line EF, making in breadth the line HM. Wherefore the line HM is rational and incommensurable in length to the line EF. And forasmuch as the squares of the lines AC and CB are incommensurable to that which is contained under the lines AC and CB twice, therefore the parallelogram EG is incommensurable to the parallelogram HG. Wherefore the line EM is incommensurable in length to the line MH, and they are both rational. Wherefore the lines EM and MH are rational commensurable in power only, Wherefore EH is a residual line. And the line joined unto it is HM. And in like sort may we prove, that the line EH is a residual line, and that the line HN is joined unto it. Wherefore unto a residual line is joined two sundry lines, being each commensurable in power only so the whole line● which (by the 79. of the tenth) is impossible. Wherefore unto the line AB can not be joined any other right line besides the line BC, which shall be incommensurable in power to the whole line, & have together with the whole line that which is made of their squares added together medial, and that which is contained under them medial, and moreover incommensurable to that which is made of their squares added together. Wherefore unto a line making with a medial superficies the whole superficies medial, can be joined only one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squares added together medial, and that which is contained under them medial, and moreover making that which is made of the squares of them added together incommensurable to that which is contained under them which was required to be proved.  

¶ Third Definitions.  
AS of binomial lines, there are 6. divers kinds, so also of residual lines which are correspondent unto them and depend of them (for a residual line is nothing else (as was before said) but that which remaineth when the less part of a binomial line is taken from the greater part or name thereof,) there are likewise six several kinds. All which are known and considered in comparison to a rational line set forth & appointed, and these residual line have the self same order of production that the binomials had. 
Six kinds of re●iduall lines.  For as the three first kinds of binomial lines, namely, the first, second, and third, were produced when the square of the greater part of the part of the binomiall exceedeth the square of the less part thereof by the square of a line commensurable unto it in length: so in likewise, the first three kinds of residual lines, namely, the first, second, and third, are produced, when the square of the whole, namely, of that which is made of the residual line, and the line joined unto it added together, exceedeth the square of the line joined to the residual, by the square of a line which is commensurable unto it in length. And as the three last kinds of binomials, namely, the fourth, fifth, and sixth were produced when the square of the greater part exceedeth the square of the less, by the square of a line incommensurable in length unto it, even so the three last kinds of residual lines are produced when the square of the whole exceedeth the square of the line adjoined, by the square of a line incommensurable unto it in length. As ye may perceive by their definitions following.  


First definition.  A first residual line is, when the square of the whole exceedeth the square of the line adjoined, by the square of a line commensurable unto it in length, and also the whole is commensurable in length to the rational line first set.  
As let AB be a rational line, whose parts are certain, distinct, & to be expressed by number. And let the residual line be CD, and let the line joined unto it be EC● and let the whole being composed of the residual CD, and the line adjoined EC, be the line ED: let moreover the square of the whole line ED, exceed the square of the line adjoined EC by the square of the line F, which line F let be commensurable in length to the whole line ED, and let the whole line ED be also commensurable in length to the rational line AB: then is the residual line CD by this definition a first residual line.   


Second definition.  A second residual line is, when the square of the whole exceedeth the square of the line adjoined, by the square of a line commensurable unto it in length, and also the line adjoined is commensurable in length to the rational line.  
As suppose the line CD to be a residual, and let the line adjoined unto it be EC, and the whole made of them both, let be the line ED: & let the square of ED the whole line exceed the square of the line adjoined EC, by the square of the line F, and let the line F be commensurable in length to the whole line ED, moreover let the line adjoined EC be commensurable in length to the rational line AB: then by this definition, the residual line CD is a second residual line.   


Third definition.  A third residual line is, when the square of the whole exceedeth the square of the line adjoined, by the square of a line commensurable unto it in length and neither the whole line, nor the line adjoined is commensurable in length to the rational line.  
As (the former supposition standing) suppose that the square of the whole line ED exceed the square of the line adjoined EC by the square of the line F, and let the line F be commensurable in length to the whole line EC, and let neither the whole line ED, nor the line adjoined EC be commensurable in length to the rational line AB, then by this definition the residual line CD is a third residual line.   

A fourth residual line is, 
Fourth definition.  when the square of the whole line exceedeth the square of the line adjoined, by the square of a line incommensurable unto it in length, and the whole line is also commensurable in length to the rational line.  
As the residual line being as before CD, & the line adjoined EC, and the whole ED, let the square of the whole line ED exceed the square of the line adjoined EC by the square of the line F, and let the line F be incommensurable in length to the whole line ED, and let ED the whole line be commensurable in length to the rational line AB, then is the residual line CD by this declaration a fourth residual line.   

A fifth residual line is, 
Fifth definition.  when the square of the whole line exceedeth the square of the line adjoined, by the square of a line incommensurable unto it in length, and the line adjoined is commensurable in length to the rational line.  
As the residual line being CD, the line adjoined EC, and the whole line ED, let the square of the whole line ED exceed the square of the line adjoined EC by the square of the line F, and let the line F be incommensurable in length to the whole line ED, and let also EC the line adjoined be commensurable in length to the rational line AE, then shall the residual CD be by this definition a fifth residual line.   

A sixth residual line is when the square of the whole line, 
Sixth definition.  exceedeth the square of the line adjoined, by the square of a line incommensurable unto it in length, and neither the whole line nor the line adjoined is commensurable in length to the rational line.  
As suppose the residual line to be CD, and the line adjoined to be ●C, and the whole line composed of them let be ED, and let the square of the whole line ●D exceed the square of the line adjoined by the square of the line F, which line F let be incommensurable in length to the whole line ED: moreover let neither the whole line CD nor the line adjoined EC, be commensurable in length to the rational line A●, then shall the residual line CD be by this explication a sixth residual line, and the last.     

¶ The 19 Problem. The 85. Proposition. To find out a first residual line.  
TAke a rational line and let the same be A, 
Third Senary.  and unto it let the line BG be commensurable in length. Wherefore the line BG also is rational. And take two square numbers DE and EF which let be such, 
Construction.  that the excess of the greater, namely, of DE, above the less EF (which excess let be the number DF) be no square number (by the corollary of the first assumpt of the 28. of the tenth). Wherefore the number ED hath not to the number DF that proportion that a square number hath to a square number (by the 24. of the eight). And as the number ED is to the number DF, so let the square of the line BG, be to the square of the line GC (by the corollary of the sixth of the tenth). 
Demonstration  Wherefore the square of the line BG is commensurable to the square of the line GC. But the square of the line BG is rational, wherefore also the square of the line GC is rational. Wherefore the line GC is also rational. And forasmuch as the number ED hath not to the number DF, that proportion that a square number hath to a square number, therefore neither also hath the square o● the line BG to the square of the line GC that proportion that a square number hath to a square number. Wherefore (by the 9 of the tenth) the line BG is incommensurable in length to the line GC. And they are both rational. Wherefore the lines BG and GC are rational commensurable in power only. Wherefore the line BC is a residual line. I say moreover that it is a first residual line. For forasmuch as the square of the line BG is greater than the square of the line GC (that it is greater it is manifest, for by supposition the square of the line BG is to the square of the line GC, as the greater number, namely, ED is to the number DF) unto the square of the line BG let the squares of the lines GC and H be equal. And for that as the number DE is to the number DF, so is the square of the line BG to the square of the line GC, therefore by conversion of proportion (by the corollary of the 9 of the fifth) as the number DE is to the number EF, so is the square of the line BG to the square of the line H. But the number DE hath to the number EF that proportion that a square number hath to a square number, for either of them is a square number, wherefore also the square of the line BG hath to the square of the line H that proportion that a square numbe● hath to a square number. Wherefore the line GB is commensurable in length to the line H. Wherefore the line GB is in power more than the line GC by the square of a line commensurable in length to the line GB: and the whole line, namely, GB is commensurable in length to the rational line A. Wherefore the line BC is a first residual line. Wherefore there is found out a first residual line which was required to be done.   

¶ The 20. Problem. The 86. Proposition. To find out a second residual line.  

Construction.  TAke a rational line, and let the same be A, and unto it let the line GC be commensurable in length. And take two square numbers DE and EF, and let them be such that the excess of the greater, namely, DF, be no square number. And as the number DF is to the number DE, so let the square of the line GC be to the square of the line GB. Wherefore both the squares are commensurable. And forasmuch as the square of the line GC is rational, 
Demo●strati●●.  therefore the square of the line BG is also rational: Wherefore also the line BG is rational. And forasmuch as the squares of the lines BG & GC have not that proportion the one to the other that a square number hath to a square number, therefore the lines BG and GC are incommensurable in length, and they are both rational. Wherefore the lines BG and GC are rational commensurable in power only. Wherefore the line BC is a residual line. I say moreover, that it is a second residual line. For forasmuch as the square of the line BG is greater than the square of the line GC, unto the square of the line BG let the squares of the lines G● & H be equal. And for that as the number DE is to the number DF, so is the square of the line GB to the square of the line GC, therefore (by conversion of proportion) as the number DE is to the number EF, so is the square of the line BG to the square of the line H. But either of these numbers DE and EF is a square number. Wherefore the line GB is commensurable in length to the line H. Wherefore the line BG is in power more than the line GC, by the square of a line commensurable in length to the line BG: and the line GC that is joined to the residual line is commensurable in length to the rational line A. Wherefore the line BC is a second residual line. Wherefore there is found out a second residual line: which was required to be done.   

¶ The 21. Problem. The 87. Proposition. To find out a third residual line.  
TAke rational line, & let the same be A: and take three numbers E, B, C, 
Construction.  and CD, not having the one to the other that proportion that a square number hath to a square number: and let the number BC have to the number BD that proportion that a square number hath to a square number. And let the number BC be greater than the number CD. And as the number E is to the number BC, so let the square of the line A be to the square of the line FG: and as the number BC is to the number CD, so let the square of the line FG be to the square of the line HG. 
Demonstration.  Wherefore the square of the line A is commensurable to the square of the line FG. But the square of the line A is rational. Wherefore also the square of the line FG is rational: wherefore the line FG is also rational. And forasmuch as the number E hath not to the number BC that proportion that a square number● hath to a square number, therefore neither also hath the square of the line A to the square of the line FG that proportion that a square number hath to a square number. Wherefore the line A is incommensurable in length to the line FG. Again for that as the number BC is to the number CD, so is the square of the line FG to the square of the line HG, therefore the square of the line FG is commensurable to the square of the line HG. But the square of the line FG is rational. Wherefore also the square of the line HG is rational. Wherefore also the line HG is rational. And for that the number BC hath not to the number CD, that proportion that a square number hath to a square number, therefore neither also hath the square of the line ●G to the square of the line HG, that proportion that a square number hath to a square number. Wherefore the line FG is incommensurable in length to the line HG: and they are both rational. Wherefore the lines FG & HG are rational commensurable in power only. Wherefore the line FH is a residual line. I say moreover, that it is a third residual line. For for that as the number E is to the number BC, so is the square of the line A to the square of the line FG: and as the number ●C is to the number CD, so is the square of the line FG to the square of the line HG● therefore by equality of proportion, as the number E is to the number CD, so is the square of the line A to the square of the line HG● but the number E hath ●●● to the number CD that proportion that a square num●●r hath to a square number, therefore neither also hath the square of the line A to the square of the line HG that proportion that a square number hath to a square number, therefore the line A is incommensurable in length to the line HG. Wherefore neither of the lines FG and HG is commensurable in length to the rational line A. And forasmuch as the square of the line FG is greater than the square of the line HG (that the line FG is greater than the line HG it is manifest, for by supposition the number BC is greater than the number CD) unto the square of the line FG let the squares of the lines HG & K● be equal. And for that as the number BC is to the number CD, so is the square of the line FG to the square of the line H●●●erfore (by conversion of proportion) as the number BC is to the number BD, so is the square of the line FG to the square of the line K. But the number BC hath to the number BD that proportion that a square number hath to a square number. Wherefore the square of the line FG hath to the square of the line K that proportion that a square number hath to a square number. Wherefore the line FG is commensurable in length to the line K. Wherefore the line FG is in power more than the line HG, by the square of a line commensurable in length to the line FG, and neither of the lines FG and GH is commensurable in length to the rational line A: when yet notwithstanding either of the lines FG and GH is rational. Wherefore the line FH is a third residual line. Wherefore there is found out a third residual line: which was required to be done.   

¶ The 22. Problem. The 88 Proposition. To find out a fourth residual line.  
TAke a rational line and let the same be A: and unto it let the line BG be commensurable in length. Wherefore the line BG is rational. And take two numbers DF and FE, 
Construction.  and let them be such that the whole number, namely, DE have to neither of the numbers DF and FE that proportion that a square number hath to a square number. And as the number DE is to the number EF, so let the square of the line BG be to the square of the line GC: wherefore the square of the line BG is commensurable to the square of the line GC, wherefore also the square of the line GC is rational, and the line GC is also rational. And for that the number DE hath not the number EF that proportion that a square number hath to a square number, 
Demonstration  therefore the line BG is incommensurable in length to the line GC. And they are both rational: wherefore the line BC is a residual line. I say moreover that it is a fourth residual line. For forasmuch as the square of the line BG is greater than the square of the line GC, unto the square of the line BG let the squares of the lines CG and H be equal. And for that as the number DE is to the number EF, so is the square of the line BG to the square of the line GC, therefore by conversion of proportion as the number DE is to the number DF, so is the square of the line BG to the square of the line H. But the numbers DE and DF have not the one to the other that porportion that a square number hath to a square number. Wherefore the line BG is incommensurable in length to the line H. Wherefore the line BG is in power more than the line GC by the square of a line incommensurable in length so the line BG: and the whole line BG is commensurable in length to the rational line A, Wherefore the line BC is a fourth residual line. Wherefore there is found out a fourth residual line: which was required to be done.   

¶ The 23. Problem. The 89. Proposition. To find out a fift residual line.  
TAke a rational line and let the same be A, 
Construction.  and unto it let the line CG be commensurable in length. Wherefore the line CG is rational. And take two numbers DF and FE, which let be such, that the number DE have to neither of these numbers DF nor FE that proportion that a square number hath to a square number. And as the number FE is to the number DE, so let the square of the line CG be to the square of the line BG. 
Demonstration.  Wherefore the square of the line CG is commensurable to the square of the line BG● Wherefore the square of the line BG is rational, and the line BG is also rational. But the numbers DE and EF have not that proportion the one to the other that a square number hath to a square number. Wherefore the lines BG and GC are rational commensurable in power only. Wher●fore the line BC is a residual line. I say moreover that it is a fift residual line. For forasmuch as the square of the line BG is greater than the square of the line GC, unto the square of the line BG let the squares of the lines GC and H be equal. Now therefore for that as the number DE is to the number EF, so is the square of the line BG to the square of the line GC, therefore by conversion of proportion, at the number DE is to the number DF, so is the square of the line BG to the square of the line H. But the numbers DE & DF have not that proportion the one to the other that a square number hath to a square number. Wherefore the line BG is incommensurable in length to the line H. Wherefore the line BG is in power more than the line CG by the square of a line incommensurable in length to the line BG, and the line CG which is joined to the residual line is commensurable in length to the rational line A. Wherefore the line BC is a fift residual line. Wherefore there is found out a fift residual line: which was required to be done.   

¶ The 24. Problem. The 90. Proposition. To find out a sixth residual line.  
TAke a rational line and let the same be A, 
Construction.  And take three numbers E, BC, and CD, not having the one to the other that proportion that a square number hath to a square number. And let not the number BC have to the number BD that proportion that a square number hath to a square number. And let the number BC be greater than the number CD, & as the number E is to the number BC, so let the square of the line A be to the square of the line FG. And as the number BC is to the number CD, so let the square of the line FG be to the square of the line GH. Now therefore for that as the number E is to the number BC, 
Demonstration.  so is the square of the line A to the square of the line FG, therefore the square of the line A is commensurable to the square of the line F G. Wherefore the square of the line FG is rational, and the line FG is also rational. And for that the number E hath not to the number BC that proportion that a square number hath to a square number, therefore the line A is incommensurable in length to the line FG. Again for that as the number BC is to the number CD, so is the square of the line FG to the square of the line GH, therefore the square of the line FG is commensurable to the square of the line GH. But the square of the line FG is rational, wherefore the square also of the line GH is rational, wherefore the line GH is also rational. And for that the number B● hath not to the number CD that proportion that a square number hath to a square number, therefore the line FG is incommensurable in length to the line GH, and they are both rational. Wherefore the lines FG and GH are rational commensurable in power only. Wherefore the line FH is a residual line. I say moreover that it is a sixth residual line. For for that as the number E is to the number BC, so is the square of the line A to the square of the line FG, and as the number BC is to the number CD, so is the square of the line FG to the square of the line GH, therefore by equality of proportion as the number E is to the number CD, so is the square of the line A to the square of the line GH. But the number E hath not to the number CD that proportion that a square number hath to a square number. Wherefore the line A is incommensurable in length to the line GH, and neither of these lines FG nor G● is commensurable in length to the rational line A. And forasmuch as the square of the line FG is greater than the square of the line GH, unto the square of the line FG let the the squares of the lines GH and K be equal. Now therefore for that as the number B● is to the number CD, so is the square of the line FG to the square of the line GH, therefore by conversion of proportion as the number BC is to the number BD, so is the square of the line FG to the square of the line K. But the number BC hath not to the number BD that proportion that a square number hath to a square number, therefore the line FG is incommensurable in length to the line K. Wherefore the line FG is in power more than the line GH by the square of a line incommensurable in length to the line FG, and neither of the lines FG nor GH is commensurable in length to the rational line A. Wherefore the line FH is a sixth residual line. Wherefore there is found out a sixth residual line: which was required to be done.  

An other more ready way to find out the six residual lines.  There is also a certain other readier way to find out every one of the foresaid six residual lines which is after this manner. Suppose that it were required to find out a first residual line: Take a first binomial line AC, & let the greater name thereof be AB. And unto the line BC let the line BD be equal. Wherefore the lines AB and BC, that is the lines AB and BD are rational commensurable in power only, and the line AB is in power more than the line BC, that is, than the line BD by the square of a line commensurable in length to the line AB. And the line AB is commensurable in length to the rational line given. For the line AC is put to be a first binomial line. Wherefore the line AD is a first residual line. And in like manner may ye find out a second, a third, a fourth, a fift, and a sixth residual line, if ye take for each a binomial line of the same order.   

¶ The 67. Theorem. The 91. Proposition. If a superficies be contained under a rational line & a first residual line: the line which containeth in power that superficies, is a residual line.  
SVppose that there be a rectangle superficies AB contained under a rational line AC and a first residual line AD. 
Fourth Senary.  Then I say that the line which containeth in power the superficies AB is a residual line. For forasmuch as AD is a first residual line, 
The ●irst par● of the Construction.  let the line joined unto it be DG (by the line joined unto it understand such a line as was spoken of in the end of the 79. proposition). Wherefore the lines AG and GD are rational commensurable in power only, & the whole line AG is commensurable in length to the rational line AC, and the line AG is in power more than the line GD by the square of a line commensurable in length unto AG, by the definition of a first residual line. Divide the line GD into two equal parts in the point E. And upon the line AG apply a parallelogram equal to the square of the line EG and wanting in figure by a square, and let the said parallelogram be that which is contained under the lines AF and FG. 
The first part of the demonstration.  Wherefore the line AF is commensurable in length to the line FG (by the 17. of the tenth) ● And by the points E, F and G, draw unto the line AC these parallel lines EH, FI, and GK. And make perfect the parallelogram AK. And for●as much as the line AF is commensurable in length to the line FG, therefore also the whole line AG is commensurable i● length to either of the lines AF and FG (by the 15. of the tenth). But the line AG is commensurable in length to the line AC. Wherefore either of the lines AF and FG is commensurable in length to the line AC. But the line AC is rational, wherefore either of the lines AF and FG is also rational. Wherefore (by the 19 of the tenth) either of the parallelograms AI and FK is also rational. 
Note. AI and FK concluded rational parallelogram.  And forasmuch as the line DE is commensurable in length to the line EG, therefore also (by the 15. of the tenth) the line DG is commensurable in length to either of the lines DE and EG. But the line DG is rational, wherefore either of the lines DE and EG is rational, and the self same line DG is incommensurable in length to the line AC (by the definition of a first residual line, or by the 13. of the tenth●). For the line DG is incommensurable in length to the line AG, which line AG is commensurable in length to the line AC: 

Note.  
DH and FK parallelograms medial.   wherefore either of the lines DE and EG is rational and incommensurable in length to the line AC ● Wherefore (by the 21. of the tenth) either of these parallelograms DH and EK is medial. Unto the parallelogram AI let the square LM be equal, and unto the parallelogram FK let the square NX be equal, being taken away from the square LM ● and ha●ing the angle LOM common to them both. 
Second part of the construction.  (And to do this, there must be found out the mean proportional between the lines FI and FG. For the square of the mean proportional is equal to the parallelogram contained under the lines FI and FG. And from the line LO cut of a line equal to the mean proportional so found out, and descri●e the square thereof). Wherefore both the squares LM and NX are about one and the self same diameter (by the 20. of the sixth) let their diameter be OR and describe the figure as it is h●●e s●t forth● Now then forasmuch as the parallelogram contained under the lines AF & FG is equal to the square of the line EG, 
Second part of the demonstration.  therefore (by the 17. of the sixth) as the line AF is to the line EG, so is the line EG to the line FG. But as the line AF is to the line EG, so is th● parallelogram AI to the parallelogram EK. And as the line EG is to the line FG, so is the parallelogram EK to the parallelogram FK. Wherefore between the pagrammes AI and FK the parallelogram EK is the mean proportional. But (by the second part of the assumpt going before the 54. of the tenth) between the squares LM and NX the parallelogram MN is the mean proportional. And unto the parallelogram AI is equal the square LM, and unto the parallelogram FK is equal the square NX by construction. Wherefore the parallelogram MN is equal to the parallelogram EK (by the 2. assumpt going before the 54. of the tenth). But the parallelogram EK is (by the first of the sixth) equal to the parallelogram DH, and the parallelogram MN is (by the 43. of the first) equal to the parallelogram LX. Wherefore the whole parallelogram DK is equal to the gnomon VTZ (which gnomon consisteth of those parallelograms by which ye see in the figure passeth a portion of a circle greater than a semicircle) and moreover to the square NX: and the parallelogram AK is equal to the squares LM and NX by construction: and it is now proved, that the parallelogram DK is equal to the gnomon VTZ, and moreover to the square NX. Wherefore the residue namely the parallelogram AB is equal to the square SQ which is the square of the line LN. 
LN, is the only li●e ●hat we sought & consider.  Wherefore the square of the line LN is equal to the parallelogram AB. Wherefore the line LN containeth in power the parallelogram AB. I say moreover that the line LN is a residual line. For forasmuch as either of these parallelograms AI and FK is rationally as it is before said, therefore the squares LM and NX which are equal unto them, that is, the squares of the lines LO and ON are rational. Wherefore the lines LO and ON are also rational. Again forasmuch as the parallelogram DH that is LX is medial, therefore the parallelogram LX is incommensurable to the square NX. Wherefore (by the 1. of the sixth, and 10. of the tenth) the line LO is incommensurable in length to the line ON ● and they are both rational. Wherefore they are lines rational commensurable in power only. Wherefore LN is a residual line by the definition, and it containeth in power the paralleloparallelogramme AB. If therefore a superficies be contained under a rational line and a first residual line, the line which containeth in power that superficies● is a residual line: which was required to be demonstrated.   

¶ The 68 Theorem. The 92. Proposition. If a superficies be contained under a rational line and a second residual line: the line which containeth in power that superficies, is a first medial residual line.  
SVppose that AB be a superficies contained under ●ra●●onall line AC, and a second residual li●● AD. 
First part of the construction.  Then I say● that the line th●● containeth in power the superficies AB is a first medial residual line. For let the line joined to the line AD be DG. Wherefore the lines AG and GD are rational commensurable in power only, and the line that is joined to the residual line, namely, the line DG is commensurable in length to the rational line AC: and the line AG is in power more than the line DG, by the square of a line commensurable in length to the line AG. Divide the line DG into two equal parts in the point E. And unto the line AG apply a parallelogram equal to the fourth part of the square of the line DG, that is, equal to the square of the line EG, and wanting in figure by a square, and let that parallelogram be that which is contained under the lines AF and FG. Wherefore (by the 1ST of the tenth) the line AF is commensurable in length to the line FG. And by the points E, F, and G, draw unto the line AC these parallel lines EH, FI, and GK: and forasmuch as the line AF is commensurable in length to the line FG, 
The first part of th● demonstration.  therefore the whole line AG is commensurable in length to either of these lines AF and FG. But the line AG is rational and incommensurable in length to the line AC. Wherefore either of these lines AF and FG are rational and incommensurable in length to the line AC. Wherefore either of th●se parallelograms AI and FK is (by the 21. of the tenth) medial. 
AI and FK concluded parallelograms medial.  Again, forasmuch as the line DE is commensurable in length to the line EG, therefore the line DG is commensurable in length to either of these lines DE and EG. But the line DG is commensurable in length to the rational line AC. Wherefore either of these lines DE and EG is rational and commensurable in length to the line AC. Wherefore (by the 19 of the tenth) either of these parallelograms DH and EK is rational. 
DH & EK, rational.  Unto the parallelogram AI describe an equal square LM, 
The second part of the construction.  and unto the parallelogram FK let the square NX be equal, as in the Proposition going before. Wherefore the squares LM and NX are both about one and the same diameter. 
The second part of the demonstration.  Let the diameter be OR, and describe the figure as is in the former Proposition expressed. Now therefore forasmuch as the parallelograms AI and FK are medial are 
* Analytically the pro●e hereof followeth among● other things.  commensurable the one to the other, and the squares of the lines LO & ON which are equal to those parallelograms, are medial, therefore the lines LO and ON are also medial commensurable in power. (And it is manifest, that the lines LO and ON are commensurable in power, for their squares are commensurable, and those squares, namely, the squares of the lines LO & ON are commensurable, for they are equal to the parallelograms AI and FK, which are commensurable the one to the other: and that those parallelograms AI and FK are commensurable the one to the other, hereby it is manifest, for that it was before proved that the lines AF and FG are commensurable in length. Wherefore (by the 1. of the sixth, and 10. of the tenth) the parallelograms AI and FK are commensurable the one to the other. Wherefore it is now manifest by the way of resolution, that the lines LO & ON are commensurable in power) ● And forasmuch as the parallelogram contained under the lines AF and FG is equal to the square of the line EG, therefore as the line AF is to the line EG, so is the line EG to the line FG. But as the line AF is to the line EG, so is the parallelogram AI to the parallelogram EK: and as the line EG is to the line FG, so is the parallelogram EK to the parallelogram FK. Wherefore the parallelogram ●K is the mean● proportional between the parallelograms AI and FK: and the parallelogram MN i● also the mean proportional between the squares LM & NX: and the parallelogram AI is equal to the square LM: and the parallelogram FK is equal to the square NX. Wherefore the parallelogram MN is equal to the parallelogram EK. But the parallelogram EK is equal to the parallelogram DH: and the parallelogram LX is equal to the parallelogram MN. Wherefore the whole parallelogram DK is equal to the Gnomon VTZ, and to the square NX. Wherefore the residue, namely, the parallelogram AB is equal to the square SQ, that is, to the square of the line LN. Wherefore the line LN containeth in power the superficies AB. 
The line LN found, which is the principal drift of all the former discourse.  I say moreover, that the line LN is a first medial residual line. For forasmuch as the parallelogram EK is rational, and is equal to the parallelogram MN, that is, LX, therefore LX, that is, the parallelogram contained under the lines LO and ON is rational. But the square NX is medial, for it is already proved that the parallelogram FK which is equal to the square NX, is medial. Wherefore the parallelogram LX is incommensurable to the square NX. But as the parallelogram LX is to the square NX, so is the line LO to the line ON (by the 1. of the sixth). Wherefore (by the 10. of the tenth) the lines LO and ON are incommensurable in length. And it is already proved, that they are medial commensurable in power. Wherefore the lines LO & ON are medial commensurable in power only containing a rational superficies. Wherefore the line LN is a first medial residual line, and containeth in power the superficies AB, which is contained under a rational line and a second residual line. If therefore a superficies be contained under a rational line and a second residual line: the line which containeth in power that superficies, is a first medial residual line: which was required to be demonstrated.   

¶ The 69. Theorem. The 93. Proposition. If a superficies be contained under a rational line and a third residual line: the line that containeth in power that superficies is a second medial residual line.  
SVppose that AB be a superficies contained under a rational line AC, & a third residual line AD. Then I say, that the line which containeth in power the superficies AB is a second medial residual line. Let the line joined unto AD, be DG. Wherefore the lines AG and GD are rational commensurable in power only, 
The first part of the Construction.  and neither of the lines AG nor GD is commensurable in length to the rational line AC, and the whole line AC is in power more than the line GD, by the square of a line commensurable in length to the line AG. Let the rest of the construction be as it was in the former Propositions. 
The fi●st part of the demonstration.  Wherefore the lines AF and FG are commensurable in length: and the parallelogram AI is commensurable to the parallelogram FK. And forasmuch as the lines AF and FG are commensurable in length, therefore the whole line AG is commensurable in length to either of these lines AF and FG. B●● the lin● AG is rational and incommensurable in length to the line AG. Wherefore either of these lines AF and FG is rational and incommensurable in length to the line AG. 

Note.  
AI and FK medial.   Wherefore (by the 21. of the tenth) either of these parallelograms AI and FK is medial. Against, forasmuch as the line DB is commensurable in length to the line EG, therefore also the whole line DG is commensurable in length to either of these lines DE and EG. But the line DG is rational commensurable in power only to the line AC. Wherefore also either of the lines DE and EG is rational and commensurable in power only to the line AC. 

Note.  
DH and EK medial.   Wherefore either of these parallelograms DH and EK is medial. Again forasmuch as the lines AG and DG are commensurable in power only, therefore they are incommensurable in length. But the line AG is commensurable in length to the line AF: and the line DG is commensurable in length to the line GOE Wherefore the line AF is incommensurable in length to the line EG. But as the line AF is to the line EG, so is the parallelogram AI to the parallelogram EK. 

Note.  
AI incommensurable to EK. Second part of the Construction.   Wherefore the parallelogram AI is incommensurable to the parallelogram EK. Unto the parallelogram AI describe an equal square LM: and unto the parallelogram FK describe an equal square NX: and describe the figure as you did in the former Proposition. Now forasmuch as the parallelogram contained under the lines AF and FG is equal to the square of the line EG, therefore as the line AF is to the line EG, so is the line EG to the line FG. But as the line AF is to the line EG, so is the parallelogram AI to the parallelogram EK: and as the line EG is to the line FG, so is the parallelogram EK to the parallelogram FK. Wherefore as the parallelogram AI is to the parallelogram EK, so is the parallelogram EK to the parallelogram FK. Wherefore the parallelogram EK is the mean proportional between the parallelograms AI and FK. But the parallelogram MN is the mean proportional between the squares LM and NX. Wherefore the parallelogram EK is equal to the parallelogram MN. Wherefore the whole parallelogram DK is equal to the Gnomon VTZ, & to the square NX. And the parallelogram AK is equal to the squares LM and NX. Wherefore the residue, namely, the parallelogram AB is equal to the square QS, that is, 
The principal line, LN found.  to the square of the line LN. Wherefore the line LN containeth in power the superficies AB. I say moreover, that the line LN is a second medial residual line. For for that as it is proved, the parallelogramme● AI and FK are medial, therefore the squares that are equal unto them, namely, the squares of the lines LO and ON are also medial. Wherefore either of these lines LO and ON is medial. And forasmuch as the parallelogram AI is 
* Because the lines AF and ●G are proved commensurable in length.  commensurable to the parallelogram FK, therefore the squares that are equal to them, namely, the squares of the lines LO and ON are also commensurable. Again, forasmuch as it is proved, that the parallelogram AI is incommensurable to the parallelogram EK, therefore the square LM is incommensurable to the parallelogram MN, that is, the square of the line LO to the parallelogram contained under the lines LO & ON. Wherefore also the 
* By the first o● the sixth and tenth of the tenth.  line LO is incommensurable in length to the line ON. Wherefore the lines LO and ON are medial commensurable in power only. I say moreover, that they contain a medial superficies. For forasmuch as it is proved, that the parallelogram EK is medial, therefore the parallelogram which is equal unto it, namely, the parallelogram contained under the lines LO and ON is also medial. Wherefore the line LN is a second medial residual line, and containeth in power the superficies AB. Wherefore the line that containeth in power the superficies AB is a second medial residual line. If therefore a superficies be contained under a rational line and a third residual line, the line that containeth i● power that superficies, is a second medial residual line: which was required to be demonstrated.   

The 70. Theorem. The 94. Proposition. If a superficies be contained under a rational line, and a fourth residual line: the line which containeth in power that superficies, is a less line.  
SVppose that there be a superficies AB contained under a rational line AC, and a forth residual line AD. Then I say that the line which containeth in power the superficies AB is a less line. For let the line joined unto it be DG. Wherefore the lines AG and DG are rational commensurable in power only, and the line AG is in power more than the line DG, by the square of a line incommensurable in length to the line AG, and the line AG is commensurable in length to the line AC. Divide the line DG into two equal parts in the point E. 
The first part of the construction.  And unto the line AG apply a parallelogram equal to the square of the line EG, and wanting in figure by a square, and let that parallelogram be that which is contained under the lines AF and FG. Wherefore (by the 18. of the tenth) the line AF is incommensurable in length to the line FG. Draw by the points E, F, & G, unto the lines AC and DB these parallel lines EH, FI, and GK. 
The first part of the demonstration.  Now forasmuch as the line AG is rational, and commensurable in length to the line AC, therefore the whole parallelogram AK is (by the 19 of the tenth) rational. 

Note.  
AK rational.   Again forasmuch as the line DG is incommensurable in length to the line AC (for if the line DG were commensurable in length to the line AC, then forasmuch as the line AG is commensurable in length to the same line AC, the lines AG and DG should be commensurable in length the one to the other, when yet they are put to be commensurable in power only) and both these lines AC and DG are rational. 

Note.  
DK medial. AI and FK incommensurable.  
The second part of the construction.  
The second part of the demonstration.   Wherefore the parallelogram DK is medial. Again forasmuch as the line AF is incommensurable in length to the line FG, therefore the parallelogram AI is incommensurable to the parallelogram FK. Unto the parallelogram AI describe an equal square LM, and unto the parallelogram FK describe an equal square NX, and let the angle LOM be common to both those squares. Wherefore the squares LM and NX are about one and the self same diameter. Let their diameter be OR, and describe the figure. And forasmuch as the parallelogram contained under the lines AF and FG is equal to the square of the line EG, therefore proportionally as the line AF is to the line EG, so is the line EG to the line FG, but as the line AF is to the line EG, so is the parallelogram AI to the parallelogram EK (by the 1. of the sixth) And as the line EG is to the line FG, so is the parallelogram EK to the parallelogram FK. Wherefore the parallelogram EK is the mean proportional between the parallelograms AI and FK: wherefore as i● was said in the former propositions, the parallelogram MN is equal to the parallelogram EK: but the parallelogram DH is equal to the parallelogram EK, and the parallelogram MN to the parallelogram LX. Wherefore the whole parallelogram DK is equal to the gnomon VTZ, and to the square NX. Wherefore the residue, namely, the parallelogram AB is equal to the residue, namely, to the square SQ, 
LN (the chief line of this theorem) found.  that is to the square of the line LN. I say moreover that LN is that irrational line which is called a less line. For forasmuch as the parallelogram AK is rational, and is equal to the squares of the lines LO and ON, therefore that which is made of the squares of the lines LO and ON added together is rational. Again forasmuch as the parallelogram DK is medial, and is equal to that which is contained under the lines LO and ON twice, therefore that which is contained under the lines LO and ON twice, is also medial. And forasmuch as the parallelogram AI is incommensurable to the parallelogram FK, therefore the squares which are equal unto them, namely, the squares of the lines LO and ON are incommensurable the one to the other. Wherefore the lines LO and ON are incommensurable in power, having that which is made of their squares added together rational, and that which is contained under them twice medial, which is commensurable to that which is contained under them once. Wherefore that which is contained under them once is also medial. Wherefore LN is that irrational line which is called a less line, and it containeth in power the superficies AB. If therefore a superficies be contained under a rational line and a fourth residual line, the line which containeth in power that superficies is a less line: which was required to be demonstrated.   

¶ The 71. Theorem. The 95. Proposition. If a superficies be contained under a rational line and a fift residual line: the line that containeth in power the same superficies, is a line making with a rational superficies, the whole superficies medial.  
SVppose that there be a superficies AB contained under a rational line AG and a fift residual line AD. Then I say that the line that containeth in power the superficies AB, is a line making with a rational superficies the whole superficies medial. For unto the line AD let the line DG be joined, which shall be commensurable in length to the rational line AC. And let the rest of the construction be as in the proposition next going before. 
Demonstration.  And forasmuch as the line AG is incommensurable in length to the line AC and they are both rational, therefore the parallelogram AK is medial. Again forasmuch as the line DG is rational and commensurable in length to the line AC, therefore the parallelogram DK is rational. Unto the parallel●gramme AI describe an equal square LM; and unto the parallelogram ●● describe an equal square N●, and as in 〈◊〉 proposition next going before, so also in this may we prove, that the line LN containeth in power the superficies AB. 
The line LN.  I say moreover that that line LN is a line making with a rational superficies the whole superficies medial. For forasmuch as the parallelogram AK is medial, therefore that which is equal unto it, namely, that which is made of the squares of the lines LO and ON added together is also medial. Again forasmuch as the parallelogram DK is rational, therefore that which is equal unto it, namely, that which is contained under the lines LO and ON twice, is also rational. And forasmuch as the line AF is incommensurable in length to the line FG, therefore (by the 1. of the sixth, & 10. of the tenth) the parallelogram A● is incommensurable to the parallelogram FK, wherefore also the square of the line LO is incommensurable to the square of the line ON. Wherefore the lines LO and ON are incommensurable in power having that which is made of their squares added together medial, and that which is contained under them twice rational. Wherefore the line LN is that irrational line which is called a line making with a rational superficies the whole superficies medial, and it containeth in power the superficies AB. Wherefore the line containing in power the superficies AB, is a line making with a rational superficies the whole superficies medial. If therefore a superficies be contained under a rational line & a fift residual line, the line that containeth in power the same superficies, is a line making with a rational superficies the whole superficies medial: which was required to be proved.   

¶ The 72. Theorem. The 96. Proposition. If a superficies be contained under a rational line and a sixth residual line, the line which containeth in power the same superficies is a line making with a medial superficies the whole superficies medial.  
SVppose that AB be a superficies contained under a rational line AC, & a sixth residual line AD. Then I say shall the line which containeth in power the superficies AB, is a line making with a medial superficies the whole superficies medial. For unto the line AD let the line DG be joined. 
Demonstration.  And let the rest be as in the propositions going before. And forasmuch as the line AF is incommensurable in length to the line FG, therefore the parallelogram AI is incommensurable to the parallelogram FK● And forasmuch as the lines AG and AC are rational commensurable in power only, therefore the parallelogram AK is medial, and in like manner the parallelogram DK is medial. Now forasmuch as the lines AG and GD are commensurable in power only, therefore they are incommensurable in length the one to the other. But as the line AG is to the line GD, so is the parallelogram AK to the parallelogram DK, therefore the parallelogram AK is incommensurable to the parallelogram DK. Describe the like figure that was described in the former propositions, and we may in like sort prove that the line LN containeth in power the superficies AB. I say moreover that it is a line making w●th a medial superficies the whole superficies medial. For the parallelogram AK is medial, wherefore that which is equal vn●● it, namely, that which is made of the sq●ares of the lines LO and ON added together is also medial. And forasmuch as the parallelogram ●K is medial, therefore that which is equal unto ●t, namely, that which is contained under the lines LO and ON twice is also medial. And forasmuch as the parallogramme AK is incommensurable to the parallelogram DK, therefore the squares of the lines LO and ON are incommensurable to that which is contained under the lines LO and ON twice. And forasmuch as the parallelogram AI is incommensurable to the parallelogram FK, therefore also the square of the line LO is incommensurable to the square of the line ON. Wherefore the lines LO and ON are incommensurable in power, having that which is made of the squares of the lines LO and ON medial, and that which is contained under them twice medial, and moreover that which is made of the squares of them is incommensurable to thate which is contained under them twice. Wherefore the line LN is that irrational line which is called a line making with a medial superficies the whole superficies medial, and it containeth in power the superficies AB. Wherefore the line which containeth in power the superficies AB is a line making with a medial superficies the whole superficies medial. If therefore a superficies be contained under a rational line and a sixth residual line, the line which containeth in power the same superficies is a line making which a medial superficies the whole superficies mediall● which was required to be demonstrated.   

¶ The 73. Theorem. The 97. Proposition. The square of a residual line applied unto a rational line, 
The fifth Senary.  maketh the breadth or other side a first re●iduall line.  
SVppose that AB be a residual line, 
These six propositions following are the conu●rses of the six former propositions.  and let CD be a rational line. And unto the line CD apply the parallelogram CE equal to the square of the line AB, and making the breadth the line CF. Then I say that the line CF is a first residual line● 
Construction.  For unto the line AB let the line conveniently joined be supposed to be B● (which self line is also called a line joined, as we declared in the end of the 79. proposition). Wherefore the ●ines AG and GB are rational commensurable in power only. And unto the line CD apply the parallelogram CH equal to the square of the line AG, and unto the line KH (which is equal to the line CD) apply the parallelogram KL equal to the square of the line BG. 
Demonstration.  Wherefore the whole parallelogram CL is equal to the squares of the lin●s A● and GB● And the parallelogr●mm● CE is equal to the square of the line AB, wherefore the parallelogram remaining, namely, the parallelogram FL is equal to that which is contained under the lines AG and GB twice. For (by the 7. of the second) the squares of the lines AG and GB are equal to that which is contained under the lines AG & GB twice, and to the square of the line AB. Divide the line FM into two equal parts in the point N. And by the point N draw unto the line CD a parallel line NX. Wherefore either of the parallelograms FX and NL is equal to that which is contained under the lines AG and GB once. And forasmuch as the squares of the lines AG and GB are rational, unto which squares the parallelogram CL is equal, therefore the parallelogram CL also is rational: wherefore the line GM is rational and 
* By the 20. of the tenth.  commensurable in length to the line CD. Again forasmuch as that which is contained under the lines AG and GB twice, is 
** By the 21. of the tenth.  medial, therefore the parallelogram equal unto it, namely the paral●elogramme FL is also medial. Wherefore the line FM is rational and 
* By the 22. of the tenth.  incommensurable in length to the line CD. And forasmuch as the squares of the lines AG and GB are rational, and that which is contained under the lines AG and GB twice, is medial, therefore the squares of the lines AG and GB are incommensurable to that which is contained under the lines AG and GB twice. But unto the squares of the lines AG and GB is equal the parallelogram CL, and to that which is contained under the lines AG and GB twice is equal the parallelogram FL, wherefore the parallelogram CL is incommensurable to the parallelogram FL. Wherefore also the line CM is incommensurable in length to the line FM: and they are both rational. Wherefore the lines CM and FM are rational commensurable in power only: and therefore the line CF is a residual line (by the 73. of the tenth). I say moreover that it is a first residual line. 
●F concluded a residual line.  For forasmuch as that which is contained under the lines AG and GB is the mean proportional between the squares of the lines AG and GB (by the assumpt going before the 54. of the tenth). And unto the square of the line AG is equal the parallelogram CH, and unto that which is contained under the lines AG and GB is equal the parallelogram NL, and unto the square of the line GB is equal the parallelogram KL. Wherefore the parallelogram NL is the mean proportional between the parallelograms CH and KL. Wherefore as CH is to NL, so i● NL to KL. But as CH is to NL, so is the line CK to the line NM & as NL is to KL, so is the line NM to the line KM. Wherefore as the line CK is to the line NM, so is the line NM to the line KM. Wherefore the parallelogram contained under the lines CK and KM is equal to the square of the line NM, that is to the fourth part of the square of the line FM. And forasmuch as the square of the line AG is commensurable to the square of the line GB, therefore the parallelogram CH is commensurable to the parallelogram KL. But as CH is to KL, so is the line CK to the line KM: wherefore the line CK is commensurable in length to the line KM. Wherefore (by the 17. of the tenth) the line CM is in power more than the line FM by the square of a line commensurable in length to the line CM. But the line CM is commensurable in length to the rational line CD. Wherefore the line CF is a fairest residual line. Wherefore the square of a residual line applied unto a rational line, maketh the breadth or other side a first residual line: which was required to be demonstrated.   

¶ The ●4. Theorem. The 98. Proposition. The square of a first medial residual line applied to a rational line, maketh the breadth or other side a second residual line.  
SVppose that AB be a first medial residual line, and let CD be a rational line. And unto the line CD apply the parallelogram CE equal to the square of the line AB, 
Construction.  and making in breadth the line CF. Then I say that the line CF is a second residual line. For unto the line AB, let the line conveniently joined be supposed to be BG. Wherefore the lines AG and BG are medial commensurable in power only, comprehending a rational superficies. And unto the line CD apply the parallelogram CH equal to the square of the line AG, and making in breadth the line CK, and un-the line KH (which is equal to the line CD) apply the parallelogram KL equal to the square of the line GB and making in breadth the line KM. 
Demonstration.  Wherefore the whole parallelogram CL is equal to both the squares of the lines AG and GB which are medial & commensurable the one to the other. Wherefore the parallelograms CH and KL are medial and commensurable the one to the other. Wherefore (by the 15. of the tenth) the whole parallelogram CL is commensurable to either of these parallelograms CH and KL. Wherefore (by the corollary of the 23. of the tenth) the whole parallelogram CL is also medial. Wherefore (by the 22. of the tenth) the line CM is rational and incommensurable in length to the line CD. And forasmuch as the parallelogram CL is equal to the squares of the lines AG and GB● and the squares of the lines AG and GB are equal to that which is contained under the lines AG and GB twice, together with the square of the line AB (by the 7. of the second): and unto the square of the line AB is equal the parallelogram CE. Wherefore the residue, namely, that which is contained under the lines AG and GB twice, is equal to the residue, namely, to the parallelogram FL. But that which is contained under the lines AG & GB twice is rational. Wherefore the parallelogram FL is also rational. Wherefore the line FM is rational and commensurable in length to the line CD (by the 20. of the tenth.) Now forasmuch as the parallelogram CL is medial, and the parallelogram FL is rational, therefore they are incommensurable the one to the other. Wherefore also the line CM is incommensurable in length to the line FM, and they are both rational. Wherefore the line CF is a residual line. 
CF concluded a residual line.  I say moreover that it is a second residual line. For divide the line FM into two equal parts in the point N, from which point draw unto the line CD a parallel line NX. Wherefore either of these parallelograms FX and NL is equal to the parallelogram contained under the lines AG and GB. And forasmuch as the parallelogram contained under the lines AG and GB is the mean proportional between the squares of the lines AG and BG. Therefore the parallelogram NL is the mean proportional between the parallelograms CH and KL. But as CH is to NL, so is the line CK to the line NM, and as NL is to KL, so is the line NM to the line KM. Wherefore as the line CK is to the line NM, so is the line NM to the line KM. Wherefore the parallelogram contained under the lines CK and KM is equal to the square of the line NM, that is, to the fourth part of the square of the line FM. But the parallelogram CH is commensurable to the parallelogram KL. Wherefore also the line CK is commensurable in length to the line KM. Wherefore (by the 17. of the tenth) the line CM is in power more than the line FM, by the square of a line commensurable in length to the line CM. And the line FM which is the line conveniently joined, is commensurable in length to the rational line CD. Wherefore the lin● CF is a second residual line. Wherefore the square of a first medial residual line applied to a rational line, maketh the breadth or other side a second residual line: which was required to be proved.   

¶ The 75. Theorem. The 99 Proposition. The square of a second medial residual line applied unto a rational line, maketh the breadth or other side a third residual line.  
SVppose that AB be a second medial residual line, and let CD be a rational line. And unto the line CD apply the parallelogram CE equal to the square of the line AB, and making in breadth the line CF. 
Construction.  Then I say, that the line CF is a third residual line. For unto the line AB let the line conveniently joined be supposed to be BG. Wherefore the lines AG & GB are medial commensurable in power only containing a medial superficies. And let the rest of the construction be as in the Proposition next going before. 
Demonstration.  Wherefore the line CM is rational and incommensurable in length to the rational line CD. And either of the parallelograms FX and NL is equal to that which is contained under the lines AG and GB. But that which is contained under the lines AG & GB is medial. Wherefore that which is contained under the lines AG and GB twice is also medial. Wherefore the whole parallelogram FL is also medial. Wherefore the line FM is rational and incommensurable in length to the line CD. And forasmuch as the lines AG and GB are incommensurable in length, therefore also the square of the line AG is incommensurable to the parallelogram contained under the lines AG and GB. But unto the square of the line AG are commensurable the squares of the lines AG and GB: and unto the parallelogram contained under the lines AG and GB, is commensurable that which is contained under the lines AG and GB twice. Wherefore the squares of the lines AG and GB are incommensurable to that which is contained under the lines AG and GB twice. Wherefore the parallelograms which are equal unto them, namely, the parallelograms CL and FL are incommensurable the one to the other. Wherefore also the line CM is incommensurable in length to the line FM: and they are both rational. Wherefore the line CF is a residual line. 
CF concluded a residual line.  I say moreover, that it is a third residual line. For forasmuch as the square of the line AG, that is, the parallelogram CH is commensurable to the square of the line BG, that is, to the parallelogram KL, therefore the line CK is commensurable in length to the line KM. And in like sort as in the former Proposition, so also in this may we prove, that the parallelogram contained under the lines CK and KM, is equal to the square of the line NM, that is, to the fourth part of the square of the line FM. Wherefore the line CM is in power more than the line FM, by the square of a line commensurable in length to the line CM● and neither of the lines CM nor FM is commensurable in length to the rational line ●D. Wherefore the line CF is a third residual line. Wherefore the square of a second medial residual line applied unto a rational line, maketh the breadth or other side a third residual line: which was required to be demonstrated.   

¶ The 76. Theorem. The 100 Proposition. The square of a less line applied unto a rational line, maketh the breadth or other side a fourth residual line.  
SVppose that AB be a less line, and let CD be a rational line. And unto the line CD apply the parallelogram CE equal to the square of the line AB, and making in breadth the line CF. Then I say, that the line CF is a fourth residual line. For unto the line AB let the line conveniently joined be supposed to be BG. 
Construction.  Wherefore the lines AG & GB are incommensurable in power, having that which is made of their squares added together rational, and that which is contained under them medial. And let the rest of the construction be as in the Propositions going before. Wherefore the whole parallelogram CL is rational. 
Demonstration.  Wherefore the line CM is also rational, and commensurable in length to the line CD. And forasmuch as that which is contained under the lines AG and GB twice is medial, therefore the parallelogram which is equal unto it, namely, the parallelogram FL, is also medial. Wherefore the line FM is rational and incommensurable in length to the line CD. But the line CM is commensurable in length to the line CD. Wherefore (by the 13. of the tenth) the line CM is incommensurable in length to the line FM: and they are both rational. Wherefore the lines CM and FM are rational commensurable in power only. 
CF proved a residual line.  Wherefore the line CF is a residual line. I say moreover, that it is a fourth residual line. For forasmuch as the lines AG and GB are incommensurable in power, therefore the squares of them, that is, the parallelograms, which are equal unto them, namely, the parallelograms CH and KL, are incommensurable the one to the other. Wherefore also the line CK is incommensurable in length to the line KM. And in like sort may we prove, that the parallelogram contained under the lines CK and KM, is equal to the square of the line NM, that is, to the fourth part of the square of the line FM. Wherefore (by the 18. of the tenth) the line CM is in power more than the line FM, by the square of a line incommensurable in length to the line CM. And the whole line CM is commensurable in length to the rational line CD. Wherefore the line CF is a fourth residual line. Wherefore the square of a less line applied unto a rational line, maketh the breadth or other side a fourth residual line: which was required to be proved.   

¶ The 77. Theorem. The 101. Proposition. The square of a line making with a rational superficies the whole superficies medial applied unto a rational line, maketh the breadth or other side a fift residual line.  
SVppose that AB be a line making with a rational superficies the whole superficies medial, and let CD be a rational line. And unto the line CD apply the parallelogram CE equal to the square of the line AB, and making in breadth the line CF. Then I say that the line CF is a fift residual line. For unto the line AB let the line conveniently joined be supposed to ●e BG. Wherefore the lines AG and GB are incommensurable in power, having that which is made of their squares added together medial, and that which is contained under them rational. Let the rest of the construction be in this as it was in the former propositions. Wherefore the whole parallelogram CL is medial. Wherefore the line CM is rational and incommensurable in length to the line CD. And either of the parallelogram FX & NL is rationally Wherefore the whole parallelogram FL is also rational. Wherefore also the line FM is rational and commensurable in length to the line CD. And forasmuch as the parallelogram CL is medial, and the parallelogram FL is rational, therefore CL and FL are incommensurable the one to the other, and the line CM is incommensurable in length to the line FM, and they are both rational. Wherefore the lines CM and FM are rational commensurable in power only. Wherefore the line GF is a residual line. I say moreover that it is a ●i●t residual line. 
CF proved a residual line.  For we may in like sort prove, that the parallelogram contained under the lines CK and KM, is equal to the square of the line NM, that is, to the fourth part of the square of the line FM. And forasmuch as the square of the line AG, that is, the parallelogram CH is incommensurable to the square of the line BG, that is to the parallelogram KL, therefore the line CK is incommensurable in length to the line KM. Wherefore (by the 18. of the tenth) the line CM is in power more than the line FM, by the square of a line incommensurable in length to the line CM. And the line conveniently joined, namely, the line FM is commensurable in length to the rational line CD. Wherefore the line CF is a ●i●t residual line. Wherefore the line CF is a fift residual line. Wherefore the square of a line making with a rational superficies the whole superficies medial, applied unto a rational line maketh the breadth or other side a fift residual line: which was required to be demonstrated.   

¶ The 78. Theorem. The 102. Proposition. The square of a line making with a medial superficies, the whole superficies medial applied to a rational line, maketh the breadth or other side, a sixth residual line.  
SVppose that AB be a line making with a medial superficies, the whole superficies medial, 
Construction.  and let CD be a rational line. And unto the line CD apply the parallelogram CE equal to the square of the line AB and making in breadth the line CF. Then I say that the line CF is a sixth residual line. For unto the line AB let the line conveniently joined be BG. Wherefore the lines AG and BG are incommensurable in power having that which is made of their squares added together medial, & that which is contained under them medial, and moreover that which is made of their squares added together is incommensurable to that which is contained under them. Let the rest of the construction be in this, as it was in the propositions going before. 
Demonstration.  Wherefore the whole parallelogram CL is medial, (for it is equal to that which is made of the squares of the lines AG & GB added together, which is supposed to be medial). Wherefore the line CM is rational and incommensurable in length to the line CD: and in like manner the parallelogram FL is medial. Wherefore also the line FM is rational and incommensurable ●n length to the line CD. And forasmuch as that which is made of the squares of the lines AG and GB added together, is incommensurable to that which is contained under the lines AG and GB twice, therefore the parallelograms equal to them, namely, the parallelograms CL and FL are incommensurable the one to the other. Wherefore also the lines GM and FM are incommensurable in length, and they are both rational. Wherefore they are rational commensurable in power only: Wherefore the line CF is a residual line. I say moreover that it is a sixth residual line. 
CF ●roued ● residual.  Let the rest of the demonstration be as it was in the former propositions. And forasmuch as the lines AG and BG are incommensurable in power, therefore their squares, that is, the parallelograms which are equal unto them, namely, the parallelograms CH and KL are incommensurable the one to the other. Wherefore also the line CK is incommensurable in length to the line KM. Wherefore (by the 18. of the tenth) the line CM is in power more than the line FM by the square of a line incommensurable in length to the line CM. And neither of the lines CM nor FM is commensurable in length to the rational line CD. Wherefore the line CF is a sixth residual line. Wherefore the square of a line making with a medial superficies the whole superficies medial applied to a rational line, maketh the breadth or other side a sixth residual line: which was required to be demonstrated.   

¶ The 79. Theorem. The 103. Proposition. A line commensurable in length to a residual line: is itself also a residual line of the self same order.  
SVppose that AB be a residual line, unto which let the line CD be commensurable in length. 
The sixth Senary.  Then I say, that the line CD is also a residual line, and of the self same order of residual lines that the line AB is. For forasmuch as the line AB is a residual line, 
Construction.  let the line conveniently joined unto it be supposed to be BE. Wherefore the lines AE and BE are rational commensurable in power only. As the line AB is to the line CD, so (by the 12. of the sixth) let the line BE be to the line DF. 
Demonstration.  Wherefore (by the 12. of the fift) as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherefore as the line AB is to the line CD, so is the whole line AE to the whole line CF, and the line BE to the line DF. Wherefore (by the 10. of the tenth) the line AE is commensurable in length to the line CF, and the line BE to the line DF. But the line AE is rational. Wherefore the line CF is also rational. And in like sort the line DF is rational, for that the line BE, to whom it is commensurable, is also rational. And for that as the line BE is to the line AE, so is the line DF to the line CF. But the lines BE and AE are commensurable in power only: Wherefore the lines CD and DF are commensurable in power only. Wherefore the line CD is a residual line. 
CD concluded a residual line.  I say moreovers that it is a residual line of the self same order that the line AB is. For for that as we have before said, as the line A● is to the line CF, so is the line BE to the line DF● therefore alternately, as the line AE is to the line BE, so is the line CF to the line DF. But the line AE is in power more than the line EB, either by the square of a line commensurable in length to the line AE, or by the square of a line incommensurable in length to the line AE. If AE be in power 〈◊〉 then BE, by the square of a line commensurable in length to AE, than the line ●F● shall also (by the 14. of the tenth) be in power more than the line DF, by the square of a line commensurable in length to the line CF, and so if the line AE be commensurable in length to the rational line put, forasmuch as the line AE is commensurable in length to the line CF, therefore (by the 12. of the tenth) the line CF shall also be commensurable in length to the same rational line. Wherefore either of the lines AB and CD is a first residual line. And if the line BE be commensurable in length to the rational line put, forasmuch as the line BE is commensurable in length to the line DF, therefore the line DF shall also be commensurable in length to the rational line put● and then either of the lines AB and CD is a second residual line. And if neither of the lines AE nor BE be commensurable in length to the rational line put, then neither of the lines CF nor DF shall be commensurable in length to the same rational line (by the 13. of the tenth). And so either of the lines AB & CD is a third residual line. But if the line AE be in power more than the line BE, by the square of a line incommensurable in length to the line AE, the line CF shall in like sort (by the 14. of the tenth) be in power more than the line DF, by the square of a line incommensurable in length to the line CF: and then if the line AE be commensurable in length to the rational line, the line CF shall also in like sort be commensurable in length to the same rational line: and so either of the lines AB and CD is a fourth residual line. And if the line BE be commensurable in length to the rational line, the line DF shall also be commensurable in length to the same line: and so either of the lines AB & CD is a ●i●t residual line. And if neither of the lines AE nor BE be commensurable in length to the rational line, in like sort neither of the lines CF nor DF shall be commensurable in length to the same rational line. And so either of the lines AB & CD is a sixth residual line. Wherefore the line CD is a residual line of the self same order that the line AB is. A line therefore commensurable in length to a residual line, is itself also a residual line of the self same order: which was required to be proved.  
As before touching binomial lines, so also touching residual lines, this is to be noted, that a line commensurable in length to a residual line, is always a residual line of the self same order that the residual line is, unto whom it is commensurable, 
Note.  as hath before in this 103. proposition been proved. But if a line be commensurable in power only to a residual line● than followeth it not, yea it is impossible, that that line should be a residual of the self same order that the residual line, is unto whom it is commensurable in power only. Howbeit those two lines shall of necessity be both either of the three first orders of resid●●ll lines, or of the three last orders: which is not hard to prove, if ye mark diligently the former demonstration, and that which was spoken of binomial lines as touching this matter.   

¶ The 80. Theorem. The 104. Proposition. A line commensurable to a medial residual line, is itself also a medial residual line, and of the self same order.  
SVppose that AB be a medial residual line, unto whom let the line CD be commensurable in length and in power, or in power only. Then I say that CD is also a medial residual line, and of the self same order. For forasmuch as the line AB is a medial residual line, 
Construction.  let the line conveniently joined unto i● 〈◊〉 BE: wherefore the lines AE and BE are medial commensurable in power only. As AB is to CD, so by the 22. of the sixth) let BE be to DF. And in like sort as in the former so also in this may we prove, 
Demonstration.  that the line AE is commensurable in length and in power, or in power only unto the line CF, & the line BE 〈◊〉 the line DF. Wherefore (by the 23. of the tenth 〈◊〉 line CF is a medial line, and the line DF is also a medial line, for that it is commensurable to the medial line BE. And in like sort the lines CF and DF are commensurable in power only: for that they have the self same proportion the one to the other, that the lines AE and EB have, which are commensurable in power only. Wherefore the line CD is a medial residual line. 
CD proved a medial.  I say moreover that it is of the self same order that the line AB is. For for that as the line AE is to the line BE, so is the line CF to the line DF. But as the line AE is to the line BE, so is the square of the line AE to the parallelogram contained under the lines AE and BE (by the first of the sixth): and as the line CF is to the line DF, so is the square of the line CF to the parallelogram contained under the lines CF and DF. Wherefore as the square of the line AE is to the parallelogram contained under the lines AE and BE, so is the square of the line CF to the parallelogram contained under the lines CF and DF. Wherefore alternately as the square of the line AE is to the square of the line CF, so is the parallelogram contained under the lines AE and BE, to the parallelogram contained under the ●ines CF and DF. But the square of the line AE is commensurable to the square of the line CF (for the line AE is commensurable to the line CF). Wherefore also the parallelogram contained under the lines AE and BE, is commensurable to the parallelogram contained under the lines CF and DF. Wherefore if the parallelogram contained under the lines AE and EB be rational, the parallelogram also contained under the lines CF and FD shall be rational. And then either of the lines AB and CD is a first medial residual line. But if the parallelogram contained under the lines AE and BE be medial, the parallelogram also contained under the lines CF and FD shall be also medial (by the corollary of the 23. of the tenth): and so either of the lines AB and CD is a second medial residual line. Wherefore the line CD is a medial residual line of the self same order that the line AB is. A line therefore commensurable to a medial residual line, is itself also a medial residual line of the self same order: which was required to be demonstrated-  
This Theorem is understanded generally, that whether a line be commensurable in length & in power, or in power only to a medial residual line, it is itself also a medial residual line, and of the self same order, which thing also is to be understanded of the three Theorems which follow.  

another demonstration after Campane.  
Suppose that A be a medial residual line, unto whom let the line B be commensurable in length, or in power only. And take a rational line CD, unto which apply the parallelogram CE equal to the square of the line A, 
Construction.  and unto the line FE (which is equal to the line CD) apply the parallelogram F● equal to the square of the line B. Now then the parallelograms CE and FG shall be commensurable, 
Demonstration.  for that the lines A, B are commensurable in power: wherefore by the 1. of the sixth and 10. of this book, th● lines DE and FG are commensurable in length. Now than if A be a first medial residual line, then is the line DE a second residual line by the 98. of this book: and if the line A be a s●cond medial residual line, then is the line ● ● a third residual line by the 99 of this book. But if DE be a second residual line, G● also shall be a second residual line (by the ●03. of this book). And if DE be a third residual line, GE also shall (by the same) be also a third residual line. Wherefore it followeth by the 9● and 93. of this book, that B is either a first medial residual line or a second medial residual line, according as the line A is supposed to be: which was required to be proved.    

¶ The 81. Theorem. The 105. Proposition. A line commensurable to a less line: is itself also a less line.  

Construction.  SVppose that AB be a less line, unto whom let the line CD be commensurable. Then I say, that the line CD is also a less line. For let the same construction be in this, that was in the former Propositions. 
Demonstration.  And forasmuch as the lines AE and EB are incommensurable in power, therefore (by the 22. of the sixth, and 10. of the tenth) the lines CF & FD are incommensurable in power. Again (by the 22. of the sixth) as the square of the line AE is to the square of the line BE, so is the square of the line CF to the square of the line DF. Wherefore by composition as the squares of the lines AE and BE are to the square of the line BE, so are the squares of the lines CF and DF, to the square of the line DF: and alternately, as the squares of the lines AE and BE are to the squares of the lines CF and DF, so is the square of the line BE to the square of the line DF. But the square of the line BE is commensurable to the square of the line DF (for the lines BE and DF are commensurable). Wherefore that which is made of the squares of the lines AE and BE added together, is commensurable to that which is made of the squares of the lines CF and DF added together. But that which is made of the squares of the lines AE and BE added together, is rational. Wherefore that which is made of the squares of the lines CF and DF added together, is also rational. Again, for that as the square of the line AE is to the parallelogram contained under the lines AE and BE, so is the square of the line CF to the parallelogram contained under the lines CF and DF (as we declared in the Proposition next going before): therefore alternately, as the square of the line AE is to the square of the line CF, so is the parallelogram contained under the lines AE and BE, to the parallelogram contained under the lines CF and DF. But the square of the line AE is commensurable to the square of the line CF, for the lines AE & CF are commensurable. Wherefore the parallelogram contained under the lines AE and BE, is commensurable to the parallelogram contained under the lines CF and DF. But the parallelogram contained under the lines AE and BE is medial. Wherefore the parallelogram contained under the lines CF and DF is also medial. Wherefore the lines CF and DF are incommensurable in power, having that which is made of their squares added together rational, and the parallelogram contained under them medial. Wherefore the line CD is a less line. A line therefore commensurable to a less line, is itself also a less line: which was required to be proved.  

another demonstration.  
Suppose that A be a less line, and unto A let the line B be commensurable whether in length and power, or in power only. Then I say that B is a less line. Take a rational line CD. 
Construction.  And unto the line CD apply (by the 44 of the first) the parallelogram CE equal to the square of the line A, and making in breadth the line CF. Wherefore (by the 100 proposition) the line CF is a fourth residual line. Unto the line FE apply (by the same) the parallelogram EH equal to the square of the line B, and making in breadth the line FH. Now forasmuch as the line A is commensurable to the line B, 
Demonstration.  therefore also the square of the line A is commensurable to the square of the line B. But unto the square of the line A is equal the parallelogram CE, & unto the square of the line B is equal the parallelogram EH. Wherefore the parallelogram CE is commensurable to the parallelogram EH. But as the parallelogram CE is to the parallelogram EH, so is the line CF to the line FH. Wherefore the line CF is commensurable in length to the line FH. But the line CF is a fourth residual line. Wherefore the line FH is also a fourth residual line (by the 103. of the tenth): and the line FE is rational. But if a superficies be contained under a rational line, and a fourth residual line, the line that containeth in power that superficies is (by the 94. of the tenth) a less line. But the line B containeth in power the superficies EH. Wherefore the line B is a less line: which was required to be proved.    

¶ The 82. Theorem. The 106. Proposition. A line commensurable to a line making with a rational superficies the whole superficies medial, is itself also a line making with a rational superficies the whole superficies medial.  
SVppose that AB be a line making with a rational superficies the whole superficies medial, unto whom let the line CD be commensurable. Then I say that the line CD is a line making with a rational superficies the whole superficies medial. 
Construction.  Unto the line AB let the line conveniently joined be BE. Wherefore the lines AE and EB are incommensurable in power, having that which is made of their squares added together medial, and the parallelogram contained under them rational. Let the construction be in this as it was in the former propositions. 
Demonstration.  And in like sort may we prove that as the line AE is to the line BE, so is the line CF to the line DF, and that that which is made of the squares of the lines AE and BE added together is commensurable to that which is made of the squares of the lines CF and DF added together, and that that which is contained under the lines AE and EB, is in like sort commensurable to that which is contained under the lines CF and DF. Wherefore also the lines CF and DF are commensurable in power, having that which is made of their squares added together medial, and that which is contained under them rational. Wherefore the line CD is a line making with a rational superficies the whole superficies medial. Wherefore a line commensurable to a line making with a rational superficies the whole superficies medial, is itself also a line making with a rational superficies the whole superficies medial: which was required to be demonstrated.  

another demonstration.  
Suppose that A be a line making with a rational superficies the whole superficies medial, and unto it let the line B be commensurable either in length and in power, or in power only. Then I say that B is a line making with a rational superficies the whole superficies medial. 
Construction.  Take a rational line CD, and unto the line CD apply the parallelogram CE equal to the square of the line A and making in breadth the line GF. Wherefore (by the 101. proposition) the line CF is a fift residual line. Again unto the line FE apply the parallelogram FG equal to the square of the line B, and making in breadth the line FH. Now forasmuch as the line A is commensurable to the line B, 
Demonstration.  therefore the square of the line A is commensurable to the square of the line B. But unto the square of the line A is equal the parallelogram CE, and unto the square of the line B is equal the parallelogram FG. Wherefore the parallelogram CE is commensurable to the parallelogram FG. Wherefore the line CF is also commensurable in length to the line FH. But the line CF is a fift residual line. Wherefore also the line FH is a fift residual line. And the line FE is rational. But if a supersicies be contained under a rational line and a ●ift residual line, the line that containeth in power that superficies, is (by the 95. of the tenth) a line making with a rational superficies the whole superficies medial. But the line B containeth in power the parallelogram FG. Wherefore the line B is a line making with a rational superficies, the whole superficies mediall● which was required to be demonstrated.    

¶ The 83. Theorem. The 107. Proposition. A line commensurable to a line, making with a medial superficies, the whole superficies medial, is itself also a line making with a medial superficies the whole superficies medial.  
SVppose that AB be a line making with a medial superficies the whole superficies medial, unto whom let the line CD be commensurable. Then I say that the line CD is also a line making with a medial superficies the whole superficies medial. For unto the line AB let the line coveniently joined be BE. And let the rest of the construction be in this as it was in the former propositions. 
Construction.  Wherefore the lines AE and BE are incommensurable in power, having that which is made of their squares added together medial, 
Demonstration.  and that which is contained under them also medial, and moreover that which is made of their squares added together is incommensurable to that which is contained under them. But the lines AE and BE (as we have before proved) are commensurable to the lines CF & DF, and that which is made of the squares of the lines AE and BE added together, is commensurable to that which is made of the squares of the lines CF and FD added together, and the parallelogram contained under the lines AE and BE is commensurable to the parallelogram contained under the lines CF and DF. Wherefore the lines CF and DF are incommensurable in power having that which is made of their squares added together medial, and that which is contained under them also medial, and moreover that which is made of their squares added together, is incommensurable to that which is contained under them. Wherefore the line CD is a line making with a medial superficies the whole superficies medial. A line therefore commensurable to a line making with a medial superficies the whole superficies medial, is itself also a line making with a medial superficies the whole superficies medial: which was required to be proved.  
This proposition may also be an other way demonstrated, as the three former propositions were. If upon a rational line you apply parallelograms equal to the squares of the lines AB and CD, the breadthes of which parallelograms shall be each a sixth residual line, and therefore the lines which contain them in power, namely, the lines AB and CD shall be both such lines as is required in the proposition, which is easy to conclude marking the order of the demonstration in the three former propositions.   

¶ The 84. Theorem. The 108. Proposition. If from a rational superficies be taken away a medialt superficies, the line which containeth in power the superficies remaining, is one of these two irrational lines, namely, either a residual line, or a less line.  
SVppose that BC be a rational superficies, 
Seventh Senary.  and from it take away a medial superficies, namely, BD. Then I say, that the line which containeth in power the superficies remaining, namely, the superficies EC, is one of ●hese two irrational lines, namely, either a residual line, or a less line. Take a rational line FG. And upon FG describe (by the 44. of the first) a rectangle parallelogram GH equal to the superficies BC. 
Constraction.  And from the parallelogram GH take away the parallelogram GK equal to the superficies BD. 
Demonstration.  Wherefore (by the third common sentence) the superficies remaining, namely, EC, is equal to the parallelogram remaining, namely, to LH. And forasmuch as BC is rational, and BD is medial, and BC is equal to the parallelogram GH, and BD to the parallelogram GK: therefore GH is rational, and GK is medial: and the parallelogram GH is applied unto the rational line FG. Wherefore (by the ●0. of the tenth) the line FH is rational and commensurable in length to the line FG. And the parallelogram GK is also applied unto the rational line FG. Wherefore (by the 22. of the tenth) the line FK is rational and incommensurable in length to the line FG. Wherefore (by the Assumpt of the 12. of the tenth) the line FH is incommensurable in length to the line FK. And they are both rational. Wherefore the lines FH and FK are rational commensurable in power only. Wherefore the line KH is a residual line: and the line conveniently soyned unto it is KF. Now the line FH is in power more th●n the line KF, either by the square of a line commensurable in length to the line FH, or by the square of a line incommensurable in length to the line FH. First let it be in power more than the line FK, by the square of a line commensurable in length to the line FH, and the whole line FH is commensurable in length to the rational line put, namely, to FG. Wherefore the line KH is a first residual line. But if a superficies be contained under a rational line, and a first residual line, the line that containeth in power that superficies, is (by the 91. of the tenth) a residual line. Wherefore the line which containeth in power LH, that is, the superficies EC, is a residual line. But if the line HF be in power more than the line FK, by the square of a line incommensurable in length to the line FH, and the whole line FH is commensurable in length to the rational line given FG. Wherefore the line KH is a fourth residual line. But a line containing in power a superficies contained under a rational line and a fourth residual line, as a less line (by the 94. of the tenth). Wherefore the line that containeth in power the superficies LH, that is, the superficies EC, is a less line. If therefore from a rational superficies be taken away a medial superficies, the line which containeth in power the superficies remaining, is one of these two irrational lines, namely, either a residual line, or a less line: which was required to be proved.   

¶ The 85. Theorem. The 109. Proposition. If from a medial superficies be taken away a rational superficies, the line which containeth in power the superficies remaining is one of these two irrational lines, namely either a first medial residual line, or a line making with a rational superficies the whole superficies medial.  
SVppose that BC be a medial superficies and from it take away a rational superficies, namely, BD. Then I say that the line which containeth in power the superficies remaining, namely, the superficies EC is one of these two irrational lines, either a first medial residual line, or a line making with a rational superficies the whole superficies medial. Take a rational line FG, and let the rest of the construction be in this as it was in the former proposition. 
Construction.  Wherefore it followeth that the line F● is rational and incommensurable in length to the line FG (by the 22. of the tenth). 
Demonstration.  And that the line KF is (by the 20. of the tenth) rational and commensurable in length to the line FG. Wherefore the lines FH and FK are rational commensurable in power only. Wherefore KH is a residual line. And the line conveniently joined unto it is FK. Now the line FH is in power more than the line FK, either by the square of a line commensurable in length to the line FH, or by the square of a line incommensurable in length unto it. If the line FH be in power more than the line FK by the square of a line commensurable in length to the line FH and the line conveniently joined unto it, namely, FK, is commensurable in length to the rational line FG. Wherefore the line KH is a second residual line. And the line FG is a rational line. But a line containing in power a superficies comprehended under a rational line and a second residual line is (by the 92. of the tenth) a first medial residual line. Wherefore the line thus containeth in power the superficies LH, that is, the superficies CE is a first medial residual lne. But if the line HF be in power more than the line FK by the square of a line incommensurable in length to the line FH, and the line conveniently joined, namely, the line FK is commensurable in length to the rational line put, namely, to FG: wherefore the line KH is a ●ift residual line. Wherefore (by the 95. of the tenth) the line that containeth in power the superficies LH, that is, the superficies EC, is a line making with a rational superficies, the whole superficies medial which was required to be proved.   

¶ The 86. Theorem. The 110. Proposition. If from a medial superficies be taken away a medial superficies incommensurable to the whole superficies, the line which containeth in power the superficies which remaineth, is one of these two irrational lines, namely, either a second medial residual line, or a line making with a medial superficies the whole superficies medial.  
AS in the former descriptions, s● 〈◊〉 also ●ake away from the medial superficies BE medial superficies BD● and let ●D be incommensurable to the whole superficies BC. Then I say, that the line which containeth in power the superficies EC, is one of th●se two irrational lines, namely, either a second medial residual line, or a li●e making with a medial superficies the whole superficies medial. 
Demonstration.  For forasmuch as either of these superficie●●s BC and BD is medial, and BC is incommensurable to BD, it followeth (by the 22. of the tenth) tha● either of these lines FH and FK is rational and incommensurable in length to the line FG. And forasmuch as the superficies BC is incommensurable to the superficies BD, that is, the superficies GH, to the superficies GK, therefore (by the first of the sixth, & 100LS of the tenth) the line FH is incommensurable in length to the line FK. Wherefore the lines HF and FK are rational commensurable in power only. Wherefore (by the 73. of the tenth) the line KH is a residual line, and the line conveniently joined unto it is FK. Now the line HF is in power more than the line FK, either by the square of a line commensurable in length to the line HF, or by the square of a line incommensurable in length unto it. If the line HF be in power more than the line FK, by the square of a line commensurable in length to the line FH, and neither of the lines HF nor FK i● commensurable to the rational line put FG. Wherefore the line KH is a third residual. But the line CF, that is, the line KL, is rational. And a rectangle superficies contained under a rational line and a third residual line, is irrational, and the line which containeth in power that superficies, is (by the 93. of the tenth) a second medial residual line. Wherefore the line that containeth in power the superficies LH, that is, the superficies EC is a second medial residual line. But if the line HF be in power more than the line FK, by the square of a line incommensurable in length to the line FH, and neither of the lines HF nor FK is commensurable in length to the line FG. Wherefore the line HK is a sixth residual line. But a line containing in power a superficies contained under a rational line and a sixth residual line, is (by the 96. of the tenth) a line making with a medial superficies the whole superficies medial. Wherefore the line that containeth in power the superficies LH, that is, the superficies EC, is a line making with a medial superficies the whole superficies medial. If therefore from a medial superficies be taken away a medial superficies, incommensurable to the whole superficies, the line that containeth in power the superficies which remaineth, is one of the two irrational lines remaining, namely, either a second medial residual line, or a line making with a medial superficies the whole superficies medial: which was required to be proved.   

¶ The 87. Theorem. The 111. Proposition. A residual line, is ●ot one and the same with a binomial line.  
SVppose that AB be a residual line. Then I say that AB is not one and the same with a binomial line. For if it be possible let it be a binomial line. And take a rational line DC. 
Construction.  And (by the 44. of the first) unto the line CD apply a rectangle parallelogram CE equal to the square of the line AB, and making in breadth the line DE. 
Demonstration on leading to an impossibility.  And forasmuch as AB is a residual line, therefore (by the 97. of the tenth) the line DE is a first residual line. Let the line conveniently joined unto it be E●. Wherefore the lines D F and FE are rational commensurable in power only, and the line DF is in power more than the line FE by the square of a line commensurable in length to the line DE & the line DF, is commensurable in length to the rational line put DC. Again forasmuch as AB is by position a binomial line, therefore (by the 60. of the tenth) the line DE is a first binomial line. Divide it into his names in the point G. And let DG be the greater name. Wherefore the lines DG and ●E are rational commensurable in power only. And the line DG is in power more than the line GE by the square of a line commensurable in length to the line DG, and the line DG is commensurable in length to the rational line put DC. Wherefore the line DF is commensurable in length to the line DG. Wherefore (by the 13. of the tenth) the whole line DF is commensurable in length to the line remaining, namely, ●o the line GF. And forasmuch as the line DF is commensurable to the line FG, but the line FD is rational. Wherefore the line FG is also rational. And forasmuch as the line FD is commensurable in length to the line FG, but the line DF is incommensurable in length to the line FE. Wherefore the line FG is incommensurable in length to the line FE (by the 13. of the tenth) and they are both rational lines. Wherefore the lines GF and FE are rational commensurable in power only. Wherefore (by the 73. of the tenth) the line EG is a residual line, but it is also rational (as before hath been proved): which is impossible, namely, that one & the same line should be both rational and irrational. Wherefore a residual line is not one and the same with a binomial line, that is, is not a binomial line: which was required to be demonstrated.  


A Corollary.  ¶ A Corollary.  
A residual line and the other five irrational lines following it, are neither medial lines, nor one and the same between themselues● that is, one is utterly of a divers kind from an other. For the square of a medial line applied to a rational line, maketh the breadth rational and incommensurable in length to the rational line, whereunto it is applied (by the 22. of the tenth) The square of a residual line applied to a rational line, maketh the breadth a first residual line (by the 97. of the tenth). The square of a first medial residual line applied to a rational line, maketh the breadth a second residual line (by the 98. of the tenth) The square of a second medial residual line applied unto a rational line, maketh the breadth a third residual line (by the 99 of the tenth) The square of a less line applied to a rational line, maketh the breadth a fourth residual line (by the 100 of the tenth) The square of a line making with a rational superficies the whole superficies medial applied to a rational line, maketh the breadth a ●ift residual line (by the 101. of the tenth) And the square of a line making with a medial superficies the whole superficies medial applied to a rational line, maketh the breadth a sixth residual line (by the 102. of the tenth) Now forasmuch as these foresaid sides which are the breadthes differ both from the first breadth, sore that it is rational, and differ also the one from the other, for that they are residuals of divers orders and kinds, it is manifest that those irrational lines differ also the one from the other. And forasmuch as it hath been proved in the 111. proposition, that 〈◊〉 residual 〈◊〉 is not one and the same with a binomial line, and it hath also been proved that the 〈…〉 of a residual line and of the five irrational lines that follow it being applied to a rational line do make their breadthes one of the residuals of that order of which they were, whose square● were applied to the rational line, likewise also the squares of a binomial line, and of the five irrational lines which follow it, being applied to a rational line, do make the breadthes one of the binomials of that order of which they were, whose squares were applied to the rational line. Wherefore the irrational lines which follow the binomial line, and the irrational lines which follow the residual line, differ the one from the other, so that all the irrational lines are 13. in number, namely, these.  

1 A medial line.  
2 A binomial line.  
3 A first bimedial line.  
4 A second bimedial line.  
5 A greater line.  
6 A line containing in power a rational superficies and a medial superficies.  
7 A line containing in power two medial superficieces.  
8 A residual line.  
9 A first medial residual line.  
10 A second medial residual line.  
11 A less line.  
12 A line making with a rational superficies the whole superficies medial.  
13 A line making with a medial superficies the whole superficies medial.     

¶ The 88 Theorem. The 112. Proposition. The square of a rational line applied unto a binomial line, maketh the breadth or other side a residual line, whose names are commensurable to the names of the binomial line, & in the self same proportion: & moreover that residual line is in the self same order of residual lines, that the binomial line is of binomial lines.  
SVppose that A be a rational line, 
The determination hath sundry parts orderly to be proved.  and BC a binomial line whose greater name let be CD. And unto the square of the line A let the parallelogram contained under the lines BC and EF (so that EF be the breadth) be equal. Then I say that EF is a residual line, whose names are commensurable to the names of the binomial line BC, which names let be CD and DB, and are in the same proportion with them: and moreover the line EF is in the self same order of residual lines, that the line BC is of binomial lines. Unto the square of the line A let the parallelogram contained under the lines BD and G be equal. 
Construction.  Now forasmuch as that which is contained under the lines BC & EF is equal to that which is contained under the lines BD and G, therefore reciprocally (by the 14. of the sixth) as the line CB is to the BD, so is the line G to the line EF. But the line BC is greater than the line BD, wherefore the line G is greater than the line EF. 
Demonstration.  Unto the line G let the line EH be equal. Wherefore (by the 11. of the fift) as the line CB is to the line BD, so is the line HE to the line FE. Wherefore by division (by the 17. of the fifth) as the line CD is to the line BD, so is the line HF to the line FE. 
This is an Assumpt, problematical, artificially used and demonstrated.  As the line HF is to the FE, so let the line FK be to the line KE (how this is to be done we will declare at the end of this demonstration). Wherefore (by the 12. of the fift) the whole line HK 
* Therefore those three lines are in continual proportion.  is to the whole line KF as the line FK is to the line KE. For as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. But as the line FK is the line KE, so is the line CD to the line DB (for FK is to EK as HF is to FE, and HF is to FE as CD is DB). Wherefore (by the 11. of the fift) as the line HK is to the line KF, so is the line CD to the line DB. But the square of the line CD is commensurable to the square of the line DB: wherefore (by the 10. of the tenth) the square of the line HK is commensurable to the square of the line FK. But these three lines HK, FK, and EK are proportional in continual proportion (as it hath already been proved). Wherefore (by the second corollary of the 20. of the sixth) the square of the line HK is to the square of the line FK as the line HK is to the line EK: wherefore the line HK is commensurable in length to the line EK. Wherefore (by the 15. of the tenth) the line HE is commensurable in length to the line FK. And forasmuch as the square of the line A is equal to that which is contained under the lines EH and BD, but the square of the line A is rational, wherefore that which is contained under the lines EH and BD is rational. And it is applied unto the rational line BD. Wherefore (by the 20. of the tenth) the line EH is rational and commensurable in length to the line BD. Wherefore also the line EK which is commensurable in length to the line HE is rational and commensurable in length to the line BD. Now for that as the line CD is to the line DB, so is t●e line FK to the line EK (for it was before proved, that as CD is to DB, so is HF to FE, and as HF is to FE, so is FK to EK) but the lines CD and DB are commensurable in power only, wherefore (by the 10. of the tenth) the lines FK and KE are also commensurable in power only. And for that as the line CD is to the line DB, so is the line FK to the line EK, therefore by contrary proportion as DB is to CD, so is EK to FK, and alternately as DB is to EK, so is CD to FK: but the lines BD and EK are commensurable in length (as it hath already been proved). Wherefore also the lines CD and FK are commensurable in length. But the line CD is rational: wherefore also the line FK is rational. Wherefore the lines FK and EK are rational commensurable in power only. Wherefore the line FE is a residual line: whose names FK and KE are commensurable to the names CD and BD of the binomial line BC and in the same proportion as is proved. 
FE concluded a residual li●●, which is sumwhat prep●●icro●sly, in respect o● the ●●der propounded, both in the proposition, and also in the determination.  I say moreover that it is a residual line of the self same order that the binomial line is. For the line CD is in power more than the line BD either by the square of a line commensurable in length to the line CD, or by the square of a line incommensurable in length. Now if the line CD be in power more than the line BD by the square of a line commensurable in length unto the line CD, than (by the 13. of the tenth) the line FK is in power more than the line EK by the square of a line commensurable in length to the line FK. And so if the line CD be commensurable in length to the rational line put, the line FK also shallbe commensurable in length to the same rational line: wherefore then the line BC is a first binomial line, & the line FE is likewise a first residual line. And if the line BD be commensurable in length to the rational line, the line ●K is also commensurable in length to the same, and then the line BC is a second binomial line, and the line FE a second residual line. And if neither of the lines CD nor DB be commensurable in length unto the rational line, neither of the lines FK nor EK are commensurable in length unto the same, and then the line BC is a third binomial line, & the line FE is a third residual line. And if the line CD be in power more than the line BD, by the square of a line incommensurable in length to the line CD, the line FK is also (by the 14. of the tenth) in power more than the line EK by the square of a line incommensurable in length to the line FK. And so if the line CD be commensurable in length to a rational line put, the line FK also is commensurable in length to the same, wherefore the line BC is a fourth binomial line, and the line FE is a fourth residual line. And if the line BD be commensurable in length to the rational line, the line EK is likewise commensurable in length to the same, and then the line BC is a fifth binomial line and the line EF a fifth residual line. And if neither of the lines CD nor DB be commensurable in length to the rational line, neither also of the lines FK nor EK is commensurable in length to the same, and then the line BC is a sixth binomial line, and the line FE a sixth residual line. Wherefore the line FE is a residual line, whose names, namely, FK and EK are commensurable to the names of the binomial line, namely, to the names CD and DB, and are in the self same proportion, and the residual line EF is in the self same order of residual lines, that the binomial line BC is of binomial lines. Wherefore the square of a rational line applied unto a binomial line maketh the breadth (or other side) a residual line, whose names are commensurable to the names of the binomial line, and in the self same proportion, and moreover that residual line is in the self same order of residual lines, that the binomial line is of binomial lines: which was required to be demonstrated.  

Here is the Assumpt (of the foregoing Proposition) confirmed.  
Now let us declare how as the line HF is to the line FE, so to make the line FK to the line EK. The line CD is greater than the line BD by supposition. Wherefore also the line HF is greater than the line FE (by alternate proportion, and the 14. of the fifth). 
Construction.  From the line HF take away the line FL equal to the line FE. Wherefore the line remaining, namely, HL, is less than the line HF, for the line HF is equal to the lines HL & LF. As HL is to HF, so (by the 12. of the sixth) let FE be to FK. 
Demonstration.  Wherefore by contrary proportion (by the Corollary of the 4. of the fifth) as HF is to HL, so is FK to FE. Wherefore by conversion of proportion (by the Corollary of the 19 of the fifth) as HF is to LF, that is, to the line equal unto it, namely, to FE, so is the line FK to the line EK.  

M. Dee, of this Assumpt, maketh (〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉, that is, Acquisively,) a Problem universal, thus:  
Two unequal right lines being propounded, to adjoin unto the less, a right line, which taken with the less (as one right line) shall have the same proportion, to the line adjoined, which, the greater of the two propounded, hath to the less.  
The construction and demonstration hereof, is word for word to be taken, as it standeth here before: after these words: The line HF is greater than the line FE.   

¶ A Corollary also noted by I Dee.  
It is therefore evident, that thus are three right lines (in our handling) in continual proportion: it is to weet, the greater, the less and the adjoined, make the first, the less with the adjoined, make the second: and the adjoined line is the third.  
This is proved in the beginning of the demonstration, after the Assumpt used.    

another demonstration after Flussas.  
Take a rational line A, and let GB be a binomial line, whose greater line let be GD: 
Construction.  and upon the line GB apply (by the 45. of the first) the parallelogram BZ equal to the square of the line A, and making in breadth the line GZ. Likewise upon the line DB (by the same) apply the parallelogram BY equal also to the square of the line A, and making in breadth the line DI: and put the line GZT equal to the line DI. Then I say, that GZ is such a residual line as is required in the Proposition. Forasmuch as the parallelograms BZ & BY are equal, therefore (by the 14. of the sixth) reciprocally as the line GB is to the line BD, so is the line DIEGO or the line GT, (which is equal unto it) unto the line GZ. Wherefore by division, as the line GD is to the line DB, so is the line TZ to the line ZG (by the 17. of the fifth). Wherefore the line TZ is greater than the line ZG. (For the line GD is the greater name of the binomial line GB). Unto the line ZG put the line ZC equal. And as the line TC is to the line TZ, so (by the 11. of the sixth) let the line ZG be to line ZK. Wherefore contrary wise (by the Corollary of the 4. of the fifth) the line TZ is to the line TC, 
Demonstration.  as the line ZK is to the line ZG. Wherefore by conversion of proportion (by the 19 of the fifth) as the line TZ is to the line ZC (that is, to ZG, which is equal unto it) so is the line ZK to the line KG. But the line TZ is to the line ZG, as the line GD is to the line DB. Wherefore (by the 11. of the fifth) the line ZK is to the line KG, as the line GD is to the line DB. But the lines GD and DB are commensurable in power only. Wherefore also the lines ZK and KG are commensurable in power only, by the 10. of this book. Farther, forasmuch as the line TZ is to the line ZG, as the line ZK is to the line KG, therefore by the 12. of the fifth, all the antecedentes, namely, the whole line TK are to all the consequentes, namely, to the line KZ, as one of the antecedentes, namely, the line ZK is to one of the consequentes, namely, to the line KG. Wherefore the line ZK is the mean proportional between the lines TK and KG. And therefore (by the Corollary of the 20. of the sixth) as the first, namely, the line TK, is to the third, namely, to the line KG: so is the square of the line TK to the square of the second, namely, of the line KZ. And forasmuch as the parallelogram BY (which is equal to the square of the rational line A) is applied upon the rational line DB, it maketh the breadth DIEGO rational and commensurable in length unto the line DB, by the 20. of the tenth. And therefore the line GT (which is equal unto the line DIEGO) is commensurable in length to the same line DB. And for that as the line GD is to the line DB, so is the line KZ to the line KG, but as the line KZ is to the line KG, so is the line TK to the line KZ, therefore (by the 11. of the fifth) as the line GD is to the line DB, so is the line T K to the line KZ. Wherefore (by the 22. of the sixth) as the square of the line GD is to the square of the line DB, so is the square of the line T K to the square of the line KZ. But the square of the line GD is commensurable to the square of the line DB (for the names GD and DB of the binomial line GB are commensurable in power). Wherefore the square of the line T K shall be commensurable to the square of the line KZ, by the 10. of this book. But as the square of the line T K is to the square of the line KZ, so is it proved, that the right line T K is to the right line KG. Wherefore the right line T K is commensurable in length to the right line KG. Wherefore it is also commensurable in length to the line TG (by the 15. of the tenth). Which line TG is (as it hath been proved a ●ationall line, and equal to the line DI. Wherefore the lines T K and KG are rational commensurable in length. And forasmuch as it hath been proved, that the line Z K is commensurable in power only unto the rational line KG, therefore the lines Z K and KG are rational commensurable in power only. 
Here are the ●ower parts of the propositi● more orderly huddled the● in the former demöstration.  Wherefore the line GZ is a residual line. And forasmuch as the rational line TG is commensurable in length to either of these lines DB and KG. Wherefore the lines DB & KG shall be commensurable in length, by the 12. of the tenth. But the line Z K is to the line KG, as the line GD is to the line DB. Wherefore alternately, by the 16. of the fifth, the line KZ is to the line GD, as the line KG is to the line DB. Wherefore the line Z K is commensurable in length unto the line GD. Wherefore the lines Z K and KG (the names of the residual line GZ) are commensurable in length to the lines GD and DB, which are the names of the binomial line GB: and the line Z K is to the line KG in the same proportion, that the line GD is to the line DB. Wherefore if the whole line Z K be in power more than the line conveniently joined KG, by the square of a line commensurable in length to the line Z K, than the greater name G D shall be in power more than the less name DB, by the square of a line commensurable in length to the line GD, by the 14. of the tenth. And if the line Z K be in power more than the line KG, by the square of a line incommensurable in length to the line Z K, the line also GD shall be in power more than the line DB, by the square of a line incommensurable in length unto the line GD (by the same Proposition). And if the greater or less name of the one be commensurable in length to the rational line put, the greater of l●sse name also of the other shall be commensurable in length to the same rational line, (by the 12. of this book). But if neither name of the one be commensurable in length to the rational line put, neither name of the other also shall be commensurable in length to the same rational line put (by the 13. of the same). Wherefore the residual line GZ shall be in the self same order of residual lines, that the binomial line GB is of binomial lines (by the definitions of residual and binomial lines. The square therefore of a rational line applied to a binomial line● & ●: which was required 〈◊〉 be proved.    

¶ The 89. Theorem. The 113. Proposition. The square of a rational line applied unto a residual, maketh the breadth or other side a binomial line, whose names are commensurable to the names of the residual line, and in the self same proportion: and moreover that binomial line is in the self same order of binomial lines, that the residual line is of residual lines.  
SVppose that A be a rational line, and BD a residual line. And unto the square of the line A let that which is contained under the lines BD and KH be equal. Wherefore the square of the rational line A applied unto the residual line BD maketh the breadth or other side KH. Then I say that the line KH is a binomial line, whose names are commensurable to the names of the residual line BD, and in the self same proportion, and that the line KH is in the self same order of binomial lines, that the line BD is of residual lines. Unto the line BD let the line conveniently joined be DC. 
Construction.  Wherefore the lines BC and DC are rational commensurable in power only. And unto the square of the line A let the parallelogram contained under the lines BC and G be equal. But the square of the line A is rational. 
Demonstration.  Wherefore the parallelogram contained under the lines BC and G is also rational. Wherefore also the line G is rational and commensurable in length to the line BC (by the 20. of the tenth). Now forasmuch as the parallelogram contained under the lines BC and G is equal to that which is contained under the lines BD and KH, therefore (by the 16. of the sixth) as the line BC is to the line BD, so i● the line KH to the line G. But the line BC is greater than the line BD. Wherefore also the line KH is greater than the line G. Unto the line G l●t the line KE be equal. Wherefore the line KE is rational and commensurable in length to the line BC, as also the line G was (by the 12. of the tenth) And for that as BC is to BD, so is KH to KE. Wherefore by ●duersion of proportion (by the corollary of the 19 of the fift) as BC is to DC, so is KH, to EH, 
An Assumpt.  ●KH into EH so let the line FH be to the line EF (how this is to be done, we will decare at the end of this demonstration). Wherefore the residue KF is to the residue FH, as the whole KH is to the whole HE (by the 19 of the fift) that is, as the line BC is to the line CD. But the lines BC and CD are commensurable in power only. Wherefore also the lines KF and FH are commensurable in power only. And for that as KH is to HE, so is KF to FH, but as KH is to HE, so is also HF to FE, therefore as KF is to FH, so is FH to FE. Wherefore (by the corollary of the 19 of the sixth) as the first is to the third, so is the square of the first, to the square of the second. Wherefore as KF is to FE, so is the square of the line KF to the square of the line FH, but these squares are commensurable, for the lines KF and FH are commensurable in power. Wherefore the lines KF and FE are commensurable in length. Wherefore (by the second part of the 15. of the tenth) the lines KE and EF are commensurable in length. Wherefore (by the same) the lines KF and FE are commensurable in length. But the line KE is rational and commensurable in length to the line BC: wherefore the line KF is also rational and commensurable in length to the line BC. And for that as the line BC is to the line CD, so it KF to EH, therefore alternately. (by the 16. of the fift) as BC is to KF, so is CD to FH. But the line BC is commensurable in length to the line KF. Wherefore the line CD is commensurable in length to the line FH. But the line CD is rational. Wherefore also the line FH is rational. And the lines BC and CD are rational commensurable in power only. Wherefore the lines KF and FH are rational commensurable in power only. Wherefore the line KH is a binomial line, whose names are commensurable to the names of the residual line, and in the same proportion. I say moreover that it is a binomial of the self same order of binomial lines, that the line BD is of residual lines. For if the line BC be in power more than the line CD by the square of a line commensurable in length to the line BC, the line KF is also in power more than the line FH by the square of a line commensurable in length to the line KF (by the 14. of the tenth). And if the line BC be commensurable in length to the rational line put, the line KF is also (by the 12. of the tenth) commensurable in length to the rational line, and so the line BD is a first residual line, and the line KH is in like sort a first binomial line. If the line CD be commensurable in length to the rational line, the line FH is also commensurable in length to the same line, and so the line BD is a second residual line, and the line KH a second binomial line. And if neither of the lines BC nor CD be commensurable in length to the rational line, neither also of the lines KF nor FH is commensurable in length to the same, and so the line BD is a third residual line, and the line KH a third binomial line. But if the line BC be in power more than the line CD by the square of a line incommensurable in length to the line BC, the line KF is in power more than the line FH by the square of a line incommensurable in length to the line KF (by the 14. of the tenth) And if the line BC be commensurable in length to the rational line put, the line KF is also commensurable in length to the same line, and so the line BD is a fourth residual line, and the line KH a fourth binomial line. And if the line CD be commensurable in length to the rational line, the line FH is also commensurable in length to the same, & so the line BD is a fift residual line, & the line KH a fift binomial line. And if neither of the lines BC nor CD be commensurable in length to the rational line, neither also of the lines KF nor FH is commensurable in length to the same, and so the line BD is a sixth residual line, and the line KH is a sixth binomial line. Wherefore KH is a binomial line, whose names KF and FH are commensurable to the names of the residual line BD, namely, to BC and CD, and in the self same proportion, and the binomial line KH is in the self same order of binomial lines, that the residual BD, is of residual lines. Wherefore the square of a rational line applied unto a residual line, maketh the breadth or other side a binomial line, whose names are commensurable to the names of the residual line, and in the self same proportion, and moreover the binomial line is in the self same order of binomial lines, that the residual line is of residual lines: which was required to be demonstrated.  

The Assumpt confirmed.  
Now let us declare how, as the line KH is to the line EH, so to make the line HF to the line FE. Add unto the line KH directly a line equal to HE, and let the whole line be KL, and (by the tenth of the sixth) let the line HE be divided as the whole line KL is divided in the point H: let the line HE be so divided in the point F. Wherefore as the line KH is to the line HL, that is, to the line HE, so is the line HF to the line FE.   

another demonstration after Flussas.  
Suppose that A be a rational line, and let BD be a residual line. And upon the line BD apply the parallelogram DT equal to the square of the line A (by the 45. of the first) making in breadth the line BT. 
another demonstration after Flussas.  Then I say that BT is a binominall line such a one as is required in the proposition. Forasmuch as BD is a residual line, let the line conveniently joined unto it be GD. Wherefore the lines BG and GD are rational commensurable in power only. 
Construction.  Upon the rational line BG apply the parallelogram BY equal to the square of the line A and making in breadth the line BE. Wherefore the line BE is rational and commensurable in length to the line BG (by the 20. of the tenth). Now forasmuch as the parallelograms BY and TD are equal (for that they are each equal to the square of the line A): 
Demonstration.  therefore reciprocally (by the 14. of the sixth) as the line BT is to the line BE, so is the line BG to the line BD. Wherefore by conversion of proportion (by the corollary of the 19 of the fifth) as the line BT is to the line TE, so is the line BG to the line GD. As the line BG is to the line GD, so let the line TZ be to the line ZE by the corollary of the 10. of the sixth. Wherefore by the 11. of the fifth the line BT is to the line TE, as the line TZ is to the line ZE. For either of them are as the line BG is to the line GD. Wherefore the residue BZ is to the residue ZT, as the whole BT is to the whole TE by the 19 of the fifth. Wherefore by the 11. of the fifth the line BZ is to the line ZT as the line ZT is to the line ZE. Wherefore the line TZ is the mean proportional between the lines BZ and ZE. Wherefore the square of the first, namely, of the line BZ, is to the square of the second, namely, of the line ZT, as the first, namely, the line BZ, is to the third, namely, to the line ZE (by the corollary of the 20. of the sixth). And for that as the line BG is to the line GD, so is the line TZ to the line ZE: but as the line TZ is to the line ZE, so is the line BZ to the line ZT. Wherefore as the line BG is to the line GD, so is the line BZ to the line ZT (by the 11. of the fifth). Wherefore the lines BZ and ZT are commensurable in power only, as also are the lines BG and GD (which are the names of the residual line BD) by the 10. of this book. Wherefore the right lines BZ and ZE are commensurable in length, for we have proved that they are in the same proportion that the squares of the lines BZ and ZT are. And therefore (by the corollary of the 15. of this book) the residue BE (which is a rational line) is commensurable in length unto the same line BZ. Wherefore also the line BG (which is commensurable in length unto the line BE) shall also be commensurable in length unto the same line EZ (by the 12. of the tenth). And it is proved that the line RZ is to the line ZT commensurable in power only. Wherefore the right lines BZ and ZT are rational commensurable in power only. Wherefore the whole line BT is a binomial line (by the 36. of this book). And for that as the line BG is to the line GD, so is the line BZ to the line ZT: therefore alternately (by the 16. of the fifth) the line BG is to the line BZ, as the line GD is to the line ZT. But the line BG is commensurable in length unto the line BZ. Wherefore (by the 10. of this book) the line GD is commensurable in length unto the line ZT. Wherefore the names BG and GD of the residual line BD are commensurable in length unto the names BZ and ZT of the binomial line BT: and the line BZ is to the line ZT in the same proportion that the line BG is to the line GD as before it was more manifest. And that they are of one and the self same order is thus proved. If the greater or less name of the residual line, namely, the right lines BG or GD be commensurable in length to any rational line put: the greater name also or less, namely, BZ or ZT shallbe commensurable in length to the same rational line put by the 12. of this book. And if neither of the names of the residual line be commensurable in length unto the rational line put, neither of the names of the binomial line shallbe commensurable in length unto the same rational line put (by the 13. of the tenth). And if the greater name BG be in power more than the less name by the square of a line commensurable in length unto the line BG, the greater name also BZ shallbe in power more than the less by the square of a line commensurable in length unto the line BZ. And if the one be in power more by the square of a line incommensurable in length, the other also shallbe in power more by the square of a line incommensurable in length by the 14. of this book. The square therefore of a rational line. etc. which was required to be proved.    

¶ The 90. Theorem. The 114. Proposition. 
This is in a manner the converse of both the former propositions, jointly.  If a parallelogram be contained under a residual line & a binomial line, whose names are commensurable to the names of the residual line, and in the sel●e same proportion: the line which containeth in power that superficies is rational.  
SVppose that a parallelogram be contained under a residual line AB and a binomial line CD, and let the greater name of the binomial line be CE, and the less name be ED, and let the names of the binomial line, namely, CE and ED be commensurable to the names of the residual line, namely, to AF and F●, and in the self same proportion. And let the line which containeth in power that parallelogram be G. Then I say that the line G is rational. Take a rational line, namely, H. And unto the line CD apply a parallelogram equal to the square of the line H, 
Construction.  and making in breadth the line KL. Wherefore (by the 112. of the tenth) KL is a residual line, 
Demonstration.  whose names let be KM and ML, which are (by the same) commensurable to the names of the binomial line, that is to CE and ED, and are in the self same proportion. But by position the lines CE and ED are commensurable to the lines AF and FB, and are in the self same proportion. Wherefore (by the 12. of the tenth) as the line AF is to the line FB● so is the line KM to the line ML. Wherefore alternately (by the 16. of the fift) as the line AF is to the line KM, so is the line BF to the line LM. Wherefore the residue AB is to the residue KL, as the whole AF is to the whole KM. But the line AF is commensurable to the line KM, for either of the lines AF and KM is commensurable to the line CE. Wherefore also the line AB is commensurable to the line KL. And as the line AB is to the line KL, so (by the first of the sixth) is the parallelogram contained under the lines CD and AB to the parallelogram contained under the lines CD and KL. Wherefore the parallelogram contained under the lines CD and AB is commensurable to the parallelogram contained under the lines CD and KL. But the parallelogram contained under the lines CD and KL is equal to the square of the line H. Wherefore the parallelogram contained under the lines CD & AB is commensurable to the square of the line H. But the parallelogram contained under the lines CD and AB is equal to the square of the line G. Wherefore the square of the line H is commensurable to the square of the line G. But the square of the line H is rational. Wherefore the square of the line G is also rational. Wherefore also the line G is rational, and it containeth in power the parallelogram contained under the lines AB and CD. If therefore a parallelogram be contained under a residual line and a binomial line, whose names are commensurable to the names of the residual line, and in the self same proportion, the line which containeth in power that superficies, is rational: which was required to be proved.  

¶ Corollary.  
Hereby it is manifest, that a rational parallelogram may be contained under irrational lines.   

¶ An ot●●r 〈…〉 Flussas'.  
〈…〉 line ●D: whos● names A● and ●D let be commensurable in length unto the names of the residual line A●, which let be AF and FB. And let the li●e AE● be to the line ED● in the same proportion that the line AF is to the line F●. And let the right line ● contain in power the superficies D●. Then I say, tha● the li●e ● is a rational lin● 〈…〉 l●ne, 
Construction.  which l●● b●●: And upon the line ●● describe (by the 4●. of the first) a parallelogram equal to the square of the line ●● and making in breadth the line DC. Wherefore (by the ●12. of this book) CD is a residu●ll line● whose names (Which let be ●● and OD) shall be co●mensurabl● in length unto the names A● and ●D, and the line C o shall be unto the line OD, 
Demonstration.  in the same proportion that the line AE is to the line ED● But as the line A● is to the line ●D, so by supposition, is the line AF to the line FE. Wherefore as the line CO is to the line OD, so is the line AF to the line F●● Wherefore the lines CO and OD are commensurable with the lines A● and ●● (by the ●●. of this book). Wherefore the residue, namely, the line CD is to the residue, namely, to the line A●, as the line CO is to the line AF (by the 19 of the fifth). But it is proved, that the line CO is commensurable unto the line AF. Wherefore the line CD is commensurable unto the line AB. Wherefore (by the first of the sixth) the parallelogram CA is commensurable to the parallelogram D●. But the parallelogram ●● i● (by construction) rational (for it is equal to the square of the rational line ●). Wherefore the parallelogram ●D ●s also rationally Wher●fore the line ● which by supposition containeth in power the superficies ●D● is also rational. If therefore a parallelogram be contained etc.: which was required to be proved.    

¶ The 91. Theorem. The 115. Proposition. Of a medial line are produced infinite irrational lines, of which none is of the self same kind with any of those that were before.  
SVppose that A be a medial line. Then I say, that of the line A may be produced infinite irrational lines, of which none shall be of the self same kind with any of those that were before. Take a rational line B. And unto that which is contained under the lines A and B let the square of the line C be equal (by the 14. of the second) ● 
Demonstration.  Wherefore the line C is irrational. For a superficies contained under a rational line and an irrational line, is (by the Assumpt following the 38. of the tenth) irrational: and the line which containeth in power an irrational superficies, is (by the Assumpt going before the 21. of the tenth) irrational. And it is not one and the self same with any of those thirteen that were before. For none of the lines that were before applied to a rational line maketh the breadth medial. Again unto that which is contained under the lines B and C, let the square of D be equal. Wherefore the square of D is irrational. Wherefore also the line D is irrational and not of the self same kind with any of those that were before. For the square of none of the lines which were before, applied to a rational line, maketh the breadth the line C. In like sort also shall it so follow, if a man proceed infinitely. Wherefore it is manifest, that of a medial line are produced infinite irrational lines, of which none is of the self same kind with any of those that were before: which was required to be proved.  

another demonstration  

another demonstration.  Suppose that AC be a medial line. Then I say, that of the line AC may be produced infinite irrational lines, of which none shall be of the self same kind with any of those irrational lines before named. Unto the line AC and from the point A, draw (by the 11. of the firs●) a perpendicular line AB, and let AB be a rational line, and make perfect● the parallelogram BC. Wherefore BG is irrational, by that which was declared and proved (in manner of an Assumpt) in the end of the demonstration of the 38: and the line that containeth it i● power is also irrational. Let the line CD contain in power the superficies BC. Wherefore CD is irrational & not of the self same kind with any of those that were before: for the square of the line CD applied to a rational line, namely, AB, maketh the breadth a medial line, namely, AC. But the square of none of the foresaid lines applied to a rational line maketh the breadth a medial line. Again, make perfect the parallelogram ED. Wherefore the parallelogram ED is also irrational (by the said Assumpt in the end of the 98. his demonstration brie●ly proved) and the line which containeth it in power is irrationally let the line which containeth it in power be DF. Wherefore DF is irrational and not of the self same kind with any of the foresaid irrational lines. For the square of none of the foresaid irrational lines applied unto a rational line, maketh the breadth the line CD. Wherefore of a medial line are produced infinite irrational lines, of which none is of the self same kind with any of those that were before: which was required to be demonstrated.    

¶ The 92. Theorem. The 116. Proposition. Now let us prove that in square figures, the diameter is incommensurable in length to the side.  
SVppose that ABCD be a square, and let the diameter thereof be AC. Then I say that the diameter AC is incommensurable in length to the side AB. For if it be possible, 
Demonstration leading to an impossibility.  let it be commensurable in length. I say that then this will follow, that one and the self same number shall be both an even number & an odd number. It is manifest (by the 47. of the first) that the square of the line AC is double to the square of the line AB. And for that the line AC is commensurable in length to the line AB (by supposition), therefore the line AC hath unto the line AB that proportion that a number hath to a number (by the 5. of the tenth). Let the line AC have unto the line AB that proportion that the number EF hath to the number G. And let EF and G be the lest numbers that have one and the same proportion with them. Wherefore EF is not unity. For if EF be unity, and it hath to the number G that proportion that the line AC hath to the line AB, and the line AC is greater than the line AB. Wherefore unity EF is greater than the number G, which is impossible. Wherefore FE is not unity, wherefore it is a number. And for that as the square of the line AC is to the square of the line AB, so is the square number of the number EF, to the square number of the number G, for in each is the proportion of their sides doubled (by the corollary of the 20 of the sixth and 11. of the eight): and the proportion of the line AC to the line AB doubled, is equal to the proportion of the number EF to the number G, doubled, for as the line AC is to the line AB, so is the number EF to the number G. But the square of the line AC is double to the square of the line AB. Wherefore the square number produced of the number EF is double to the square number produced of the number G. Wherefore the square number produced of EF is an even number. Wherefore EF is also an even number. For if EF were an odd number, the square number also produced of it, should (by the 23. and 29. of the ninth) be an odd number. For if odd numbers how many soever be added together, and if the multitude of them be odd, the whole also shall be odd. Wherefore EF is an even number. Divide the number EF into two equal parts in H. And forasmuch as the numbers EF and G are the least numbers in that proportion, therefore (by the 24. of the seventh) they are prime numbers the one to the other. And EF is an even number. Wherefore G is an odd number. For if G were an even number, the number two should measure both the number EF and the number G (for every even number hath an half part by the definition) but these numbers EF & G are prime the one to the other. Wherefore it is impossible that they should be measured by two or by any other number besides unity. Wherefore G is an odd number. And forasmuch as the number EF is double to the number EH, therefore the square number produced of EF is quadruple to the square number produced of EH. And the square number produced of EF is double to the square number produced of G. Wherefore the square number produced of G is double to the square number produced of EH. Wherefore the square number produced of G is an even number. Wherefore also by those things which have been before spoken, the number G is an even number, but it is proved that it is an odd number, which is impossible. Wherefore the line AC is not commensurable in length to the line AB, wherefore it is incommensurable.  

another demonstration.  
We may by an other demonstration prove, that the diameter of a square is incommensurable to the side thereof. 
another demonstration leading to an impossibility.  Suppose that there be a square, whose diameter let be A and let the side thereof be B. Then I say that the line A is incommensurable in length to the line B. For if it be possible let it be commensurable in length. And again as the line A is to the line B so let the number EF be to the number G: and let them be the lest that have one and the same proportion with them: wherefore the numbers EF and G, are prime the one to the other. First I say that G is not unity. For if it be possible let it be unity. And for that the square of the line A is to the square of the line B as the square number produced of EF is to the square number produced of G (as it was proved in the ●ormer demonstration) but the square of the line A is double to the square of the line B. Wherefore the square number produced of EF is double to the square number produced of G. And by your supposition G is unity. Wherefore the square number produced of EF is the number two which is impossible. Wherefore G is not unity. Wherefore it is a number. And for that as the square of the line A is to the square of the line B, so is the square number produced of EF to the square number produced of G. Wherefore the square number produced of EF is double to the square number produced of G. Wherefore the square number produced of G. measureth the square number produced of EF. Wherefore also (by the 14. of the eight) the number G measureth the number EF: and the number G also measureth itself. Wherefore the number G measureth these numbers EF and G, when yet they are prime the one to the other: which is impossible. Wherefore the diameter A is not commensurable in length to the side B. Wherefore it is incommensurable: which was required to be demonstrated.   

another demonstration after Flussas.  
Suppose that upon the line AB be described a square whose diameter let be the line AC. Then I say that the side AB is incommensurable in length unto the diameter AC. Forasmuch as the lines AB and BC are equal, therefore the square of the line AC is double to the square of the line AB by the 47. of the first. Take by the 2. of the eight numbers how many soever in continual proportion from unity, and in the proportion of the squares of the lines AB and AC. Which let be the numbers D, E, F, G. And forasmuch as the first from unity namely E is no square number, for that it is a prime number, neither is also any other of the said numbers a square number except the third from unity and so all the rest leving one between, by the 10. of the ninth. Wherefore D is to E, or E to F, or F to G, in that proportion that a square number is to a number not square. Wherefore by the corollary of the 25. of the eight, they are not in that proportion the one to the other that a square number is to a square number. Wherefore neither also have the squares of the lines AB and AC (which are in the same proportion) that porportion that a square number hath to a square number. Wherefore by the 9 of this book their sides, namely, the side AB and the diameter AC are incommensurable in length the one to the other which was required to be proved.  
This demonstration I thought good to add, for that the former demonstrations seem not so full, and they are thought of some to be none of Theons, as also the proposition to be none of Euclides.   

Here followeth an instruction by some studious and skilful Graecian (perchance Theon) which teacheth us of farther use and fruit of these irrational lines.  
Seeing that there are found out right lines incommensurable in length the one to the ●ther, as the lines A and B, there may also be found out many other magnitudes having length and breadth (such as are plain superficieces) which shallbe incommensurable the one to the other. For if (by the 13. of the sixth) between the lines A and B there be taken the mean proportional line, namely C, then (by the second corollary of the 20. of the sixth) as the line A is to the line B, so is the figure described upon the line A to the figure described upon the line C, being both like and in like sort described, that is, whether they be squares (which are always like the one to the other), or whether they be any other like rectiline figures, or whether they be circles about the diameters A and C. For circles have that proportion the one to the other, that the squares of their diameters have (by the 2. of the twelfth). Wherefore (by the second part of the 10. of the tenth) the ●igures described upon the lines A and C being like and in like sort described are incommensurable the one to the other. Wherefore by this means there are found out superficieces incommensurable the one to the other. In like sort there may be found out figures commensurable the one to the other, if ye put the lines A and B to be commensurable in length the one to the other. And seeing that it is so, now let us also prove that even in soli●es also or bodies there are some commensurable the one to the other, and other some incommensurable the one to the other. For if from each of the squares of the lines A and B, or from any other rectiline figures equal to these squares be erected solids of equal altitude, whether those solids be composed of equidistant supersicieces, or whether they be pyramids or prisms, thos● solids s● erected shallbe in that proportion the one to the other that their bases are (by the 32. o● the eleventh and 5. and 6. of the twelfth) Howbeit there is no such proposition concerning prisms, And so if the bases of the solids b● commensurable the one to the other, the solids also shall be commensurable the one to the other, and if the bases be incommensurable the one to the other, the solids also shall be incommensurable the one to the other (by the 10. of the tenth). And if there be two circles A and B: and upon each of the circles be erected Cones or Cilinders of equal altitude, those Cones & Cilinders s●all be in that proportion the one to the other that the circles are, which are their bases (by the 11. of the twelfth): and so if the circles be commensurable the one to the other, the Cones and Cilinders also shall be commensurable the one to the other. But if the circles be incommensurable the one to the other, the Cones also and Cilinders shallbe incommensurable the one to the other, (by the 10. of the tenth). Wherefore it is manifest that not only in lines and super●icieces, but also in solids or bodies is found commensurabilitie or incommensurability.   

An advertisement by john Dee.  
Although this proposition were by Euclid to this book allotted, (as by the ancient Graecian published under the name of Aristoteles 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉, it would seem to be, and also the property of the same, agreeable to the matter of this book, and the proposition itself, so famous in Philosophy and Logic, as it was, would in manner crave his elemental place, in this tenth book) yet the dignity & perfection● of Mathematical Method can not allow it here: as in due order following: But most aptly after the 9 proposition of this book, as a corollary of the last part thereof. And undoubtedly the proposition hath for this 2000 years been notably regarded among the greek Philosophers: and before Aristotle's time was concluded with the very same inconvenience to the gaynesayer, that the first demonstration here induceth, namely, Odd number to be equal to even: as may appear● in Aristotle's work, named Analitica prima, the first book and 40. chapter. But else in very many places of his works he maketh mention of the proposition. Evident also it is that Euclid was about Aristotle's time, and in that age the most excellent Geometrician among the Greeks. Wherefore, seeing it was so public in his time, so famous, and so appertaining to the property of this book: it is most likely, both to be known to Euclid, and also to have been by him in apt order placed. But of the disordering of it, can remain no doubt, if ye consider in Zamberts' translation, two other propositions going next before it, so far misplaced, that where they are, word for word, before du●ly placed, being the 105. and 106. yet here (after the book ended), they are repeated with the numbers of 116. and 117. proposition. Zambert therein was more faithful to follow as he found in his greek example, than he was skilful or careful to do what was necessary. Yea and some greek written ancient copies have them not so: Though in deed they be well demonstrated, yet truth disorded, is half disgraced● especially where the pattern of good order, by profession is avouched to be. But through ignorance, arrogancy and ●emerltie of unskilful Method Masters, many things remain yet, in these Geometrical Elements, unduly tumbled in: though true, yet with disgrace: which by help of so many wits and ability of such, as now may have good cause to be skilful herein, will (I hope) ere long be taken away: and things of importance (wanting) supplied.    
The end of the tenth book of Euclides Elements   

¶ The eleventh book of Euclides Elements.  

The argument of the eleventh book.  HITHERTO HATH ●VCLID● IN TH●S● former books with a wonderful Method and order entreated of such kinds of figures superficial, which are or may be described in a superficies or plain. And hath taught and set forth their properties, natures, generations, and productions even from the first root, ground, and beginning of them: namely, from a point, which although it be indivisible, yet is it the beginning of all quantity, 
A point the beginning of all quantity continual.  and of it and of the motion and slowing thereof is produced a line, and consequently all quantity continual, as all figures plain and solid what so ever. Euclid therefore in his first book began with it, 
The method used by Euclid in the ten ●●●mer boots.  and from thence went he to a line, as to a thing most simple next unto a point, then to a superficies, and to angles, and so through the whole first book, 
●irst bo●●e.  he entreated of these most simple and plain grounds. In the second book he entreated further, 
Second ●●o●e.  and went unto more harder matter, and taught of divisions of lines, and of the multiplication of lines, and of their parts, and of their passions and properties. And for that rightlined ●igures are far distant in nature and property from round and circular figures, in the third book he instructeth the reader of the nature and condition of circles. 
Third boo●e.  In the fourth book he compareth figures of right lines and circles together, 
forth b●o●e.  and teacheth how to describe a figure of right lines with in or about a circle: and contrariwise a circle with in or about a rectiline figure. In the fifth book he searcheth out the nature of proportion (a matter of wonderful use and deep consideration), 
●iueth bo●●e.  for that otherwise he could not compare ●igure with figure, or the sides of figures together. For whatsoever is compared to any other thing, is compared unto it undoubtedly under some kind of proportion. Wherefore in the sixth book he compareth figures together, 
Sixth boo●e.  one to an other, likewise their sides. And for that the nature of proportion, can not be fully and clearly seen without the knowledge of number, wherein it is first and chiefly found: in the seventh, 
Seventh book●  eight, 
●ight boo●●.  and ninth books, 
Ninth book.  he entreat●th of number, & of the kinds and properties thereof. And because that the sides of solid bodies, for the most part are of such sort, that compared together, they have such proportion the one to the other, 
Tenth boo●e.  which can not be expressed by any number certain, and therefore are called irrational lines, he in the tenth book hath written & taught which line● are commensurable or incommensurable the one to the other, and of the diversity of kinds of irrational lines, with all the conditions & proprieties of them. And thus hath Euclid in these ten foresaid books, fully & most plenteously in a marvelous order taught, whatsoever seemed necessary, and requisite to the knowledge of all superficial figures, of what sort & form so ever they be. Now in these books following he entreateth of figures of an other kind, namely, of bodily figures: 
What is entrea●ea of in the fi●e boots following.  as of Cubes, Pyramids, Cones, Columns, Cilinders, Parallelipipedons. Spheres and such others● and showeth the diversity of them, the generation, and production of them, and demonstrateth with great and wonderful art, their proprieties and passions, with all their natures and conditions. He also compareth one o● them to an other, whereby to know the reason and proportion of the one to the other, chiefly of the five bodies which are called regular bodies. 
〈◊〉 ●●●ular bodies● the ●●all end 〈…〉 o● I u●●●●es ●eome●●●all elements.  And these are the things of all other entreated of in Geometry, most worthy and of greatest dignity, and as it were the end and final intent of the whole are of Geometry, and for whose cause hath been written, and spoken whatsoever hath hitherto in the former books been said or written. As the first book was a ground, and a necessary entry to all the r●st ●ollowing, so is this eleventh book a necessary entry and ground to the rest which follow. 
Comparison ●● the 〈◊〉 ●●o●e and 〈◊〉 book 〈◊〉.  And as that contained the declaration of words, and definitions of thinger requisite to the knowledge of superficial figures, and entreated of lines (and of their divisions and sections) which are the terms and limits of superficial figures: so in this book is set forth the declaration of words and definitions of things pertaining to solid and corporal figures: and also of superficieces which are the terms & limits of solids: moreover of the division and intersection of them, and divers other things, without which the knowledge of bodily and solid forms can not be attained unto. And first is set the definitions as followeth.  

Definitions. A solid or body is that which hath length, breadth, and thickness, 
First definition  and the term or limit of a solid is a superficies.  
There are three kinds of continual quantity, a line, a superficies, and a solid or body: the beginning of all which (as before hath been said) is a point, which is indivisible. Two of these quantities, namely, a line and a superficies, were defined of Euclid before in his first book. But the third kind, namely, a solid or body he there defined not, as a thing which pertained not then to his purpose: but here in this place he setteth the definition thereof, as that which chiefly now pertaineth to his purpose, and without which nothing in these things can profitably be taught. A solid (saith he) is that which hath length, breadth, and thickness, or depth. There are (as before hath been taught) three reasons or means of measuring, which are called commonly dimensions, namely, length, breadth, and thickness. These dimensions are ascribed unto quantities only. By these are all kinds of quantity defined, ●● are counted perfect or imperfect, according as they are partaker of fewer or more of them. As Euclid defined a line, ascribing unto it only one of these dimensions, namely, length: Wherefore a line is the imperfectest kind of quantity. In defining of a superficies, he ascribed unto it two dimensions, namely, length, and breadth: whereby a superficies is a quantity of greater perfection than is a line, but here in the definition of a solid or body. Euclid attributeth unto it all the three dimensions, length, breadth, and thickness. Wherefore a solid is the most perfectest quantity, 
A solid the most perfectest quantity.  which wanteth no dimension at all, passing a line by two dimensions, and passing a superficies by one. This definition of a solid is without any designation of ●orme or figure easily understanded, only conceiving in mind, or beholding with the eye a piece of timber or stone, or what matter so ever else, whose dimensions let be equal or unequal. For example let the length thereof be 5. inches, the breadth 4. and the thickness 2. if the dimensions were equal, the reason is like, and all one, as it is in a Sphere and in ● cube. For in that respect and consideration only, that it is long, broad, and thick, it beareth the name of a solid or body, ●nd hath the nature and properties thereof. There is added to the end● of the definition of a solid, that the term and limit of a solid ●s a superficies. Of things infinity there i● no Art or Science. All quantities therefore in this Art entreated of, are imagined to be finite, 
No science of things infinite.  and to have their ends and borders as hath been showed in the first book, that the limits and ends of a line are points, and the limits or borders of a superficies are lines, so now he saith tha● the ends, limits, or borders of a solide● are superficieces. As the side of any square piece of timber, or of a table, or die, or any other like, are the terms and limits of them.   

2 A right line is then erected perpendicularly to a pl 〈…〉 erficies, when the right line maketh right angles with all the lines 〈…〉 it, 
Second definition.  and are drawn upon the ground plain superficies.  
Suppose that upon the ground plain superficies, CDEF from the point B be erected a right line, namely, ●A, so that let the point A be a lo●e in the air. Draw also from the point ● in the plain superficies CDBF, as many right lines as ye list, as the lines BC, BD, ●●, BF, BG, HK, BH, and BL. If the erected line BA with all these lines drawn in the superficies CDEF make a right angle, so that all those ●ngles A●●, A●D, A●E, ABF● A●G, A●K, ABH, ABL, and so of others, be right angles, then by this definition, the line AB, i● a line ●●●cted upon the superficies CDEF: it is also called commonly a perpendicular line or a plumb line, unto or upon a superficies.   


Third definition.  3 A plain superficies is then upright or erected perpendicularly to a plain superficies, when all the right lines drawn in one of the plain superficieces unto the common section of those two plain superficieces, making therewith right angles, do also make right angles to the other plain superficies. Inclination or leaning of a right line, to a plain superficies, is an acute angle, contained under a right line falling from a point above to the plain superficies, and under an other right line, from the lower end of the said line (let down) drawn in the same plain superficies, by a certain point assigned, where a right line from the first point above, to the same plain superficies falling perpendicularly, toucheth.  
In this third definition are included two definitions: the first is of a plain superficies erected perpendicularly upon a plain superficies. 
Two dif●initions included in this definition.  The second is of the inclination or leaning of a right line unto a superficies: of the first take this example. Suppose ye have two super●icieces ABCD and CDEF. Of which let the superficies CDEF be a ground plain superficies, and let the superficies ABCD be erected unto it, and let the line CD be a common term or intersection to them both, that is, let it be the end or bound of either of them, 
Declaration of the first part.  & be drawn in either of them: in which line note at pleasure certain points, as the point G, H. From which points unto the line CD, draw perpendicular lines in the superficies ABCD, which let be GL and HK, which falling upon the superficies CDEF, if they 'cause right angles with it, that is, with lines drawn in it from the same points G and H, as if the angle LGM or the angle LGN contained under the line ●G drawn in the superficies erected, and under the GM or GN drawn in the ground superficies CDEF lying flat, be a right angle, then by this definition, the superficies ABCD is upright or erected upon the superficies CDEF. It is also commonly called a superficies perpendicular upon or unto a superficies.  

Declaration of the second part.  For the second part of this definition, which is of the inclination of a right line unto a plain superficies, take this example. Let ABCD be a ground plain superficies, upon which from a point being a fit, namely, the point E, suppose a right line to fall, which let be the line EG, touching the plain superficies ABCD at the point G. Again, from the point E, being the top or higher limit and end of the inclining line EG, let a perpendicular line fall unto the plain superficies ABCD, which let be the line EF, and let F be the point where EF toucheth the plain superficies ABCD. Then from the point of the fall of the line inclining upon the superficies unto the point of the falling of the perpendicular line upon the same superficies, that is, from the point G to the point F, draw a right line GF. Now by this definition, the acute angle EGF is the inclination of the line EG unto the superficies ABCD. Because it is contained of the inclining line, and of the right line drawn in the superficies, from the point of the fall of the line inclining to the point of the fall of the perpendicular line: which angle must of necessity be an acute angle. For the angle EFG is by construction a right angle, and three angles in a triangle are equ● 〈…〉 ●ight angles. Wherefore the other two angles, namely, the angles EGF, and GEF, are equ● 〈…〉 right angle. Wherefore either of them is less than a right angle. Wherefore the angle EGF is an 〈…〉 gle.   


Fourth definition.  4 Inclination of a plain superficies to a plain superficies, is an acute angle contained under the right lines, which being drawn in either of the plain superficieces to one & the self same point of the common section, make with the section right angles.  
Suppose that there be two superficieces ABCD & EFGH, and let the superficies ABCD be supposed to be erected not perpendicularly, but somewhat leaning and inclining unto the plain superficies EFGH, as much or as little as ye will: the common term or section of which two superficieces let be the line CD. From some one point, a● from the point M assigned in the common section of the two superficieces, namely, in the line CD, draw a perpendicular line in either superficies. In the ground superficies EFGH draw the line MK, and in the superficies ABCD draw the line ML. Now if the angle LMK be an acute angle, then is that angle the inclination of the superficies ABCD unto the superficies EFGH, by this definition, because it is contained of perpendicular lines drawn in either of the superficieces to one and the self same point being the common section of them both.   

5 Plain superficieces are in like sort inclined the on● 〈…〉 her, 
Fifth definition.  when the said angles of inclination are equal the one to the oh 〈…〉  
This definition needeth no declaration at all, but is most manifest by the definition last going before. For in considering the inclinations of divers superficieces to others, if the acute angles contained under the perpendicular lines drawn in them from one point assigned in each of their common sections be equal, as if to the angle LMK in the former example be given an other angle in the inclination of two other superficieces equal, then is the inclination of these superficieces like, and are by this definition said in like sort to incline the one to the other.  
Now also let there be an other ground plain superficies, namely, the superficies MNOP, unto whom also let lean and incline the superficies Q●●T, and let the common section or segment of them be the line QR. And draw in the superficies MNOP to some one point of the common section as to the point X the line VX, making with the common section right angles, namely, the angle VXR, or the angle VXQ: also in the superficies STQR draw the right line YX to the same point X in the common section, making therewith right angles, as the angle YX●, or the angle YXQ. Now (as saith the definition) if the angles contained under the right lines drawn in these superficieces & making right angles with the common section, be in the points, that is, in the points of their meeting in the common section, equal: then is the inclination of the superficieces equal. As in this example, if the angle LGH contained under the line LG being in the inclining superficies ●KEF and under the line HG being in the ground superficies ABCD, been equal, to the angle YXV contained under the line VX being in the ground superficies MNOP and under the line YX being in the inclining superficies STQR: then is the inclination of the superficies IKEF unto the superficies ABCD, like unto the inclination of the superficies STQR unto the superficies MNOP. And so by this definition these two superficieces are said to be in like sort inclined.   


Sixth definition.  6 Parallel plain superficieces are those, which being produced or extended any way never touch or concur together.  
Neither needeth this definition any declaration, but is very easy to be understanded by the definition of parallel lines: ●or as they being drawn on any part, never touch or come together: so parallel plain super●icieces are such, which admit no touch, that is, being produced any way infinitely never meet or come together.   


Seventh def●inition.  7 Like solid or bodily figures are such, which are contained under like plain superficieces, and equal in multitude.  
What plain super●icieces are called like, hath in the beginning of the sixth book, been sufficiently declared. Now when solid figures or bodies be contained under such like plain superficieces as there are defined, and equal in number, that is, that the one solid have as many in number as the other, in their sides and limits: they are called like solid figures, or like bodies.   


Eighth definition.  8 Equal and like solid (or bodily) figures are those which are contained under like superficieces, and equal both in multitude and in magnitude.  
In like solid figures it is sufficient, that the superficieces which contain them be like and equal in number only, but in like solid figures and equal, it is necessary that the like superficieces containing them, be also equal in magnitude. So that besides the likeness between them, they be (each being compared to his correspondent superficies) o● one greatness, and that their areas or fields be equal. When such super●icieces contain bodies or solids, then are such bodies equal and like solids or bodies.   


Ninth dissilition.  9 A solid or bodily angle, is an inclination of more than two lines to all the lines which touch themselves mutually, and are not in one and the self same superficies.  
Or else thus: A solid or bodily angle is that which is contained under more than two plain angles, not being in one and the self same plain superficies, but consisting all at one point.  
Of a solid angle doth Euclid here give two several definitions. The first is given by the concourse and touch of many lines. The second by the touch & concourse of many superficial angles. And both these definitions tend to one, and are not much different, for that lines are the limits and terms of superficieces. But the second given by super●iciall angles is the more natural definition, because that supe●ficieces a●e the next and immediate limits of bodies, and so are not lines. An example of a solid angle cannot well and at ●ully be given or described in a pla●●e superficies. But touching this first definition, lay before you a cube or a die, and consider any of the corners or angles thereof so shall ye see that at every angle there concur three lines (for two lines concurring cannot make a solid angle) namely, the line or edge of his breadth, of his length, and of his thickness, which their so inclining & concurring tovether, make a solid angle, and so of others. And now concerning the second definition, what superficial or plain angles be hath been taught before in the first book, namely, that it is the touch of two right lines. And as a super●iciall or plain angle is caused & contained of right lines, so si a solid angle caused & contained of plain superficial angles. Two right lines touching together, make a plain angle, but two plain angles joined together can not make a solid angle, but according to the definition, they must be more than two, as three, ●oure, ●iue, or mo●: which also must not be in one & the self same superficies, but must be in divers superficieces, ●eeting at one point. This definition is not hard, but may easily be conceived in a cube or a die, where ye see three angles of any three superficieces or sides of the die concur and meet together in one point, which three plain angles so joined together, make a solid angle. Likewise in a Pyrami● or a spi●e of a steeple or any other such thing, all the sides thereof tending upward narrower and narrower, at length end their angles (at the height or top thereof) in one point. So all their angles there joined together, make a solid angle. And for the better ●ig●t thereof, I have set here a figure whereby ye shall more easily conceive ●● the base of the figure is a triangle, namely, ABC, if on every side of the triangle ABC, ye raise up a triangle, as upon the side AB, ye raise up the triangle AFB, and upon the side AC the triangle AFC, and upon the side BC, the triangle BFC, and so bowing the triangles raised up, that their tops, namely, the points F meet and join together in one point, ye shall easily and plainly see how these three superficial angles AFBBFC, CFA, join and close together, touching the one the other in the point F, and so make a solid angle.   

10 A Pyramid is a solid figure contained under many plain superficieces set upon one plain superficies, and gathered together to one point. 
Tenth definition.   
Two superficieces raised upon any ground can not make a Pyramid, for that two superficial angles joined together in the top, cannot (as before is said) make a solid angle. Wherefore when three, four, five, or more (how many soever) superficieces are raised up from one superficies being the ground, or base, and ever ascending diminish their breadth, till at the length all their angles concur in one point, making there a solid angle: the solid enclosed, bounded, and terminated by these superficieces is called a Pyramid, as ye see in a taper of four sides, and in a spire of a tower which containeth many sides, either of which is a Pyramid.  
And because that all the superficieces of every Pyramid ascend from one plain superficies as from the base, and tend to one point, it must of necessity come to pass; that all the superficieces of a Pyramid are trianguler, except the base, which may be of any form or figure except a circle. For if the base be a circle, than it ascendeth not with sides, or divers superficieces, but with one round superficies, and hath not the name of a Pyramid, but is called (as hereafter shall appear) a Cone.   


Eleventh definition.  11 A prism is a solid or a bodily figure contained under many plain superficieces, of which the two superficieces which are opposite, are equal and like, and parallels, & all the other superficieces are parallelograms.  
Flussas' here noteth that Theon and Campane disagree in defining a Prism, and he preferreth the definition given of Campane before the definition given of Euclid (which because he may seem with out less offence to reject, he calleth it Theons definition) and following Campane he giveth an other definition, which is this.  

another definition of a prism, which is a special definition of a prism: as it is commonly called and used.  A Prism is a solid figure, which is contained under five plain superficieces, of which two are triangles, like, equal, and parallels, and the rest are parallelograms.  
The example before set agreeth likewise with this definition, and manifestly declareth the same. For in it were ●iue superficieces, the base, the two erected superficieces, and the two ends: of which the two ends are triangles like, equal and parallels, and all the other are parallelograms as this definition requireth. The cause why he preferreth the difinion of Campane before the definition of Theon (as he calleth it, but in very deed it is Euclides definition, as certainly, as are all those which are given of him in the former books, neither is there any cause at all, why it should be doubted in this one definition more than in any of the other) as he himself allegeth, is, for that it is (as he saith) to large, and comprehendeth many more kinds of solid figures besides prisms, as Columns having sides, and all Parallelipipedons, which a definition should not do: but should be convertible with the thing defined, and declare the nature of it only, and stretch no farther.  
Me ●hinketh Flussas aught not to have made so much a do in this matter, nor to have been so sharp in sight and so quick as to see and espy out such faults, which can of no man that will see rightly without affection be espied for such great faults. For it may well be answered that these faults which he noteth (if yet they be faults) are not to be found in this definion. It may be said that it extendeth itself not ●arther than it should, but declareth only the thing defined, namely, a Prism. Neither doth it agreed (as ●lussas cavilleth) with all Parallelipipedons and Columns having sides. All Parallelipipedons what so ever right angled, or not right angled which are described of equidistant sides or superficieces, have their sides opposite. So that in any of them there is no one side, but it hath a side opposite unto it. So likewise is it of even sided Columns, each hath his opposite side directly against it, which agreeth not with this definition of Euclid. Here it is evidently said, that of all the superficieces, the two which are opposite are equal, like, and parallels, meaning undoubtedly only two & no more. Which is manifest by that which followeth. The other (saith he) are parallelograms, signifying most evidently that none of the rest besides the two aforesaid, which are equal, like, and parallels, are opposite: but two of necessity are raised up, and concur in one common line, and the other is the base. So that it containeth not under it the figures aforesaid, that is sided Columns, & all Parallelipipedons, as Flussas hath not so advisedly noted.  
Again where Flussas setteth in his definition, as an essential part thereof, that of the five superficieces, of which a Prism is contained, two of them must be triangles, that undoubtedly is not of necessity, they may be of some other figure. Suppose that in the figure before given that in the place of the two opposite figures, which there were two triangl●s, were placed two pentagons: yet should the figure remain a Prism still, and agreed with the definition of Euclid, and falleth not under the definition of Flussas. So that his definition seemeth to be to narrow and stretcheth not so far as it aught to do, nor declareth the whole nature of the thing defined. Wherefore it is not to be preferred before Euclides definition, as he would have it. This figure of Euclid called a Prism, is called of Campane and certain others Figura Serr●tilis, for that it repres●teth in some manner the form of a Saw. 
This body called Figura Serratilis.  And of some others it is called Cuneus, that is, a Wedge, because it beareth the figure of a wedge.  
Moreover although it were so, that the definition of a Prism should be so large, that it should contain all these figures noted of Flussas as sided Columns, & all Parallelipipedons: yet should not Flussas have so great a cause to find so notably a fault, so utterly to reject it. It is no rare thing in all learnings, chiefly in the mathematicals, to have one thing more general than an other. Is it not true that every Isosceles is a triangle, but not every triangle is an Isosceles? And why may not likewise a Prism be more general, than a Parallelepipedon, or a Column having sides (and contain them under it as a triangle containeth under it an Isosceles and other kinds of triangles). So that every Prallelipipedon, or every sided Column be a Prism, but not every Prism a Parallelipipedom or a sided Column. This aught not to be so much offensive. And indeed it seemeth manifestly of many, yea & of the learned so to be taken, as clearly appeareth by the words of Psellus in his Epitome of Geometry, where he entreateth of the production and constitution of these bodies. His words are these. 
Psellus.  All rectiline figures being erected upon their plains or bases by right angles, make prisms. Who perceiveth not but that a Pentagon erected upon his base of ●iue sides maketh by his motion a sided Column of five sides? Likewise an Hexagon erected at right angles produceth a Column having six sides: and so of all other rectillne figures. All which solids or bodies so produced, whether they be sided Columns or Parallelipipedons, be here in most plain words (of this excellent and ancient Greek author Psellus) called prisms. Wherefore if the definition of a Prism given of Euclid should extend itself so largely as Flussas imagineth, and should include such figures or bodies, as he noted: he aught not yet for all that so much to be offended, and so narrowly to have sought faults. For Euclid in so defining might have that meaning & sense of a Prism which Psellus had. So ye see that Euclid may be defended either of these two ways, either by that that the definition extendeth not to these figures, and so not to be over general nor stretch farther than it aught: or else by that that if it should stretch so far it is not so heinous. For that as ye see many have tak● it in that sense. In deed commonly a Prism is taken in that signification and meaning in which Campa●●● Flussas and others take it. In which sense it seemeth also that in divers propositions in these books following it aught of necessity to be taken.   

12 A Sphere is a figure which is made, 
twelfth definition.  when the diameter of a semicircle abiding fixed, the semicircle is turned round about, until it return unto the self same place from whence it began to be moved.  
To the end we may fully and perfectly understand this definition, how a Sphere is produced of the motion of a semicircle, it shall be expedient to consider how quantities Mathematically are by imagination conceived to be produced, by flowing and motion, as was somewhat touched in the beginning of the first book. Ever the less quantity by his motion bringeth for●h the quantity next above it. As a point moving, flowing, or gliding, bringeth forth a line, which is the first quantity, and next to a point. A line moving produceth a superficies, which is the second quantity, and next unto a line. And last of all, a superficies moving bringeth forth a solid or body, which is the third & last quantity. These things well marked, it shall not be very hard to attain to the right understanding of this definition. Upon the line AB being the diameter, describe a semicircle ACB, whose centre let be D: the diameter AB being fixed on his ends or● points, imagine the whole superficies of the semicircle to move round from some one point assigned, till it return to the same point again. So shall it produce a perfect Sphere or Globe, the form whereof you see in a ball or bowl. And it is fully round and solid, for that it is described of a semicircle which is perfectly round, as our country man johannes de Sacro Busco in his book of the Sphere, 
What is to be ta●●n heed of in the definition of a sphere given by johannes de Sacro Busco.  of this definition which he taketh out of Euclid, doth well collect. But it is to be noted and taken heed of, that none be deceived by the definition of a Sphere given by johannes de Sacro Busco: A Sphere (saith he) is the passage or moving of the circumference of a semicircle, till it return unto the place where it began, which agreeth not with Euclid. Euclid plainly saith, that a Sphere is the passage or motion of a semicircle, and not the passage or motion of the circumference of a semicircle: neither can it be true that the circumference of a semicircle, which is a line, should describe a body. It was before noted that every quantity moved, describeth and produceth the quantity next unto it. Wherefore a line moved can not bring forth a body, but a superficies only. As if ye imagine a right line fastened at one of his ends to move about from some one point till it return to the same again, it shall describe a plain superficies, namely, a circle. So also if ye likewise conceive of a crooked line, such as is the circumference of a semicircle, that his diameter fastened on both the ends it should move from a point assigned till it return to the same again, it should describe & produce a ●ound superficies only, which is the superficies and limit of the Sphere, and should not produce the body and solidity of the Sphere. But the whole semicircle, which is a superficies, by his motion, as is before said, produceth a body, that is, a perfect Sphere. So see you the error of this definition of the author of the Sphere: which whether it happened by the author himself, which I think not: or that that particle was thrust in by some one after him, which is more likely, it it not certain. But it is certain, that it is unaptly put in, and maketh an untrue definition: which thing is not here spoken, any thing to derogate the author of the book, which assuredly was a man of excellent knowledged neither to the hindrance or diminishing of the worthiness of the book, which undoubtedly is a very necessary book, than which I know none more mere to be taught and read in schools touching the grounds and principles of Astronomy and Geography: but only to admonish the young and unskilful reader of not falling into error. 
Theodosiu● definition of a sphere.  Theodosius in his book De Sphericis (a book very necessary for all those which will see the grounds and principles of Geometry and Astronomy, which also I have translated into our vulgar tongue, ready to the press) defineth a Sphere after this manner: A Sphere is a solid or body contained under one superficies, in the middle whereof there is a point, from which all lines drawn to the circumference are equal. This definition of Theodosius is more essential and natural, then is the other given by Euclid. The other did not so much declare the inward nature and substance of a Sphere, as it showed the industry and knowledge of the producing of a Sphere, and therefore is a causal definition given by the cause efficient, or rather a description then a definition. But this definition is very essential, declaring the nature and substance of a Sphere. As if a circle should be thus defined, as it well may: A circle is the passage or moving of a line from a point till it return to the same point against it is a causal definition, showing the efficient cause whereof a circle is produced, namely, of the motion of a line. And it is a very good description fully showing what a circle is. Such like description is the definition of a Sphere given o● Euclide ● by the motion of a semicircle. But when a circle is defined to be a plain superficies, in the midst whereof is a point, from which all lines drawn to the circumference thereof, are equal: this definition is essential and formal, and declareth the very nature of a circle. And unto this definition of a circle, is correspondent the definition of a Sphere given by Theodosius, saying: that it is a solid o● body, in the midst, whereof there is a point, from which all the lines drawn to the circumference are equal. So see you the affinity between a circle and a Sphere. For what a circle is in a plain, that is a Sphere in a Solid. The fullness and content of a circle is described by the motion of a line moved about: but the circumference thereof, which is the limit and border thereof, is described of the end and point of the same line moved about. So the fullness, content, and body of a Sphere or Globe is described of a semicircle moved about. But the Spherical superficies, which is the limit and border of a Sphere, 
The circumference of a sphere.  is described of the circumference of the same semicircle moved about. And this is the superficies meant in the definition, when it is said, that it is contained under one superficies, which superficies is called of johannes de ●acro Busco & others, the circumference of the Sphere.  

Galens' definition 〈◊〉 a sph●r●.  Galene in his book de diffinitionibus medici● ● giveth yet an other definition of a Sphere, by his property or common accidence of moving, which is thus. A Sphere is a figure most apt to all motion, as having no base whereon th' stay. This is a very plain and witty definition, declaring the dignity thereof above all figures generally. 
The dignity of a sphere.  All other bodies or solids, as Cubes, Pyramids, and others have sides, bases, and angles, all which are stays to rest upon, or impediments and lets to motion. But the Sphere having no side or base to stay one, nor angle to let the course thereof, but only in a point touching the plain wherein 〈◊〉 standeth, moveth freely and fully with out let. And for the dignity and worthiness thereof, this circular and Spherical motion is attributed to the heavens, which are the most worthy bodies. Wherefore there is ascribed unto them this chief kind of motion. This solid or bodily figure is also commonly called a Globe. 
A sphere called a Globe.    

13 The axe of a Sphere is that right line which abideth fixed, about which the semicircle was moved. 
thirteenth definition.   
As in the example before given in the definition of a Sphere, the line AB, about which his ends being fixed, the semicircle was moved (which line also yet remaineth after the motion ended) is the axe of the Sphere described of that semicircle. Theodosius defineth the axe of a Sphere after this manner. 
Theodosius definition of the axe of a sphere.  The axe of a Sphere is a certain right line drawn by the centre, ending on either side in the superficies of the Sphere, about which being fixed the Sphere is turned. As the line AB in the former example. There needeth to this definition no other declaration, but only to consider, that the whole Sphere turneth upon that line AB, which passeth by the centre D, and is extended one either side to the superficies of the Sphere, wherefore by this definition of Theodosius it is the axe of the Sphere.   

14 The centre of a Sphere is that point which is also the centre of the semicircle. 
fourteenth definition.   
This definition of the centre of a Sphere is given as was the other definition of the axe, namely, having a relation to the definition of a Sphere here given of Euclid: where it was said that a Sphere is made by the revolution of a semicircle, whose diameter abideth fixed. The diameter of a circle and of a semicrcle is all one. And in the diameter either of a circle or of a semicircle is contained the centre of either of them, for that they diameter of each ever passeth by the centre. Now (saith Euclid) the point which is the centre of the semicircle, by whose motion the Sphere was described, is also the centre of the Sphere. As in the example there given, the point D is the centre both of the semicircle & also of the Sphere. Theodosius giveth as other definition of the centre of a Sphere which is thus. 
Theodosius definition of the centre of a sphere.  The centre of a Sphere is a point with in the Sphere, from which all lines drawn to the superficies of the Sphere are equal. As in a circle being a plain figure there is a point in the midst, from which all lines drawn to the circumfrence are equal, which is the centre of the circle: so in like manner with in a Sphere which is a solid and bodily figure, there must be conceived a point in the midst thereof, from which all lines drawn to the superficies thereof are equal. And this point is the centre of the Sphere by this definition of Theodosius. Flussas in defining the centre of a Sphere comprehendeth both those definitions in one, after this sort. The centre of a Sphere is a point assigned in a Sphere, from which all the lines drawn to the superficies are equal, and it is the same which was also the centre of the semicircle which described the Sphere. 
Flussas' definition of the centre of a sphere.  This definition is superfluous and containeth more than needeth. For either part thereof is a full and sufficient definition, as before hath been showed. Or else had Euclid been insufficient for leaving out the one part, or Theodosius for leaving out the other. Peradventure Flussas did it for the more explication of either, that the one part might open the other.   

15 The diameter of a Sphere is a certain right line drawn by the centre, and one each side ending at the superficies of the same Sphere. 
fifteenth definition   
This definition also is not hard, but may easily be couceaved by the definition of the diameter of a circle. For as the diameter of a circle is a right line drawn from one side of the circumfrence of a circle to the other, passing by the centre of the circle: so imagine you a right line to be drawn from one side of the superficies of a Sphere to the other, passing by the centre of the Sphere, 
Difference between the diameter & axe of a sphere.  and that line is the diameter of the Sphere. So it is not all one to say, the axe of a Sphere, and the diameter of a Sphere. Any line in a Sphere drawn from side to side by the centre is a diameter. But not every line so drawn by the centre is the axe of the Sphere, but only one right line about which the Sphere is imagined to be moued● So that the name of a diameter of a Sphere is more general, then is the name of an axe. For every axe in a Sphere is a diameter of the same: but not every diameter of a Sphere is an axe of the same. And therefore Flussas setteth a diameter in the definition of an axe as a more general word ●n this manner. The axe of a Sphere, is that fixed diameter above which the Sphere is moved. A Sphere (as also a circle) may have infinite diameters, but it can have but only one axe.   


seventeenth definition.  16 A cone is a solid or bodily figure which is made, when one of the sides of a rectangle triangle, namely, one of the sides which contain the right angle, abiding fixed, the triangle is moved about, until it return unto the self same place from whence it began first to be moved. Now if the right line which abideth fixed be equal to the other side which is moved about and containeth the right angle: then the cone is a rectangle cone. But if it be less, then is it an obtuse angle cone. And if it be greater, than is it an a cuteangle cone,  
If the right line which abideth fixed, be equal to the other side which moveth ro●●d about, and containeth the right angle, than the Cone is a rectangle Cone.  

First kind of Cones.  As suppose in the former example, that the line AB which is fixed, and about which the triangle was moved, and after the motion yet remaineth, be equal to the line BC, which is the other line containing the right angle, which also is moved about together with the whole triangle● than is the Cone described, as the Cone ADC in this example, a right angled Cone: so called for that the angle at the top of the Cone is a right angle. For forasmuch as the lines AB and BC of the triangle ABC are equal, the angle BAC is equal to the angle BCA (by the 5. of the first). And each of them is the half of the right angle ABC (by the 32. of the first). In like sort may it be showed in the triangle ABD, that the angle ●DA is equal to the angle ●AD, and that each of them is the half of a right angle. Wherefore the whole angle CAD, which is composed of the two half right angles, namely, DA● and CA● is a right angle. And so have ye what is a right angled Cone.  
This figure of a Cone is of Campane, of Vitellio, and of others which have written in these latter times, called a round Pyramid, 
A Cone called of Campane a ro●●de pyramis.  which is not so aptly. For a Pyramid, and a Cone, are far distant, & of sundry natures. A Cone is a regular body produced of one circumvolution of a rectangle triangle, and limited and bordered with one only round superficies. But a Pyramid is terminated and bordered with divers superficieces. Therefore can not a Cone by any just reason bear the name of a Pyramid. This solid of many is called Turbo, which to our purpose may be Englished a Top or Ghyg: and moreover, peculiarly Campane calleth a Cone the Pyramid of a round Column, namely, of that Column which is produced of the motion of a parallelogram (contained of the lines AB and BC) moved about, the line AB being fixed. Of which Columns shall be showed hereafter.   

17 The axe of a Cone is that line, which abideth fixed, about which the triangle is moved. 
seventeenth definition.  And the base of the Cone is the circle which is described by the right line which is moved about.  
As in the example the line AB is supposed to be the line about which the right angled triangle ABC (to the production of the Cone) was moved: and that line is here of Euclid called the axe of the Cone described. The base of the Cone is the circle which is described by the right line which is moved about. As the line AB was fixed and slayed, so was the line BC (together with the whole triangle ABC) moved and turned about. A line moved, as hath been said before, produceth a superficies: and because the line BC is moved about a point, namely, the point B, being the end of the axe of the Cone AB, it produceth by his motion, and revolution a circle, which circle is the base of the Cone: as in this example, the circle CDE.  
The line which produceth the base of the Cone, is the line of the triangle which together with the axe of the Cone containeth the right angle. The other side also of the triangle, namely, the line AC, is moved about also with the motion of the triangle, which with his revolution describeth also a superficies, 
A conical superficies.  which is a round superficies, & is erected upon the base of the Cone, & endeth in a point, namely, in the higher part or top of the Cone. And it is commonly called a conical superficies.   


eighteenth definition.  18 A cylinder is a solid or bodily figure which is made, when one of the sides of a rectangle parallelogram, abiding fixed, the parallelogram is moved about, until it return to the self same place from whence it began to be moved.  
This definition also is of the same sort and condition, that the two definition● before geum were, namely, the definition of a Sphere and the definition of a Cone. For all are given by moving of a superficies about a right line fixed, the one of a semicircle about his diameter, the other of a rectangle triangle about one of his sides And this solid or body here defined is caused of the motion of a rectangle parallelogram having one of his sides containing the right angle fixed from some one point till it return to the same again where it began. As suppose ABCD to be a rectangle parallelogram, having his side AB fastened, about which imagine the whole parallelogram to be turned, till it return to the point where it began, then is that solid or body, by this motion described, a Cylinder: which because of his roundness can not at full be described in a plain superficies, yet have you for an example thereof a sufficient designation thereof in the margin 〈◊〉 as in a plain may be. If you will perfectly behold the form of a cilinder. Consider a round pillar that is perfectly round.   


Ninetenth definition.  19 The axe of a cilinder is that right line which abideth fixed, about which the parallelogram is moved. And the bases of the cilinder are the circles described of the two opposite sides which are moved about.  
Even as in the description of a Sphere the line fastened was the axe of the Sphere produced: and in the description of a cone, the line fastened was the axe of the cone brought forth: so in this description of a cilinder the line abiding, which was fixed, about which the rectangle parallelogram was moved is the axe of that cilinder. As in this example is the line AB. The bases of the cilinder ●●c. In the revolution of a parallelogram only one side is fixed, therefore the three other sides are moved about: of which the two sides which with the axe make right angles, and which also are opposite sides, in their motion describe each of them a circle, which two circles are called the bases of the cilinder. As ye see in the figure before put two circles described of the motion of the two opposite lines AD and BC, which are the bases of the Cilinder.  

A cillindricall superficies.  The other line of the rectangle parallelogram moved, by his motion describeth the round superficies about the Cilinder. As the third line or side of a rectangle triangle by his motion described the round Conical superficies about the Cone. And as the circumference of the semicircle described the round spherical superficies about the Sphere. In this example it is the superficies described of the line DC.  

Corollary.  By this definition it is plain that the two circles, or bases of a cilinder are ever equal and parallels: for that the lines moved which produced them remained always equal and parallels. Also the axe of a cilinder is ever an erected line unto either of the bases. For with all the lines described in the bases, and touching it, it maketh right angles,  

A round● Column or sphere.  Campane, Vitell●o, with other later writers, call this solid or body a round Column● or pillar. And Campane addeth unto this definition this, as a corollary. That of a round Column, of a Sphere, and of a circle the centre is one and the self same. 
A Corollary added by Campane.  That is (as he himself declareth it & proveth the same) where the Column, the Sphere, and the circle have one diameter.   

20 Like cones and cilinders are those, 
Twenty definition.  whose axes and diameters of their bases are proportional  
The similitude of cones and cilinders standeth in the proportion of those right lines, of which they have their original and spring. For by the diameters of their bases is had their length and breadth, and by their axe is had their height or deepness. Wherefore to see whether they be like or unlike, ye must compare their axes together, which is their depth, and also their diameters together, which is their length & breadth. As if the axe ●G of the cone ABC be to to the axe EI of the cone DEF, as the diameter AC of the cone ABC is to the diameter DF of the cone DEF, then a●e the cones ABC and DEF like cones. Likewise in the cilinders. If the axe LN of the cilinder LHMN have that proportion to the axe OQ of the cilinder ROPQ, which the diameter HM hath to the diameter RP: then are the cilinders HLMN and ROPQ like cilinders, and so of all others.   

21 A Cube is a solid or bodily figure contained under six equal squares. 
Twenty one definition.   
As is a die which hath six sides, and each of them is a full and perfect square, as limits or borders under which it is contained. And as ye may conceive in a piece of timber containing a foot square every way, or in any such like. So that a Cube is such a solid whose three dimensions are equal, the length is equal to the breadth thereof, and each of them equal to the depth. Here is as it may be in a plain superficies set an image thereof, in these two figures whereof the first is as it is commonly described in a plain, the second (which is in the beginning of the other side of this leaf) is drawn as it is described by art upon a plain superficies to show somewhat bodilike. And in deed the latter description is for the sight better than the first. But the first for the demonstrations of Euclides propositions in the five books following is of more use, for that in it may be considered and seen all the fix sides of the Cube. And so any lines or sections drawn in any one of the six sides. Which can not be so well seen in the other figure described upon a playnd. And as touching the first figure (which is set at the end of the other side of this leaf) ye see that there are six parallelograms which ye must conceive to be both equilater and rectangle, although in deed there can be in this description only two of them rectangle, they may in deed be described all equilater. Now if ye imagine one of the six parallelograms, as in this example, the parallelogram ABCD to be the base lying upon a ground plain superfices. And so conceive the parallelogram EFGH to be in the top over it, in such sort, that the lines AE, CG, DH, & BF may be erected perpendicularly from the points A, C, B, D, to the ground plain superficies or square ABCD. For by this imagination this figure will show unto you bodilike. And this imagination perfectly had, will make many of the propositions in these five books following, in which are required to be described such like solids (although not all cubes) to be more plainly and easily conceived.  
In many examples of the Greek and also of the Latin, there is in this place set the definition of a Tetrahedron, which is thus.   


Twenty two definition.  22 A Tetrahedron is a solid which is contained under four triangles equal and equilater.  
A form of this solid ye may see in these two examples here set, whereof one is as it is commonly described in a plain. Neither is it hard to conceive. For (as we before taught in a Pyramid) if ye imagine the triangle BCD to lie upon a ground plain superficies, and the point A to be pulled up together with the lines AB, AC, and AD, ye shall perceive the form of the Tetrahedron to be contained under 4. triangles, which ye must imagine to be all four equilater and equiangle, though they can not so be drawn in a plain. And a Tetrahedron thus described, is of more use in these five books following, then is the other, although the other appear in form to the eye more bodilike. 
A Tetrahedron one of the five regular bodies.   
Why this definition is here left out both of Campane and of Flussas, I can not but marvel, considering that a Tetrahedron, is of all Philosophers counted one of the five chief solids which are here defined of Euclid, which are called commonly regular bodies, without mention of which, the entreaty of these should seem much maimed: unless they thought it sufficiently defined under the definition of a Pyramid, 
Difference between a Tetrahedron and a pyramis.  which plainly and generally taken, includeth in deed a Tetrahedron, although a Tetrahedron properly much differeth from a Pyramid, as a thing special or a particular, from a more general. For so taking it, every Tetrahedron is a Pyramid, but not every Pyramid is a Tetrahedron. By the general definition of a Pyramid, the superficieces of the sides may be as many in number as ye list, as 3.4. 5.6. or more, according to the form of the base, whereon it is set, whereof before in the definition of a Pyramid were examples given. But in a Tetrahedron the superficieces erected can be but three in number according to the base thereof, which is ever a triangle. Again, by the general definition of a Pyrami●, the superficieces erected may ascend as high as ye list, but in a Tetrahedron they must all be equal to the base. Wherefore a Pyramid may seem to be more general than a Tetrahedron, as before a Prism seemed to be more general than a Parallelipipedon, or a sided Column: so that every Parallelipipedon is a Prism, but not every Prism is a Parallelipipedon. And every axe in a Sphere is a diameter: but not every diameter of a Sphere is the axe thereof. So also noting well the definition of a Pyramid, every Tetrahedron may be called a Pyramid, but not every Pyramid a Tetrahedron. And in deed Psellus in numbering of these five solids or bodies, calleth a Tetrahedron a Pyramid in manifest words. 
Psellus calleth a Tetrahedron a pyramis.  This I say might make Flussas & others (as I think it did) to omit the definition of a Tetrahedron in this place, as sufficiently comprehended within the definition of a Pyramid given before. But why then did he not count that definition of a Pyramid faulty, for that it extendeth itself to large, and comprehendeth under it a Tetrahedron (which differeth from a Pyramid by that it is contained of equal triangles) as he not so advisedly did before the definition of a Prism.   


Twenty three definition.  23 An Octohedron is a solid or bodily figure contained under eight equal and equilater triangles.  
As a Cube is a solid figure contained under six superficial figures of four sides or squares which are equilater, equiangle, and equal the one to the other: so is an Octohedron a solid figure contained under eight triangles which are equilater and equal the one to the other. As ye may in these two figures here set behold. Whereof the first is drawn according as this solid is commonly described upon a plain superficies. The second is drawn as it is described by art upon a plain, to show bodilike. And in deed although the second appear to the eye more bodilike, yet as I before noted in a Cube, for the understanding of divers Propositions in these five books following, is the first description of more use yea & of necessity. For without it, ye can not conceive the draft of lines and sections in any one of the eight sides which are sometimes in the descriptions of some of those Propositions required. Wherefore to the consideration of this first description, imagine first that upon the upper face of the superficies of the parallelogram ABCD, be described a Pyramid, having his four triangles AFB, AFC, CFD, and DFB, equilater and equiangle, and concurring in the point F. Then conceive that on the lower face of the superficies of the former parallelogram be described an other Pyramid, having his four triangles AEB, AEC, CED, & DEB, equilater and equiangle, and concurring in the point E. For so although somewhat grossly by reason the triangles can not be described equilater, you may in a plain perceive the form of this solid, and by that means conceive any lines or sections required to be drawn in any of the said eight triangles which are the sides of that body.   

24 A Dodecahedron is a solid or bodily figure contained under twelve equal, equilater, 
Twenty ●o●er definition.  and equiangle Pentagons'.  
As a Cube, a Tetrahedron, and an Octohedron, are contained under equal plain figures, a Cube under squares, the other two under triangles: so is this solid figure contained under twelve equilater, equiangle, and equal Pentagons', or figures of five sides. As in these two figures here set you may perceive. Of which the first (which thing also was before noted of a Cube, a Tetrahedron, and an Octohedron) is the common description of it in a plain, the other is the description of it by art upon a plain to make it to appear somewhat bodilike. The first description in deed is very obscure to conceive, but yet of necessity it must so, neither can it otherwise, be in a plain described to understand those Propositions of Euclid in these five books a following which concern the same. For in it although rudely, may you see all the twelve Pentagons', which should in deed be all equal, equilater, and equiangle. And now how you may somewhat conceive the first figure described in the plain to be a body. Imagine first the Pentagon ABCDE ●o be upon a ground plain superficies, then imagine the Pentagon FGHKL to be on high opposite unto the Pentagon ABCDE. And between those two Pentagons' there will be ten Pentagons' pulled up, five from the five sides of the ground Pentagon, namely, from the side AB the Pentagon ABONM, from the side BC the Pentagon BCQPO, from the side CD the Pentagon CDSRQ, from the side DE, the Pentagon DEVTS, from the side EA the Pentagon EAMXV, the other five Pentagons' have each one of their sides common with one of the sides of the Pentagon FGHKL, which is opposite unto the Pentagon in the ground superficies: namely, these are the other five Pentagons' FGNMX, GHPON, HKRQP, KLRST, LFXVT. So here you may behold twelve Pentagons', which if you imagine to be equal, equilater, & equiangle, and to be lifted up, ye shall (although somewhat rudely) conceive the bodily form of a Pentagon. And some light it will give to the understanding of certain Propositions of the five books following concerning the same.   


Twenty five definition.  25 An Icosahedron is a solid or bodily figure contained under twenty equal and equilater triangles.  
These ●iue solids now last defined, namely, a Cube, a Tetrahedron, an Octohedron, a Dodecahedron and an Icosahedron are called regular bodies. 
Five regular bodies.  As in plain superficieces, those are called regular figures, whose sides and angles are equal, as are equilater triangles, equilater pentagons, hexagons, & such like, so in solids such only are counted and called regular, which are comprehended under equal plain superficieces, which have equal sides and equal angles, as all these five foresaid have, as manifestly appeareth by their definitions, which were all given by this propriety of equality of their superficieces, which have also their sides and angles equal. And in all the course of nature there are no other bodies of this condition and perfection, but only these five. Wherefore they have ever of the ancient Philosophers been had in great estimation and admiration, and have been thought worthy of much contemplation, about which they have bestowed most diligent study and endeavour to search out the natures & properties of them. They are as it were the end and perfection of all Geometry, for whose sake is written whatsoever is written in Geometry. They were (as men say) first invented by the most witty Pythagoras then afterward set forth by the divine Plato, and last of all marvelously taught and declared by the most excellent Philosopher Euclid in these books following, and ever since wonderfully embraced of all learned Philosophers. 
The dignity of these bodies.  The knowledge of them containeth infinite secrets of nature. Pithag●ras, Timeus and Plato, by them searched out the composition of the world, with the harmony and preservation thereof, and applied these ●iue solids to the simple parts thereof, the Pyramid, or Tetrahedron they ascribed to the ●ire, 
A Tetrahedron ascribed unto the fire.  for that it ascendeth upward according to the figure of the Pyramid. To the air they ascribed the Octohedron, 
An octohedron ascribed unto the air.  for that through the subtle moisture which it hath, it extendeth itself every way to the one side, and to the other, according as that figure doth. Unto the water they assigned the Ikosahedron, for that it is continually flowing and moving, 
An Ikosahedron assigned unto the water.  and as it were making angle● 〈…〉 ●ide according to that figure. And to the earth they attributed a Cube, 
A cube assigned unto the earth.  as to a thing stable● 〈◊〉 and sure as the figure signifieth. Last of all a Dodecahedron, 
A dodecahedron assigned to heaven.  for that it is made of P●ntago●, whose angles are more ample and large than the angles of the other bodies, and by that ●ea●●● draw more ●● roundness, 〈◊〉 & to the form and nature of a sphere, they assigned to a sphere, namely, 〈…〉. Who so will 〈…〉 in his Tineus, shall ●ead of these figures, and of their mutual proportion●●●raunge ma●ter●, which h●re are not to be entreated of, this which is said, shall be sufficient for the 〈◊〉 of them and for th● declaration of their definitions.  
After all these definitions here set of Euclid, Flussas hath added an other definition, which 〈◊〉 of a Parallelipipedon, which because it hath not hitherto of Euclid in any place been defined, and because it is very good and necessary to be had, I thought good not to omit it, thus it is.   

A parallelipipedon is a solid figure comprehended under four plain quadrangle figures, 
Definition of a parallelipipedon.  of which those which are opposite are parallels.  
Because these five regular bodies here defined are not by these figures here set, so fully and lively expressed, that the studious beholder can thoroughly according to their definitions conceive them. I have here given of them other descriptions drawn in a plain, by which ye may easily attain to the knowledge of them. For if ye draw the like forms in matter that will bow and give place, as most aptly ye may do in fine pasted paper, such as pastwives make women's pastes of, & then with a knife cut every line finely, not through, but half way only, if then ye bow and bend them accordingly, ye shall most plainly and manifestly see the forms and shapes of these bodies, even as their definitions show. And it shall be very necessary for you to had●●tore of that pasted paper by you, for so shall yo● upon it 〈…〉 the forms of other bodies, as prisms and Parallelipopedons, 〈…〉 set forth in these five books following, and see the very 〈◊〉 of th●se bodies there mentioned: which will make these books concerning bodies, as easy unto you as were the other books, whose figures you might plainly see upon a plain superficies.  
Describe thi● figur●, which consist●th of tw●lu●●quil●●●● and equiangle P●nt●●●●●, vpo● the foresaid matter, and finely cut as before was ●●ught t●●●l●u●n lines contained within th● figur●, and bow and fold the Pen●●gon● accordingly. And they will so close together, tha● th●y will ●●k● th● very form of a Dodecahedron. 
A D●d●●●●edron.   

An Icosa●edron.  If ye describe this figure which consisteth of twenty equilater and equiangle triangles upon the foresaid matter, and finely cut as before was showed the nin●t●ne lines which are contained within the figure, and then bow and fold them accordingly, they will in such sort close together, that ther● will be made a perfect form of an Icosahedron.  
Because in these five books there are sometimes required other bodies besides the foresaid five regular bodies, as Pyramids of divers forms, prisms, and others, I have here set forth three figures of three sundry Pyramids, one having to his base a triangle, an other a quadrangle figure, the other ● Pentagon● which if ye describe upon the foresaid matter & finely cut as it was before taught the lines contained within each figure, namely, in the first, three lines, in the second, four lines, and in the third, five lines, and so bend and fold them accordingly, they will so close together at the tops, that they will ●ake Pyramids of that form that their bases are of. And if ye conceive well the describing of these, ye may most easily describe the body of a Pyramid of what form so ever ye william.  
Because these five books following are somewhat hard for young beginners, by reason they must in the figures described in a plain imagine lines and superficieces to be elevated and erected, the one to the other, and also conceive solids or bodies, which, for that they have not hitherto been acquainted with, will at the first sight be somewhat strange unto them, I have for their more ●ase, in this eleventh book, at the end of the demonstration of every Proposition either set new figures, if they concern the elevating or erecting of lines or superficieces, or else if they concern bodies, I have showed how they shall describe bodies to be compared with the constructions and demonstrations of the Propositions to them belonging. And if they diligently weigh the manner observed in this eleventh book touching the description of new figures agreeing with the figures described in the plain, it shall not be hard for them of themselves to do the like in the other books following, when they come to a Proposition which concerneth either the elevating or erecting of lines and superficieces, or any kinds of bodies to be imagined.   

¶ The 1. Theorem. The 1. Proposition. That part of a right line should be in a ground plain superficies, & part elevated upward is impossible.  
FOr if it be possible, let part of the right line ABC, namely, the part AB be in a ground plain superficies, and the other part thereof, namely, BC be elevated upward. And produce directly upon the ground plain superficies the right line AB beyond the point B unto the point D. Wherefore unto two right lines given ABC, and ABD, the line AB is a common section or part, which is impossible. 
Demonstration leading to an impossibility.  For a right line can not touch a right line in 〈◊〉 points then one, v●lesse those right be exactly agreeing and laid the one upon the other. Wherefore that part of a right line should be in a ground plain superficies, and part elevated upward is impossible: which was required to be proved.  
This figure more plainly setteth forth the foresaid demonstration, if ye elevate the superficies wherein the line BC.  

another demonstration after Fl●s●●s.  
If it be possible let there be a right line ABG, whose part AB let be in the ground plain superficies AED; 
another demonstration after Flussas.  and let the rest thereof BG be elevated on high, that is, without the plain AED. Then I say that ABG is not one right line. For forasmuch as AED is a plain superficies, produce directly & equally upon the said plain AED the right line AB towards D, which by the 4. definition of the first shall be a right line. And from some one point of the right line ABD, namely, from C, dra● unto the point G a right line CG. Wherefore in the triangle 〈…〉 the outward ang●● AB● is equal to the two inward and opposite angles (by the 32. of the first) and therefore it is less than two right angles (by the 17. of the same) Wherefore the line ABG forasmuch as it maketh an angle, is not ● right line. Wherefore that part of a right line should be in a ground plain superficies, and part elevated upward is impossible.  
If ye mark well the figure before added for the plainer declaration of Euclides demonstration, i● will not be hard for you to co●●●●e this figure which ulysses putteth for his demonstration ● wherein is no difference but only the draft of the line GC.    

¶ The 2. Theorem. The 2. Proposition. If two right line cut the ou● to the other, they are ●●●ne and the self same plain superficies: & every triangle is in one & the self same superficies.  
SVppose that these two right lines AB and CD do cut the one the other in the point E. Then I say that these lines AB and CD are in one and the self same superficies, and that every triangle is in one & self same plain superficies. 
Construction.  Take in the lines EC and EB points at all adventures, and let the same be F and G, and draw a right line from the point B to the point C, and an other from the point F to the point G. And draw the lines FH and GK. First I say that the triangle EBC is in one and the same ground superficies. 
Demonstration leading to an impossibility.  For if part of the triangle EBC, namely the triangle FCH, or the triangle GBK be in the ground superficies, and the residue be in an other, then also part of one of the right lines EC or EB shall be in the ground superficies, and part in an other. So also if part of the triangle EBC, namely, the part EFG be in the ground superficies and the residue be in an other, then also one part of each of the right lines EC and EB shall be in the ground superficies, & an other part in an other superficies, which (by the first of the eleventh) is proved to be impossible. Wherefore the triangle EBC is in one and the self same plain superficies. For in what superficies the triangle BCE is, in the same also is either of the lines EC and EB, and in what superficies either of the lines EC and EB is, in the self same also are the lines AB and CD. Wherefore the right lines lines AB and CD are in one & the self same plain superficies, and every triangle is in one & the self same plain superficies: which was required to be proved.  
In this figure here set may ye more plainly conceive the demonstration of the former proposition where 〈◊〉 may ele●●●● what part of the triangle ECB ye will, namely the part FCH or the part GBK, or finally the part FCGB as is required in the demonstration.   

¶ The 3. Theorem. The 3. Proposition. If two plain superficieces cut the one the other: their common section is a right line.  
SVppose that these two superficieces AB & BC do cut the one the other, and let their common secti●●●e the line DB. Then I say that DB is a right line. For if not, draw from the point D to the point B a right line DFB in the plain superficies AB, 
Demonstration leading to an impossibility.  and likewise from the same points draw an other right line DEB in the plain superficies BC. Now therefore two right lines DEB and DFB shall ●aue the self sa●● e●de●, and therefore do include a superficies which (by the last common sentence) is impossible● Wherefore the lines DEB and DFB are not right lines. In like sort also may we prove that no other right line can be drawn from the point D to the point B besides the line DB which is the common section of the two superficieces AB and BC. If therefore two plain superficieces cut the one the other, their common section is a right line: which was required to be demonstrated.  
 
This figure here set, showeth most plainly not only this third proposition, but also the demonstration thereof, if ye elevate the superficies AB, and so compare it with the demonstration.   

¶ The 4. Theorem. The 4. Proposition. If from two right lines, cutting the one the other, at their common section, a right line be perpendicularly erected: the same shall also be perpendicularly erected from the plain superficies by the said two lines passing.  
SVppose that there be two right lines AB and CD cutting the one the other in the point E. And from the point E let there be erected a right line EF perpendicularly to the said two right lines AB and CD: then I say that the right line EF, 
Construction.  is also erected perpendicular to the plain superficies which passeth by the lines A B and CD. Let these lines AE, EB, EC, and ED be put equal the one to the other. And by the point E extend a right line at all adventures, and let the same be GEH. And draw these right lines AD, CB, FA, FG, FD, FC, FH, and FB. 
Demonstration.  And forasmuch as these two right lines AE & ED are equal to these two lines CE and EB, and they comprehend equal angles (by the 15. of the first): therefore (by the 4. of the first) the base AD is equal to the base CB, and the triangle AED is equal to the triangle CEB. Wherefore also the angle DAE, is equal to the angle EBC. But the angle AEG is equal to the angle BEH (by the 15 of the first). Wherefore there are two triangles AGE, and BEH having two angles of the one equal to two angles of the other, each to his correspondent angle, and one side of the one equal to one side of the other, namely one of the sides which lie between the equal angles, namely, the side AE is equal to the side EB. Wherefore (by the 26. of the first) the sides remaining are equal to the sides remaining. Wherefore the side GE is equal to the side EH, and the side AG to the side BH. And forasmuch as the line AE is equal to the line EB, and the line FE is common to them both, and maketh with them right angles, wherefore (by the fourth of the first) the base FA it equal to the base FB. And (by the same reason) the base FC is equal to the base FD. And forasmuch as the line AD is equal to the line BC, and the line FA is equal to the line FB as it hath been proved. Therefore these two lines FA and AD are equal to these two lines FB & BC, the one to the other, & the base FD is equal to the base FC. Wherefore also the angle FAD is equal to the angle FBC. And again forasmuch as it hath been proved, that the line AG is equal to the line BH, but the line FA is equal to the line FB. Wherefore there are two lines FA and AG equal to two lines FB and BH and it is proved that the angle FAG is equal to the angle FBH: wherefore (by the 4. of the first) the base FG is equal to the base FH. Again forasmuch as it hath been proved that the line GE is equal to the line EH, and the line EF is common to them both: wherefore these two lines GE and EF are equal to these two lines HE and EF, and the base FH is equal to the base FG: wherefore the angle GEF is equal to the angle HEF. Wherefore either of the angles GEF, and HEF is a right angle. Wherefore the line EF is erected, from the point E, perpendicularly to the line GH. In like sort may we prove, that the same line FE maketh right angles with all the right lines which are drawn upon the ground plain superficies and touch the point B. But a right line is then erected perpendicularly to a plain superficies, when it maketh right angles with all the lines which touch it, and are drawn upon the ground plain superficies (by the 2. definition of the eleventh). Wherefore the right line FE is erected perpendicularly to the ground plain superficies. And the ground plain superficies is that which passeth by these right lines AB and CD. Wherefore the right line FE is erected perpendicularly to the plain superficies which passeth by the right lines AB and CD. If therefore from two right lines cutting the one the other and at their common section a right line be perpendicularly erected: it shall also be erected perpendicularly to the plain superficies by the said two lines passing: which was required to be proved.  
In this figure you may most evidently conceive the former proposition and demonstration, if ye erect perpendicularly unto the ground plain superficies ACBD the● triangle AFB: and elevate the triangles AFD, & CFB in such sort, that the line AF of the triangle AFB may join & make one line with the line AF of the triangle AFD: and likewise that the line BF of the triangle AFB may join & make one right line with the line BF of the triangle BFC.   

¶ The 5. Theorem. The 5. Proposition. If unto three right lines which touch the one the other, be erected a perpendicular line from the common point where those three lines touch: those three right lines are in one and the self same plain superficies.  
SVppose that unto these three right lines BC, BD, and BE, touching the one the other in the point B, be erected perpendicularly from the point B, the line AB. Then I say, that those three right lines BC, BD and BE, are in one & the self same plain superficies. For if not, then if it be possible, let the lines BD & BE be in the ground superficies, and let the line BC be erected upward (now the lines AB and BC are in one and the same plain superficies (by the 2. of the eleventh) for they touch the one the other in the point B). Extend the plain superficies wherein the lines AB and BC are, 
Demonstration leading to an impossibility.  and it shall make at the length a common section with the ground superficies, which common section shall be a right line (by the 3. of the eleventh): let that common section be the line BF. Wherefore the three right lines AB, BC, and BF are in one and the self same superficies, namely, in the superficies wherein the lines AB and BC are. And forasmuch as the right line AB is erected perpendicularly to either of these lines BD and BE, therefore the line AB is also (by the 4. of the eleventh) erected perpendicularly to the plain superficies, wherein the lines BD and BE are. But the superficies wherein the lines BD and BE are is the ground superficies. Wherefore the line AB is erected perpendicularly to the ground plain superficies. Wherefore (by the 2. definition of the eleventh) the line AB maketh right angles with all the lines which are drawn upon the ground superficies and touch it. But the line BF which is in the ground superficies doth touch it. Wherefore the angle ABF is a right angle. And it is supposed that the angle ABC is a right angle. Wherefore the angle ABF is equal to the angle ABC, and they are in one and the self same plain superficies which is impossible. Wherefore the right line BC is not in an higher superficies. Wherefore the right lines BC, BD, BE are in one and the self same plain superficies. If therefore unto three right lines touching the one the one the other, be erected a perpendicular line from the common point where those three lines touch: those three right lines are in one and the self same plain superficies: which was required to be demonstrated.  
This figure here set more plainly declareth the demonstration of the former proposition, if ye erect perpendicularly unto the ground superficies, the s●●perficies wherein is drawn the line 〈◊〉 and so compare it with the said defenestration.   

The 6. Theorem. The 6. Proposition. If two right lines be erected perpendicularly to one & the self same plain superficies: those right lines are parallels the one to the other.  
SVppose that these two right lines AB and CD be erected perpendicularly to a ground plain superficies. Then I say that the line AB is a parallel to the line CD. Let the points which those two right lines touch in the plain superficies be B and D. 
Construction.  And draw a right line from the point B to the point D. And (by the 11. of the first) from the point D, draw unto the line BD in the ground superficies a perpendicular line DE. And (by the 2. of the first) 
* An Assumpt as M. Dee proveth it. Demonstration.  put the line DE equal to the line AB. And draw these right lines BE, AE, and AD. And forasmuch as the line AB is erected perpendicularly to the ground superficies, therefore (by the 2. definition of the eleventh) the line AB maketh right angles with all the lines which are drawn upon the ground plain superficies and touch it. But either of t●ese lines BD and BE which are in the ground superficies, touch the line AB, wherefore either of these angles ABD and ABE is a right angle● and by the same reason also either of the angles CDB, & CDE is a right angle. And forasmuch as the line AB is equal to the line DE, and the line BD is common to them both, therefore these two lines AB and BD, are equal to these two lines ED and DB, and they contain right angles: wherefore (by the 4. of the first) the base AD is equal to the base BE. And forasmuch as the line AB is equal to the line DE, and the line AD to the line BE, therefore these two lines AB and BE are equal to these two lines ED and DA, and the line AE is a common base to them both. Wherefore the angle ABE is (by the 8. of the first) equal to the angle EDA. But the angle ABE is a right angle: wherefore also the angle EDA is a right angle: wherefore the line ED is erected perpendicularly to the line DA: and it is also erected perpendicularly to either of these lines BD and DC, wherefore the line ED is unto these three right lines BD, DA, and DC erected perpendicularly from the point where these three right lines touch the one the other: wherefore (by the 5. of the eleventh) these three right lines BD, DA, and DC are in one and the self same superficies. And in what superficies the lines BD and DA are, in the self same also is the line BA: for every triangle is (by the 2. of the eleventh) in one and the self same superficies. Wherefore these right lines AB, BD, and DC are in one and the self same superficies, and either of these angles ABD and BDC is a right angle (by supposition). Wherefore (by the 28. of the first) the line AB is a parallel to the line CD. If therefore two right lines be erected perpendicularly to one and the self same plain superficies, those right lines are parallels the one to the other: which was required to be proved.  
Here for the better understanding of this 6. proposition I have described an other figure: as touching which if ye erect the superficies ABD perpendicularly to the superficies BDE, and imagine only a line to be drawn from the point A to the point E (if ye will ye may extend a thread from the said point A to the point E) and so compare it with the demonstration, it will make both the proposition, and also the demonstration most clear unto you.  

¶ An other demonstration of the sixth proposition by M. Dee.  
Suppose that the two right lines AB & CD be perpendicularly erected to one & the same plain superficies, namely the plain superficies OP. Then I say that ●● and CD are parallels. Let the end points of the right lines AB and CD, which touch the plain superficies O●, be the points ● and D, frō● to D let a strait line be drawn (by the first petition) and (by the second petition) let the strait line ●D be extended, as to the points M & N. Now forasmuch as the right line AB, from the point ● produced, doth cut the line MN (by construction). Therefore (by the second proposition of this eleventh book) the right lines AB & MN are in one plain● superficies. Which let be QR, cutting the superficies OP in the right line MN. By the same means may we conclude the right line CD to be in one plain superficies with the right line MN. But the right line MN (by supposition) is in the plain superficies QR: wherefore CD is in the plain superficies QR. And A● the right line was proved to be in the same plain superficies QR. Therefore AB and CD are in one plain superficies, namely QR. And forasmuch as the lines A● and CD (by supposition) are perpendicular upon the plain superficies OP, therefore (by the second definition of this book) with all the right lines drawn in the superficies OP and touching AB and CD, the same perpendiculars A● and CD, do make right angles. But (by construction) MN, being drawn in the plain superficies OP toucheth the perpendiculars AB and CD at the points ● and D. Therefore the perpendiculars A● and CD, make with the right line MN two right angles namely ABN, and CDM: and MN the right line is proved to be in the one and the same plain superficies, with the right lines AB & CD: namely in the plain superficies QR. Wherefore by the second part of the 28. proposition of the first book, the right line● AB and CD are parallels. If therefore two right lines be erected perpendicularly to one and the self same plain superficies those right lines are parallels the one to the other: which was required to be demonstrated.   

A Corollary added by M. Dee.  
Hereby it is evident that any two right lines perpendicularly erected to one and the self same plain superficies, are also themselves in one and the same plain superficies, which is likewise perpendicularly erected to the same plain superficies, unto which the two right lines are perpendicular.  
The first part hereof is proved by the former construction and demonstration, that the right lines AB and CD are in one and the same plain superficies Q●. The second part is also manifest (that is, that the plain superficies QR is perpendicularly erected upon the plain superficies OP) for that A● and CD being in the plain superficies QR, are by supposition perpendicular to the plain superficies OP: wherefore by the third definition of this book QR is perpendicularly erected to, or upon OP: which was required to be proved.   

Io. Dee his advise upon the Assumpt of the 6.  
As concerning the making of the line DE, equal to the right line AB, verily the second of the first, without some farther consideration, is not properly enough alleged. And no wonder it is, for that in the former booke●, whatsoe●●●●a●h of lines been spoken, the same hath always been imagined to be in one only plain superficies considered or executed. But here the perpendicular line AB, is not in the same plaint superficies, that the right line DB is. Therefore some other help must be put into the hands of young beginners, how to bring this problem to execution: which is this, most plain and brief. Understand that BD the right line, is the common section of the plain superficies, wherein the perpendiculars AB and CD are, & of the other plain superficies, to which they are perpendiculars. The first of these (in my former demonstration of the 6) ● I noted by the plain superficies QR: and the other, I noted by the plain superficies OP. Wherefore BD being a right line common to both the plain sup●rficieces QR & OP, thereby the ponits B and D are common to the plains QR and OP. Now from BD (sufficiently extended) cut a right line equal to AB, (which suppose to be BF) by the third of the first, and orderly to BF make DE equal, by the 3. o● the first, if DE be greater than BF. (Which always you may 'cause so to be, by producing of DE sufficiently). Now forasmuch as BF by construction is cut equal to AB, and DE also, by construction, put equal to BF, therefore by the 1. common sentence, DE is put equal to AB: which was required to be done.  
In like sort, if DE were a line given to whom AB were to be cut and made equal, first out of the line DB (sufficiently produced) cutting of DG, equal to DE by the third of the first: and by the same 3. cutting from BA (sufficiently produced) BA, equal to DG: then is it evident, that to the right line DE, the perpendicular line AB is put equal. And though this be easy to conceive, yet I have designed the figure accordingly, whereby you may instruct your imagination. Many such helps are in this book requisite, as well to inform the young students therewith, as also to master the froward gaynesayer of our conclusion, or interrupter of our demonstrations course.    

¶ The 7. Theorem. The 7. Proposition. If there be two parallel right lines, and in either of them be taken a point at all adventures: a right line drawn by the said points is in the self same superficies with the parallel right lines.  
SVppose that these two right lines AB and CD be parallels, and in either of them take a point at all adventures, namely, E and F. Then I say, that a right line drawn from the point E to the point F, is in the self same plain superficies that the parallel lines are. For if not, then if it be possible, let it be in an higher superficies, 
Demonstration leading to an impossibility.  as the line EGF is, and draw the superficies wherein the line EGF is, & extend it, and it shall make a common section with the ground superficies, which section shall (by the 3. of the eleventh) be a right line: let that section be the right line EF. Wherefore two right lines EGF and EF include a superficies: which (by the last common sentence) is impossible. Wherefore a right line drawn from the point E to the point F, is not in an higher superficies. Wherefore a right line drawn from the point E to the point F, is in the self same superficies wherein are the parallel right lines AB and CD. If therefore there be two parallel right lines, and in either of them be taken a point at all adventures, a right line drawn by th●se points is in the self same plain superficies with the parallel right lines: which was required to be demonstrated.  
By this figure it is easy to see the former demonstration, if ye elevate the superficies wherein is drawn the line EGF.   

The 8. Theorem. The 8. Proposition. If there be two parallel right lines, of which one is erected perpendicularly to a round plain superficies: the other also is erected perpendicularly to the self same ground plain superficies.  
SVppose that there be two parallel right lines AB and CD, and let one of them, namely, AB be erected perpendicularly to a ground superficies. 
This proposition is as it were the converse of the sixth Construction.  Then I say that the line CD is also erected perpendicularly, to the self same ground superficies. Let the lines AB and CD fall upon the ground superficies in t●e points B and D, and (by the first petition) draw a righ● line from the point B to the point D. And draw (by the 11. of the first) in the ground superficies from the point D unto the line BD a perpendicular line DE, and (by the 2. of the first) put the line DE equal to the line AB, and draw a right line from the point B to the point E, and an other from the point A to the point E, and an other from the ●oint A to the point D. 
Demonstration.  And forasmuch as the line AB is erested perpendicularly to the ground superficieces, therefore (by the 2. definition of the eleventh) the line AB is erected perpendicularly to all the right lines that are in the ground superficies and touch it. Wherefore either of these angles ABD & ABE is a right angle. And forasmuch as upon these parallel lines AB and CD falleth a certain right line BD, therefore (by the 29. of the first) the angles ABD and CDB are equal to two right angles. But the angle ABD is a right angle, wherefore also the angle CDB is a right angle. Wherefore the line CD is erected perpendicularly to the line BD. And forasmuch as the line AB is equal to the line DE, and the line ●D is common to them both, therefore these two lines AB and BD are equal to these two lines ED and DB, and the angle ABD is equal to the angle EDB for either of them is a right angle. Wherefore (by the 4. of the first) the base AD is equal to the base BE. And forasmuch as the line AB is equal to the line DE, and the line BE to the lin● AD, therefore these two lines AB and BE are equal to these two lines AD & DE, the on● to the other, and the line AE is a common base to them both. Wherefore (by the 8. of the first) the angle ABE is equal to the angle ADE: but the angle A●E is a right angle, wherefore th●●ngle EDA, is also a right angle. Wherefore the line ED is erected perpendicularly to the line AD, and it is also erected perpendicularly to th● line DB. Wherefore the line ED is erected perpendicularly to the plain superficies wherein th● l●n●s BD and BA are (by the 4. of ●his book) Wherefore (by the 2. definition of the eleventh) the line ED is erected perpendicularly to all the right lines that touch it and are in the superficies wherein the lines BD and AD are. But in what superficies the lines BD and DA are, in the self same superficies is the line DC. For the line AD being drawn from two points taken in the parallel lines AB and CD is by the former proposition in the self same superficies with them. Now forasmuch as the lines AB and BD ar● in the superficies wherein the lines BD and DA are, but in what superficies the lines AB & BD are, in the same is the line DC. Wherefore the line ED is erected perpendicularly to the line DC. Wherefore also the line CD is erected perpendicularly to the line DE. And the line CD is erected perpendicularly to the line DB. For by the 29. of the first, the angle CDB being equal to the angle ABD is a right angle. Wherefore the line CD is from the point D erected perpendicularly to two right lines DE and DB cutting the one the other in the point D. Wherefore by the 4. of the eleventh, the line CD is erected perpendiculaaly to the plain superficies, wherein are the lines DE and DB. But the ground plain superficies is that wherein are the lines DE and DB, to which superficies also the line AB is supposed to be erected perpendicularly. Wherefore the line CD is erected perpendicularly to the ground plain superficies, whereunto the line AB is erected perpendicularly. If therefore there be two parallel right lines, of which one is erected perpendicularly to a ground plain superficies, the other also is erected perpendicularly to the self same ground plain superficies: which was required to be demonstrated.  
This figure will more clearly set forth the former demonstration, if ye erect perpendicularly the superficies ABD to the superficies BDE, and imagine a line to be drawn from the point A to the point D, in stead whereof, as in the 6. proposition ye may extend a thread.   

¶ The 9 Theorem. The 9 Pro 〈◊〉 Right lines which are parallels to one and the self same right line, and are not in the self same superficies that it is in: are also parallels the one to the other.  
SVppose that either of these right lines AB and CD be a parallel to the line EF not being in the self same superficies with it. Then I say that the line AB is a parallel to the line CD. Take in the line EF a point at all adventures, and let the same be G. 
Construction.  And from the point G raise up in the superficies wherein are the lines EF and AB, unto the line EF a perpendicular line GH, and again in the superficies wherein are the lines EF and CD, raise up from the same point G to the line EF a perpendicular line GK. 
Demonstration.  And forasmuch as the line EF is erected perpendicularly to either of the lines GH and GK, therefore (by the 4. of the eleventh) the line EF is erected perpendicularly to the superficies wherein the lines GH and GK are, but the line EF is a parallel line to the line AB. Wherefore (by t●e 8. of the eleventh) the line AB is erected perpendicularly to the plain superficies, wherein are the lines GH and GK. And by the same reason also the line CD is erected perpendicularly to the plain superficies wherein are the lines GH & GK. Wherefore either of these lines AB and CD is erected perpendicularly to the plain superficies, wherein the lines GH and GK are. But if two right lines be erected perpendicularly to one and the self same plain superficies, those right lines are parallels the one to the other (by the 6. of the eleventh) Wherefore the line AB is a parallel to the line CD. Wherefore right lines which are parallels to one & the self same right line, and are not in the self same superficies with it are also parallels the one to the other: which was required to be proved.  
This figure more clearly manifesteth the former proposition and demonstration, if ye elevate the superficieces ABEF and CDEF that they may incline and concur in the line EF.   

¶ The 10. Theorem. The 1 〈…〉 If two right lines touching the one the other 〈…〉 her right lines touching the one the other, and no 〈…〉 lfe same superficies with the two first: those right lines contain equal angles.  
SVppose that these two right lines AB and BC touching the one the other, be parallels to these two lines DE and EF touching also the one the other, and not being in the self same superficies that the lines AB and BC are. Then I say, that the angle ABC is equal to the angle DEF. 
Construction.  For let the lines BA, BC, ED, EF, be put equal the one to the other: and draw these right lines AD, CF, BE, AC, and DF. 
Demonstration.  And forasmuch as the line BA is equal to the line ED, and also parallel unto it, therefore (by the 33. of the first) the line AD is equal and parallel to the line BE: and by the same reason also the line CF is equal & parallel to the line BE. Wherefore either of these lines AD and CF is equal & parallel to the line EB. But right lines which are parallels to one and the self same right line, and are not in the self same superficies with it, are also (by the 9 of the eleventh) parallels the one to the other. Wherefore the line AD is a parallel line to the line CF. And the lines AC and DF join them together. Wherefore (by the 33. of the first) the line AC is equal and parallel to the line DF. And forasmuch as these two right lines AB & BC are equal to these two right lines DE and EF, and the base AC also is equal to the base DF: therefore (by the 8. of the first) the angle ABC is equal to the angle DEF. If therefore two right lines touching the one the other be parallels to two other right lines touching the one the other, and not being in one and the self same superficies with the two first: those righ● lines contain equal angles: which was required to be demonstrated.  
This figure here set more plainly declareth the former Proposition and demonstration, if ye elevate the superficieces DABE, and FCBE, till they concur in the line FE.   

¶ The 1. Problem. The 11. Proposition. From a point given on high, to draw unto a ground plain superficies a perpendicular right line.  


Construction.  
Two cases in this proposition.  
The first case.   SVppose that the point given on high be A, and suppose a ground plain superficies, namely, BCGH. It is required from the point A to draw unto the ground superficies a perpendicular line. Draw in the ground superficies a right line at adventures, and let the same be BC. 
john Dee.  And (by the 12. of the first) from the point A draw unto the line BC a perpendicular line AD. 
* This requireth the imagination of a plain superficies passing by the point A, and the strait line BC. And so help yourself in the like cases either Mathematically imagining, or Mechanically practising.  Now if AD be a perpendicular line to the ground superficies, then is that done which was sought for. But if not, then (by the 11. of the first) from the point D raise up in the ground superficies unto the line BC a perpendicular line DE. And (by the 12. of the first) from the point A draw unto the line DE a perpendicular line AF. And by the point F draw (by the 31. of the ●irst) unto the line BC a parallel line FH: And extend the line FH from the point F to the point G. 

Second cas●.  
Demonstration.   And forasmuch as the line BC is erected perpendicularly to either of these lines DE and DA, therefore (by the 4. of the eleventh) the line BC is erected perpendicularly to the superficies wherein the lines ED and AD are: and to the line BC the line GH is a parallel. But i● there be two parallel right lines, of which one is erected perpendicularly to a certain plain superficies, the other also (by the 8. of the eleventh) is erected perpendicularly to the self same superficies. Wherefore the line GH is erected perpendicularly to the plain● superficies wherein the lines ED and DA are. Wherefore also (by the 2. definition of the eleventh) the line GH is erected perpendicularly to all the right lines which touch it, and are in the plain superficies wherein the lines ED and AD are. But the line AF toucheth it being in the superficies wherein the lines ED and AD are (by the ●. of this book). Wherefore the line GH is erected perpendicularly to the line FA. Wherefore also the line FA is erected perpendicularly to the line GH: and the line AF is also erected perpendicularly to the line DE. Wherefore AF is erected perpendicularly to either of these lines HG and DE. But if a right line be erected perpendicularly from the common section of two right lines cutting the one the other, it shall also be erected perpendicularly to the plain superficies of the said two lines (by the 4. of the eleventh). Wherefore the line AF is erected perpendicularly to that superficies wherein the lines ED and GH are. But the superficies wherein the lines ED and GH are, is the ground superficies. Wherefore the line AF is erected perpendicularly to the ground superficies. Wherefore from a point given on high, namely, from the point A, is drawn to the ground superficies a perpendicular line: which was required to be done.  
In this figure shall ye much more plainly see both the cases of this former demonstration. For as touching the first case, ye must erect perpendicularly to the ground superficies, the superficies wherein is drawn the line AD, and compare it with the demonstration, and it will be clear unto you. For the second case ye must erect perpendicularly unto the ground superficies the superficies wherein is drawn the line AF, and unto it let the other superficies wherein is drawn the line AD, incline, so that the point A of the one may concur with the point A of the other: and so with your figure thus ordered, compare it with the demonstration, and there will be in it no hardness at all.   

¶ The 2. Problem. The 12. Proposition. Unto a plain superficies given, and from a point in it given, to raise up a perpendicular line.  
SVppose that there be a ground plain superficies given, and let the point in it given be A. It is required from the point A to raise up unto the ground plain superficies a perpendicular line. Understand some certain point on high, and let the same be B. 
Construction.  And from the point B draw (by the 11. of the eleventh) a perpendicular line to the ground superficies, and let the same be BC. And (by the 31. of the first) by the point A draw unto the line BC a parallel line DA. Now forasmuch as there are two parallel right lines AD and CB, the one of them, namely, 
Demonstration.  CB is erected perpendicularly to the 〈◊〉 superficies: wherefore the other line also, namely, AD, is 〈◊〉 perpendicularly to the same ground superficies (by the eight of 〈…〉 eleventh). Wherefore unto a plain superficies given, and 〈◊〉 point in it given, namely, A, is raised up a perpendicular lyn●●required to be done.  
In this second figure ye may consider plainly the demonstration of the former proposition if ye erect perpendicularly the superficies wherein are drawn the lines AD and CB.   

¶ The 11. Theorem. The 13. Pr●position. From one and the self point, and to one and the self same plain superficies, can not be erected two perpendicular right lines on one and the self same side.  
FOr if it be possible from the point A let there be erected perpendicularly to one and the self same plain superficies two righ● lines AB and AC on one and the self same side. 
Demonstration leading to an impossibility.  And extend the superficies wherein are the lines AB and AC: 
Note this manner of imagination Mathematical.  and it shall make at length a common section in the ground superficies which common section shall be a right line, and shall pass by the point A: let that common section be the line DAE. Wherefore (by the 3. of the eleventh) the lines AB, AC, and DAE are in one and the self same plain superficies. And forasmuch as the line CA is erected perpendicularly to the ground superficies, therefore (by the 2. definition of the eleventh) it maketh right angles with all the right lines that touch it, and are in the ground superficies. But the line DAE toucheth it, being in the ground superficies. Wherefore the angle CAE is a right angle, and by the same reason also the angle BAE is a right angle. Wherefore (by the 4 petition) the angle CAE is equal to the angle BAE the less to the more, both angles being in one & the self same plain superficies: which is impossible. Wherefore from one and the self same point, and to one and the self same plain superficies can not be erected two perpendicular right lines on one & the self same side: which was required to be demonstrated.  
In this figure if ye erect perpendicularly the superficies wherein are drawn the lines ●A and CA to the ground superficies wherein is drawn the line DAE, and so compare it with the the demonstration of the former proposition it will be clear unto you.  

M. Dee his annotation.  
Euclides words in this 13. proposition admit two cases: one, if th● 〈…〉 in the plain superficies, (as commonly the demonstrations suppose) the other, if the point assigned be any where without the said plain superficies, to which, the perpendiculars fall, is considered. Contrary to either of which, if the adversary affirm, admitting from one point two right lines, perpendiculars to one and the self same plain superficies, and on one and the same side thereof, by the 6. of the eleventh he may be bridled: which will ●ore him to confess his two perpendiculars to be also parallels. But by supposition agreed one, they concur at one and the same poyn●, which (by the definition of parallels) i● impossible. Therefore our adversary must recant a●d yield to out proposition.    

¶ The 12. Theorem. The 14. Proposition. To whatsoever plain superficieces one and the self same right line is erected perpendicularly: those superficieces are parallels the one to the other.  
SVppose that a right line AB be erected perpendicularly to either of these plain superficieces CD and EF. Then I say, that these superficieces CD and EF are parallels the one to the other. 
Demonstration leading to an impossibility.  For if not, then if they be extended they will at the length meet. Let them meet, if it be possible. Now than their common section shall (by the 3. of the eleventh) be a right line. Let that common section be GH. And in the line GH take a point at all adventures and let the same be K. And draw a right line from the point A to the point K, and an other from the point B to the point K. And forasmuch as the line AB is erected perpendicularly to the plain superficies EF, therefore the line AB is also erected perpendicularly to the line BK which is in the extended superficies EF. Wherefore the angle ABK is a right angle. And by the same reason also the angle BAK is a right angle. Wherefore in the triangle ABK, these two angles ABK & BAK, are equal to two right angles: which (by the 17. of the first) is impossible. Wherefore these superficieces CD and EF being extended meet not together. Wherefore the superficieces CD and EF are parallels. Wherefore to what soever plain superficieces one and the self same right line i● erected perpendicularly: those superficies are parallels the one to the other: which was required to be proved.  
In this figure may ye plainly see the former demonstration if ye erect the three superficieces, GD, GE, and KLM perpendiculary to the ground plain superficies: but yet in such sort that the two superfici●● 〈…〉 may concur in the common line G 〈…〉 the demonstration.  

A corollary added by Campane.  
If a right line be erected perpendicularly to one of those superficies, it 〈…〉 erected perpendicularly to the other.  
For if it should not be erected perpendicularly to the other, than it falling upon that other shall make with some one line thereof an angle less than a right angle: which line should (by the 5. petition of the first) concur with some one line of that superficies whereunto it is perpendicular. So that those superficieces should not be parallels: which is contrary to the supposition. For they are suppsed to be parallels.    

¶ The 13. Theorem. The 15. Proposition. If two right lines touching the one the other be parallels to two other right lines touching also the one the other and not being in the self same plain superficies with the two first: the plain superficieces extended by those right lines, are also parallels the one to the other.  
SVppose that these two right lines AB and BC touching the one the other be parallels to these two right lines DE & EF touching also the one the other, and not being in the self same plain superficies with the right lines AB and BC. Then I say, that the plain superficieces by the lines AB and BC, and the lines DE and EF being extended, shall not meet together, that is, they are equedistant and parallels: 
Construction.  From the point B draw (by the 11. of the eleventh) a perpendicular line to the superficies wherein are the lines DE and EF, and let that perpendicular line be BG. And by the point G in the plain superficies passing by DE, and EF, draw (by the 31. of the first) unto the line ED a parallel line GH: and likewise by that point G draw in the same superficies unto the line EF a parallel line GK. 
Demonstration.  And forasmuch as the line BG is erected perpendicularly to the superficies wherein are the lines DE and EF, therefore (by the 2. definition of the eleventh) it is also erected perpendicularly to all the right lines which touch it, and are in the self same superficies wherein are the lines DE and EF. But either of these lines GH and GK touch it, and are also in the superficies wherein are the lines DE and EF, therefore either of these angles BGH, and BGK, is a right angle. And forasmuch as the line BA is a parallel to the line GH (that the lines GH and GK are parallels unto the lines AB and BC it is manifest by the 9 of this book): therefore (by the 29. of the first) the angles GBA and BGH are equal to two right angles. But the angle BGH is (by construction) a right angle, therefore also the angle GBA is a right angle: therefore the line GB is erected perpendicularly to the line BA. And by the same reason also may it be proved, that the line BG is erected perpendicularly to the line BC. Now forasmuch as the right line BG is erected perpendicularly to these two right lines BA and BC touching the one the other, therefore (by the 4. of the eleventh) the line BG is erected perpendicularly to the superficies wherein are the lines BA and BE And it is also erected perpendicularly to the superficies wherein are the lines GH and GK. But the superficies wherein are the lines GH and GK, is that superficies wherein are the lines DE and EF: wherefore the line BG is erected perpendicularly to the superficies wherein are the lines DE and EF. Wherefore the line BG is erected perpendicularly to the superficies wherein are the lines DE and EF, and to the superficies wherein are the lines AB and BC. But if one and the self same right line be erected perpendicularly to plain superficieces, those superficieces are (by the 14. of the eleventh) parallels the one to the other. Wherefore the superficies wherein are the lines AB and BC is a parallel to the superficies wherein are the lines DE and EF. If therefore two right lines touching the one the other be parallels to two other right lines touching also the one the other, and not being in the self same plain superficies with the two first, the plain superficieces extended by those right lines are also parallels the one to the other which was required to be demonstrated.  
By this figure here put, ye may more clearly see both the former 15. Proposition and also the demonstration thereof: if ye erect perpendicularly unto the ground superficies, the three superficieces ABC, KHE, and LHBM, and so compare it with the demonstration.  

¶ A Corollary added by Flussas.  
Unto a plain superficies being given, to draw by a point given without it, a parallel plain superficies. Suppose as in the former description that the superficies given be ABC, & let the point given without it be G. Now then by the point G draw (by the 31. of the first) unto the lines AB and BC parallel lines GH and HK. And the superficies extended by the lines GH and GK shall be parallel unto the superficies ABC, by this 15. Proposition.    

The 14. Theorem. The 16. Proposition. If two parallel plain superficieces be cut by some one plain superficies: their common sections are parallel lines.  
SVppose that these two plain superficieces AB and CD be cut by this plain superficies EFGH, ●nd let their common sections be the right lines EF and GH. Then I say that the line EF is a parallel to the line GH. For if not, than the lines EF and GH being produced, shall at the length meet together either on the side that the points FH, are, or on the side that the points E, G be. First let them be produced on that side that the points F, H are, and let them meet in the point K. And forasmuch as the line EFK is in the superficies AB, therefore all the points which are in the line EF are in the superficies AB (by the first of this book) But one of the points which are in the right line EFK is the point K, 
Demonstration leading to an absurdities  therefore the point K is in the superficies AB. And by the same reason also the point K is in the superficies CD. Wherefore the two superficieces AB and CD being produced do meet together, but by supposition they meet not together, for they are supposed to be parallels. Wherefore the right lines EF and GH produced shall not meet together on that side that the points F, H are. In like sort also may we prove that the right lines EF and GH produced meet not together on that side that the points E, G be. But right lines which being produced on no side meet together, are parallels (by the last definition of the first). Wherefore the line EF is a parallel to the line GH. If therefore two parallel plain superficieces be cut by some one plain superficies their common sections are parallel lines: which was required to be proved.  
This figure here set more plainly 〈…〉 demonstration, if ye erect perpendicular 〈…〉 superficies the three superficieces A● 〈…〉 and so compare it with the demonstr● 〈…〉  

A Corollary added by Flussas.  
If two plain superficieces be parallels to one and the s●lfe same plain 〈…〉 also be parallels the one to the other, or they shall make one and the self same plain sup 〈…〉  
In this figure here set, ye may more plainly see the former demonstration, if ye elevate to the ground super●icieces ACDI, the three super●icieces AB, DG, & GI', and ●o compare it with the demonstration.    

The 15. Theorem. The 17. Proposition. I● two right lines be cut by plain superficieces being parallels: the parts o● the lines divided shall be proportional.  


In t●is ●rono●●●o●●● must understand the proportional ●artes or s●●●ions to be th●se which are contained 〈◊〉 the parallel superficies.  
Construction.  
Demonstration.   Suppose that these two right lines AB and CD be divided by these plain superfi●i●ces being parallels, namely, GH, KL, MN in the points A, E, B, C, F, D. Then I say that as the right line AE is to the right line EB, so is the right line CF to the right line FD. Draw these right lines AC, BD and AD. And let the line AD and the superficies KL concur in the point X. And draw a right line from the point E to the point X and an other from the point X to the point F. And forasmuch as these two parallel superficieces KL and MN are cut by the superficies EBDX, therefore their common sections which are the lines EX and BD, are (by the 16. of the eleventh) parallels the one to the other. And by the same reason also forasmuch as the two parallel superficies GH and KL be cut by the superficies AXFC, their common sections AC and XF are (by the 16. of the eleventh) parallels. And forasmuch as to one of the sides of the triangle ABD● namely, to the side BD is drawn a parallel line EX, therefore (by the 2. of the sixth) proportionally as the line AE is to the line EB, so is the line AX to the line XD. Again forasmuch as to one of the sides of the triangle ADC, namely, to the side AC is drawn a parallel line XF, therefore by the 2. of the sixth, proportionally as the line AX is to the line XD, so is the line CF to the line FD. And it was proved that as the line AX is to the line XD, so is the line AE to the line EB, therefore also (by the 11. of the fift) as the line AE is to the line EB, so is the line CF to the line FD. If therefore two right lines ●e divided by plain super●icieces being parallels, the parts of the lines divided shall be proportional: which was required to be demonstrated.  
In this figure it is more easy to see the former demonstration, if ye erect perpendicularly unto the ground superficies ACBD, the three superficieces, GH, KL, and MN, or if ye so ●r●ct them that th●y be equedistant one to the other.   

¶ The 16. Theorem. The 18. Proposition. If a right line be erected perpendicularly to a plain superficies: all the superficieces extended by that right line, are erected perpendicularly to the self same plain superficies.  
SVppose that a right line AB be erected perpendicularly to a ground superficies. Then I say, that all the superficieces passing by the line AB, are erected perpendicularly to the ground superficies. Extend a superficies by the line AB, and let the same be ED, & let the common section of the plain superficies and of the ground superficies be the right line CE. And take in the line CE a point at all adventures, 
Construction.  and let the same be F: and (by the 11. of the first) from the point F draw unto the line CE a perpendicular line in the superficies DE, and let the same be FG. And forasmuch as the line AB is erected perpendicularly to the ground superficies, therefore (by the 2. definition of the eleventh) the line AB is erected perpendicularly to all the right lines that are in the ground plain superficies, 
Demonstration.  and which touch it. Wherefore it is erected perpendicularly to the line CE. Wherefore the angle ABE is a right angle. And the angle GFB is also a right angle (by construction). Wherefore (by the 28. of the first) the line AB is a parallel to the line FG. But the line AB is erected perpendicularly to the ground superficies: wherefore (by the 8. of the eleventh) the line FG is also erected perpendicularly to the ground superficies. And forasmuch as (by the 3. definition of the eleventh) a plain superficies is then erected perpendicularly to a plain superficies, when all the right lines drawn in one of the plain superficieces unto the common section of those two plain superficieces making therewith right angles, do also make right angles with the other plain superficies and it is proved that the line FG drawn in one of the plain superficieces, namely, in DE, perpendicularly to the common section of the plain superficieces, namely, to the line CE, is erected perpendicularly to the ground superficies: wherefore the plain superficies DE is erected perpendicularly to the ground superficies. In like sort also may we prove, that all the plain superficieces which pass by the line AB, are erected perpendicularly to the ground superficies. If therefore a right line be erected perpendicularly to a plain superficies all the superficieces passing by the right line, are erected perpendicularly to the self same plain superficies: which was required to be demonstrated.  
In this figure here set ye may erect perpendicularly at your pleasure the superficies wherein are drawn the lines DC, GF, AB, and HE, to the ground superficies wherein is drawn the line CFBE, and so plainly compare it with the demonstration before put.   

¶ The 17. Theorem. The 19 Proposition. If two plain superficieces cutting the one the other be erected perpendicularly to any plain superficies: their common section is also erected perpendicularly to the self same plain superficies.  
SVppose that these two plain super●icieces AB & BC cutting the one the other be erected perpendicularly to a ground superficies, and let their common section be the line BD. Then I say, that the line BD is erected perpendicularly to the ground superficies. 
Demonstration leading to an impossibility.  For if not, then (by the 11. of the first) from the point D draw in the superficies AB unto the right line DA a perpendicular line DE. And in the superficies CB draw unto the line DC a perpendicular line DF. And forasmuch as the superficies AB is erected perpendicularly to the ground superficies, and in the plain superficies AB unto the common section of the plain superficies and of the ground superficies, namely, to the line DA is erected a perpendicular line DE, therefore (by the converse of the 3. definition of this book) the line DE is erected perpendicularly to the ground superficies. And in like sort may we prove, that the line DF is erected perpendicularly to the ground superficies. Wherefore from one and the self same point, namely, from D, are erected perpendicularly to the ground superficies two right lines both on one and the self same side: which is (by the 15. of the eleventh) impossible. Wherefore from the point D can not be erected perpendicularly to the ground superficies any other right lines beside BD, which is the common section of the two superficieces AB and BC. If therefore two plain super●icieces cutting the one the other be erected perpendicularly to any plain superficies, their common section is also erected perpendicularly to the self same plain superficies: which was required to be proved.  
Here have I set an other figure which will more plainly show unto you the former demonstration, if ye erect perpendicularly to the ground superficies AC the two superficieces AB and BC which cut the one the other in the line BD.   

The 18. Theorem. The 20. proposition. If a solid angle be contained under three plain superficial angles: every two of those three angles, which two so ever be taken, are greater than the third.  
SVppose that the solid angle A be contained under three plain superficial angles, that is, under BAC, CAD, and DAB. Then I say that two of these superficial angles how so ever they be taken, are greater than the third. If the angles BAC, CAD, & DAB be equal the one to the other, then is it manifest that two of them which two so ever be taken are greater than the third. But if not, let the angle BAC be the greater of the three angles. And unto the right line AB and from the point A make in the plain superficies BAC unto the angle DAB an equal angle BAE. And (by the 2. of the first) make the line AE equal to the line AD. Now a right line BEC drawn by the point E, shall cut the right lines AB and AC in the points B and C: draw a right line from D to B, and a other from D to C. And forasmuch as the line DA is equal to the line AE, 
Demonstration.  and the line AB is common to them both, therefore these two lines DA and AB are equal to these two lines AB and AE and the angle DAB is equal to the angle BAE. Wherefore (by the 4. of the first) the base DB is equal to the base BE. And forasmuch as these two lines DB and DC are greater than the line BC, of which the line DB is proved to be equal to the line BE. Wherefore the residue, namely, the line DC is greater than the residue, namely, than the line EC. And forasmuch as the line DA is equal to the line AE, and the line AC is common to them both, and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EAC. And it is proved that the angle DAB is equal to the angle BAE: wherefore the angles DAB and DAC are greater than the angle BAC. If therefore a solid angle be contained under three plain superficial angles every two of those three angles, which two so ever be taken are greater than the third: which was required to be proved.  
In this figure ye may plainly behold the former demonstration, if ye elevate the three triangles ABD, A●C and ACD in such ●or●that they may all meet together in the point A.   

The 19 Theorem. The 21. Proposition. Every solid angle is comprehended under plain angles less than four right angles.  
SVppose that A be a solid angle contained under these superficial angles BAC, DAC and DAB. Then I say that the angles BAC, DAC and DAB are less than four right angles. 
Construction  Take in every one of these right lines ACAB and AD a point at all adventures and let the same be B, C, D. And draw these right lines BC, CD and DB. 
Demonstration.  And forasmuch as the angle B is a solid angle, for it is contained under three superficial angles, that is, under CBA, ABD and CBD, therefore (by the 20. of the eleventh) two of them which two so ever be taken are greater than the third. Wherefore the angles CBA and ABD are greater than the angle CBD: and by the same reason the angles BCA and ACD are greater than the angle BCD● and moreover the angles CDA and ADB are greater than the angle CDB. Wherefore these six angles CBA, ABD, BCA, ACD, CDA, and ADB are greater than these three angles, namely, CBD, BCD, & CDB. But the three angles CBD, BDC, and BCD are equal to two right angles. Wherefore the six angles CBA, ABD, BCA, ACD, CDA, and ADB are greater than two right angles. And forasmuch as in every one of these triangles ABC, and ABD and ACD three angles are equal two right angles (by the 32. of the first). Wherefore the nine angles of the three triangles, that is, the angles CBA, ACB, BAC, ACD, DAC, CDA, ADB, DBA and BAD are equal to six right angles. Of which angles the six angles ABC, BCA, ACD, CDA, ADB and DBA are greater than two right angles. Wherefore the angles remaining, namely, the angles BAC, CAD and DAB which contain the solid angle are less than sour right angles. Wherefore every solid angle is comprehended under plain angles less than four right angles: which was required to be proved.  
If ye will more fully see this demonstration compare it with the figure which I put for the better sight of the demonstration of the proposition next going before. Only here is not required the draft of the line AE.  
Although this demonstration of Euclid be here put for solid angles contained under three superficial angles, yet after the like manner may you proceed if the solid angle be contained under superficial angles how many so ever. As for example if it be contained under four superficial angles, if ye follow the former construction, the base will be a quadrangled figure, whose four angles are equal to four right angles: but the 8. angles at the bases of the 4. triangles set upon this quadrangled figure may by the 20. proposition of this book be proved to be greater than those 4. angles of the quadrangled figure: As we saw by the discourse of the former demonstration. Wherefore those 8. angles are greater than four right angles: but the 12. angles of those four triangles are equal to 8. right angles. Wherefore the four angles remaining at the top which make the solid angle are less than four right angles. And observing this course ye may proceed infinitely.   

¶ The 20. Theorem. The 22. Proposition. If there be three superficial plain angles of which two how soever they be taken, be greater than the third, and if the right lines also which contain those angles be equal: then of the lines coupling those equal right lines together, it is possible to make a triangle.  
SVppose that there be three superficial angles ABC, DEF, and GHK, of which let two, which two soever be taken, be greater than the third, that is, let the angles ABC, and DEF be greater than the angle GHK, and let the angles DEF and GHK be greater than the angle ABC: and moreover let the angles GHK and ABC be greater than the angle DEF. And let the right lines AB, BC, DE, EF GH, and HK be equal the one to the other, and draw a right line from the point A to the point C, and an other from the point D to the point F, and moreover an other from the point G to the point K. 
Two cases in this proposition.  Then I say that it is possible of three right lines equal to the lines AC, DF and GK, to make a triangle, that is, that two of the right lines AC, DF, and GK, which two soever be taken, are greater than the third. Now if the angles ABC, DEF, 
The first case.  and GHK be equal the one to the other, it is manifest that these right lines AC, DF, and GK being also (by the 4. of the first) equal the one to the other, it is possible of three right lines equal to the lines AC, DF, and GK to make a triangle. 
Second case.  But if they be not equal, let them be unequal. And (by the 23. of the first) unto the right line HK, 
Constructions  and at the point in it H, make unto the angle ABC an equal angle KHL. And by the ●. of the first) to one of the lines AB, BG, DE, EF, GH, 
Demonstration.  or HK make the line HL equal, & draw these right lines KL and GL. And forasmuch as these two lines AB and DC, are equal to these two lines KH and HL, and the angle B is equal to the angle KHL, ther●●●● (by the 4. of the first) the base AC is equal to the base KL. And forasmuch as the angles ABC, and GHK are greater than the angle DEF, but the angle GHL is equal to the angles ABC, & GHK● therefore the angle GHL is greater than the angle DEF. And forasmuch as these two lines GH and HL are equal to these two lines DE and EF, and the angle GHL is grea●er than the angle DEF, therefore (by the 25. of the first) the base GD is greater than the base DF. But the lines GK and KL are greater than the line GL. Wherefore then lines GK & KL are much greater than the line DF. But the line KL is equal to the line AC. Wherefore the lines AC and GK are greater than the line DF. In like sort also may we pro●●● tha● the lines AC and DF are greater than the line GK, and that the lines GK and DF are greater than the line AC. Wherefore it is possible to make a triangle of three lines equal to the lines AC, DF, and GK: which was required to be demonstrated.  

another demonstration.  
Suppose that the three superficial angles be ABC, DEF, and GHK, of which angles, 
another demonstration.  two howsoever they be taken, are greater than the third. And let them be contained under these equal right lines AB, BC, DE, EF, GH, HK, which equal right lines let th●se lines AC, DF, and GK join together. Then I say that it is possible of three right lines equal to the lines AC, DF, and GK to make a triangle, whi●h again is as much to say, as that two of those lines which two soever be taken, are greater than the third. Now again if the angles B, E, H, be equal, the lines also AC DF, and GK are equal, and so two of them shall be greater than the third. But if not, l●t the angles B, E, H, be unequal, and let the angle B be greater than ●ither of the angles E and H. Therefore (by the 24. of the first) the right line AC is greater than either of the lines DF & GK. And it is manifest that the line AC with either of the lines DF or GK is greater than the third. I say also that the lines DF and GK are greater than the line AC. Unto the right line AB, 
Construction.  and to the point in it B, make (by the 22. of the first) unto the angle GHK an equal angle AB L, and unto one of the lines AB, BC, DE, EF, GH or HK, make by the 2. of the first) an equal line BL. And draw a right line from the point A to the point L, and an other from the point L to the point C. 
Demonstration.  And forasmuch as these two lines AB & BL are equal to these two lines GH & HK the one to the other, and they contain equal angles: therefore (by the 4. of the first) the base AL is equal to the base GK. And forasmuch as the angles E and H be greater than the angle ABC, of which the angle GHK is equal to the angle ABL, therefore the angle remaining, namely, the angle E is greater than the angle LBC. And forasmuch as these two lines LB and BC, are equal to these two lines DE and EF the one to the other, and the angle DEF is greater than the angle LBC, therefore (by the 25. of the first) the base DF is greater than the base LC: and it is proved that the line GK is equal to the line AL. Wherefore the lines DF & GK are greater than the lin●s AL & LC. B●t the li●es AL and LC are greater than the line AC. Wherefore the lines DF & GK are much greater than th● line AC. Wherefore two of these right lines AC, DF & GK which two soever be taken are greater than th● third. Wherefore it is possible of three right lines equal to the lines AC, DF and GK to make a triangle: which was required to be demonstrated.    

¶ The 3. Problem. The 23. Proposition. Of three plain superficial angles, two of which how soever they be taken, are greater than the third, to make a solid angle: Now it is necessary that those three superficial angles be less than four right angles.  
SVppose that the superficial angles given be ABC, DEF, GHK: of which let two how soever they be taken, be greater than the third: and moreover, let those three angles be l●sse than four right angles. It is required of three superficial angles equal to the angles ABC, DEF, and GHK, to make a solid or bodily angle. Let the lines AB, BC, DE, EF, GH, and HK, be made equal: 
Construction.  and draw a right line from the point A to the point C, & an other from the point D to the point F, and an other from the point G to the point K. Now (by the 22. of the eleventh) it is possible of three right lines equal to the right lines AC, DF, and GK, to make a triangle. Make such a triangle, and let the same be LMN, so that let the line AC be equal to the line LM, and the line DF to the line MN, and the line GK to the line LN. And (by the 5. of the fourth) about the triangle LMN describe a circle LMN, and take (by the 1. of the third) the centre of the same circle, 

Three cases in this proposition.  
The first case.   which centre shall either be within the triangle LMN, or in one of the sides thereof, or without it. First let it be within the triangle, and let the same be the point X, & draw these right lines LX, MX, and NX. Now I say, that the line AB is greater than the line LX. 
A necessary thing to be proved before he proceed any ●arther in the construction of the Problems.  For if not, than the line AB is either equal to the line LX, or else it is less than it. First let it be equal. And forasmuch as the line AB is equal to the line LX, but the line AB is equal to the line BC, therefore the line LX is equal to the line BC: and unto the line LX the line XM is (by the 15. definition of the first) equal: wherefore these two lines AB and BC, are equal to these two lines LX and XM the one to the other: and the base AC is supposed to be equal to the base LM. Wherefore (by the 8. of the first) the angle ABC is equal to the angle LXM. And by the same reason also the angle DEF is equal to the angle MXN, and moreover the angle GHK to the angle NXL. Wherefore these three angles ABC, DEF, and GHK, are equal to these three angles LXM, MXN, & NXL. But the three angles LXM, MXN, and NXL, are equal to four right angles (as it is manifest to see by the 13. of the first, if any one of these lines MX, LX, or NX, be extended on the side that the point X is). Wherefore the three angles ABC, DEF, and GHK, are also equal to four right angles. But they are supposed to be less than four right angles: which is impossible. Wherefore the line AB is not equal to the line LX. I say also that the line AB is not less than the line LX. For if it be possible, let it be less: and (by the 2. of the first) unto the line AB put an equal line XO: and to the line BC put an equal line XP, and draw a right line from the point O to the point P. And forasmuch as the line AB is equal to the line BC, therefore also the line XO is equal to the line XP. Wherefore the residue OL is equal to the residue MP. Wherefore (by the 2. of the sixth) the line LM is a parallel to the line OP● and the triangle LMX is equiangle to the triangle OPX. Wherefore as the line XL is to the line LM, so is the line XO to the line OP. Wherefore alternately (by the 16. of the fift) as the line LX is to the line XO, so is the line LM to the line OP. But the line LX is greater than the line XO. Wherefore also the line LM is greater than the line OP. But the line LM is put to be equal to the line AC: wherefore also the line AC is greater than the line OP. Now forasmuch as these two right lines AB and BC are equal to these two right lines OX and XP, and the base AC is greater than the base OP, therefore (by the 25. of the first) the angle ABC is greater than the angle OXP. In like sort also may we prove, that the angle DEF is greater than the angle MXN, and that the angle GHK is greater than the angle NXL. Wherefore the three angles ABC, DEF, and GHK, are greater than the three angles LXM, MXN, and NXL. But the angles ABC, DEF, and GHK, are supposed to be less than four right angles: wherefore much more are the angles LXM, MXN, & NXL less than four right angles. But they are also equal to four right angles: which is impossible. Wherefore the line AB is not less than the line LX: and it is also proved that it is not equal unto it. Wherefore the line AB is greater than the line LX.  
Now from the point X raise up unto the plain superficies of the circle LMN a perpendicular line XR (by the 12. of the eleventh). And unto that which the square of the line AB exceedeth the square of the line XL 
* Which how to find out is taught at the end of this demonstration, and also was taught in the assumpt put before the 14. proposition of the tenth book.  let the square of the line XR be equal. And draw a right line from the point R to the point L, and an other from the point R to the point M, and an other from the point R to the point N. And forasmuch as the line RX is erected perpendicularly to the plain superficies of the circle LMN, therefore (by conversion of the second definition of the eleventh) the line RX is erected perpendicularly to every one of these lines LX, MX, and NX. 
Demonstration of the first case.  And forasmuch as the line LX is equal to the line XM, & the line XR i● common to them both, and is also erected perpendicularly to them both, therefore (by the 4. of the first) the base RL is equal to the base RM. And by the same reason also the line RN is equal to either of these lines RL and RM. Wherefore these three lines RL, RM, and RN, are equal the one to the other. And forasmuch as unto that which the square of the line AB exceedeth the square of the line LX, the square of the line RX is supposed to be equal, therefore the square of the line AB is equal to the squares of the lines LX and RX. But unto the squares of the lines LX and XR, the square of the line LR is (by the 47. of the first) equal, for the angle LXR is a right angle. Wherefore the square of the line AB is equal to the square of the line RL. Wherefore also the line AB is equal to the line RL. But unto the line AB is equal every one of these lines BC, DE, EF, GH, and HK, and unto the line RL is equal either of these lines RM and RN. Wherefore every one of these lines AB, BC, DE, EF, GH, and HK, is equal to every one of these lines RL, RM, and RN. And forasmuch as these two lines RL and RM are equal to these two lines AB and BC, and the base LM is supposed to be equal to the base AC, therefore (by the 8. of the first) the angle LRM is equal to the angle ABC. And by the same reason also the MRN is equal to the angle DEF, and the angle LRN to the angle GHK. Wherefore of three superficial angles LRM, MRN, and LRN, which are equal to three superficial angles given, namely, to the angles ABC, DEF, & GHK, is made a solid angle R, comprehended under the superficial angles LRM, MRN, and LRN: which was required to be done.  

Second case.  But now let the centre of the circle be in one of the sides of the triangle, let it be in th● side MN, and let the centre be X. And draw a right line from the point L to the point X. I say again, that the line AB is greater than the line LX. For if not, then AB is either equal to LX, or else it is less than it. First let it be equal. Now then these two lines AB and BC, that is, DE & EF are equal to these two lines MX and XL, that is, to the line MN. But the line MN is supposed to be equal to the line DF. Wherefore also the lines DE and EF are equal to the line DF: which (by the 20. of the first) is impossible. Wherefore the line AB is not equal to the line LX. In like sort also may we prove, that it is not less. Wherefore the line AB is greater than the line LX. And now if as before unto the plain superficies of the circle be erected from the point X a perpendicular line RX whose square let be equal unto that which the square of the line AB exceedeth the square of the line LX, and if the rest of the construction and demonstration be observed in this that was in the forme● case, then shall the Problem be finished.  
But now let the centre of the circle be without the triangle LMN, and let it be in th● point X. And draw these right lines LX, MX, and NX. 
Third case.  I say that in this case also the line AB is greater than the line LX. For if not, then is it either equal or less. First let it be equal. Wherefore these two lines AB and BC are equal to these two lines MX and XL the one to th● other, and the base AC is equal to the base ML. Wherefore (by the 8. of the first) the angle ABC is equal to the angle MXL: and by the same reason also the angle GHK is equal to the angle LXN. Wherefore the whole angle MXN is equal to these two angles ABC and GHK. But the angles ABC and GHK are greater than the angle DEF. Wherefore the angle MXN is greater than the angle DEF. And forasmuch as these two lines DE and EF are equal to these two lines MX and XN, and the base DF is equal to the base MN, therefore (by the 8. of the first) the angle MXN is equal to the angle DEF. And it is proved that it is also greater: which is impossible. Wherefore the line AB is not equal to the line LX. In like sort also may we prove, that it is not less. Wherefore the line AB is greater than the line LX. And again if unto the plain superficies of the circle be erected perpendicularly from the point X a line XR, whose square is equal to that which the square of the line AB exceedeth the square of the line LX, and the rest of the construction be done in this that was in the former cases, then shall the Problem be finished.  
I say moreover, that the line AB is not less than the line LX. 
another demonstration to prove that th● line AB is not less than the line LX.  For if it be possibl●, let it be less. And unto the line AB, put (by the 2. of the first) the line XO equal: and unto the line BC put the line XP equal: And draw a right line from the point O to the point P. And forasmuch as the line AB is equal to the line BC, therefore the line XO is equal to the line XP. Wherefore the residue OL is equal to the residue MP. Wherefore (by the 2. of the sixth) the line LM, is a parallel to the line PO. And the triangle LXM is equiangle to the triangle PXO. Wherefore (by the 6. of the sixth) as the line LX is to the line LM, so is the line XO to the line PO. Wherefore alternately (by the 16. of the fift) as the line LX is to the line XO, so is the line LM to the line OP. But the line LX is greater than the line XO. Wherefore also the line LM is greater than the line OP. But the line LM is equal to the line AC. Wherefore also the line AC is greater than the line OP. Now forasmuch as these two lines AB and BC are equal to these two lines OX & XP the one to the other, and the base AC is greater than the base OP, therefore (by the 25. of the first) the angle ABC is greater than the angle OXP. And in like sort if we put the line XR equal to either of these lines XO or XP, and draw a right line from the point O to the point R, we may prove that the angle GHK is greater than the angle OXR. Unto the right line LX, and unto the point in it X, make (by the 23. of the first) unto the angle ABC an equal angle LXS: and unto the angle GHK make an equal angle LXT. And (by the second of the first) let either of these lines SX, and XT, be equal to the line OX. And draw these lines OS, OT, and ST. And forasmuch as these two lines AB and BC are equal to these two lines OX and XS, and the angle ABC is equal to the angle OXS, therefore (by the 4. of the first) the base AC, that is, LM, is equal to the base OS. And by the same reason also the line LN is equal to the line OT. And forasmuch as these two lines LM and LN are equal to these two lines SO and OT, and the angle MLN is greater than the angle SOT, therefore (by the 25. of the first) the base MN is greater than the base ST. But the line MN is equal to the line DF. Wherefore the line DF is greater than the line ST. Now forasmuch as these two lines DE and EF are equal to these two lines SX and XT, and the base DF is greater than the base ST, therefore (by the 25. of the first) the angle DEF is greater than the angle SXT. But the angle SXT is equal to the angles ABC and GHK. Wher●fore also the angle DEF is greater than the angles ABC and GHK. But it is also less: which is impossible.  

This was before ta●ght in the tenth book in the assumpt put before the 14. proposition.  But now let us declare how to find out the line XR, whose square shall be equal ●o that which the square of the line AB exceedeth the square of the line LX. Take the two right lines AB and LX, and let AB be the greater, and upon AB describe a semicircle ACB, and from the point A apply into the semicircle a right line AC equal to the right line LX: and draw a right line from the point C to the point B. And forasmuch as in the semicircle ACB is an angle ACB, therefore (by the 31. of the third) the angle ACB is a right angle. Wherefore (by the 47. of the first) the square of the line AB is equal to the squares of the lines AC and CB: wherefore the square of the line AB is in power more than the square of the line AC by the square of the line CB: but the line AC is equal to the line LX, wherefore the square of the line AB is in power more than the square of the line LX by the square of the line CB. If therefore unto the line CB we make the line XR equal, then is the square of the line AB greater than the square of the line LX by the square of the line XR: which was required to be done.  
In this figure may ye more fully s●● the construction and demonstration of the ●●rst case of the former 23. Proposition, if ye erect perpendicularly the triangle ● RN, and unto it bend the triangle LMR, that the angles R of each may join together in the point R. And so fully understanding this case, the other cases will not be hard to conceive.   

¶ The 21. Theorem. The 24. Proposition. If a solid or body be contained under 
* 
M. Dee (to avoid cavillation) addeth to Euclides proposition this word six: whom I have followed accordingly, and not Zamberts, in this.  
This kind of body mentioned in the proposition is called a Parallelipipedom according to the definition before given thereof.   six parallel plain superficieces, the opposite plain superficieces of the same body are equal and parallelograms.  
SVppose that this solid body CDHG be contained under these 6. parallel plain superficieces, namely, AC, GF, BG, CE, FB, and AE. Then I say that the opposite superficieces of the same body, are equal and parallelograms, it is to weet, the two opposites AC and GF, and the two opposites BG and CE, and the two opposites FB and AE to be equal, and all to be parallelograms. 
Demonstration that the opposite sides are parallelograms.  For forasmuch as two parallel plain superficieces, that is, BG, and CE are divided by the plain superficies AC, their common sections are (by the 16. of the eleventh) parallels. Wherefore the line AB is a parallel to the line CD. Against forasmuch as two parallel plain superficieces FB and AE are divided by the plain superficies AC their common sections are by the same proposition, parallels. Wherefore the line AD is a parallel to the line BC. And it is also proved, that the line AB is a parallel to the line DC. Wherefore the superficies AC is a parallelogram. In like sort also may we prove, that every one of these superficices CE, GF, BG, FB, and AE are parallelograms. Draw a right line from the point A, to the point H, and an other from the point D to the point F. 
Demonstration that the opposite superficies are equal.  And forasmuch as the line AB is proved a parallel to the line CD, and the line BH to the line CF, therefore these two right lines AB and BH touching the one the other, are parallels to these two right lines DC and CF touching also the one the other, and not being in one and the self same plain superficies. Wherefore (by the 10. of the eleventh) they comprehend equal angles. Wherefore the angle ABH is equal to the angle DCF. And forasmuch as these two lines AB and BH are 
* AB is equal to DC, because the superficies AC is proved a parallelogram, and by the same reason, is BH equal to CF, because the superficies FB is proved a parallelogram: therefore the 34. of the first is our proof.  equal to these two lines DC and CF, and the angle ABH is proved equal to the angle DCF● therefore (by the 4. of the first) the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DCF. And forasmuch as (by the 41. of the first) the parallelogram BG is double to the triangle ABH, and the parallelogram CE is also double to the triangle DCF, therefore the parallelogram BG is equal to the parallelogram CE. In like sort also may we prove that the parallelogram AC is equal to the parallelogram GF, and the parallelogram AE to the parallelogram FB. If therefore a solid or body be contained under six parallel plain superficieces, the opposite plain superficieces of the same body are equal & parallelograms which was required to be demonstrated.  
I have for the better help of young beginners, described here an other figure whose form if it be described upon pasted paper with the letters placed in the same order that it is here, and then if ye cut finely these lines AG, DE and CF not through the paper, and fold it accordingly, compare it with the demonstration, and it will shake of all hardness from it.   

The 22. Theorem. The 25. Proposition. If a Parallelipipedom be cut of a plain superficies being a parallel to the two opposite plain superficieces of the same body: then, as the base is to the base, so is the one solid to the other solid.  
I have for the better sight of the construction & demonstration of the former 25. proposition, here set another figure, whose form if ye describe upon pasted paper, and finely cut the three lines XI, BS, and TY, not through the paper but half way, and then fold it accordingly, and compare it with the construction and demonstration, you shall see that it will give great light thereunto.  
Here Flussas addeth three Corollaries.  

First Corollary.  
If a Prism be cut of a plain superficies parallel to the opposite superficieces, 
F●●●● Corollary.  the session's of the Prism shall be the one to the other in that proportion, that the sections of the base are the one to the other.  
For the sections of the bases, which bases (by the 11. definition of this book) are parallelograms, shall likewise be parallelograms, by the 16. of this book (when as the superficies which cutteth is parallelel to the opposite super●icieces) and shall also be equiangle. Wherefore if unto the bases (by producing the right lines) be added like and equal bases, as was before showed in a parallelipipedon, of those sections shallbe made as many like prisms as ye william. And so by the same reason, namely, by the common excess, equality, or want of the multiplices of the bases & of the sections by the 5. definition of the fifth may be proved, that every section of the Prism multiplied by any multiplication whatsoever, shall have to any other section that proportion that the sections of the bases have.   

Second Corollary.  
Solides whose two opposite superficie●es are poligonon figures like equal and parallels, the other superficies, which of necessity are parallelograms, 

Second Corolry.  
These solids which he speaketh of in this Corollary are of some called sided columns.   being cut of a plain superficies parallel to the two opposite superficies: the sections of the base are the one to the other, as the sections of the solid are th● one to the other.  
Which thing is manifest, for such solids are divided into prisms, which have one common side, namely, the axe or right line, which is drawn by the centres of the opposite bases. Wherefore how many pa●allelogrāmes or bases are set upon the opposite poligonon figures, of so many prisms shall the whole solid be composed. For those polygonon figures are divided into so many like triangles by the 20. of the sixth, which describe prisms. Which prisms being cut by a superficies parallel to the opposite superficieces, the sections of the bases shall (by the former Corollary) be proportional with the sections of the prisms. Wherefore by the ●● of the fifth, as the sections of the one are the one to the other, so are the sections of the whole the one to the other.  
Of these solids there are infinite kinds, according to the variety of the opposite and parallel poligonon figures, which poligonon figures do altar the angles of the parallelograms set upon them according to the diversity o● their situation. But this is no let at all to this corollary, for that which we have proved will always follow. When as the superficieces which are set upon the opposite like & equal polygonon and parallel superficieces are always parallelograms.   

Third Corollary.  

Third Corollary.  T●e foresaid solids composed of prisms, being cut by a plain superficies parallel to the opposite superficieces, are the one to the other as the heads or higher parts cut are.  
For it is proved that they are the one to the other as the bases are. Which bases forasmuch as they are par●llelogrammes, are the one to the other as the right lines are upon which they are set by, the first of the sixth, which right lines are the heads or higher parts of the prisms.    

The 4. Problem. The 26. Proposition. Upon a right line given, and at a point in it given, to make a solid angle equal to a solid angle given.  
In these two 〈…〉 here put, you may in 〈◊〉 clearly concern the ●●●●mer construction and d●●monstratiō, if ye erect pe●●pendicularly unto the ground superficies the triangles ALB and DCE, & elevate the triangles ALH and DCF that the lines LA and CD of them may exactly agreed with the line● LA and CD of the ●riangles erec●ed● For so ordering them, if ye compare the former construction and demonstration with them, they will be plaint unto you.  
Although Euclides demonstration be touching solid angles which are contained only under three superficial angles, that is, whose bases are triangles: yet by it may ye perform the Problem touching solid angles contained under superficial angles how many soever, that is, having to their bases any kind of Poligonon figures. For every Poligonon figure may by the 20. of the sixth, be resolved into like tringles. And so also shall the solid angle be divided into so many solid angles as there be triangles in the base. Unto every one of which solid angles you may by this proposition describe 〈◊〉 equal solid angle upon a line given, and at a point in it given. And so at the length the whole solid angle after this manner described shall be equal to the solid angle given.   

The 5. Theorem. The 27. Proposition. Upon a right line given to describe a parallelipipedon like and in like sort situate to a parallelipipedon given.  
●Vppose that the right line given be AB, and let the parallelipipedon given be CD. It is required upon the right line given AB to describe a parallelipipedon like and in like sort situate to the parallelipipedon given, namely, to CD. 
Constr●ction.  Unto the right line AB and at the point in it A describe (by the 26. of the ele●enth (a solid angle equal to the solid angle C, and let it be contained under these superficial angles BAH, HAK, and KAB, so that let the angle BAH be equal to the angle ECF, and the angle BAK to the angle ECG, and moreover the angle KAH to the angle GCF. And as the line EC is to the line CG, so let the AB be to the line AK (by the 12. of the sixth) and as the line GC is to the line CF, so let the KA be 〈◊〉 the line AH. Wherefore of equality (by the 22. of the fift) as the line EC is to the line CF, so is the line BA to the line AH. 
Demonstration.  Make perfect the parallelogram BH, and also the solid AL. Now for that as the line EC is to the line CG, so is the line BA to the line AK, therefore the sides which contain the equal angles, namely, the angles ECG and BAK are proportional: wherefore (by the first definition of the sixth) the parallelogram GE is like to the parallelogram KB. And by the same reason the parallelogram KH is like to the parallelogram GF, and moreover the parallelogram FE to the parallelogram HB ● Wherefore there are three parallelograms of the solid CD like to the three parallelograms of the solid AL. But the three other sides in each of these solids are equal and like to their opposite sides. Wherefore the whole solid CD is like to the whole solid AL. Wherefore upon the right line given AB is described the solid AL contained under parallel plain superficieces like and in like sort situate to the solid given CD contained also under parallel plain superficieces: which was required to be done.  
This demonstration is not hard to conceive by the former figure as it is described in a plain, if ye that imagination of parallelipipedons described in a plain which we before taught in the definition of a cube. Howbeit I have here for the more ease of such as are not yet acquainted with solids, described two figures, whose forms first describe upon pasted paper with the like letters noted in them, and then finely cut the three middle lines of each figure, and so fold them accordingly, and that done compare them with the construction and demonstration of this 27. proposition, and they will be very easy to conceive.   

The 23. Theorem. The 28. Proposition. If a parallelipipedom be cut by a plain superficies drawn by the diagonal lines of those plain superficieces which are opposite: that solid is by this plain superficies cut into two equal parts.  
A diagonal line is a right line which in angular figures is drawn from one angle and extended to his contrary angle as you see in the figure AB.  
Describe for the better sight of this demonstration a figure upon pasted paper like unto that which you described for the 24. proposition only altering the letters: namely, in stead of the letter A put the letter F, and in stead of the letter H the letter C: moreover in stead of the letter C put the letter H, and the letter E for the letter D, and the letter A for the letter E, and finally put the letter D for the letter F. And your figure thus ordered compare it with the demonstration, only imagining a superficies to pass through the body by the diagonal lines CF and DE.   

¶ The 24. Theorem. The 29. Proposition. Parallelipipedons consisting upon one and the self same base, and under one and the self same altitude, whose 
* Look at the end of the demonstration what is understanded by standing lines.  standing lines are in the self same right lines, are equal the one to the other.  
SVppose that that these parallelipipedons CM and CN do consist upon one and the self same base, namely AB, and let them be under one and the self same altitude, whose standing lines, that is, the four sides of each solid which fall upon the base, as the lines AF, CD, BH, LM of the solid CM, and the lines CE, BK, AG, and LN of the solid CN, let be in the self same right lines or parallel lines FN, DK. 

john Dee his figure.  
By this figure it appears why ●uch prisms were called pledges: of 〈◊〉 v●ry shape of a wedge, as is the solid DEFGAC. etc.   Then I say that the solid CM is equal to the solid CN. For forasmuch as either of these superficieces CBDH, CBEK is a parallelogram, therefore (by the 34. of the first) the line CB is equal to either of these lines DH and EK. Wherefore also the line DH is equal to the line EK. Take away EH which is common to them both, wherefore the residue namely DE is equal to the residue HK. Wherefore also the triangle DCE is equal to the triangle HKB. And the parallelogram DG is equal to the parallelogram HN. And by the same reason the triangle AGF is equal to the triangle MLN, and the parallelogram CF is equal to the parallelogram BM. But the parallelogram CG is equal to the parallelogram BN, by the 24. of the tenth for they are opposite the one to the other. Wherefore the prism contained under the two triangles FAG and DCE and under the three parrallelogrammes AD, DG, and CG is equal to the prism contained under the two triangles MLN and HBK, and under the three parallelograms, that is, BM, NH, and BN. Put that solid common to them both, whose base is the parallelogram AB, and the opposite side unto the base is the superficies GEHM. Wherefore the whole parallelipipedon CM is equal to the whole parallelipipedon CN: Wherefore parallelipipedons consisting upon one and the self same base, and under one and the self same altitude, whose standing lines are in the self same right lines, are equal the one to the other: which was requi●ed to be demonstrated.  
Although this demonstration be not hard to a good imagination to conceive by the former figure (which yet by M. Dee● reforming is much better than the figure of this proposition commonly described in other copies both greek and latin): yet for the ease of those which are young beginners in this matter of solids, I have here set an other figure whose form if it be described upon pasted paper, with the like letters to every line as they be here put, and then if ye finely cut not through but as it were half way the three lines LA, NMGF, and KHED, & so fold it accordingly, & compare it with the demonstration, it will give great light thereunto.  
Standing lines are called those four right lines of every parallelipipedon which join together the angles of the upper and neither bases of the same body. 
Standing lines.  Which according to the diversity of the angles of the solids, may either be perpendicular upon the base, or fall obliquely. And forasmuch as in this proposition and in the next proposition following, the solids compared together are supposed to have one and the self same base, it is manifest that the standing lines are in respect of the lower base in the self same parallel lines, namely, in the two sides of the lower base. But because there are put two solids upon one and the self same base, and under one and the self same altitude, the two upper bases of the solids may be diversly placed. For forasmuch as they are equal and like (by the 24. of this book) either they may be placed between the self same parallel lines: and then the standing lines are in either solid said to be in the self same parallel lines, or right lines: namely, when the two sides of each of the upper bases are contained in the self same parallel lines: but contrariwise if those two sides of the upper bases be not contained in the self same parallel or right lines, neither shall the standing lines which are joined to those sides be said to be in the self same parallel or right lines. And therefore the standing lines are said to be in the self same right lines, when the sides of the upper bases, at the lest two of the sides are contained in the self same right lines: which thing is required in the supposition of this 29, proposition. But the standing lines are said not to be in the self same right lines, when none of the two sides of the upper bases are contained in the self same right lines, which thing the next proposition following supposeth.   

¶ The 25. Theorem. The 30. Proposition. Parallelipipedons consisting upon one and the self same base, and under the self same altitude, whose standing lines are not in the self same right lines, are equal the one to the other.  
SVppose that these Parallelipipedons CM and CN, do consist upon one and the self same base, namely, AB, and under one and the self same altitude, whose standing lines, namely, the lines AF, CD, BH, and LM, of the Parallelipipedon CM, and the standing lines AG, CH, BK, and LN, of the Parallelipipedon CN, let not be in the self same right lines. Then I say, that the Parallelipipedon CM, is equal to the Parallelipipedon CN. Forasmuch as the upper superficieces FH & ●K, of the former Parallelipipedons, are in one and the self same superficies (by reason they are supposed to be of one and the self same altitude): 
Construction.  Extend the lines FD and MH, till they concur with the lines N● and KE (sufficiently both ways extended: for in divers cases their concourse is divers). Let ●D extended, meet with NG, and cut it in the point X: and with KE in the point P. Let likewise MH extended, meet with NG (sufficiently produced) in the point O, and with KE in the point R. And draw these right lines AX, LO, CP, and BR. Now (by the 29. of the eleventh) the solid CM, whose base is the parallelogram ACBL, and the parallelogram opposite unto it is FDHM, is equal to the solid CO, whose base is ACBL and the opposite side the parallelogram XPRO, 
Demonstration.  for they consist upon one and the self same base, namely, upon the parallelogram ACBL, whose standing lines AF, AX, LM, LO, CD, CP, BH, and ●K, 
I Dees figure.  are in the self same right lines FP and MR. But the solid CO, whose base is the parallelogram ACBL, and the opposite superficies unto it is XPRO, is equal to the solid CN, whose base is the parallelogram ACBL, and the opposite superficies unto it is the superficies G●KN, for they are upon one and the self same base, namely, ACBL, and their standing lines AG, AX, CF, CP, LN, LO, BK, and BR, are in the self same right lines NX, and PK. Wherefore also the solid CM, is equal to the solid CN. Wherefore Parallelipipedons consisting upon one and the self same base, and under the self same altitude, whose standing lines are not in the self same right lines, are equal the one to the other: which was required to be proved.  
This demonstration is somewhat harder than the former to conceive by the figure described in the plain (and yet before M. Dee invented that figure which is placed for it, it was much harder) by reason one solid is contained in an other. And therefore for the clearer light thereof, describe upon pasted paper this figure here put with the like letters and finely cut the lines AC, CB, EG, BL, EPK, ROHM, and fold it accordingly that every line may exactly agreed with his correspondent line (which observing the letters of every line ye shall easily do) and so compare your figure with the demonstration, and it will make it very plain unto you.   

The 26. Theorem. The 31. Proposit. Parallelipipedons consisting upon equal bases, and being under one and the self same altitude, are equal the one to the other.  
SVppose that upon these equal bases AB and CD do consist these parallelipipedons AE and CF, being under one & the self same altitude. 

Two cases in this proposition.  
Th● first case.  
Construction.   Then I say, that the solid AE is equal to the solid CF. First let the standing lines, namely, HK, BE, AG, LM, OP, DF, CX, and RS, be erected perpendicularly to the bases AB and CD, and let the angle ALB not be equal to the angle CRD. Extend the line CR to the point T. And (by the 23. of the first) upon the right line RT, and at the point in it R, describe unto 
We are beholding to M. Dee for inventing this figure, with other, which till his reforming were as much misshapen as this was, and so both in the Greek and Latin copies remain.  the angle ALB an equal angle TRU. And (by the third of the first) put the line RT equal to the line LB, and the line RV equal to the line AL. And (by the 31. of the first) by the point V draw unto the line RT a parallel line VW: 
Demonstration.  and make perfect the base RW, and the solid YU. Now forasmuch as these two lines TR and RV are equal to these two lines BL and LA, and they contain equal angles, therefore the parallelogram RW is equal and like to the parallelogram AB. Again, forasmuch as the line LB is equal to the line RT, and the line LM to the line RS (for the lines LM and RS are the altitudes of the Parallelipipedons AE and CF, which altitudes are supposed to be equal) and they contain right angles by supposition, therefore the parallelogram RY is equal and like to the parallelogram BM. And by the same reason also the parallelogram LG is equal and like to the parallelogram SU. Wherefore three parallelograms of the solid AE are equal and like to the three parallelograms of the solid YU. But these three parallelograms are equal and like to the three opposite sides. Wherefore the whole Parallelipipedon AE is equal and like to the whole Parallelipipedon YU. Extend (by the second petition) the lines DR and WV, until they concur, and let them concur in the point Q. And (by the 31. of the first) by the point T draw unto the line RQ a parallel line T 4, and extend duly the lines Ta and DO until they concur, and let them concur in the point ✚. And make perfect the solids QY and RI. Now the solid QY, whose 
* Note now, how the base respectively is takend for so may alteration of respects altar the name of the bounds either of solids or plains.  base is the parallelogram RY, and the opposite side unto the base the parallelogram Qb, is equal to the solid YV, whose base is the parallelogram RY, and the opposite side unto the base the parallelogram VZ. For they consist upon one and the self fame base, namely, RY, and are under one and the self same altitude, and their standing lines, namely, RQ, RV, Ta, TW, SN, Sd, Yb, and YZ, are in the self same right lines, namely, QW, and NZ. But the solid YV is proved equal to the solid AE. Wherefore also the solid YQ is equal to the solid AE. Now forasmuch as the parallelogram RVWT is equal to the parallelogram QT (by the 35. of the first) and the parallelogram AB is equal to the parallelogram RW: therefore the parallelogram QT is equal to the parallelogram AB, and the parallelogram CD is equal to the parallelogram AB (by supposition). Wherefore the parallelogram CD is equal to the parallelogram QT. And there is a certain other superficies, namely, DT. Wherefore (by the 7. of the fift) as the base CD is to the base DT, so is the base QT to the base DT. And forasmuch as the whole Parallelipipedon CI is cut by the plain superficies RF, which is a parallel to either of the opposite plain superficieces, therefore as the base CD is to the base DT, so is the solid CF to the solid RI (by the 25. of the eleventh). And by the same reason also, forasmuch as the whole Parallelipipedon QI is cut by the plain superficies RY, which is a parallel to either of the opposite plain superficieces, therefore as the base QT is to the base DT, so is the solid QY to the solid RI. But as the base CD is to the base DT, so is the base QT to the base TD. Wherefore (by the 11. of the fift) as the solid CF is to the solid RI, so is the solid QY to the solid RI. Wherefore either of these solids CF and QY, have to the solid RI one and the same proportion. Wherefore the solid CF is equal to the solid QY. But it is proved that the solid QY is equal to the solid AE. Wherefore also the solid CF is equal to the solid AE.  
But now suppose that the standing lines, namely, AG, HK, 
Second case.  BE, LM, CX, OP, DF, and RS, be not erected perpendicularly to the bases AB and CD. Then also I say, that the solid AE is equal to the solid CF. Draw (by the 11. of the eleventh) unto the ground plain superficieces AB and CD from these points K, E, C, M, P, F, X, S, these perpendicular lines KN, ET, GV, MZ, PW, FY, XQ, and SI. And draw these right lines NT, NV, ZV, ZT, WY, WQ, IQ, and JY. Now (by that which hath before been proved in this 31. Proposition) the solid KZ is equal to the solid PI, for they consist upon equal 
* There you perceive how the base is diversly considered & chosen: as before we advertised you.  bases, namely, KM, and PS, and are under one and the self same altitude, whose standing lines also are erected perpendicularly to the bases. But the solid KZ is equal to the solid AE (by the 29. of the eleventh): and the solid PI is (by the same) equal to the solid CF, for they consist upon one and the self same base, and are under one & the self same altitude, whose standing lines are upon the self same right lines. Wherefore also the solid AE is equal to the solid CE. Wherefore Parallelipipedons consisting upon equal bases and being under one and the self same altitude, are equal the one to the other: which was required to be demonstrated.  
The demonstration of the first case of this proposition is very hard to conceive by the figure described for it in a plain. And yet before M. Dee invented that figure which we have there placed for it, it was much harder. For both in the Greek and Latin Euclid, it is very ill made, and it is in a manner impossible to conceive by it the construction and demonstration thereto appertaining. Wherefore I have here described other figures, which first describe upon pasted paper, or such like matter and then order them in manner following.  
As touching the solid AE in the first case, I need not to make any new description. For it is plain enough to conceive as it is there drawn. Although you may for your more ease of imagination describe of pasted paper a parallelipipedo● having his sides equal with the sides of the parallelipipedon AE before described, and having also the six parallelograms thereof (contained under those sides) equiangle with the six parallelograms of that figure, each side and each angle equal to his correspondent side, and to his correspondent angle.  
But concerning the other solid. When ye have described these three figures upon pasted paper: Where note for the proportion of each line, to make your figure of pasted paper equal with the figure before described upon the plain, let your lines OP, CX, RS, DF, etc. namely, the rest of the standing lines, of these figures, be equal to the standing lines OP, CX, RS, DF, etc. of that figure. Likewise let the lines OC, CR, RD, DO, etc. namely, the sides which contain the bases of these figures be equal to the lines OC, CR, RD, DO, etc. namely, to the sides which contain the bases of that figure. Moreover let the lines PX, X●, SF, FP, etc. namely, the rest of the lines which contain the upper superficieces of these figures, be equal to the lines PX, XS, SF, F●, etc. namely, to the rest of the lines which contain the upper superficieces of that figure (to have described all those foresaid lines of these figures equal to all the lines of that figure, would have required much more space than here can be spared: I have made them equal only to the halves of those lines, but by the example of these ye may, if ye will describe the like figures having their lines equal to the whole lines of the figure in the plain, each line to his correspondent line). When I say ye have as before is taught described these three figures, cut finely the lines XC, SR, FD of the first figure, and the lines SR, YT, and I ✚ of the second figure: likewise the lines ●R, NQ, ZVV, and YT, of the third figure, and fold these figures accordingly, which ye can not choose but do if ye mark well the letters of every line.  
As touching the second case ye need no new figures, for it is plain to see by the figures (now reformed by M. Dee) described for it in the plain, especially if ye remember the form of the figure of the 29. proposition of this book. Only that which there ye conceive to be the base, imagine here in both the figures of this second case to be the upper superficies opposite to the base, and that which was there supposed to be the upper superficies conceive here to be the base. You may describe them upon pasted paper for your better sight, taking heed ye note the letters rightly according as the construction requireth.  

Flussas' demonstrateth this proposition an otherway taking only the bases of the solids, and that after this manner.  
Take equal bases (which yet for the surer understanding let be utterly unlike) namely, AEBF and ADCH, and let one of the sides of each concur in one & the same right line AED, & the bases being upon one and the self same plain let there be supposed to be set upon them parallelipipedons under one & the self same altitude. Then I say that the solid set upon the base AB is equal to the solid set upon the base AH. By the point E draw unto the line AC a parallel line EG, which if it fall without the base AB, produce the right line HC to the point I Now forasmuch as AB and AH are parallelogrmaes, therefore by the 24. of this book, the triangles ACI and EGL shall be equaliter the one to the other: and by the 4. of the first, they shall be equiangle and equal: and by the first definition of the sixth, and fourth Proposition of the same, they shall be like. Wherefore prisms erected upon those triangles and under the same altitude that the solids AB and AH a●e, shall be equal and like, by the 8. definition of this book. For they are contained under like plain superficieces equal both in multitude and magnitude. Add the solid set upon the base ACLE common to them both. Wherefore the solid set upon the base AEGC, is equal to the solid set upon the base AELI. And forasmuch as the superficieces AEBF, and ADHC are equal (by supposition): and the part taken away AG is equal to the part taken away AL: therefore the residue BY shall be equal to the residue GD. Wherefore as AG is to GD as AL is to BY (namely, equals to equals). But as AG is to GD, so i● the solid set upon AG to the solid set upon GD by the 25. of this book, for it is cut by a plain superficies set upon the line GE, which superficies is parallel to the opposite superficieces. Wherefore as AL is to BY, so is the solid set upon AL to the solid set upon BY. Wherefore (by the 11. of the fifth) as the solid set upon AG (or upon AL which is equal unto it) is to the solid set upon GD, so is the same solid set upon AG or AL to the solid set upon BY. Wherefore (by the 2. part of the 9 of the fifth) the solids set upon GD and BY shall be equal. Unto which solids if ye add equal solids, namely, the solid set upon AG to the solid set upon GD, and the solid set upon AL to the solid set upon BY: the whole solids set upon the base AH and upon the base AB ●hall be equal. Wherefore Parallelipedons consisting upon equal bases and being under one and the self same altitude are equal the one to the other: which was required to be proved.    

¶ The 27. Theorem. The 32. Proposition. Parallelipipedons being under one and the self same altitude, are in that proportion the one to the other that their bases are.  
SVppose that these parallelipipedons AB and CD be under one & the self same altitude. Then I say that those parallelipipedons AB and CD are in that proportion the one to the other, that their bases are, that is, that as the base AE is to the base CF, so is the parallelipipedon AB to the parallelipipedon CD. 
Construction.  Upon the line FG describe (by the 45. of the first) the parallelogram FH equal to the parallelogram AE and equiangle with the parallelogram CF. And upon the base FH describe a parallelipipedom of the self same altitude that the parallelipipedom CD is, & let the same be GK. 
Demonstration.  Now (by the 31. of the eleventh) the parallelipipedon AB is equal to the parallelipipedon GK, for they consist upon equal bases, namely, AE and FH, and are under one and the self same altitude. And forasmuch as the parallelipipedon CK is cut by a plain superficies DG, being parallel to either of the opposite plain super●icieces, therefore (by the 25. of the eleventh) as the base HF is to the base FC, so is the parallelipipedon GK to parallelipipedon CD: but the base HF is equal to the base AE, and the parallelipipedon GK is proved equal to the parallelipipedon AB. Wherefore as the base A●E is to the base CF, so is the parallelipedon AB, to the parallelipipedon CD. Wherefore parallelipipedons being under one and the self same altitude, are in that proportion the one to the other that their bases are: which was required to be demonstrated.  
I need not to put any other figure for the declaration of this demonstration, for it is easy to see by the figure there described. Howbeit ye may for the more full sight thereof, describe solids of pasted paper according to the construction there set forth, which will not be hard for you to do, if ye remember the descriptions of such bodies before taught.  

A Corollary added by Flussas.  
Equal parallelipipedons contained under one and the self same altitude, have also their bases equal. For if the bases should be unequal, the parallelipipedons also should be unequal by this 32 proposition. And equal parallelipipedons having equal bases, have also one and the self same altitude. For if they should have a greater altitude, they should exceed the equal parallelipipedons which have the self same altitude: But if they should have a less they should want so much of those self same equal parallelipipedons.    

The 28. Theorem. The 33. Proposition. Like parallelipipedons are in triple proportion the one to the other of that in which their sides of like proportion are.  
SVppose that these parallelipipedons AB and CD be like, & let the sides AE and CF be sides of like proportion. Then I say, the parallelipipedon AB is unto the parallelipipedon CD in triple proportion of that in which the side AE is to the side CF. Extend the right lines AE, GE and HE to the points K, L, M. 
Construction.  And (by the 2. of the first) unto the line CF put the line EK equal, and unto the line FN put the line EL equal, and moreover unto the line FR put the line EM equal, and make perfect the parallelogram KL, and the parallelipipedon KO. 
Demonstration.  Now forasmuch as these two lines EK and EL are equal to these two lines CF and FN, but the angle KEL is equal to the angle CFN (for the angle AEG is equal to the angle CFM by reason that the solids AB and CD are like). Wherefore the parallelogram KL is equal and like to the parallelogram CN, and by the same reason also the parallelogram KM is equal and like to the parallelogram CR, and moreover the parallelogram OE to the parallelogram FD. Wherefore three parallelograms of the parallelipipedon KO are like and equal to three parallelograms of the parallelipipedon CD: but those three sides are equal and like to the three opposite sides: wherefore the whole parallelipipedon KO is equal and like to the whole parallelipipedon CD. Make perfect the parallelogram GK. And upon the bases GK and KL make perfect to the altitude of the parallelipipedon AB, the parallelipipedons EX & LP. And forasmuch as by reason that the parallelipipedons AB & CD are like, as the line AE is to the line CF, so is the line EG, to the line FN, and the line EH to the line FR. But the line CF is equal to the line EK, and the line FN to the line EL, and the line FR to the line EM, therefore as the line AE is to the line EK, so is the line GE to the line EL, and the line HE to the line EM (by construction). But as the line AE is to the line EK, so is the parallelogram AG to the parallelogram GK (by the first of the sixth). And as the line GE is to the line EL, so is the parallelogram GK to the parallelogram KL. And moreover as the line HE is to the line EM, so is the parallelogramm● PE to the parallelogram KM. Wherefore (by the 11. of the fift) as the parallelogram AG is to the parallelogram GK, so is the parallelogram GK to the parallelogram KL, and the parallelogram PE to the parallelogram KM. But as the parallelogram AG is to the parallelogram GK, so is the parallelipipedon AB to the parallellpipedon EX, by the former proposition, and as the parallelogram GK is to the parallelogram KL, so by the same is the parallelipipedon XE to the parallelipipedon PL: and again by the same, as the parallelogram PE is to the parallelogram KM, so is the parallelipipedon PL to the parallelipipedon KO. Wherefore as the parallelipipedom AB is to the parallelipipedon EX, so is the parallelipipedon EX to the parallelipipedon PL, and the parallelipipedon PL to the parallelipipedon KO. But if there be four magnitudes in* continual proportion, the first shallbe unto the fourth in triple proportion that it is to the second (by the 10. definition of the fift). Wherefore the parallelipipedon AB is unto the parallelipedon KO in triple proportion that the parallelipipedon AB is to the parallelipipedon EX. But as the parallelipipedon AB is to the parallelipipedom EX, so is the parallelogram AG to the parallelogram GK, and the right line AE to the right line EK. Wherefore also the parallelipipedon AB is to the parallelipipedon KO in triple proportion of that which the line AE hath to the line EK. But the parallelipipedon KO is equal to the parallelipipedon CD, and the right line EK to the right line CF. Wherefore the parallelipipedon AB is to the parallelipipedon CD in triple proportion that the side of like proportion, namely, AE is to the side of like proportion, namely, to CF. Wherefore like parallelipipedons are in triple proportion the one to the other of that in which their sides of like proportion are: which was required to be demonstrated.  

¶ Corellary.  
Hereby it is manifest, that if there be four right lines in 
*  continual proportion, as the first is to the fourth, so shall the Parallelipipedon described of the first line, be to the Parallelipipedon described of the second, both the Parallelipipedons being like and in like sort described. For the first line is to the fourth in triple proportion that it is to the second: and it hath before been proved that the Parallelipipedon described of the first, is to the Parallelipipedon described of the second, in the same proportion that the first line is to the fourth.  
Because the one of the figures before, described in a plain, pertaining to the demonstration of this 33. Proposition, is not altogether so easy to a young beginner to conceive, I have here for the same described an other figure, which if ye first draw upon pasted paper, and afterward cut the lines & fold the sides accordingly, will agreed with the construction & demonstration of the said Proposition. Howbeit this ye must note that ye must cut the lines OQ & MR on the contrary side ●o that which ye cut the other lines. For the solids which have to their base the parallelogram LK are set on upward and the other downward: You may if ye think good describe after the s●me manner of pasted paper a body equal to the solid CD: though that be easy enough to conceive by the figure thereof described in the plain.   

¶ Certain most profitable Corollaries, Annotations, Theorems, and Problems, with other practices, logistical, and Mechanical, partly upon this 33. and partly upon the 34. 36. and other following, added by Master john Dee. ¶ A Corollary. 1.  
1. Hereby it is manifest, that two right lines may be found, which shall have that proportion, the one to the other, that any two like Parallelipipedons, and in like sort described, given, have the one to the other.  
Suppose Q and X to be two like Parallelipipedons, and in like sort described. Of Q take any of the three lines, of which it is produced: as namely, RG. Of X, take that right line of his production, which line ●s answerable to R G in proportion (which most aptly, after the Greek name, may be called Omologall to RG) & let that be TU. By the 11. of the sixth, to RG and TV, let the third line in proportion with them be found, and let that be Y. By the same 11. of the sixth, to TV and Y, let the third right line be found, in the said proportion of TV to Y: & let that be Z. I say now that RG hath that proportion to Z, which Q hath to X. For by construction, we have four right lines in continual propiotion, namely, RG, TV, Y, and Z. Wherefore by Euclides Corollary, here before, RG is to Z, as Q is to X. Wherefore we have found two right lines having that proportion the one to the other, which any two like Parallelipipedons of like description, given, have the one to the other: which was required to be done.   

¶ A Corollary.  
As a Converse, of my ●ormer Corollary, doth it follow: To find two like Parallelipipedons of like description, which shall have that proportion the one to the other, that any two right lines, given, have the one to the other.  
Suppose the two right lines given to be A and B: Imagine of four right lines in continual proportion, A. to be the first, and B to be the fourth: (or contrariwise, B to be first, & A to be the fourth). The second and third are to be found, which may, between A & B, be two means in continual proportion: as now, 
* Note this famous Lemma.  suppose such two lines, found: and let them be C and D. Wherefore by Euclides Corollary, as A is to B (if A were taken as first) so shall the Parallelipipedon described of A, be to the like Parallelipipedon and in like sort described of C: being the second of the four lines in continual proportion: it is to we●e, A, C, D, and B. Or, if B shall be taken as first, (and that thus they are orderly in continual proportion, B, D, C, A,) then, by the said Corollary, as B is to A, so shall the Parallelipipedon described of B, be unto the like Parallelipipedon and in like sort described o● D. And unto a Parallelipipedon of A or B, at pleasure described, may an other of C or D be made like, and in like sort situated or described, by the 27. of this eleventh book. Wherefore any two right lines being given, etc.: which was required to be done.  
Thus have I most briefly brought to your understanding, if (first) B were double to A, than what Parallelipipedon soever, were described of A, the like Parallelipipedon and in like sort described of C, shall be double to the Parallelipipedon described of A. And so likewise (secondly) if A were double to B, the Parallelipipedon of D, should be double to the like, of B described, both being like situated. Wherefore if of A or B, were Cubes made, the Cubes of C and D are proved double to them: 
The doubling o●●he Cube.  as that of C, to the Cube of A: and the Cube of D to the Cube of B: in the second case. 
* Note.  And so of any proportion else between A and B.  
Now also do you most clearly perceive the Mathematical occasion, whereby (first of all men) Hypocrates, to double any Cube given, was led to the former Lemma: Between any two right lines given, to find two other right lines, which shall be with the two first lines, in continual proportion: After whose time (many years) divine Plato, Heron, Philo, Appollonius, Di●●l●●, Pappus, Sporus, Menech●us, Archytas Tarentinus (who made the wooden dove to sly) Erato●●hene, Nicomedes, with many other (to their immortal fame and renown) published, 
* 〈…〉 Lemma.  divers their witty devices, methods, and engines (which yet are extant) whereby to execute this problematical Lemma. But not withstanding all the travails of the foresaid Philosophers and mathematicians, yea and all others doings and contrivinge (unto this day) about the said Lemma, yet there remaineth sufficient matter, Mathematically so to demonstrate the same, that most exactly & readily, it may also be Mechanically practised: that who soever shall achieve that feat, shall not be counted a second Archimedes, but rather a perales Mathematicien, and Mathematicorum Princeps. I will sundry ways (in my brief additions and annotations upon Euclid) excite you thereto, 
Note what i● yet lacking requisite to the doubling of the Cube.  yea and bring before your eyes sundry new ways, by meinuented: and in this book so placed, as matter thereof, to my inventions appertaining, may give occasion: Leaving the farther, full, & absolute my concluding of the Lemma, to an other place and time: which will, now, more compendiously be done: so great a part thereof, being before hand in this book published.   

¶ A Corollary added by Flussas.  
Parallelipipedons consisting upon equal bases, are in proportion the one to the other as their altitudes are. For if those altitudes be cut by a plain superficies parallel to the bases: the sections shall be in proportion the one to the other as the sections of the bases cut, by the 25. of this book. Which sections of the bases are the one to the other in that proportion that their sides or the altitudes of the solids are, by the ●irst of the sixth. Wherefore the solids are the one to the other as their altitudes are. But if the bases be unlike, the self same thing may be proved by the Corollary of the 25. of this book, which by the 25. Proposition was proved in like bases.    

¶ The 29. Theorem. The 34. Proposition. In equal Parallelipipedons the bases are reciprocal to their altitudes. And Parallelipipedons whose bases are reciprocal to their altitudes, are equal the one to the other.  
But now again suppose that the bases of the Parallelipipedons AB and CD be reciprocal to their altitudes, 
The converse of both the parts of the first case.  that is, as the base EH is to the base NP, so let the altitude of the solid CD be to the altitude of the solid AB. Then I say, that the solid AB is equal to the solid CD. For again let the standing lines be erected perpendicularly to their bases.  
And now if the base EH be equal to the base NP: but as the base EH is to the base NP, so is the altitude of the solid CD to the altitude of the solid AB. Wherefore the altitude o● the solid CD is equal to the altitude of the solid AB. But Parallelipipedons consisting upon equal bases and under one and the self same altitude, are (by the 31. of the eleventh) equal the one to the other. Wherefore the solid AB is equal to the solid CD.  
But now suppose that the base EH be not equal to the base NP: but let the base EH be the greater. Wherefore also the altitude of the solid CD, that is, the line CM is greater than the altitude of the solid AB, that is, than the line AG. Put again (by the 3. of the first) the line CT equal to the line AG, and make perfect the solid CZ. Now for that as the base EH is to the base NP, so is the line MC to the line AG. But the line AG is equal to the line CT. Wherefore as the base EH is to the base NP, so is the line CM to the line CT. But as the base EH is to the base NP, so (by the 32. of the eleventh) is the solid AB to the solid CZ, ●or the solides AB and CZ are under equal altitudes. And as the line CM is to the line CT, so (by the 1. of the sixth) is the base MP to the base P●T, and (by the 32. of the eleventh) the solid CD to the solid CZ. Wherefore also (by the 11. and 9 of the fift) as the solid AB is to the solid CZ, so is the solid CD to the solid CZ. Wherefore either of these solids AB and CD have to the solid CZ one and the same proportion. Wherefore (by the 7. of the fift) the solid AB is equal to the solid CD: which was required to be demonstrated.  
Again suppose that the bases of the Parallelipipedons AB and CD be reciprocal to their altitudes, 
The converse of the second case.  that is, as the base EH is to the base NP, so let the altitude of the solid CD be to the altitude of the solid AB. Then I say, that the solid AB is equal to the solid CD. For the same order of construction remaining, for that as the base EH is to the base NP, so is the altitude of the solid CD to the altitude of the solid AB: but the base EH is equal to the parallelogram FK, and the base NP to the parallelogram XR: wherefore as the base FK is to the base XR, so is the altitude of the solid CD to the altitude of the solid AB But the altitudes of the solids AB and BT are equal, and so also are the altitudes of the solids DC and DY. Wherefore as the base FK is to the base XR, so is the altitude of the solid DY to the altitude of the solid BT. Wherefore the bases of the Parallelipipedons BT and DY are reciprocal to their altitudes. But Parallelipipedons whose altitudes are erected perpendicularly to their bases, and whose bases are reciprocal to their altitudes, are equal the one to the other (by this Proposition). Wherefore the solid BT is equal to the solid DY. But the solid BT is equal to the solid BA (by the 29. of the eleventh) for they consist upon one and the self same base, namely, FK, and are under one and the self same altitude, whose standing lines are in the self same right lines. And the solid DY is equal to the solid DC, for they consist upon one and the self same base, namely, XR, and are under one and the self same altitude, whose standing lines are in the self same right lines. Wherefore also the solid AB is equal to the solid CD. 
The general conclusion.  Wherefore in equal Parallelipipedons the bases are reciprocal to their altitudes. And Parallelipipedons whose bases are reciprocal with their altitudes, are equal the one to the other: which was required to be proved.  
The demonstration of the first case of this Proposition is easy to conceive by the figure as it is described in the plain. But ye may for your more full sight describe Parallelipipedons of pasted paper, according as the construction teacheth you.  
And for the second case, if ye remember well the form of the figure● which you made for the second case of the 31. Proposition: and describe the like for this, taking ●eede to the letters that ye place them like as the construction in this case requireth, ye shall most easily by them come to the full understanding of the construction and demonstration of the said case.  

M. john Dee, his sundry Inventions and Annotations, very necessary, here to be added and considered.  

A Theorem.  
If four right lines be in continual proportion, and upon the square of the first, as a base, be erected a rectangle parallelipipedon, whose heith is the fourth line: that rectangle parallelipipedon is equal to the Cube made of the second line. And if upon the square of the fourth line, as a base, be erected a rectangle parallelipipedon, whose heith is the first line, that parallelipipedon is equal to the Cube made of the third line.  
Suppose AB, CD, E●, and ●H to be four right lines in continual proportion: 
Construction.  and upon the square of AB (which let be AI) as a base, l●t be erected a rectangle parallelipipedom, having his heith IK, equal to GH, the fourth line. And let that parallelipipedon be AK. Of the second line CD, let a Cube be made: whose square base, let be noted wi●h CQ: and let his heith b● noted by QL: & let the whole Cube be signified by CL. I say that AK is equal to CL. Let the like construction be for the cube of the third line: that is, upon the square of ●H (which suppose to be GN) let a rectangle parallelipipedon be erected, having his heith NO, equal to A●, the first line: which parallelipipedon let be noted with GO. And suppose the cube of the third line (●F) to be ●M● whose square base, let be noted by ●●● and his heith by RM. I say now (secondly) that GO is equal to ●M. For the first part consider, 
Demonstration of the first part.  that AI (the square base of AK) is to CQ, the square base of CL, as A● is to the third line EF, by the ●. Corollary of the 20. of the sixth. But as A●, is to EF, so (by alternate proportion) is CD to GH, to CD. The cubes root, is QL, the same cubes heith equal: and to GH is IK (by construction) equal: wherefore, as AI is to CQ, so is QL to IK. The bases therefore and heithes of AK and CL, are reciprocally in proportion: wherefore by the second part of this 34. proposition, AK and CL are equal. 
Demonstration of the second part.  For proof of the second part of my theorem, I say, that as AB, CD, ●F, and GH, are in continual proportion forward, so are they backward in continual proportion, as by the fourth of the fift may be proved. Wherefore now considering GH to be as first, and so A● to be the fourth: the square base GN, is to the square base ●K, as GH is to CD, by the 2. corollary of the 20. of the sixth: But as GH is to CD, so is ●● to A●, by alternate proportion: to the Cubik root ●F, is RM (the heith of the same Cube ●M) equal. And to AB, is the heith NO equal, by construction: wherefore as GN is to ●R, so is RM to NO. Therefore by the second part of this 34. proposition, GO is equal to EN. If four right lines (therefore) be in continual proportion etc. as in the proposition: which was required to be demonstrated.    

A Corollary logistical.  
Of my former Theorem it followeth: Any two numbers being given, between which two we would have two other numbers, middle, in continual proportion: 
To find two middle proportionals between two numbers given.  That if we multiply the square of the first number given, by the other given number (as if it were the fourth): the root Cubik of that of come or product, shall be the second number sought: And farther, if we multiply the square of the other number given, by the first given number, the root Cubike of that of come shall be the third number sought.  
For (by my Theorem) those rectangle parallelipipedons made of the squares of the first & fourth, multiplied by the fourth & the first, accordingly, are equal to the Cubes made of the second & third numbers: which we make our two ●iddl● proportionals● Wherefore of those parallelipipedo●s (as Cubes) the Cubik roots, by good and usual art sought and found, give the very two middle numbers desired. And where those numbers, are not by logistical consideration accounted Cubik numbers, ye may use the logistical secret of approaching near to the precise verity: 
Note the practice of approaching to precie●es in Cubik roots.  so that thereof most easily you shall perceive, that your fail is of the s●nce never to be perceived: it is to weet, as in a line of an inch long, not to want or exceed the thousand thousand part: or farther you may (infinitely approach at pleasure. O Mechanical friend, be of good comfort, put to thy hand: Labour improbus, omnia vincit.   

A Problem. 1.  
Upon a right lined plain superficies given, to apply a rectangle parallelipipedon given.   

Or we may thus express the same thing.  
Upon a right lined plain superficies given, to erect a rectangle parallelipipedon, equal to a rectangle parallelipipedon given.  
Suppose the right lined plain superficies given to be ●: and the rectangle parallelipipedon given to be AM. Upon ●, as a base must AM be applied: that is, a rectangle par●llipipedon must be erected upon ●, as a base, which shall be equal to AM. By the last of the second, to the right lined figure ●, let an equal square be made: which suppose to be FRX. 
* This is the way to apply any square given, to a line also given, sufficienty extended.  produce one side of the base of the parallelipipedom AM, which let be AC, extended to the point P. Let the other side of the said base, concurring with AC, be CG. As CG is to FR (the side of the square FRX) so let the same FR be to a line cut of from CP sufficiently extended: by the 11. of the sixth: and let that third proportional line be CP. Let the rectangle parallelogram be made perfect, as CD. It is evident, that CD, is equal to the square FRX by the 17. of the sixth: and by construction FRX, is equal to B. Wherefore CD, is equal to ●. By the 12. of the sixth, as CP, is to AC, so let AN (the heith of AM) be to the right line O. I say that a solid perpendicularly erected upon the base ●, having the heith of the line O, is equal to the parallelipipedon AM. For CD is to AG, as CP is to A● by the first of the sixth, and ● is proved equal to CD: Wherefore by the 7. of the fifth, B is to AG as CP is to AC: But as CP is to AC, so is AN to O, by construction: Wherefore B is to AG as AN is to O. So than the bases ● and AG are reciprocally in proportion with the heithes AN and O. By this 34 therefore, a solid erected perpendicularly upon ● as a base, having the height O, is equal to AM. Wherefore upon a right lined plain superficies given, we have applied a rectangle parallelipipedon given: Which was requisite to be done.   

A Problem 2.  
A rectangle parallelipipedon being given to make an other equal to it of any heith assigned.  
Suppose the rectangle parallelipipedon given to be A, and the heith assigned to be the right line ●: Now must we make a rectangle parallelipipedon, equal to A: Whose heith must be equal to ●. According to the manner before used, we must frame our construction to a reciprocal proportion between the bases and heithes. Which will be done if, as the heith assigned beareth itself in proportion to the heith of the parallelipipedon given: so, one of the sides of the base of the parallelipipedon given, be to a fourth line, by the 12. of the sixth found. For that line found, and the other side of the base of the given parallelipipedon, contain a parallelogram, which doth serve for the base, (which only, we wanted) to use with our given heith: and so is the Problem to be executed.   

Note.  
Euclid in the 27. of this eleventh hath taught, how, of a right line given, to describe a parallepipedom, like, & likewise situated, to a parallelipipedom given: I have also added, How, to a parallepipedon given, an other may be made equal, upon any right lined base given, or of any heith assigned: But if either Euclid, or any other before our time (answerably to the 25. of the sixth, in plains) had among solids invented this proposition: 
A problem worth the searching for.  Two unequal and unlike parallelipipedons being given, to describe a parallelipipedon equal to the one, and like to the other, we would have given them their deserved praise: and I would also have been right glad to have been eased of my great travails and discourses about the inventing thereof.  
Here end I Dee his additions upon this 34. Proposition.    

The 30. Theorem. The 35. Proposition. If there be two superficial angles equal, and from the points of those angles be elevated on high right lines, comprehending together with those right lines which contain the superficial angles, equal angles, each to his corespondent angle, and if in each of the elevated lines be taken a point at all adventures, and from those points be drawn perpendicular lines to the ground plain superficieces in which are the angles given at the beginning, and from the points which are by those perpendicular lines made in the two plain superficieces be joined to those angles which were put at the beginning right lines: those right lines together with the lines elevated on high shall contain equal angles.  
SVppose that these two rectiline superficial angles BAC, and EDF be equal the one to the other: 
Construction.  and from the points A and D let there be elevated upward these right lines AG and DM, comprehending together with the lines put at the beginning equal angles, each to his correspondent angle, that is, the angle MDE to the angle GAB, and the angle MDF to the angle GAC, and take in the lines AG and DM points at all adventures and let the same be G and M. And (by the 11. of the eleventh) from the points G and M draw unto the ground plain superficieces wherein are the angles BAC and E DF perpendicular lines GL and MN and let them fall in the said plain super●icieces in the points N and L, and draw a right line from the point L to the point A and an other from the point N to the point D. Then I say that the angle GAL is equal to the angle MDN. Fron the greater of the two lines AG and DM, (which let be AG) cut of by the 3. of the first the line AH equal unto the line DM. And (by the 31. of the first) by the point H, draw unto the line GL a parallel line, and let the same be HK. Now the line GL is erected perpendicularly to the ground plain superficies BAL: Wherefore also (by the 8. of the eleventh) the line HK is erected perpendicularly to the same ground plain superficies BAC. Draw (by the 12. of the first) from the points K and N unto the right lines AB, AC, DF, & DE perpendicular right lines, and let the same be KC, NF, KB, NE. And draw these right lines HC, CB, MF, FE. Now forasmuch a● (by the 47. of the first) the square of the line HA is equal to the squares of the lines HK and KA, 
Demonstration.  but unto the square of the line KA are equal the squares of the lines KC and CA: Wherefore the square of the line HA is equal to the squares of the lines HK, KC and CA But by the same unto the squares of the lines HK and KC is equal the square of the line HC: Wherefore the square of the line HA is equal to the squares of the lines HC and CA: wherefore the angle HCA is (by the 48. of the first) a right angle. And by the same reason also the angle MFD is a right angle. Wherefore the angle HCA is equal to the angle MFD. But the angle HAC is (by supposition) equal to the angle MDF. Wherefore there are two triangles MDF and HAC having two angles of the one equal to two angles of the other, each to his correspondent angle, and one side of the one equal to one side of the other, namely, that side which subtendeth one of the equal angles, that is, the side HA is equal to the side DM by construction. Wherefore the sides remaining are (by the 26. of the first) equal to the sides remaining. Wherefore the side AC is equal to the side DF. In like sort may we prove that the side AB is equal to the side DE, if ye draw a right line from the point H to the point B, and an other from the point M to the point E. For forasmuch as the square of the line AH is (by the 47. of the first) equal to the squares of the lines AK and KH, and (by the same) unto the square of the line AK are equal the squares of the lines AB and BK. Wherefore the squares of the lines AB, BK, and KH are equal to the square of the line AH. But unto the squares of the lines BK and KH is equal the square of the line BH (by the 47. of the first) for the angle HKB is a right angle, for that the line HK is erected perpendicularly to the ground plain superficies: Wherefore the square of the line AH is equal to the squares of the lines AB and BH. Wherefore (by the 48. of the first) the angle ABH is a right angle. And by the same reason the angle DEM is a right angle. Now the angle BAH is equal to the angle EDM, for it is so supposed, and the line AH is equal to the line DM. Wherefore (by the 26. of the first) the line AB is equal to the line DE. Now forasmuch as the line AC is equal to the line DF, and the line AB to the line DE, therefore these two lines AC and AB are equal to these two lines FD and DE. But the angle also CAB is by supposition equal to the angle FDE. Wherefore (by the 4. of the first) the base BC is equal to the base EF, and the triangle to the triangle, and the rest of the angles to the rest of the angles. Wherefore the angle ACB is equal to the angle DFE. And the right angle ACK is equal to the right angle DFN. Wherefore the angle remaining, namely, BCK, is equal to the angle remaining, namely, to EFN. And by the same reason also the angle CBK is equal to the angle FEN. Wherefore there are two triangles BCK, & EFN, having two angles of the one equal to two angles of the other, each to his correspondent angle, and one side of the one equal to one side of the other, namely, that side that lieth between the equal angles, that is the side BC is equal to the side EF: Wherefore (by the 26. of the first) the sides remaining are equal to the sides remaining. Wherefore the side CK is equal to the side FN: but the side AC is equal to the side DF. Wherefore these two sides AC and CK are equal to these two sides DF and FN, and they contain equal angles: Wherefore (by the 4. of the first) the base AK is equal to the base DN. And forasmuch as the line AH is equal to the line DM, therefore the square of the line AH is equal to the square of the line DM. But unto the square of the line AH are equal the squares of the lines AK and KH (by the 47. of the first) for the angle AKH is a right angle. And to the square of the line DM are equal the squares of the lines DN and NM, for the angle DNM is a right angle. Wherefore the squares of the lines AK and KH are equal to the squares of the lines DN and NM: of which two, the square of the line AK is equal to the square of the line DN (for the line AK is proved equal to the line AN). Wherefore the residue, namely, the square of the line KH is equal to the residue, namely, to the square of the line NM. Wherefore the line HK is equal to the line MN. And forasmuch as these two lines HA and AK are equal to these two lines MD and DN, the one to the other, and the base HK is equal to the base MN: therefore (by the 8. of the first) the angle HAK is equal to the angle MDN. If therefore there be two superficial angles equal, and from the points of those angles be elevated on high right lines, comprehending together with those right lines which were put at the beginning, equal angles, each to his corespondent angle, and if in each of the erected lines be taken a point at all adventures, and from those points be drawn perpendicular lines to the plain superficieces in which are the angles given at the beginning, and fr●● the points which are by the perpendicular lines made in the two plain superficieces be joined right lines to those angles which were put at the beginning, those right lines shall together with the lines elevated on high make equal angles which was required to be proved.  
Because the figures of the former demonstration are somewhat hard to conceive as they are there drawn in a plain, by reason of the lines that are imagined to be elevated on high, I have here set other figures, wherein you must erect perpendicularly to the ground superficieces the two triangles BHK, and EMN, and then elevate the triangles DFM, & ACH, in such sort that the angles M and H of these triangles, may concur with the angles M and H of the other erected triangles. And then imagining only a line to be drawn from the point G of the line AG to the point L in the ground superficies, compare it with the former construction & demonstration, and it will make it very easy to conceive.  

¶ Corollary.  
By this it is manifest, that if there be two rectiline superficial angles equal, and upon those angles be elevated on high equal right lines containing together with the right lines put at the beginning equal angles: perpendicular lines drawn from those elevated lines to the ground plain superficieces wherein are the angles put at the beginning, are equal the one to the other. For it is manifest, that the perpendicular lines HK, & MN, which are drawn from the ends of the equal elevated lines AH, and DM, to the ground superficieces, are equal.    

¶ The 31. Theorem. The 36. Proposition. If there be three right lines proportional: a Parallelipipedon described of those three right lines, is equal to the Parallelipipedon described of the middle line, so that it consist of equal sides, and also be equiangle to the foresaid Parallelipipedon.  
SVppose that these three lines A, B, C, be proportional, as A is to B, so let B be to C. Then I say, that the Parallelipipedon made of the lines A, B, C, is equal to the Parallelipipedon made of the line B, so that the solid made of the line B consist of equal sides, and be also equiangle to the solid made of the lines A, B, C. Describe (by the 23. of the eleventh) a solid angle E contained under three superficial angles, 
Construction.  that is, DEG, GEF, and FED: and (by the 3. of the first) put unto the line B every one of these lines DE, GE, & EF, equal: and make perfect the solid EK. And unto the line A let the line LM be equal. And (by the 26. of the eleventh) unto the right line LM, and at the point in it L, describe unto the solid angle E an equal solid angle, contained under these plain superficial angles NLX, XLM, and NLM, and unto the line B put the line LX equal, & the line LN to the line C. Now for that as the line A is to the line B, 
Demonstration.  so is the line B to the line C: but the line A is equal to the line LM, and the line B to every one of these lines LX, EF, EG, and ED, and the line C to the line LN. Wherefore as LM is to EF, so is DE to LN: So than the sides about the equal angles MLN, & DEF, are reciprocal: Wherefore (by the 14. of the sixth) the parallelogram MN is equal to the parallelogram DF. And forasmuch as two plain superficial angles, namely, DEF and NLM are equal the one to the other, and upon them are erected upward equal right lines, LX and EG; comprehending with the right lines put at the beginning equal angles the one to the other. Wherefore 
* It is evident that those perpendiculars are all one with the standing lines of the solids, if their solid angles be made of super●ic●●il right angles only.  perpendicular lines drawn from the points X and G to the plain super●icieces wherein are the angles NLM, and DEF, are (by the Corollary of the former Proposition) equal the one to the other: and those perpendiculars are the altitudes of the Parallelipipedons LH and EK, by the 4. definition of the sixth. Wherefore the solids LH and EK, are under one and the self same altitude. But Parallelipipedons consisting upon equal bases, and being under one and the self altitude, are (by the 31. of the eleventh) equal the one to the other. Wherefore the solid LH is equal to the solid EK. But the solid LH is described of the lines A, B, C, and the solid EK is described of the line B. Wherefore the Parallelipipedon described of the lines A, B, C, is equal to the Parallelipipedon made of the line B, which consisteth of equal sides, and is also equiangle to the foresaid Parallelipipedon. If therefore there be three right lines proportional, a Parallelipipedom described of those three lines is equal to the Parallelipipedom described of the middle line, so that is consist of equal sides, and also be equiangle to the foresaid Parallelipipedon: which was required to be proved.  
The construction and demonstration of this Proposition, and of the next Proposition following, may easily be conceived and understanded by the figures described in the plain belonging to them. But ye may for the more full sight of them, describe such bodies of pasted paper, having their sides proportional, as is required in the Propositions.  

¶ New inventions (coincident) added by Master john Dee.  

A Corollary. 1.  
Hereby it is evident, that if three right lines be proportional: the Cube produced of the middle line, is equal to the rectangle Parallelipipedon made of those three lines.  
For a Cube is a Parallelipipedon of equal sides: and also rectangled: as we suppose the Parallelipipedon, made of the three lines to be likewise rectangled.  

¶ A Problem. 1.  
A Cube being given, to find three right lines proportional, in any proportion given between two right lines: of which three lines, the rectangle Parallelipipedon produced, shall be equal to the Cube given.  
Suppose AC to be the Cube given: whose root, suppose to be AB. Let the proportion given, be that which in between the two right lines D and E, I say now, three right lines are to be found, proportional, in the proportion of D to E, of which, the rectangle Parallelipipedon produced, shall be equal to AC. By the 12. of the sixth let a line be found, which to AB have that proportion that D hath to E. Let that line be F: and by the same 12. of the sixth, let an other line be found, to which, AB, hath that proportion that D hath to E: and let that line found be H. Let a rectangle Parallelipipedon mathematically be produced of the three right lines F, AB, and H, which suppose to be K: I say now, that F, AB, and H, are three right lines found proportional in the proportion of D to E, of which, the rectangle Parallelipipedon K, produced, is equal to AC the Cube given. First it is evident that F, AB, and H, are proportional in the proportion of D to E. For, by construction, as D is to E, so is F to AB: and by construction likewise, as D is to E, so is AB to H. Wherefore F is to AB, and AB is to H, as D is to E. So then it is manifest, F, AB, and H, to be proportional in the proportion of D to E, and AB to be the middle line. By my former Corollary, therefore, the rectangle parallelipipedon made of F, AB, and H, is equal to the Cnbe made of AB. But AC, is (by supposition) the Cube made of AB● and of the three lines F, AB, and H, the rectangle parallelipipedon produced, is K, by construction: Wherefore, K, is equal to AC: A Cube being given, therefore, three right lines are found, proportional in ●ny prop●●●ion given between two right lines, of which three right lines the rectangle parallelipipedon produced, is equal to the Cube given. Which aught to be done.   

A Problem. 2.  
A rectangle Parallelipipedon being given, to find three right lines proportional: of the which, the rectangle Parallelipipedon produced, is equal to the rectangle Parallelipipedon given.  

Doubling of the 〈◊〉. etc.  Listen to this new devise, you courageous mathematicians: consider, how near this creepeth to the famous Problem of doubling the Cube. What hope may (in manner) any young beginner conceive, by one means or other, at one time or other, to execute this Problem? 
* D●m●●●●ratio●●f possibility in the problem.  Seeing to a Cube may infinitely infinite Parallelipipedons be found equal: all which Parallelipipedons shall be produced of three right lines proportional, by the fo●me● Problem: but to any rectangle Parallelipipedon given, some one Cube is equal ● as is ●asie to demonstrate: We can not doubt, but unto our rectangle Parallelipipedon given, many other rectangle Parallelipipedons are also equal, having their three lines of production, proportional. 
another argument to com●o●t the studious.  In th● former Problems infinitely infinite Parallelipipedons may be found of three proportional lines produced, equal to the Cube given: it is to weet, the three lines to be of all proportions, that a man can devise between two right lines: and here any one will serve: where also i● infinite variety: though all of one quantity: as be●ore in the Cube. I leave as now, with this mark here set up to shoot at. Hit it who can.      

¶ The 32. Theorem. The 37. Proposition. If there be four right lines proportional: the Parallelipipedons described of those lines, being like and in like sort described, shall be proportional. And i● the Parallelipipedons described of them, being like and in like sort described, be proportional: those right lines also shall be proportional.  
SVppose that these four right lines AB, CD, EF, and GH, be proportional, as AB is to CD, so let EF be to GH, and upon the lines AB, CD, EF, and GH, describe these Parallelipipedons KA, LC, ME, and NG, being like and in like sort described. Then I say, that as the solid KA is ●o the solid LC, so is the solid ME to the solid NG. 
Demonstration of the first part.  For forasmuch as the Parallelipipedon KA is like to the Parallelipipedon LC: therefore (by the 33. of the eleventh) the solid KA is to the solid LC in triple proportion of that which the side AB is to the side CD●: and by the same reason the Parallelipipedon ME is to the Parallelipipedon NG in triple proportion of that which the side EF is to the side GH. Wherefore (by the 11. of the fift) as the Parallelipipedon KA is to the Parallelipipedon LC, so is the Parallelipipedon ME t● the Parallelipipedon NG.  
But now suppose, that as the Parallelipipedon KA is to the Parallelipipedon LC, so is the Parallelipipedon ME to the Parallelipipedon NG. Then I say, 
Demonstration of the second part, which is the converse of the first part.  that as the right line AB is to the right line CD, so is the right li●● EF to the right line GH. For again forasmuch as the solid KA is to the solid LC in triple proportion of that which the side AB is to the side CD, and the solid ME also is to the solid NG in triple proportion of that which the line EF is to the line GH, and as the solid KA is to the solid LC, so is the solid ME to the solid NG. Wherefore also as the line AB is to the line CD, so is the line EF to the line GH. If therefore there be four right lines proportional: the Parallelipipedons described of those lines, being like & in like sort described, shall be proportional. And if the Parallelipipedons described of them, and being like and in like sort described, be proportional: those right lines also shall be proportionally which was required to be proved.   

¶ The 33. Theorem. The 38. Proposition. If a plain superficies be erected perpendicularly to a plain superficies, and from a point taken in one of the plain superficieces be drawn to the other plain superficies, a perpendicular line: that perpendicular line shall fall upon the common section of those plain superficieces.  
SVppose that the plain superficies CD be erected perpendicularly to the plain superficies AB, and let their common section be the line DA: and in the superficies CD take a point at all adventures, and let the same be E. Then I say, that a perpendicular line drawn from the point E to the plain superficies AB, shall fall upon the right line DA. For if not, then let it fall without the line DA, as the line EF doth, and let it fall upon the plain superficies AB in the point F. 
Demonstration leading to an impossibility.  And (by the 12. of the first) from the point F draw unto the line DA, being in the superficies AB a perpendicular line FG, which line also is erected perpendicularly to the plain superficies CD: by the third definition: by reason we presuppose CD and AB to be perpendicularly erected each to other. Draw a right line from the point E to the point G. And forasmuch as the line FG is erected perpendicularly to the plain superficies CD, and the line EG toucheth it being in the superficies CD. Wherefore the angle FGE is (by the 2. definition of the eleventh) a right angle. But the line EF is also erected perpendicularly to the superficies AB: wherefore the angle EFG is a right angle. Now therefore two angles of the triangle EFG, are equal to two right angles: which (by the 17. of the first) is impossible. Wherefore a perpendicular line drawn from the point E to the superficies AB, falleth not without the line DA. Wherefore it falleth upon the line DA: which was required to be proved.  

¶ Note.  
Campane maketh this as a Corollary, following upon the 13: and very well, with small aid of other Propositions, he proveth it● whose demonstration there, Flussas hath in this place, and none other: though he saith that Campane of such a Proposition, as of Euclides, maketh no mention.  
In this figure ye may more fully see the former Proposition and demonstration if ye erect perpendicularly unto the ground plain superficies AB the superficies CD, and imagine a line to be extended from the point E to the point F, instead whereof ye may extend if ye will a thread.    

¶ The 34. Theorem. The 39 Proposition. If the opposite sides of a Parallelipipedon be divided into two equal parts, and by their common sections be extended plain superficieces: the common section of those plain superficieces, and the diameter of the Parallelipipedon shall divide the one the other into two equal parts.  
SVppose that AF be a Parallelipipedon, and let the opposite sides thereof CF and AH be divided into two equal parts in the points K, L, M, N, and likewise let the opposite sides AD and GF be divided into two equal parts in the point●s X, P, O, R, and by those sections extend these two plain superficieces KN & XR, and let the common section of those plain superficieces be the line VS, and let the diagonal line of the solid AB be the line DG. Then I say, that the lines VS and DG do divide the one the other into two equal parts, 
Construction.  that is, that the line VT is equal to the line TS, and the line DT to the line TG. Draw these right lines DV, VE, BS, and SG. 
Demonstration.  Now forasmuch as the line DX is a parallel to the line OE, therefore (by the 29. of the first) the angles DXV and VOE being alternate angles, are equal the one to the other. And forasmuch as the line DX is equal to the line OE, and the line XV to the line VO, and they comprehend equal angles: Wherefore the base DV is equal to the base VE (by the 4. of the first) and the triangle DXV is equal to the triangle VOE, and the rest of the angles to the rest of the angles. Wherefore the angle XVD is equal to the angle OVE. Wherefore DVE is one right line, and by the same reason BSG is also one right line, and the line BS is equal to the line SG. And forasmuch as the line CA is equal to the line DB, and is unto it a parallel, but the line CA is equal to the line GE, and is unto it also a parallel: wherefore (by the firs● common sentence) the line DB is equal to the line GE, & is also a parallel unto it: but the right lines DE and BG do join these parallel lines together: Wherefore (by the 33. of the first) the line DE is a parallel unto the line BG. And in either of these lines are taken points at all adventures, namely, D, V, G, S, and a right line is drawn from the point D to the point G, and an other from the point V to the point S. Wherefore (by the 7. of the eleventh) the lines DG and VS are in one and the self same plain superficies. And forasmuch as the line DE is a parallel to the line BG, therefore (by the 24. of the first) the angle EDT is equal to the angle BGT, for they are alternate angles, and likewise the angle DTV is equal to the angle GTS. Now then there are two triangles, that is, DTV and GTS, having two angles of the one equal to two angles of the other, and one side of the one equal to one side of the other, namely, the side which subtendeth the equal angles, that is, the side DV to the side GS, for they are the halves of the lines DE and BG: Wherefore the sides remaining are equal to the sides remaining. Wherefore the line DT is equal to the line TG, and the line VT to the line TS. If therefore the opposite sides of a Parallelipipedon be divided into two equal parts, and by their sections be extended plain superficieces, the common section of those plain superficieces, and the diameter of the Parallelipipedon, do divide the one the other into two equal parts: which was required to be demonstrated.  

A Corollary added by Flussas.  
Every plain superficies extended by the centre of a parallelipipedon, divideth that solid into two equal parts: and so doth not any other plain superficies not extended by the centre.  
For every plain extended by the centre, cutteth the diameter of the parallelipipedon in the centre into two equal parts. For it is proved, that plain superficieces which cut the solid into two equal parts, do cut the dimetient into two equal parts in the centre. Wherefore all the lines drawn by the centre in that plain superficies shall make angles with the dimetient. And forasmuch as the diameter falleth upon the parallel right lines of the solid, which describe the opposite sides of the said solid, or upon the parallel plain superficieces of the solid, which make angels at the ends of the diameter: the triangles contained under the diameter, and the right line extended in that plain by the centre, and the right line, which being drawn in the opposite superficieces of the solid, joineth together the ends of the foresaid right lines, namely, the end of the diameter, and the end of the line drawn by the centre in the superficies extended by the centre, shall always be equal, and equiangle, by the 26. of the first. For the opposite right lines drawn by the opposite plain superficieces of the solid do make equal angles with the diameter, forasmuch as they are parallel lines, by the 16. of this book. But the angles at the centre are equal, by the 15. of the first, for they are head angles: & one side is equal to one side, namely, half the dimetient. Wherefore the triangles contained under every right line drawn by the centre of the parallelipipedon in the superficies, which is extended also by the said centre, and the diameter thereof, whose ends are the angles of the solid, are equal, equilater, & equiangle (by the 26. of the first). Wherefore it followeth that the plain superficies which cutteth the parallelipipedon, doth make the parts of the bases on the opposite side, equal, and equiangle, and therefore like, and equal both in multitude, and in magnitude: wherefore the two solid sections of that solid, shallbe equal and like, by the 8. definition of this book. And now that no other plain superficies, besides that which is extended by the centre, divideth the parallelipipedon into two equal parts, it is manifest: if unto the plain superficies which is not extended by the centre, we extend by the centre a parallel plain superficies (by the Corollary of the 15. of this book). For forasmuch as that superficies which is extended by the centre, doth divide the parallelipipedom into two equal par●●: it is manifest, that the other plain superficies (which is parallel to the superficies which divideth the solid into two equal parts) is in one of the equal parts of the solid: wherefore seeing that the whole is ever greater than his parts, it must needs be that one of these sections is less than the half of the solid, and therefore the other is greater.  
For the better understanding of this former proposition, & also of this Corollary added by Flussas, it shallbe very needful for you to describe of pasted paper or such like matter a parallelipipedom or a Cube, and to divide all the parallelograms thereof into two equal parts, by drawing by the centres of the said parallelograms (which centres are the points made by the cutting of diagonal lines drawn from th● opposite angles of the said parallelograms) lines parallels to the sides of the parallelograms: as in the former figure described in a plain ye may see, are the six parallelograms DE, EH, HA, AD, DH, and CG, whom these parallel lines drawn by the centres of the said parallelograms, namely, XO, OR, PR, and PX, do divide into two equal parts: by which four lines ye must imagine a plain superficies to be extended, also these parallel lines KL, LN, NM, and M●, by which four lines likewise y● must imagine a plain superficies to be extended ye: may if ye will put within your body made thus of pasted paper, two superficieces made also of the said paper, having to their limits lines equal to the foresaid parallel lines: which superficieces must also be divided into two equal parts by parallel lines drawn by their centres, and must cut the one the other by these parallel lines. And for the diameter of this body, extend a thread from one angle in the base of the solid to his opposite angle, which shall pass by the centre of the parallelipipedon, as doth the line DG in the figure before described in the plain. And draw in the base and the opposite superficies unto it, diagonal lines, from the angles from which is extended the diameter of the solid: as in the former description are the lines BG and DE. And when you have thus described this body, compare it with the former demonstration, and it will make it very plain unto you, so your letters agreed with the letters of the figure described in the book. And this description will plainly set forth unto you the corollary following that proposition. For where as to the understanding of the demonstration of the proposition the superficieces put within the body were extended by parallel lines drawn by the centres of the bases of the parallelipipedon: to the understanding of the said Corollary, ye may extend a superficies by any other lines drawn in the said bases, so that yet it pass through the midst of the thread, which is supposed to be the centre of the parallelipipedon.    

¶ The 35. Theorem. The 40. Proposition. If there be two prisms under equal altitudes, & the one have to his base a parallelogram, and the other a triangle, and if the parallelogram be double to the triangle: those prisms are equal the one to the other.  
SVppose that these two prisms ABCDEF, GHKMON, be under equal altitudes, and let the one have to his base the parallelogram AC, and the other the triangle GHK, and let the parallelogram AC be double to the triangle GHK. Then I say, that the Prism ABCDEF is equal to the Prism GHKMON. 
Construction.  Make perfect the Parallelipipedons AX & GO. And forasmuch as the parallelogram AC is double to the triangle GHK, 
Demonstration.  but the parallelogram GH is also (by the 41. of the first) double to the triangle GHK, wherefore the parallelogram AC is equal to the parallelogram GH. But Parallelipipedons consisting upon equal bases and under one and the self same altitude, are equal the one to the other (by the 31. of the eleventh). Wherefore the solid AX is equal to the solid GO. But the half of the solid AX is the Prism ABCDEF, and the half of the solid GO is the Prism GHKMON. Wherefore the Prism ABCDEF is equal to the Prism GHKMON. If therefore there be two prisms under equal altitudes, and the one have to his base a parallelogram, & the other a triangle, and if the parallelogram be double to the triangle: those prisms are equal the one to the other: which was required to be proved.  
This Proposition and the demonstration thereof are not hard to conceive by the former figures: but ye may for your fuller understanding of them take two equal Parallelipipedons equilater and equiangle the one to the other described of pasted paper or such like matter, and in the base of the one Parallelipipedon draw a diagonal line, and draw an other diagonal line in the upper superficies opposite unto the said diagonal line drawn in the base. And in one of the parallelograms which are set upon the base of the other Parallelipipedon draw a diagonal line, and draw an other diagonal line in the parallelogram opposite to the same. For so if ye extend plain superficieces by those diagonal lines there will be made two prisms in each body. You must take heed that ye put for the bases of each of these Parallelipipedons equal parallelograms. And then note them with letters according to the letters of the figures before described in the plain. And compare them with the demonstration, and they will make both it and the Proposition very clear unto you. They will also give great light to the Corollary following added by Flussas.  

A Corollary added by Flussas.  
By this and the former propositions it is manifest, that prisms and solids 
* Which of some are called sided Columns.  contained under two poligo●on figures equal, like, and parallels, and the rest parallelograms: may be compared the one to the other after the self same manner that parallelipipedons are.  
For forasmuch as (by this proposition and by the second Corollary of the 2●. of this book) it is manifest, that every parallelipipedon may be resolved into two like, and equal prisms, of one and the same altitude, whose base shallbe one and the self same with the base of the parallelipipedon or the half thereof, which Pris●es also shallbe cont●yned under the self same side● with the parallelipipedom, the said side● being also sides of like proportion: I say that Prisme● may be compared together after the like manner that their Parallelipipedon● are● For if we would divide a Prism like unto his foli●e by the 25. of this book, ye shall find in the Corollaryes of the 25. proposition, that that which is set forth touching a parallelipipedon, followeth not only in a Prism, but also in any sided column whose opposite bases are equal, and like, and his sides parallelograms.  
If it be required by the 27. proposition upon a right line given to describe a Prism like and in like sort situate to a Prism given: describe ●●●st the whole parallelipipedon whereof the prism given is the half (which thing ye see by this 40. proposition may be done). And unto that parallelipipedom describe upon the right line given by the said 27. proposition an other parallelipipedon like: and the half thereof shallbe the prism which ye seek for, namely, shallbe a prism described upon the right line given, and like unto the prism given.  
In deed prisms can not be cut according to the 28. proposition. For that in their opposite sides can be drawn no diagonal lines: howbeit by that 28. proposition those prisms are manifestly confirmed to be equal and like, which are the halves of one and the self same parallelipipedon.  
And as touching the 29. proposition, and the three following it, which proveth that parallelipipedons under one and the self same altitude, and upon equal bases, or the self same bases, are equal: or if they be under one and the self same altitude they are in proportion the one to the other, as their bases are● to apply these comparisons unto 〈◊〉 it is to 〈◊〉 required, that the bases of the prisms compared together, be either all parallelograms, or all tria●gles. For so one and the self altitude remaining, the comparison of things equal 〈◊〉 one and th●● self same, and the halves of the bases are ever the one to the other in the same proportion, that their wholes are. Wherefore prisms which are the halves of the parallelipipedons, and which have the same proportion the one to the other that the whole parallelipipedons have, which are under one and th● self●●ame altitude: must needs 'cause that their bases being the halves of the base● of the parallelip●p●●●●●●e in the same proportion the one to the other, that their whole parallelipip●don● are. If therefore the w●ole parallelipipedons be in the proportion of the whole bases, their h●l●●● also (which are prisms) shallbe in thee (proportion either of the wholes if their bases be parallel●gr●mm●●● or of the hal●●●●f they be triangles, which is ever all one by the 15. of the fifth.  
And forasmuch as by the 33. proposition, like parallelipipedons which are the doubles of their prisms are in triple proportion the one to the other that their sides of like proportion are, it is manifest, that prisms being their halves (which have the one to the other the same proportion that their wholes have, by the 15 of the fifth) and having the self same sides that thei● parallelipipedons have, are the one to the other in triple proportion of that which the sides of like proportion are.  
And for that prisms are the one to the other in the same proportion that their parallelipipedons are, and the bases of the prisms (being all either triangles or parallelograms) are the one to the other in the same proportion that the bases of the parallelipipedons are, whose altitudes also are always equal, we may by the 34. proposition conclude, that the bases of the prisms and the bases of the parallelipipedons their doubles (being each the one to the other in one and the self same proportion) are to the altitudes, in the same proportion that the bases of the double solids, namely, of the parallelipipedons are. For if the bases of the equal parallelipipedons be reciprocal with their altitudes, than their halves which are prisms shall have their bases reciprocal with their altitudes.  
By the 36. proposition we may conclude, that if there be three right lines proportional, the angle of a Prism made of these three lines (being common with the angle of his parallelipipedon which is double) doth make a prism, which is equal to the Prism described of the middle line, and containing the like angle, consisting also of equal sides. For a● in the parallelipipedon, so also in the Prism, this one thing is required, namely, that the three dimensions of the proportional lines do make an angle like unto the angle contained of the middle line taken three times. Now than if the solid angle of the Prism be made of those three right lines, there shall of them be made an angle like to the angle of the parallelipipedon which is double unto it. Wherefore it followeth of necessity, that the prisms which are always the halves of the Parallelipipedons, are equiangle the one to the other, as also are their doubles, although they be not equilater: and therefore those halves of equal solids are equal the one to the other: namely, that which is described of the middle proportional line is equal to that which is described of the three proportional lines.  
By the 37. proposition also we may conclude the same touching prisms which was concluded touching Parallelipipedons. For forasmuch as prisms, described like & in like sort of the lines given, are the halves of the Parallelipipedons which are like and in like sort described, it followeth that these prisms have the one to the other the same proportion that the solids which are their doubles have. And therefore if the lines which describe them be porportionall, they shallbe proportional, and so conversedly according to the rule of the said 37. proposition.  
But forasmuch as the 39 proposition supposeth the opposite superficial sides of the solid to be parallelograms, and the same solid to have one diameter, which things a Prism can not have, therefore this proposition can by ●o means by applied to prisms.  
But as touching solids whose bases are two like, equal, and parallel poligonon figures, and their sides are parallelograms, 
Sided Columns.  forasmuch as by the second Corollary of the 25. of this book it hath been declared, that such solids are composed of prisms, it may easily be proved that their nature is such, as is the nature of the prisms, whereof they are composed. Wherefore a parallelipipedon being by the 27. proposition of this book described, there may also be described the half thereof, which is a Prism: and by the description of prisms, there may be composed a solid like unto a solid given composed of prisms. So that it is manifest, that that which the 29. 30. 31. 32. 33. 34. and 37. propositions set forth touching parallelipipedons, may well be applied also to these kinds of solids.    
The end of the eleventh book of Euclides Elements.   

¶ The twelfth book of Euclides Elements.  
IN THIS TWELVETH BOOK, EUCLID setteth forth the passions and proprieties of Pyramids, prisms, Cones, Cylinders, and Spheres. And compareth Pyramids, first to Pyramids, then to prisms: so likewise doth he Cones, and Cylinders. And lastly he compareth Spheres the one to the other. But before he goeth to the treaty of those bodies, he proveth that, like Poligonon figures inscribed in circles, and also the circles themselves are in proportion the one to the other, as the squares of the diameters of those circles are. Because that was necessary to be proved, for the confirmation of certain passions and proprieties of those bodies.  

¶ The 1. Theorem. The 1. Proposition. Like Poligonon figures described in circles: are in that proportion the one to the other, that the squares of their diameters are.  
SVppose that there be two circles ABCDE, and FGHKL, and in them let there be described like Poligonon figures, namely, ABCDE, and FGHKL, and let the diameters of the circles be BM, and GN. Then I say, that as the square of the line BM is to the square of the line GN, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. 
Construction.  Draw these right lines BE, AM, GL, and FN. And forasmuch as the Poligonon figure ABCDE is like to the Poligonon figure FGHKL, 
Demonstration.  therefore the angle BAE is equal to the angle GFL, and as the line BA is to the line AE, so is the line GF to the line FL (by the definition of like Poligonon figures). Now therefore there are two triangles BAE and GFL, having one angle of the one equal to one angle of the other, namely, the angle BAE equal to the angle GFL, and the sides about the equal angles are proportional. Wherefore (by the first definition of the sixth) the triangle ABE is equiangle to the triangle FGL. Wherefore the angle AEB is equal to the angle FLG. But (by the 21. of the third) the angle AEB is equal to the angle AMB, for they consist upon one and the self same circumference: and by the same reason the angle FLG is equal to the angle FNG. Wherefore the angle AMB is equal to the angle FNG. And the right angle BAM, is (by the 4. petition) equal to the right angle GFN. Wherefore the angle remaining, is equal to the angle remaining. Wherefore the triangle AMB is equiangle to the triangle FNG. Wherefore proportionally as the line BM is to the line GN, so is the line BA to the line GF. But the square of the line BM is to the square of the line GN in double proportion of that which the line BM is to the line GN (by the Corollary of the 20. of the sixth). And the Poligonon figure ABCDE is to the Poligonon figure FGHKL in double proportion of that which the line BA is to the line GF (by the ●0. of the sixth). Wherefore (by the 11. of the fift) as the square of the line BM is to the square of the line GN, so is the Poligonon figure ABCDE, to the Poligonon figure FGHKL. Wherefore like Poligonon figures described in circles, are in that proportion the one to the other, that the squares of the diameters are: which was required to be demonstrated.  

¶ john Dee his fruitful instructions, with certain Corollaries, and their great use.  
WHo can not easily perceive, what occasion and aid, Archimedes had, by these first & second Propositions, to ●inde the near Area, or Content of a circle: between a Poligonon figure within the circle, and the like about the same circle, described? Whose precise quantities are most easily known: being comprehended of right lines. Where also (to avoid all occasions of error) it is good in numbers, not having precise square roots, to use the logistical process, according to the rules, with √ ● 12, √ ● 19, and so, of such like. Who can not readily fall into Archimedes reckoning and account, by his method? To find the proportion of the circumference of any circle to his diameter, to be almost triple, and one seventh of the diameter: but to be more than triple and ten one & seventithes: that is to be less than 3 1/● and more than 3 10/71. And where Archimedes used a Poligonon figure of 96. sides: he that, for exercise sake, or for earnest desire of a more nearness, will use Polygonon figures of 384. sides (or more) may well travail therein, till either weariness 'cause him stay, or else he find his labour fruitless. In deed Archimedes concluded proportion, of the circumference to the diameter, hath hitherto served the vulgar and● Mechanical wor● men: wherewith, who so is not concented, let his own Methodical travail satisfy his desire: or let him procure other thereto. For, narrower terms (of greater and less) found, and appointed to the circumference, will also win to the Area of the circle a nearer quantity: seeing, it is well demonstrated of Archimed●s, that a triangle rectangle, of whose two sides (containing the right angle) one is equal to the semidiameter of the circle, and the other to the circumference of the same, is equal to the Area of that circle. Upon which two Theorems, it followeth, that the square made of the diameter, is in that proportion to the circle (very near) in which, 14, is to 11● Wherefore every circle is— eleven fowertenthes (well near) of the square about him described. The one side, then, of that square, divide into 14. equal parts: 
The squaring of the circle.  and from that point which endeth the eleventh part, draw to the opposite side, a line, parallel to the other sides, and so make perfect the parallelogram. Then, by the last Proposition of the second book, unto that parallelogram (whose one side hath those 11. equal parts), make a square equal. Then is it evident, that square to be equal to the circle, about which the first square is described. As ye may here behold in these figures.  
Gentle friend, the great desire, which I have, that both with pleasure and also profit, thou mayest spend thy time in these excellent studies, doth 'cause me here to furnish thee somewhat (extraordinarily) about the circle: not only by pointing unto thee, the wellspring of Archimedes, his so much wondered at; and justly commended travail (in the former 3. Theorems, here repeated), but also to make thee more apt, to understand and practise this and other books following, where, use of the circle may be had in any consideration: as in Cones, Cylinders, and Spheres, etc.  

¶ A Corollary. 1.  
By Archimedes second Theorem (as I have here alleged them) it is manifest, that a parallelogram contained, either under the semidiameter and half the circumference, or under the half semidiameter and whole circumference of any circle, is equal to the circle: by the 41. of the first: and first of the six●.   

¶ A Corollary. 2.  
Likewise it is evident, that the parallelogram contained under the semidiameter, and half of any portion of the circumference of a circle given, is equal to that sector of the same circle, to which the whole portion of the circumference given, doth belong. Or you may use the half semidiameter, and the whole portion of the circumference as sides of the said parallelogram.  
The farther winning, and inferring, I commi●●● to your skill, care, and ●●udy●. But in an other sort will I give you new aid, and instruction here.    

¶ A Theorem.  
Of all circles, the circumferences to their own diameters, have one and the same proportion: in what one circle soever, they are assigned.  
That is (as Archimedes hath demonstrated) almost, as 22. to 7: or nearer, if nearer be fou●d: until the very precise proportion be demonstrated. Which, what soever it be, in all circumferences to their proper diameters, will be demonstrated one and the same.  

A Corollary. 1.  
Wherefore if two circles be propounded, which suppose to be A and B, as the circumference of A is to the circumference of B, so is the diameter of A to the diameter of B.  
For by the former Theorem, as the circumference of A, is to his own diameter, so is the circumference of B, to his own diameter: Wherefore, alternately, as the circumference of A, is to the circumference of B, so is the diameter of A, to the diameter of B: Which was required to be demonstrated.   

A Corollary. 2.  
It is now then evident, that we can give two circles whose circumferences one to the other, shall have any proportion given in two right lines.  
The great Mechanical use (besides Mathematical considerations) which, these two Corollaryes may have in Wheels of Milles, Clocks, Cranes, and other engines for water works, and for wars, and many other purposes, the earnest and witty Mechanicien will soon bolt out, & gladly practise.  
● john Dee.     

¶ The 2. Theorem. The 2. Proposition. Circles are in that proportion the one to the other, that the squares of their diameters are.  
SVppose that there be two circles ABCD, and EFGH, and let their diameters be BD and FH. Then I say, that as the square of the line DB is to the square of the line FH, 

Demonstration ●eading to an impossibility.  
Two cases in this proposition.  
The first case.   so is the circle ABCD to the circle EFGH. For if the circle ABCD be not unto the circle EFGH, as the square of the line BD is to the square of the line FH: then the square of the line BD shall be to the square of the line FH, as the circle ABCD is to a superficies, either less than the circle EFGH, or greater. First let the square of the line BD be to the square of the line FH, as the circle ABCD is to a superficies less than the circle EFGH, namely, to the superficies S. Describe (by the 6. of the fourth) in the circle EFGH a square EFGH. Now this square thus described is greater than the half of the circle EFGH. For if by the points E, F, G, H, we draw right lines touching the circle, 

That a square within any circle described is bigger than half the circle.  
That the Isosceles triangles, without the square, are greater than half the segments wherein they are.   the square EFGH, is the half of the square described about the circle, but the square described about the circle, is greater than the circle. Wherefore the square EFGH, which is inscribed in the circle, is greater than the half of the circle EFGH. Divide the circumferences EF, FG, GH, and HE, into two equal parts in the points K, L, M, N. And draw these right lines EK, KF, FL, LG, GM, MH, HN, and NE. Wherefore every one of these triangles EKF, FLG, GMH, and HNE, is greater than the half of the segment of the circle which is described about it. For if by the points K, L, M, N, be drawn lines touching the circle, and then be made perfect the parallelograms made of the right lines EF, FG, GH, & HE, every one of the triangles EKF, FLG, GMH, & HNE, is the half of the parallelogragramme which is described about it (by the 41. of the first): but the segment described about it is less than the parallelogram. Wherefore every one of these triangles EKF, FLG, GMH, and HNE, is greater than the half of the segment of the circle which is described about it. Now then dividing the circumferences remaining into two equal parts, and drawing right lines from the points where those divisions are made, & so continually doing this, we shall at the length (by the 1. of the tenth) leave certain segments of the circle, which shall be less than the excess, whereby the circle EFGH exceedeth the superficies S. For it hath been proved in the first Proposition of the tenth book, that two unequal magnitudes being given, if from the greater be taken away more than the half, and likewise again from the residue more than the hal●e, and so continually, there shall at the length be left a certain magnitude which shall be less than the less magnitude given. Let there be such segments left, & let the segments of the circle EFGH, namely, which are made by the lines EK, KF, FL, LG, GM, MH, HN, and NE, be less than the excess, whereby the circle EFGH exceedeth the superficies S. Wherefore the residue, namely, the Poligonon figure EKFLGMHN, is greater than the superficies S. Inscribe in the circle ABCD a Poligonon figure like to the Poligonon figure EKFLGMHN, and let the same be AXBOCPDR. Wherefore (by the Proposition next going before) as the square of the line BD is to the square of the line FH, so is the Poligonon figure AXBOCPDR to the Poligonon figure EKFLGMHN. But as the square of the line BD is to the square of the line FG, so is the circle ABCD supposed to be to the superficies S. Wherefore (by the 11. of the fift) as the circle ABCD is to the superficies S, so is the Poligonon figure AXBOCPDR to the Poligonon figure EKFLGMHN. Wherefore alternately (by the 16. of the fift) as the circle ABCD is to the Poligonon figure described in it, so is the superficies S to the Poligonon figure EKFLGMHN. But the circle ABCD is greater than the Poligonon figure described in it. Wherefore also the superficies S is greater than the Poligonon figure EKFLGHMN: but it is also less: which is impossible. Wherefore as the square of the line BD is to the square of the line FH, so is not the circle ABCD to any superficies less than the circles EFGH.  

Second case.  In like sort also may wprove, that as the square of the line FH is to the square of the line BD, so is not the circle EFGH to any superficies less than the circle ABCD. I say, namely, that as the square of the line BD is to the square of the line FH, so is not the circle ABCD to any superficies greater than than the circle EFGH. For if it be possible, let it be to a greater, namely, to the superficies S. Wherefore by conversion, as the square of the line FH is to the square of the line BD, so is the superficies S to the circle ABCD. 
* This As 〈…〉 afterwards at the end of the dem●stra●ion proved.  But as the superficies S is to the circle ABCD, so is the circle EFGH to some superficies l●sse them the circle ABCD. Wherefore (by the 11. of the fift) as the square of the line FH is to the square of the line BD, so is the circle EFGH, to some superficies less than the circle ABCD: which is in the first case proved to be impossible. Wherefore as the square of the line BD is to the square of the line FH, so is not the circle ABCD to any superficies greater than the circle EFGH. And it is also proved that it is not, to any less. Wherefore as the square of the l●ne BD is to the square of the line FH, so is the circle ABCD to the circle EFGH. Wherefore circles are in that proportion the one to the other, that the squares of their diameters are: which was required to be proved.  

¶ An Assumpt.  
I say now, that the superficies S being greater than the circle EFGH, as the superficies S is to the circle ABCD, so is the circle EFGH to some superficies less than the circle ABCD. For, as the superficies S is to the circle ABCD, so let the circle EFGH be to the superficies T. Now I say, that the superficies T is less than the circle ABCD. For for that as the superficies S is to the circle ABCD, so is the circle EFGH to the superficies T, therefore alternately (by the 16. of the fift) as the superficies S is to the circle EFGH, so is the circle ABCD to the superficies T. But the superficies S is greater than the circle EFGH (by supposition). Wherefore also the circle ABCD is greater than the superficies T (by the 14. of the fift). Wherefore as the superficies S is to the circle ABCD, so is the circle EFGH to some superficies less than the circle ABCD: which was required to be demonstrated.   

¶ A Corollary added by Flussas.  
Circles have the one to the other, that proportion, that like Poligonon figures and in like sort described in them have. For, it was by the first Proposition proved, that the Poligonon figures have that proportion the one to the other, that the squares of the diameters have, which proportion likewise, by this Propositions the circles have.   

¶ Very needful Problems and Corollaryes by Master John Dee invented: whose wonderful use also, be partly declareth.  

A Problem. 1.  
Two circles being given: to find two right lines, which have the same proportion, one to the other, that the given circles have, o●e to the other●  
Suppose A and B, to be the diameters of two circles given: I say that two right lines are to be found, having that proportion, that the circle of A hath to the circle of B. Let to A & B (by the 11 of the sixth) a third proportional line be found, which suppose to be C. 
Construction.  I say now that A hath to C, that proportion which the circle of A hath to the circle of B. For forasmuch as A, B, and C, are (by construction) three proportional lines, 
Demonstration.  the square of A is to the square of B, as A is to C, (by the Corollary of the 20. of the sixth) ● but as the square of the line A is to the square of the line B, so is the circle whose diameter is the line A, to the circle whose diameter is the line B, by this second of the eleventh. Wherefore the circles of the line● A and B, are in the proportion of the right lines A and C. Therefore two circles being given, we have found two right lines having the same proportion between them, that the circles given, have one to the other: which aught to be done.   

A Problem. 2.  
Two circles being given, and a right line: to find an other right line, to which the line given shall have that proportion, which the one circle hath to the other.  
Suppose two circles given: which let be A & B, & a right line given, which let be C: I say that an other right line is to be ●ounde, to which the line C shall have that proportion that the circle A, 
Construction  hath to the circle B. As the diameter of the circle A, is to the diameter of the circle B, so let the line C be to a fourth line, (by the 12. of the 〈…〉 line be D. And, by the 11. of the sixth, let a third line proportional be found, to the lines C & D, which let be E● I say now, that the line C hath to the line E, that proportion which the circle A, hath to the circle B. For (by construction) the line● C, D, and E, 
Demonstration.  are proportional: therefore the square of C● is to the square of D, as C is to E, by the Corollary of the 10. of the sixth. But by construction, as the diameter of the circle A, is to the diameter of the circle B, so is C, to D: wherefore as the square of the diameter of the circle A, is to the square of the diameter of the circle B, so is the square of the line C to the square of the line D, by the 22. of the sixth. But as the square of the diameter of A, the circle, is to the square of the diameter of the circle D, so is the circle A, to the circle B, by the second of the twelfth: wherefore by 11. of the fifth, as the circle A, is to the circle B, so is the square of the line C, to the square of the li●e D. But it is proved, that ●s the square of the line C, is to the square of the line D, so is the line C to the line E. Wherefore by the 11. of the fifth, as the circle A, is to the circle B, so is the line C to the line E. Two circles being given therefore, and a right line, we have found a right line, to which the right line given, hath that proportion, which the one circle hath to the other. Which aught to be done.  

Note.  
The difference between this Problem, and that next before, is this: there, although we had two circles given, 
Di●●reence between the first problem and the second.  and two lines were found in that proportion the one to the other, in which the given circles were,: and here likewise are two circles given, and two line● also are had in the same proportion, that the given circles are: yet there we took at pleasure the first of the two lines, whereunto we framed the second proportionally, to the circles given. But here the first of the two lines, is assigned, pointed, a●d determined to us: and not our choice to be had therein, as was in the former Problem.    

A Problem. 3.  
A circle being given, to find an other circle, to which the given c●rcle is in any proportion given in two right lines.  
Suppose the circle A●C given, and therefore his semidiameter is given: whereby his diameter also is given: which diameter let be AC. Let the proportion given, be that which is between ●, F, two right lines. I say, a circle is to be found, unto which A●C hath that proportion that ● hath to ●. As ● is to ●, 
Construction.  so let AC the diameter, be to an other right line, by the 12. of the sixth. Which line suppose to be 11. Between AC and ●● find a middle proportional line, by the 13. of the sixth: which let be LN. By the 10. of the first, divide ●N, into two equal parts: and let that be done in the point ●. Now upon ●●, (oh being made the centre) describe a circle: which let be ●N●. 
Demonstration.  I say that ABC, is to LMN, as ● is to F. For seeing that AC, LN, and H, are three right lines in continual proportion (by construction) therefore (by the Corollary of the 20. of the sixth) as AC is to H, so is the square of A● to the square of LN. But AC is to H, as ● is to ●, by construction. Wherefore the square of AC is to the square of LN, as E is F: but as the square of the diameter AC, is to the square of the diameter LN, so is the circle ABC to the circle ●MN, by this 2. of the twelfth, wherefore by the 11. of the fifth, the circle ABC is to the circle LMN, as ● is to F. A circle being given (therefore) an other circle is found, to which the given circle is in any proportion given between two right lines: which aught to be done.   

A Problem. 4.  
Two circles being given, to find one circle equal to them both.  
Suppose the two circles given, have their diameters A● & CD. I say that a circle must be ●ound equal to the two circles whose diameters are A● and CD: unto the line A●, 
Construction.  at the point A, erect a perpendicular line A●: from which (sufficiently produced) cut a line equal to CD, which let be AF. By the first petition draw from F to ● a right line: so is FA● made a triangle rectangle. I say now that a circle whose diameter is F●, is equal to the two circles whose diameters are A● and ●D. 
Demonstration.  For by the 47. of the first, the square of F● is equal to the squares of A● & AF. Which AF is (by construction) equal to CD● wherefore the square of F● is equal to the squares of AB and CD. But circles are one to an other, as the squares of their diameters are one to the other, by this second of the twelfth. Therefore the circle whose diameter is ●● is equal to the circles whose diameters are A● and CD. Therefore two circles being given we have found a circle equal to them both. Which was required to be done.  

A Corollary. 1.  
Hereby it is made evident, that in all triangles rectangle, the circles, semicircles, quadrants, o● any other portions of circles described upon the subtendent line, is equal to the two circles, semicircles, quadrants, or any two other like portions of circles, described on the two lines comprehending the right angle, like to like being compared.  
For like parts have that proportion between themselves, that their whole magnitudes have, of which they are like parts, by the 15. of the fifth. But of the whole circles, in the former problem it is evident: and therefore in the forenamed like portions of circle's, it is a true consequent.   

A Corollary. 2.  
By the former problem, it is also manifest, unto circles three, four, five, or to how many so ever one will give, one circle may be given equal.  
For if first, to any two, by the former problem, you find one equal, and then unto your found circle and the third of the given circles, as two given circles, find one other circle equal, and then to that second found circle, and to the fourth of the first given circles● as two circles, one new circle be found equal, and so proceed till you have once coupled orderly, every one of your propounded circles (except the first and second already done) with the new circle thus found for so the last found circle is equal to all the first given circles. If ye doubt, or sufficiently understand me not: help yourself by the discourse and demonstration of the last proposition in the second book, and also of the 31. in the sixth book.    

¶ A Problem 5.  
Two unequal circles being given, to find a circle equal to the excess of the greater to the less.  
Suppose the two unequal circles given, to be ABC & DEF, & let ABC be the greater: 
Construction.  whose diameter suppose to be AC: & the diameter of DEF suppose to be DF. I say a circle must be found equal to that excess in magnitude, by which ABC is greater th● DEF. By the first of the fourth, in the circle ABC. Apply a right line equal to DF: whose one end let be at C, and the other let be at B. Fron B to A draw a right line. By the 30. of the third, it may appear, 
Demonstration.  that ABC is a right angle: and thereby ABC, the triangle is rectangled: wherefore by the first of the two corollaries, here before, the circle ABC is equal to the circle DEF, (For BC by construction is equal to DF) and more over to the circle whose diameter is AB. That circle therefore whose diameter is AB, is the circle containing the magnitude, by which ABC is greater than DEF. Wherefore two unequal circles being given, we have found a circle equal to the excess of the greater to the less: which aught to be done.   

A Problem. 6.  
A Circle being given to find two Circles equal to the same: which found Circles, shall have the one to the other, any proportion given in two right lines.  
Suppose ABC, a circle given: and the proportion given, let it be that, which is between the two right lines D and E. I say, that two circles are to be found equal to ABC: and with all, one to the other, 
Construction.  in the proportion of D to E. Let the diameter of ABC be AC. As D is to E, so let AC be divided, by the 10. of the sixth, in the point F. At F, to the line AC let a perpendicular be drawn FB, and let it meet the circumference at the point B. From the point B to the points A and C, let right lines be drawn: BA and BC. I say that the circles whose diameter are the lines BA and BC are equal to the circle ABC: and that those circles having to their diameters the lines BA and BC are one to the other in the proportion of the line D to the line E. For, first that they are equal, it is evident: by reason that ABC is a triangle rectangle: wherefore by the 47. of the first the squares of BA and BC are equal to the square of AC: And so by this second it is manifest, the two circles to be equal to the circle ABC. Secondly as D is to ●, so is AF to FC: by construction. And as the line AF is to the line FC, so is the square of the line ●A to the square of the line BC. [Which thing, we will briefly prove thus. The parallelogram contained under AC and AF, 
Note this well: for it i● of great use.  is equal to the square of BA: by the Lemma after the 32. of the tenth book: and by the same Lemma or Assumpt, the parallelogram contained under AC and ●C, is equal to the square of the line BC. Wherefore as the first parallelogram hath itself to the second● so hath the square of BA (equal to the first parallelogram) itself, to the square of BC, equal to the second parallelogram. But both the parallelograms have one height, namely, the line AC: and bases, the lines AF and FC: wherefore as AF is to FC, so is the parallelog●amme contained under AC, AF, to the parallelogram contained under AC, FC, by the fi●st of the sixth. And therefore as AF is to FC, so is the square of BA to the square of BC.] And as the square of BA is to the square of BC: so is the circle whose diameter is BA, to the circle whose diameter is BC, by this second of the twelfth. Wherefore the circle whose diameter is BA, is to the circle whose diameter is BC, as D is to E. And before we proved them equal to the circle ABC. Wher●fore a circle being given, we have found two circles equal to the same: which have the one to the other any proportion given in two right lines. Which aught to be done.  

Note.  

another way of demonstration of the f●●st problem of th●● addition.  He●e may you perceive an other way how to execute my first problem, for if you make a right angle contained of the diameters given, as in this figure suppose them BA and BC: and then subtend the right angle with the line AC: and from the right angle, let fall a line perpendicular to the base AC: that perpendicular at the point of his fall, divideth AC into AF and FC, of the proportion required.   

A Corollary.  
It followeth of things manifestly proved in the demonstration of this problem, that in a triangle rectangle, if from the right angle to the base, a perpendicular be let fall: the same perpendicular cutteth the base into two parts, 
Note this proper●ie of a triangle rectangle.  in that proportion, one to the other, that the squares of the righ● lines, containing the right angle, are in, one to the other: those on the one side the perpendicular, being compared to those on the other, both square and segment.    

A Problem 7.  
Between two circles given, to find a circle middle proportional.  
Let the two circles given, be ACD and BEF: I say, that a circle is to be found, which between ACD and BEF is middle proportional. 
Construction.  Let the diameter of ACD, be AD, and of BEF, let B● be the diameter: between AD and BF, find a line middle proportional, by the 13. of the sixth: which let be HK: I say that a circle, whose diameter is HK is middle proportional between ACD and BEF. To AD, HK, and BF, (three right lines in continual proportion, by construction) let a fourth line be found: to which BF shall have that proportion, that AD hath to HK: by the 12. of the sixth, & let that line be ●. 
Demonstration.  It is manifest that the ●ower lines AD, HK, BF, and L, are in continual proportion. [For by construction, as AD is to HK, so is B● to L. And by construction, on, as AD is to HK, so is HK to BF: wherefore HK is to BF, as BF is to L: by the 11. of the fifth, wherefore the 4. lines are in continual proportion.] Wherefore as the first is to the third, that is AD to BF, so is the square of the first to the square of the second: that is, the square of AD, to the square of HK: by the corollary of the 20. of the sixth. And by the same corollary, as HK is to L, so is the square of HK to the square of BF. But by alternate proportion, the line AD is to BF, as HK is to L: wherefore the square of AD is to the square of HK, as the square of HK is to the square of BF. Wherefore the square of HK, is middle proportional, between the square of AD and the square of BF. But as the squares are one to the other, so are the circles (whose diameters produce the same squares) one to the other, by this second of the twelfth: wherefore the circle whose diameter is the line HK, is middle proportional, between the circles whose diameters are the line a AD and BF. Wherefore between two circles given, we have found a circle middle proportional: which was requisite to be done.  

¶ A Corollary.  
Hereby it is manifest, three lines or more, being in continual proportion, that the circles having those lines to their diameters, are also in continual proportion.  
As of three, our demonstration hath already proved: so of four, will the proof go forward if you add a fifth line in continual proportion to the four given: as we did to the three, add the fourth: namely, the line L. And so, if you have 6, by putting to one more, the demonstration will be ●asie and plain. And so of as many as you william.    

A Problem. 8.  
To a circle being given, to find three circles equal: Which three circles shall be in continual proportion, in any proportion given between two right lines.  
Suppose the circle given to be ABC: and the proportion given to be that which is between the lines X and Y. I say, that three circles are to be given, which three, together, shall be equal to the circle ABC: and withal in continual proportion, in the same proportion which is between the right lines X and Y. Let the diameter of ABC, be AC. Of AC, make a square: by the 46. of the first: which which let be ACDE. From the point D draw a line, sufficiently long (any way, 
* 
Though I say, without the square, yet you must think, that it may be also within the square: & that diversly. Wherefore this Problem may have diverse cases, so, but briefly, to aside all, may thus be said: cut any side of that square into 3 partest in the proportion of X to Y.  
Note the manner of the drift in this demonstration and construction, mixedly and with no determination to the constructions as commonly i● in probleme●● which is here of me so vsed● for an example to young students of variety in art.   without the square): which let be DO. At the point D, and from the line DO, cut a line equal to X: which let be DM. At the point M, and from the line MO, cut a line equal to ●● which let be MN. At the point N, to the two lines DM and MN, set a third line proportional, by the 12. of the sixth: which let be NO. From E (one of the angles of the square ACDE, next to D) draw a right line to O: making perfect the triangle DEO. Now from the points M and N, draw lines, to the ●ide DE, parallel to the side EO: by the 31. of the first: which let be MF and NG. Wherefore, by the 2. of the sixth, the side DE, is proportionally cut in the points F and G, as DO is cut in the points M and N: therefore, as DM is to MN, so is DF, to FG: and as MN is to NO, so is FG to GOE Wherefore, seeing DM, MN, and NO, are, by construction, continually proportioned, in the proportion of X to Y: So likewise, are DF, FG, and GE, in continual proportion, in the proportion of X to Y, by the 11. of the fift. From the points F and G, to the opposite ●ide AC, let right lines be drawn parallel to the other sides: which lines, suppose to be FI, and GK: making thereby, three parallelograms DIEGO, ●K, and GC, equal to the whole square ACDE. Which three parallelograms, by the first of the sixth, are one to an other, as their bases, DF, FG, and GE, are. But DF, FG, and GE, were proved to be in continual proportion, in the proportion of X to Y: Wherefore, the three parallelograms DIEGO, FK, and GC, by the 11. of the fifth, are also in continual proportion, and in the same, which X is in, to Y. Let three squares be made, equal to the three parallelograms DIEGO, FK, and GC: by the last of the second: Let the sides of those squares be, orderly, S, T, and V Forasmuch as, it was last concluded that the three parallelograms, DIEGO, FK, and GC (which are equal to the square ACDE) are also in continual proportion, in the proportion of X to Y, therefore their equals, namely, the three squares of S, T, & V, are also equal to the whole square ACDE, and in continual proportion, in the proportion of X to Y. Wherefore the three circles, whose diameters are S, T, and V, are equal to the circle, whose diameter is AC, the side of the square ACDE, and also in continual proportion, in the proportion of X to Y: by this second of the twelfth. But, by construction, AC is the diameter of the circle ABC. Wherefore we have found three circles, equal to ABC: namely, the circle, whose diameter is S: and the circle, whose diameter is T: and the circle, whose diameter is V: which three circles, also, are in continual proportion, in the proportion of X to Y. Wherefore to a circle being given, we have found three circle's equal in any proportion, given, between two right lines: which was requisite to be done.  

¶ A Corollary.  
Hereby, it is evident, that a circle given, we may find circles 4, 5, 6, 10, 20, 100, 1000, or how many soever shall be appointed, being in continual proportion, in any proportion, given between two right lines: which circles, all together, shall be equal to the circle given.  
For, evermore dividing the one side of the chief square (which is made of the diameter of the circle given) into so many parts, as circles are to be made: so that between those parts be continued the proportion given between two right lines● and from the points of those divisions, draw parallels, perpendiculars to the other side of the said chief square: making so many parallelograms of the chief square, as are circles to be made: and to those parallelograms (orderly) making equal squares: it is manifest that the sides of those squares, are the diameters of the circles required to be made.    

A Problem. 9  
Three circles being given, to find three equal to them: which three found circles shall be in continual proportion, in any proportion given between two right lines.  
Suppose the three circles given to be A, B, and C, and let the proportion given, be that which is between the right lines X & Y. I say, three other circles are to be found, equal to A, B, and C, & with all, in continual proportion, in the proportion of X to Y. By the 2. Corollary of my 4. Problem, 
Construction.  make one circle equal to the three circles A, B, and C. Which one circle suppose to be D: And by the problem next before, let three circles be ●ound, equal to D, and with all, in continual proportion, in the same proportion which is between X and Y. Which three circles, suppose to be E, F, and G. I say, that E, F, G, are equal to A, B, C: and with all, in continual proportion, in the proportion of X to Y. 
Demonstration.  For, by construction, the circle D is equal to the circles A, B, & C: and by construction likewise, the circles E, F, and G, are equal to the same circle D: Wherefore the three circles E, F, & G, are equal to the three circles A, B, & C: and by construction E, F, and G, are in continual proportion, in the proportion of the line ● to the line Y. Wherefore E, F, and G, are equal to A, B, and C: and in continual proportion, in the proportion of X to Y. Three circles, therefore, being given, we have found three circles, equal to them, and also in continual proportion, in any proportion given, between two right lines. Which was requisite to be done.  

¶ A Corollary. 1.  
It is hereby very manifest, that unto 4.5.6.10.20. 100 or how many circles soever, shall be given, we may find 3.4.5.8.10. or how many soever, shall be appointed: which, all together, shall be equal to the circles given, how many soever they are: and with all, our found circles, to be in continual proportion, in any proportion assigned between two right lines given.  
For, evermore, by the Corollary of the 4. Problem, reduce all your circles to one: and by the Corollary of my 8. Problem, make as many circles as you are appointed, equal to the circles given, and continual in proportion, in the same, wherein, the two right lines given, are● And so have you performed, the thing required.   

Note.  
What incredible fruit in the Science of proportions may hereby grow, no man's tongue can sufficiently express. And sorry I am, that utterly leisure is taken from me, somewhat to specify in particular hereof.   

¶ The key of one of the chief treasure houses, belonging to the State Mathematical.  
THat● which in these 9 Problems, is said of circles● is much more said of squares, by whose means, circles, are thus handled. And therefore seeing to all Polygonon right lined figures, equal squares may be made, by the fault of the second: and contrariwise, to any square, a right lined figure may be made equal, and withal, like to any right lined figure given, by the 25 of the sixth. And fourthly, seeing upon the said plain figures● as upon base● may Pri●mes, Parallelipipedons, Pyramids, sided Columns, Cones, and Cylinders, be reared: which being 
* Note and remember one ●e●th in these solids.  all of one height, shall have that proportion, one to the other, that their bases have, one to the other. And fifthly, seeing Spheres, Cones, and Cylinders are one to other in 〈◊〉 known proportions: and so may be made, one to the other in any proportion assigned. And 〈◊〉, seeing under every one of the kinds of figures, both plain, and solid, infinite cases may chance, by the aid of these Problems, to be soluted and executed: How infinite (then) upon infinite, is the number of practices, either Mathematical, o● Mechanical, to be performed, of comparisons between divers kinds, of plains to plains, and solids to solidest  
Farthermore, to speak of plain superficial figures, in respect of the con●●r, or Area of the circle, sundry mixed line figures, Anular and Lunular figures: and also of circles to be given equal to the said ●●●sed figures● and in all proportions else: and evermore thinking of solids, (like high) set upon any of those unused figures, (O Lord) in consideration of all the premises, how infinite, how strange and ●●credible ●●●●●●ation● and practise●, may (by the aid and direction of these few problems) 〈◊〉 readily into the imagination's and hands of them, that will bring their mind and intent wholly and fixedly to such mathematical discourses? In these Elements, I intend but to give to young beginners some 〈◊〉, and courage to exercise ●heir own wits, and talon, in this most pleasant and profitable sci●nc●. All ●hinge● 〈◊〉 not, neither y●●●an, in every place be said. Opportunity, and Sufficienty, best are to be allowed.      

¶ The 3. Theorem. The 3. Proposition. Every Pyramid having a triangle to his base: may be divided into two Pyramids equal and like the one to the other, and also like to the whole, having also triangles to their bases, and into two equal prisms: and those two prisms are greater than the half of the whole Pyramid.  
And forasmuch as to one of the sides of the triangle ADB, namely, to the side AB, is drawn ● parallel line HK, therefore the whole triangle A●DB is equiangle to the triangle DHK, and their sides are proportional (by the Corollary of the ●. of the sixth). Wherefore the triangle ADB is like to the triangle DHK. And by the same reason also the triangle DBC is like to the triangle DKL, and the triangle ADC to the triangle DHL. And forasmuch as two right lines BA and AC touching the one the other, are parallels to two right lines KH and HL, touching also the one the other, but not being in one and the self same superficies with the two first lines, therefore (by the 10. of the eleventh) they contain equal angles. Wherefore the angle BAC is equal to the angle KHL. And as the line BA is to the line AC, so is the line KH to the line HERALD Wherefore the triangle ABC is like to the triangle KHL. Wherefore the whole Pyramid whose base is the triangle ABC & top the point D, is like to the pyramid whose base is the triangle HKL, and top the point D. But the pryamis whose base is the triangle HKL and top the point D, is proved to be like to the pyramid whose base is the triangle AEG and top the point H. Wherefore also the pyramid whose base is the triangle ABC, and top the point D, is like to the pyramid whose base is the triangle AEG and top the point H (by the 21. of the sixth). 
The conclusion of the first part.  Wherefore either of these pyramids AEGH and HKLD is like to the whole pyramid ABCD.  

Demonstration of the second part, namely, that it is divided moreover into two equal prisms.  [Andrea forasmuch as the lines BE, KH, are parallel lines and equal, as it hath been proved, therefore the right lines BK and EH, which join them together, are equal and parallels, by the 33. of the first. Again forasmuch as the lines BE and FG are parallel lines and equal, therefore lines EG and BF, which join them together, are also equal and parallels: and by the same reason forasmuch as the lines FG and KH are equal parallels, the lines FK and GH, which join them together, are also equal parallels. Wherefore BEHK, DEGF, and KHGF, are parallelograms. And forasmuch as their opposite sides are equal, by the 34. of the first● therefore the triangles FHG, & BKF, are equiangle, by the 8. of the first: and therefore, by the 4. of the same, they are equal: and moreover, by the 15. of the eleventh, their superficiec●s are parallels. Wherefore the solid BKFEHG is a Prism, by the 11. definition of the eleventh. Likewise forasmuch as the sides of the triangle HKL are equal and parallels to the sides of the triangle GFC, as it hath before been proved: It is manifest, that CFKL, FKHG, and CLHG, are parallelograms, by the 33. of the first. Wherefor● the whole solid KLHFCG, is a Prism, by the 11. definition of the eleventh, and is contained under the said parallelograms CFKL, FKHG, and CLHG, and the two triangles HKL and GFC, which are opposite and parallels.]  
And forasmuch as the line BF is equal to the line FC, therefore (by the 41. of the first) the parallelogram EBFG is double to the triangle GFC. And forasmuch as if there be two prisms of equal altitudes, and the one have to his base a parallelogram, and the other a triangle, and if the parallelogram be double to the triangle, those prisms are (by the 40. of the eleventh) equal the one to the other: therefore the Prism contained under the two triangles BKF and EHG, and under the three parallelograms EBFG, EBKH, and KHFG, 
Conclusion of the second part.  is equal to the Prism contained under the two triangles GFC, and HKL, and under the three parallelograms KFCL, LCGH, and HKFG.  
And it is manifest, that both these prisms, of which the base of one is the parallelogram EBFG, 
Demonstration of the last part that the two prisms are greater than the half of the whole Pyramid.  and the opposite unto it the line KH, and the base of the other is the triangle GFC, and the opposite side unto it the triangle KLH, are greater than both these Pyramids, whose bases are the triangles AGE, and HKL, and tops the points H & D. For if we draw these right lines EF and EK, the Prism whose base is the parallelogram EBFG, and the opposite unto it the right line HK, is greater than the Pyramid whose base is the triangle EBF, & top the point K. But the Pyramid whose base is the triangle EBF, and top the point K, is equal to the Pyramid whose base is the triangle AEG and top the point H, for they are contained under equal and like plain superficieces. Wherefore also the Prism whose base is the parallelogram EBFG and the opposite unto it the right line HK, is greater than the Pyramid, whose base is the triangle AEG, and top the point H. But the prism whose base is the parallelogram EBFG, and the opposite unto it the right line HK, is equal to the prism, whose base is the triangle GFC, and the opposite side unto it the triangle HKL: And the Pyramid whose base is the triangle AEG, and top the point H, is equal to the Pyramid, whose base is the triangle HKL, and top the point D. 
Conclusion of the last part.  Wherefore the foresaid two prisms are greater than the foresaid two Pyramids, whose bases are the triangles AEG, HKL, and tops the points H and D. Wherefore the whole Pyramid whose base is the triangle ABC, 
Conclusion of the whole proposition.  and top the point D, is divided into two Pyramids equal and like the one to the other, and like also unto the whole Pyramid, having also triangles to their bases, and into two equal prisms, and the two prisms are greater than half of the whole Pyramid: which was required to be demonstrated.  
If ye will with diligence read these four books following of Euclid, which concern bodies, and clearly see the demonstrations in them contained, it shall be requisite for you when you come to any proposition, which concerneth a body or bodies, whether they be regular or not, first to describe of p●s●ed paper (according as I taught you in the end of the definitions of the eleventh book) such a body or bodies, as are there required, and having your body, or bodies thus described, when you have noted it with letters according to the figure set forth upon a plain in the proposition, follow the construction required in the proposition. As for example, in this third proposition it is said that, Every pyramid having a triangle to ●is base, may be divided into two pyramids. etc. Here first describe a pyramid of pasted paper ha●ing his base triangled (it skilleth not whether it be equilater, or equiangled, or not, only in this proposition is required that the base be a triangle. Then for that the proposition supposeth the base of the pyramid to be the triangle ABC, note the base of your pyramid which you have described with the letters ABC, and the top of your pyramid with the letter D: For so is required in the proposition. And thus have you your body ordered ready to the construction. Now in the construction it is required that ye divide the lines, AB, BC, CA etc., namely, the six lines which are the sides of the four triangles containing the pyramis, into two equal parts in the poyntet ●, F, G, etc. That is, ye must divide the line AB of your pyramid into two equal parts, and note the point of the division with the letter E, and so the line BC being divided into two equal parts, note the point of the division with the letter F. And so the rest, and this order follow ye as touching the rest of the construction there put, and when ye have finished the construction, compare your body thus described with the demonstration: and it will make it very plain and easy to be understanded. Whereas without such a body described of matter, it is hard for young beginners (unless they have a very deep imagination) fully to conceive the demonstration by the sig●e as it is described in a plain. Here for the better declaration of that which I have said, have I set a figure, whose form if ye describe upon pasted paper noted with the like letters, and cut the lines ●A, DA, DC, and fold it accordingly, it will make a Pyramid described according to the construction required in the proposition. And this order follow ye as touching all other propositions which concern bodies.  

¶ An other demonstration after Campane of the 3. proposition.  
Suppose that there be a Pyramid ABCD having to his base the triangle BCD, and let his top be the solid angle A: from which let there be drawn three subtended lines AB, AC, and AD to the three angles of the base, and divide all the sides of the base into two equal parts in the three points E, F, G: divide also the three subtended lines AB, AC, and AD in two equal parts in the three points H, K, L. And draw in the base these two lines EF and EG: So shall the base of the pyramid be divided into three superficieces: whereof two are the two triangles BEF, and EGD, which are like both the one to the other, and also to the whole base, by the 2 part of the second of the sixth, & by the definition of like super●iciec●s, & they are equal the one to the other, by the 8. of the first: the third superficies is a quadrangled parallelogram, namely, EFGC: which is double to the triangle EGD, by the 40. and 41. of the first. Now then again from the point H draw unto the points E and F these two subtendent lines HE and HF: draw also a subtended line from the point K to the point G. And draw these lines HK, KL, and LH. Wherefore the whole pyramid ABCD is divided into two pyramids, which are HBEF, and AHKL, and into two prisms of which the one is EHFGKC, and is set upon the quadrangled base CFGE: the other is EGDHKL, and hath to his base the triangle EGD. Now as touching the two pyramids HBEF and AHKL, that they are equal the one to the other, and also that they are like both the one to the other and also to the whole, it is manifest by the definition of equal and like bodies, and by the 10. of the eleventh, and by 2. part of the second of the sixth. And that the two prisms are equal it is manifest by the last of the eleventh. And now that both the prisms taken together are greater than the half of the whole pyramid, hereby it is manifest, for that either of them may be divided into two pyramids, of which the one is a triangular pyramid equal to one of the two pyramids into which together with the two prisms is divided the whole pyramid, and the other is a quadrangled pyramid double to the other pyramid. Wherefore it is plain that the two prisms taken together are three quarters of the whole pyramid divided. But if ye are desirous to know the proportion between them, read the ●, of this book. But now here to this purpose it shall be sufficient to know, that the two prisms taken together do exceed in quantity the two partial pyramids taken together, into which together with the two prisms the whole pyramid was divided.    

¶ The 4. Theorem. The 4. Proposition. If there be two Pyramids under equal altitudes, having triangles to their bases, and either of those Pyramids be divided into two Pyramids equal the one to the other, and like unto the whole, and into two squall prisms, and again if in either of the Pyramids made of the two first Pyramids be still observed the same order and manner: then as the base of the one Pyramid is to the base of the other Pyramid, so are all the prisms which are in the one Pyramid to all the prisms which are in the other, being equal in multitude with them.  
SVppose that there be two Pyramids under equal altitudes, having triangles to their bases, namely, ABC, and DEF, and having to their tops the points G and H. And let either of these pyramids be divided into two pyramids equal the one to the other, and like unto the whole, and into two equal prisms (according to the method of the former Proposition). And again, let either of those pyramids so● made of the two first pyramids, be imagined to be after the same order divided, and so do continually. Then I say, that as the base ABC is to the base DEF, so are all the prisms which are in the pyramid ABCG, to all the prisms which are in the pyramid DEFH being equal in multitude with them. For forasmuch as the line BX is equal to the line XC, and the line AL to the line LC: (For as we saw in the construction pertaining to the former Proposition, all the six sides of the whole pyramids, are each divided into two equal parts, the like of which construction is in this proposition also supposed): therefore the line XL is a parallel to the line AB, & the triangle ABC, is like to the triangle LXC, (by the Corollary of the second of the sixth): and by the same reason the triangle DEF is like to the triangle RWF. And forasmuch as the line BC is double to the line CX, and the line FE to the line FW: therefore as the line BC is to the line CX, so is the line EF to the line FW. And upon the lines BC and CX are described rectiline figures like and in like sort set, namely, the triangles ABC and LXC, and upon the lines EF and FW are also described rectiline figures, like and in like sort set, namely, the triangles DEF & RWF● But if there be four right lines proportional, the rectiline figures described of them being like and in like sort set, shall also be proportional (by the 22. of the sixth). Wherefore as the triangle ABC is to the triangle LXC, so is the triangle DEF to the triangle RWF. Wherefore alternately (by the 16. of the fift) as the triangle ABC is to the triangle DEF, so is the triangle LXC to the triangle RWF. 
* An Assumpt.  But as the triangle LXC is to the triangle RWF, so is the prism whose base is the triangle LXC, and the opposite side unto it the triangle OMN, to the prism whose base is the triangle RWF, and the opposite side unto it the triangle STV (by the Corollary of the 40. of the eleventh). For these prisms are under one & the self same altitude, namely, under the half of the altitude of the whole Pyramids, which Pyramids are supposed to be under one and the self same altitude: this is also proved in the Assumpt following). Wherefore (by the 11. of the fift) as the triangle ABC is to the triangle DEF, so is the prism whose base is the triangle LXC, and the opposite side unto it the triangle OMN, to the prism whose base is the triangle RWF, and the opposite side unto it the triangle STV. And forasmuch as there are two prisms in the pyramid ABCG equal the one to the other, & two prisms also in the pyramid DEFH equal the one to the other: therefore as the prism, whose base is the parallelogram BKLX, and the opposite side unto it the line MO, is to the prism, whose base is the triangle LXC, and the opposite side unto it the triangle OMN, so is the prism, whose base is the parallelogram PERW, and the opposite unto it the line ST, to the prism, whose base is the triangle RWF, and the opposite side unto it the triangle STV. Wherefore by composition (by the 18. of the fift) as the prisms KBXLMO, & LXCMNO, are to the prism LXCMNO, so are the prisms PEWRST, and RWFSTV, to the prism RWFSTV. Wherefore alternately (by the 16. of the fift) as the two prisms KBXLMO, and LXCMNO, are to the two prisms PEWRST, and RWFSTV, so is the prism LXCMNO to the prism RWFSTV. But as the prism LXCMNO is to the prism RWFSTV, so have we proved that the base LXC is to the base RWF, and the base ABC to the base DEF. Wherefore (by the 16. of the fift) as the triangle ABC is to the triangle DEF, so are both the prisms which are in the pyramid ABCG, to both the prisms which are in the pyramid DEFH. And in like sort if we divide the other pyramids after the self same manner, namely, the pyramid OMNG, and the pyramid STVH: as the base OMN is to the base STV, so shall both the prisms that are in the pyramid OMNG, be to both the prisms which are in the pyramid STVH. But as the base OMN is to the base STV, so is the base ABC to the base DEF. Wherefore (by the 11. of the fift) as the base ABC is to the base DEF, so are the two prisms that are in the pyramid ABCG, to the two prisms that are in the pyramid DEFH, and the two prisms that are in the pyramid OMNG, to the two prisms that are in the pyramid STVH, and the four prisms to the four prisms. And so also shall it follow in the prisms made by dividing the two pyramids AKLO, and DPRS, and of all the other pyramids in general, being equal in multitude.  
[Andrea that as the triangle LXC is to the triangle RWF, so is the prism whose base is the triangle LXC and the opposite side OMN, 
An Assumpt.  to the prism whose base is the triangle RWF and the opposite side the triangle STV, may thus be proved. For in the self same construction, imagine perpendicular lines to be drawn from the points G and H to the two plain superficieces wherein are the triangles ABC and DEF. Now those perpendicular lines shall be equal the one to the other, for that the two pyramids are supposed to be of equal altitude. And forasmuch as two right lines, namely, GC, and the perpendicular line drawn from the point G, are divided by two parallel plain superficieces, namely, ABC and OMN, therefore (by the 17. of the eleventh) the parts of the lines divided are proportional. But the line GC is by the plain superficies OMN divided into two equal parts in the point N. Wherefore also the perpendicular line drawn from the point G to the plain superficies wherein is the triangle ABC, is divided into two equal parts by the superficies OMN: & by the same reason also the perpendicular line which is drawn from the point H to the plain superficies DEF is divided into two equal parts, by the plain superficies STV. And the perpendiculars drawn from the points G and H to the plain superficieces ABC and DEF are equal. Wherefore also the perpendicular lines which are drawn from the triangles OMN and STV to the plain superficieces ABC and DEF are equal the one to the other. Wherefore also the prisms whose bases are the triangles LXC and RWF and the opposite sides the triangles OMN and STV, are of equal altitude. Wherefore also the parallelipipedons which are described of the foresaid prisms and being equal in altitude with them, are the one to the other, as the base of the one is to the base of the other. Wherefore also as the half of the bases of those parallelipipedons, namely, as the base LXC is to the base RWF, so are the foresaid prisms the one to the other.] If therefore there be two pyramid under equal altitudes, having triangles to their bases and either of those pyramids be divided into two pyramids equal the one to the other and like unto the whole, and into two equal prisms, 
Conclusion of the whole.  and again if in either of the pyramids made of the two first pyramids be still observed the same order and manner: then as the base of the one pyramid is to the base of the other pyramis, so all the prisms which are in the one pyramid to all the prisms which are in the other being equal in multitude: which was required to be proved.   

¶ The 5. Theorem. The 5. Proposition. Pyramids consisting under one and the self same altitude, having triangles to their bases: are in that proportion the one to the other that their bases are.  
SVppose that these two Pyramids, whose bases are the triangles ABC & DEF, and tops the points G and H, be under equal altitudes. Then I say, that as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH. For if the pyramid ABCG be not to the pyramid DEFH, as the base ABC is to the base DEF, then as the base ABC is to the base DEF, 
Demonstration leading to an impossibility.  so is the pyramid ABCG to a solid, either less than the pyramid DEFH, or greater. First let it be to some less, and let the same be X. And (by the 3. of the twelfth) let the pyramid DEFH be divided into two pyramids equal the one to the other, and like unto the whole, and into two equal prisms. Now the two prisms are greater than the half of the whole pyramid. And again (by the same) let the pyramids which are made of the division, be in like sort divided, and do this continually, until there remain some pyramids made of the pyramid DEFH, which are less than the excess, whereby the pyramid DEFH exceedeth the solid X. Let such pyramids be taken, and for example sake, let those pyramids be DPRS, & STVH. Wherefore the prisms remaining which are in the pyramid DEFH, are greater than the solid X. Divide (by the Proposition next going before) the pyramid ABCG in like sort, & as many times as the pyramid DEFH ●s divided. Wherefore (by the same) as the base ABC is to the base DEF, so are all the prisms which are in the pyramid ABCG, to all the prisms which are in the pyramid DEFH. But as the base ABC is to the base DEF, so is the pyramid ABCG to the solid X. Wherefore (by the 11. of the fift) as the pyramid ABCG, is to the solid X, so are the prisms which are in the pyramid ABCG, to the prisms which are in the pyramid DEFH. Wherefore alternately (by the 16. of the fift) as the pyramid ABCG is to the prisms which are in it, so is the solid X, to the prisms which are i● the pyramid DEFH. But the pyramid ABCG is greater than the prisms which are in it. Wherefore also the solid X is greater than the prisms which are in the pyramid DEFH (by the 14. of the fift). But it is supposed to be less which is impossible. Wherefore as the base ABC is to the base DEF, so is not the pyramid ABCG to any solid less than the pyramid DEFH.  
I say moreover, that as the base ABC is to the base DEF, so is not the pyramid ABCG, to any solid greater than the pyramid DEFH. For if it be possible, let it be unto some greater, namely, to the solid X. Wherefore (by conversion, by the Corollary of the 4. of the ●i●●) as the base DEF is to the base ABC, so is the solid X to the pyramid ABCG. But as the solid X is to the pyramid ABCG, so is the pyramid DEFH to some solid less than the pyramid ABCG, 
* In the Assu●p●●●llowin● the second proposition of this b●●ke.  as we have before proved. Wherefore also (by the 11. of the ●ift) as th● base DEF is to the base ABC, so is the pyramid DEFH, to some solid less than the pyramid ABCG which thing we have proved to be impossible. Wherefore as the base ABC is to the base DEF, so is not the pyramid ABCG to any solid greater than the pyramid DEFH: and it is als● proved that it is not in that proportion to any less than the pyramid DEFH. Wherefore as the base ABC is to the base DEF, so is the pyramid ABCG to the pyramid DEFH. Wherefore pyramids consisting under one and the self same altitude, and having triangles to their bases, are in that proportion the one to the other, that their bases are: which was required to be demonstrated.   

¶ The 6. Theorem. The 6. Proposition. Pyramids consisting under one and the self same altitude, and having P●ligo●on figures to their bases: are in that proportion the one to the other, that their bases are.  
SVppose that there be two Pyramids, having to their bases these Polig●non figures ABCED, and FGHKL, and let their ●oppes be the points M and N, which let be of one and the self same altitude. Then I say, that as the base ABCED is to the base FGHKL, so is the pyramid ABCEDM, to the pyramid FGHKLN. 
Construction.  Divide the base ABCED into these triangles ABC, ACD, & CDE, and likewise the base FGHKL into these triangles FGH, FHL, and HKL. And imagine that upon every one of those triangles be set a pyramid of equal altitude with the two pyramids put at the beginning. 
Demonstration.  And for that as the triangle ABC is to the triangle ADC, so is the pyramid ABCM to the pyramid ADCM (by the 5. of this book). Wherefore, by composition (by the 18. of the fift) as the four sided figure ABCD is to the triangle ACD, so is the pyramid ABCDM to the pyramid ACDM. But as the triangle ACD is to the triangle CDE, so is the pyramid ACDM to the pyramid CDEM. Wherefore of equality (by the 22. of the fift) as the base ABCD is to th● base CDE, so is the pyramid ABCDM to the pyramid CDEM. Wherefore again by composition (by the 18. of the fift) as the base ABCDE is to the base CDE, so is the pyramid ABCEDM to the pyramid CDEM. And by the same reason also as the base FGHKL is to the base HKL, so is the pyramid FGHKLN to the pyramid HKLN. And forasmuch as there are two pyramids CDEM and HKLN, having triangles to their bases, and being under one and the self same altitude, therefore (by the 5. of the twelfth) as the base CDE is to the base HKL, so is the pyramid CDEM to the pyramid HKLN. Now for that as the base ABCED is to the base CDE, so is the pyramid ABCEDM to the pyramid CDEM. But as the base CDE is to the base HKL, so is the pyramid CDEM to the pyramid HKLN. Wherefore of equality (by the 22. of the fift) as the base ABCED is to the base HKL, so is the pyramid ABCEDM to the pyramid HKLN. But also as the base HKL is the base FGHKL, so is the Pyramid HKLN to to the pyramid FGHKLN. Wherefore again of equality (by the 22. of the fift) as the base ABCED is to the base FGHKL, so is the pyramid ABCEDM to the pyramid FGHKLN. Wherefore pyramids consisting under one and the self same altitude, and having Polygonon figures to their bases, are in that proportion the one to the other, that their bases are: which was required to be proved.   

The 7. Theorem. The 7. Proposition. Every prism having a triangle to his base, may be divided into three pyramids equal the one to the other, having also triangles to their bases.  
SVppose that ABCDEF be a prism, having to his base the triangle ABC, and the opposite side unto it, the triangle DEF. Then I say that the prism ABCDEF, may be divided into three pyramids equal the one to the other, and having triangles to their bases. 
Demonstration.  Draw these right lines BD, EC, and CD. And forasmuch as ABED is a parallelogram, and his diameter is the line BD, therefore the triangle ABD is equal to the triangle EDB. Wherefore also the pyramid whose base is the triangle ABD, and top the point C, is equal to the pyramid whose base is the triangle EDB, & top the point C, by the 5. of this book. But the pyramid whose base is the triangle EDB, and top the point C, is one and the same which the pyramid whose base is the triangle EBC, and top the point D, for they are comprehended of the self same plain superficieces, namely, of the triangles BDEDEC, DBC, and EBC. Wherefore also the pyramid whose base is the triangle ABD and top the point C, is equal to the pyramid whose base is the triangle EBC and top the point D. Again forasmuch as BCFE is a parallelogram, and the diameter thereof is EC, therefore the triangle ECF is equal to the triangle CBE. Wherefore also the pyramid whose base is the triangle EBC and top the point D, is equal to the pyramid, whose base is the triangle ECF and top the point D, by the 5. of this book. But the pyramid whose base is the triangle BEC and top the point D, is proved to be equal to the pyramid whose base i● the triangle ABD, and top the point C. Wherefore also the pyramid whose base is the triangle CEF and top the point D, is equal to the pyramid whose base is the triangle ABD & top the point C. Wherefore the prism ABDEF is divided into three equal pyramids having triangles to their bases. And forasmuch as the pyramid whose base is the triangle ABD and top the point C, is one & the self same with the pyramid whose base is the triangle CAB & top the point D (for they are contained under the self same plain superficieces) but it hath been proved that the pyramid whose base is the triangle ABD and top the point C, is the third pyramid of the prism whose base is the triangle ABC a●d the opposite side unto it the triangle DEF. Wherefore the pyramid whose base is the triangle ABC and top the point D is the third pyramid of the prism, whose base is the triangle ABC, and opposite side the triangle DEF. Wherefore every prism having a triangle to his base, may be divided into three pyramids equal the one to the other, having also triangles to their bases: which was required to be proved.  

¶ Corollary.  
Hereby it is manifest that every pyramid is the third part of a prism having one and the same base with it and also being under the self same altitude with it. For if the base of the prism be any other rectiline figure than a triangle, that also may be divided into prisms which shall have triangles to their bases.  
Here Campane and Flussas add certain Corollaryes.  

First Corollary.  
Every Prism is triple to the pyramis, which hath the self same triangle to his base that the Prism hath, and the self same altitude. As it is manifest by this proposition, where the Prism is divided into three equal Pyramids, of which, two are upon one and the self same base, and under one and the self same altitude. But if the Prism have to his ●ase a parallelogram, and if the Pyramid have to his base the half of the same parallelogram, and their altitudes be equal, then again the Pyramid shallbe the third part of the Prism. For it was manifest, by the 40. of the cleveth, that prisms, being under equal altitudes, and the one having to his base a triangle, and the other a parallelogram double to the same triangle, are equal the one to the other. Whereof followeth the former conclusion.   

Second Corollary.  
If there be many prisms under one and the same altitude, and having triangles to their bases, 

Note.  
Sided Columns (sometime called prisms) are triple to pyramids, having one base and equal he●th with them.   and if the triangular bases be so joined together upon one and the same plain, that they compose a Poligonon figure: A pyramid set upon that base being a Poligonon figure, and under the same altitude, is the third part of that solid, which is composed of all the prisms added together.  
For forasmuch as eu●ry one of the prisms which hath to his base a triangle, to every one of the Pyramids set upon the same base (the altitude being always one and the same) is triple, it is manifest by the 12. of the fifth, that all the prisms are to all the Pyramids triple. Wherefore Parallelipipedons are triple to Pyramids set upon the self same base with them, and under the same altitude, 
Note: ●arallelipipedons triple to pyramids of one base and heith with them.  for that they contain two prisms.   

Third Corollary.  
If two prisms being under one and the self same altitude, have to their bases either, both triangles, or both parallelograms, the prisms are the one to the other, as their bases are.  
For forasmuch as those prisms are equemultiqlices unto the Pyramids upon the self same bases, and under the same altitude, which Pyramids are in proportion as their bases, it is manifest (by the 15. of the fift), that the prisms are in the proportion of the bases. For by the former Corollary, the prisms are triple to the Pyramids s●t upon the triangular bases.   

Fourth Corollary.  
prisms are in sesquealtera proportion to Pyramids which have the self same quadrangled base that the prisms have, and are under the self same altitude.  
For, that Pyramid containeth two Pyramids set upon a triangular base of the same Prism, for it is proved, that that Prism is triple to the Pyramid which is set upon the half of his quadrangled base, unto which the other which is set upon the whole base is double, by the sixth of this book.   

Fiveth Corollary.  
Wherefore we may in like sort conclude, that solids mentioned in the second Corollary (which solids Campane calleth sided Columns) being under one and the self same altitude, are in proportion the one to the other, as their bases, which are poligonon figures.  
For they are in the proportion of the Pyramids or prisms, set upon the self same bases, and under the self same altitude, that is, they are in the proportion of the bases of the said Pyramids or prisms. For those solids may be divided into prisms having the self same altitude, when as their opposite bases may be divided into triangles, by the 20 of the sixth. Upon which triangles prisms being set, are in proportion as their bases.  
By this 7. Proposition it plainly appeareth that ●u●lide, as it was before noted in the diffinition●● under the definition of a Prism, comprehended also those kinds of solids, which Campane calleth sided Columns. For in that he saith, Every Prism having a triangle to his base, may be deuided● etc. he needed not (taking a Prism in that sense which Campane and most men take it) to have added that particle, having to his base a triangle. For by their sense, there is no Prism, but it may have to his base a triangle● and so it may seem that Euclid aught without exception have said, that, every prism whatsoever, may be divided into three pyramids equal the one to the other, having also triangles to ●heir bases. For so do Campane and Flussas put the proposition, leaving out the former particle having to his base a triangle, which yet is read in the Greek copy, & not le●t out by any other interpreters known abroad except by Campane and Flussas. Yea and the Corollary following of this proposition added by Theon or Euclid, and amended by M. Dee seemeth to confirm this sense.  
Of this, ●s 〈◊〉 made manifest, that every pyramid is the third part of the prism, having the same base with it, and equal altitude. For, and if the base of the prism have any other right lined figure (than a triangle) and also the superficies opposite to the base, the same figure: that prism may be divided into prisms, having triangled bases: and the superficieces to those bases opposite, also triangled a ●●ike and equally.  
For there, as we see are put these words, ●or and if the base of the prism be any other right lined figure● etc. whereof a man may well infer that the base may be any other rectiline figure whatsoever, & not only a triangle or a parallelogram, and it is true also in that sense, as it is plain to see by the second corollary added out of Flussas, which corollary, as also the first of his corollaries, is in a manner all one with the Corollary added by Theon or Euclid. Farther Theon in the demonstration of the 10. proposition of this book (as we shall afterwards see) most plainly calleth not only sided columns prisms, but also parallelipipedons. And although the 40. proposition of the eleventh book may seem hereunto to be a l●t. For that it can be understanded of those prisms only which have triangles to their like, equal, opposite, and parallel sides, or but of some sided columns, and not of all: yet may that let be thus removed away, to say that Euclid in that proposition used genus pro specie, that is, the general word for some special kind thereof: which thing also is not rare, not only with him, but also with other learned philosophers. Thus much I thought good by the way to note in farther defence of Euclid definition of a Prism.     

The 8. Theorem. The 8. Proposition. Pyramids being like & having triangles to their bases, are in triple proportion the one to the other, of that in which their sides of like proportion are.  
SVppose that these pyramids whose bases are the triangles GBC and HEF and tops, the points A and D be like, and in like sort described, and let AB and DE be sides of like proportion. Then I say that the pyramid ABCG is to the pyramid DEFH in triple proportion, of that in which the side AB is to the side DE. Make perfect the parallelipipedons, namely, the solids BCKL & EFXO. And forasmuch as the pyramid ABCG is like to the pyramid DEFH, 
Construction.  therefore the angle ABC is equal to the angle DEF, 
Demonstration.  & the angle GBC to the angle HEF, and moreover the angle ABG to the angle DEH, and as the line AB is to the line DE, so is the line BC to the line EF, and the line BG to the line EH. And for that as the line AB is to the line DE, so is the line BC to the line EF, and the sides about the equal angles, are proportional, therefore the parallelogram BM is like to the parallelogram EP: and by the same reason the parallelogram BN is like to the parallelogram ER, and the parellelogramme BK is like unto the parallelogram EX. Wherefore the three parallelograms BM, KB and BN are like to the three parallelograms EP, EX, and ER. But the three parallelograms BM, KB, and BN are equal and like to the three opposite parallelograms, and the three parallelograms EP, EX, and ER are equal and like to the three opposite parallelograms. Wherefore the parallelipipedons BCKL and EFXO are comprehended under plain superficieces like and equal in multitude. Wherefore the solid BCKL is like to the solid EFXO. But like parallelipipedons are (by the 33. of the eleventh) in triple proportion the one to the other of that in which side of like proportion is to side of like proportion. Wherefore the solid BCKL is to the solid EFXO in triple proportion of that in which the side of like proportion AB is to the side of like proportion DE. But as the solid BCKL is to the solid EFXO, so is the pyramid ABCG to the pyramid DEFH (by the 15. of the fifth) for that the pyramid is the sixth part of this solid: for the prism, being the half of the parallelipipedon is triple to the pyramid. Wherefore the pyramid ABCG is to the pyramid DEFH in triple proportion of that in which the side AB is to the side DE. Which was required to be proved.  

Corollary.  
Hereby it is manifest that like pyramids having to their bases poligonon figures, are in triple proportion the one to the other, of that in which side of like proportion, is to side of like proportion.  
For if they be divided into pyramids having triangles to their bases (for like poligonon figures are divided into like triangles, and equal in multitude, and the sides are of like proportion) as one of the pyramids of the one, having a triangle to his base, is to one of the pyramids of the other, having also a triangle to his base, so also are all the pyramids of the one pyramid having triangles to their bases to all the pyramids of the other pyramid having also triangles to their bases. That is, the pyramid having to his base a poligonō●igure, to the pyramid having also to his base a poligonō●igure. But a pyramid having a triangle to his base, is to a pyramid having also a triangle to his base, & being like unto it, in triple proportion of that in which side of like proportion is to side of like proportion. Wherefore a pyramid having to his base a poligonon figure, is to a pyramid having also a polygonon figure to his base, the said pyramids being like the one to the other, in triple proportion of that in which side of like proportion is to side of like proportion. Likewise prisms and sided columns, being set upon the bases of those pyramids, 
An addition by Campane and Flussas.  and under the same altitude (forasmuch as they are equemultiplices unto the pyramids, namely, triples, by the corollary of the 7. of this book) shall have the ●ormer porportion that the pyramids have, by the 15, of the fifth, and therefore they shall be in triple proportion of that in which the sides of like proportion are.    

¶ The 9 Theorem. The 9 Proposition. In equal pyramids having triangles to their bases, the bases are reciprocal to their altitudes. And pyramids having triangles to their bases, whose bases are reciprocal to their altitudes, are equal the one to the other.  
SVppose that BCGA and EFHD be equal pyramids, having to their bases the triangles BCG and EFH, and the tops the points A and D. Then I say that the bases of the two pyramids BCGA and EFHD are reciprocal to their altitudes: that is, as the base BCG is to the base EFH, so is the altitude of the pyramid EFHD to the altitude of the pyramid BCGA. Make perfect the parallelipipedons, namely, BGML and EHPO. And forasmuch as the pyramid BCGA is equal to the pyramid EFHD, 
Demonstration of the first part.  & the solid BGML is sextuple to the pyramid BCGA. (For the parallelipipedon is duple to the Prism set upon the base of the Pyramid, & the Prism is triple to the pyramid): and likewise the solid EHPO is sextuple to the pyramid EFHD. Wherefore the solid BGML is equal to the solid EHPO. But in equal parallelipipedons, the bases are (by the 34. of the eleventh) reciprocal to their altitudes. Wherefore as the base BN is to the base EQ, so is the altitude of the solid EHP●, to the altitude of the solid BGML. But as the base BN is to the base EQ, so is the triangle GBC to the triangle HEF (by the 15. of the ●ifth, for the triangles GBC & HEF are the halves of the parallelograms BN and EQ) ● Wherefore (by the 11. of the fifth) as the triangle GBC is to the triangle HEF, so is the altitude of the solid EHPO to the altitude of the solid BGML. But the altitude of the solid EHPO is one and the same with the altitude of the pyramid EFHD, and the altitude of the solid BGML is one and the same with the altitude of the pyramid BCGA. Wherefore as the base GBC is to the base HEF, so is the altitude of the pyramid EFHD to the altitude of the pyramid BCGA. Wherefore the bases of the two pyramids BCGA and EFHD are reciprocal to their altitudes.  
But now suppose that the bases of the pyramids BCGA and EFHD, be reciprocal to their altitudes, 
Demonstration ●f the second part, which i● the converse of the first.  that is, as the base GBC is to the base HEF, so let the altitude of the pyramid EFHD be to the altitude of the pyramid BCGA. Then I say that the pyramid BCGA is equal to the pyramid EFHD. For (the self same order of construction remaining), for that as the base GBC is to the base ●EF, so is the altitude of the pyramid EFHD to the altitude of the pyramid BCGA. But as the base GBC is to the base HEF, so is the parallelogram GC to the parallelogram HF. Wherefore (by the 11. of the fifth) as the parallelogram GC is to the parallegoramme HF, so is the altitude of the pyramid EFHD to the altitude of the pyramid BCGA. But the altitude of the pyramid EFND and of the solid EHPO, is one and the self same, and the altitude of the pyramid BCGA and of the solid BGML, is also one and the same. Wherefore as the base GC is to the base HF, so is the altitude of the solid EHPO to the altitude of the solid BGML. But parallelipipedons, whose bases are reciprocal to their altitudes are (by the 34. of the eleventh) equal the one to the other. Wherefore the parallelipipedon BGML is equal to the parallelipipedon EHPO. But the pyramid BCGA is the sixth part of the solid BGML, and likewise the pyramid EFHD is the sixth part of the solid EHPO. Wherefore the pyramid BCGA is equal to the pyramid EFHD. Wherefore in equal pyramids having triangles to their bases, the bases are reciprocal to their altitudes. And pyramids having triangles to their bases, whose bases are reciprocal to their altitudes, are equal the one to the other: which was required to be demonstrated.  

A corollary added by Campane and Flussas.  
Hereby it is manifest that equal pyramids having to their bases Poligonon figures, have their bases reciprocal with their altitudes. And Pyramids whose bases being poligonon figures are reciprocal with their altitudes, are equal the one to the other.  
Suppose that upon the polygonon figures A and B, be set equal pyramids. Then I say that their bases A and B are reciprocal with their altitudes. Describe by the 25. of the sixth, triangles equal to the bases A and B. Which let be C and D. Upon which let there be set pyramids equal in altitude with the pyramids A and B. Wherefore the pyramids C and D, being set upon bases equal with the bases of the pyramids A and B, and having also their altitudes equal with the altitudes of the said pyramids A and B, shall be equal by the 6. of this book. Wherefore by the first part of this proposition, the bases of the pyramids, C to D are reciprocal with the altitudes of D to C. But in what proportion are the bases C to D, in the same are the bases A to B, forasmuch as they are equal. And in what proportion are the altitudes of D to C, in the same are the altitudes of B to A, which altitudes are likewise equal. Wherefore by the 11. of the fifth, in what proportion the bases A to B are, in the same reciprocally are the altitudes of the pyramids B to A. In like sort by the second part of this proposition may be proved the converse of this corollary. The same thing followeth also in a Prism, and in a sided column, as hath before at large been declared in the corollary of the 40. proposition of the 11. book. For those solids are in proportion the one to the other, as the pyramids or parallelipipedons, for they are either parts of equemultiplices or equemultiplices to parts.    

The 10. Theorem. The 10. Proposition. Every cone is the third part of a cilinder, having one and the self same base and one and the self same altitude with it.  
SVppose that there be a cone having to his base the circle ABCD; and let there be a cilinder having the self same base, and also the same altitude that the cone hath. Then I say that the cone is the third part of the cilinder, that is, that the cilinder is in triple proportion to the cone. For if the cilinder be not in triple proportion to the cone, than the cilinder is either in greater proportions then triple to the cone, or else in less. First let it be in greater than triple. 
Constr 〈◊〉 〈◊〉  And describe (by the 6. of the fourth) in the circle ABCD a square ABCD. Now the square ABCD, is greater than the half of the circle ABCD. For if about the circle ABCD, we describe a square, the square described in the circle ABCD is the half of the square described about the circle. And let there be Parallelipipedon prisms described upon those squares, 
Parallelipipedons called prisms.  equal in altitude with the cilinder. But prisms are in that proportion the one to the other, that their bases are (by the 32. of the eleventh, and 5. Corollary of the 7. of this book). Wherefore the prism described upon the square ABCD is the half of the prism described upon the square that is described about the circle. Now the clinder is less than the prism which is made of the square described abou● the circle ABCD, being equal in altitude with it, for it containeth it. Wherefore the prism described upon the square ABCD and being equal in altitude with the cylinder, is greater than half the cylinder. Divide (by the 30. of the third) the circumferences AB, BC, CD and DA into two equal parts in the points E, F, G, H, And draw these right lines AE, EB, BF, FC, CG, GD, DH & HA. Wherefore every one of these triangles AEB, BFC, CGD and DHA is greater than half of that segment of the circle ABCD which is described about it, as we have before in the 2. proposition declared. Describe upon every one of these triangles AEB, BFC, CGD, and DHA a prism of equal altitude with the cylinder. Wherefore every one of these prisms so described is greater than the half part of the segment of the cylinder that is set upon the said segments of the circle. For if by the points E, F, G, H, be drawn parallel lines to the lines AB, BC, CD and DA, and then be made perfect the parallelograms made by those parallel lines, and moreover upon those parallelograms be erected parallelipipedons equal in altitude with the cylinder, the prisms which are described upon each of the triangles AEB, BFC, CGD, and DHA are the halves of every one of those parallelipipedons. And the segments of the cylinder are less than those parallelipipedons so described. Wherefore also every one of the prisms which are described upon the triangles AEB, BFC, CGD and DHA is greater than the half of the segment of the cylinder set upon the said segment. Now therefore dividing every one of the circumferences remaining into two equal parts, and drawing right lines, and raising up upon every one of these triangles prisms equal in altitude with the cylinder, and doing this continually, we shall at the length (by the first of the tenth) leave certain segments of the cylinder which shallbe less then the excess whereby the cylinder exceedeth the cone more than thrice. Let those segments be AE, EB, BF, FC, CG, GD, DH and HA. Wherefore the prism remaining, whose base is the polygonon ●igure AEBFCGDH, and altitude the self same that the cylinder hath, is greater than the cone taken three times. 
* By this it is manifest that Euclid comprehended sided Columns also under the name of a Prism.  But the prism whose base is the polygonon figure AEBFCGDH and altitude the self same that the cylinder hath, is triple to the pyramid whose base is the polygonon figure AEBFCGDA and altitude the self same that the cone hath, by the corollary of the 3. of this book. Wherefore also the pyramid whose base is the polygonon figure AEBFCGDH and top the self same that the cone hath, is greater than the cone which hath to his base the circle ABCD. But it is also less, for it is contained of it which is impossible. Wherefore the cylinder is not in greater proportion then triple to the cone.  
I say moreover that the cylinder is not in less proportion then triple to the cone● For if it be possible let the cylinder be in less proportion then triple to the cone. Wherefore by conversion, the cone is greater than the third part of the cylinder. Describe now (by the sixth of the fourth) in the circle ABCD a square ABCD. Wherefore the square ABCD is greater than the half of the circle ABCD upon the square ABCD describe a pyramid having one & the self same altitude with the cone. Wherefore the pyramid so described is greater than half of the cone. (For if as we have before declared we describe a square about the circle, the square ABCD is the half of the square described about the circle, and if upon the squares be described parallelipipedons equal in altitude with the cone, which solids are also called prisms, the prism or parallelipipedon described upon the square ABCD is the half of the prism which is described upon the square described about the circle, for they are the one to the other in that proportion that their bases are (by the 32. of the eleventh, & 5. corollary of the 7. of this book.) Wherefore also their third parts are in the self same proportion (by the 15. of the fift). Wherefore the pyramid whose base is the square ABCD is the half of the pyramid set upon the square described about the circle. But the pyramid set upon the square described about the circle is greater than the cone whom it comprehendeth. Wherefore the pyramid whose base is the square ABCD, and altitude the self same that the cone hath; is greater than the half of the cone.) Divide (by the 30. of the third) every one of the circumferences AB, BC, CD, and DA into two equal parts in the points E, F, G, and H: and draw these right lines AE, EB, BF, FC, CG, GD, DH, and HA. Wherefore every one of these triangles AEB, BFC, CGD, and DHA is greater than the half part of the segment of the circle described about it. Upon every one of these triangles AEB, BFC, CGD, and DHA describe a pyramid of equal altitude with the cone and after the same manner every one of those pyramids so described is greater than the half part of the segment of the cone set upon the segment of the circle. Now therefore dividing (by the 30, of the third) the circumferences remaining into two equal parts, & drawing right lines & raising up upon every one of those triangles a pyramid of equal altitude with the cone, and doing this continually, we shall at the length (by the first of the tenth) leave certain segments of the cone, which shallbe less then the excess whereby the cone exceedeth the third part of the cylinder. Let those segments be AE, EB, BF, FC, CG, GD, DH, and HA. Wherefore the pyramid remaining, whose base is the poligonon figure AEBFCGDH and altitude the self same with the cone, is greater than the third part of the cylinder. But the pyramid whose base is the polygonon figure AEBFCGDH and altitude the self same with the cone, is the third part of the prism whose base is the poligonon figure AEBFCGDH and altitude the self same with the cylinder. Wherefore 
* A prism having for his base a polygonon figure as we have often before noted unto you.  the prism whose base is the polygonon figure AEBFCGDH, and altitude the self same with the cylinder, is greater than the cylinder whose base is the circle ABCD. But it is also less, for it is contained of it, which is impossible. Wherefore the cylinder is not in less proportion to the cone then in triple proportion. And it is proved that it is not in greater proportion to the cone then in triple proportion, wherefore the cone is the third part of the cylinder. Wherefore every cone is the third part of a cylinder, having one & the self same base, and one and the self same altitude with it: which was required to be demonstrated.  

¶ Added by M. john Dee.  

¶ A Theorem. 1.  
The superficies of every upright Cylinder, except his bases, is equal to that circle whose semidiameter is middle proportional between the side of the Cylinder, and the diameter of his base.   

¶ A Theorem. 2.  
The superficies of every upright, or Isosceles Cone, except the base, is equal to that circle, whose semidiameter is middle proportional between the side of that Cone, and the semidiameter of the circle: which is the base of the Cone.  
My intent in additions is not to amend Euclide● Method, (which needeth little adding or none at all). But my desire is somewhat to furnish you, toward a more general art Mathematical than Euclides Elements, 
Note: M. Dee his chief purpose in his additions.  (remaining in the terms in which they are written) can sufficiently help you unto. And though Euclides Elements with my Additions, run not in one Methodical race toward my mark: yet in the mean space my Additions either give light, where they are annexed to Euclides matter, or give some ready aid, and show the way to dilate your discourses Mathematical; or to invent and practise things Mechanically. And (in deed) if more leysor had happened, many more strange matters Mathematical had, (according to my purpose general) been presently published to your knowledge: but want of due leasour causeth you to want, that, which my good will toward you, most heartily doth wish you.  
As concerning the two Theorems here annexed, their verity, is by Archimedes, in his book of the Sphere and Cylinder manifestly demonstrated, and at large: you may therefore boldly trust to them, and use them, as suppositions, in any your purposes: till you have also their demonstrations. But if you well remember my instructions upon the first proposition of this book, and my other addition, upon the second, with the suppositions how a Cylinder and a Cone are Mathematically produced, you will not need Archimedes demonstration: nor yet be utterly ignorant of the solid quantities of this Cylinder and Cone here compared: (the diameter of their base, and heith being known in any measure) neither can their crooked superficies remain unmeasured. Whereof undoubtedly great pleasure and commodity may grow to the sincere student, and precise practiser.     

¶ The 11. Theorem. The 11. Proposition. Cones and Cylinders being under one and the self same altitude, are in that proportion, the one other that their bases are.  
In like sort also may we proue● that as the circle EFGH, is to the circle ABCD, so is not the cone EN to any solid less than the cone AL. Now I say that as the circle ABCD, is to the circle EFGH, so is not the cone AL to any solid greater than the cone EN. For if it be possible let it be unto a greater, namely to the solid X. Wherefore by conversion, as the circle EFGH, is to the circle ABCD, so is the solid X to the cone AL: but as the solid X is to the cone AL, so is the cone EN, to some solid less than the cone AL (as we may see by the assumpt put after th● second of this book): Wherefore (by the 11. of the fift) as the circle EFGH is to the circle ABCG● so is the cone EN to some solid less than the cone AL, which we have proved to be impossible. Wherefore as the circle ABCD, is to the circle EFGH, so is not the cone AL to any solid greater than the cone EN. And it is also proved that it is not to any less. Wherefore as the circle ABCD, is to the circle EFGH, so is the cone AL to the cone EN.  
But as the cone is to the cone, so is the cylinder to the cylinder, (by the 15. of the fift) for the one is in triple proportion to the other. 
Demonstration touching cylinders.  Wherefore) by the 11. of the fift) as the circle ABCD is to the circle EFGH, so are the cylinders which are set upon them the one to the other, the said cylinders being under equal altitudes with the cones. Cones therefore and cylinders, being under one & the self same altitude, are in that proportion the one to the other, that their bases are: which was required to be demonstrated.   

¶ The 12. Theorem. The 12. Proposition. Like Cones and Cylinders, are in triple proportion of that in which the diameters of their bases are.  
Now also I say, that the cone ABCDL is not to any solid greater than the cone EFGHN, 
Second case.  in triple proportion of that in which the diameter BD is to the diameter FH. For if it be possible, let it be to a greater, namely, to the solid X. Wherefore by conversion (by the Corollary of the 4. of the fift) the solid X is to the cone ABCDL, in triple proportion of that in which the diameter FH is to the diameter BD. But as the solid X is to the cone ABCDL, so is the cone EFGHN to some solid less than the cone ABCDL (as it is easy to see by the Assumpt put after the 2. Proposition). Wherefore also the cone EFGHN is unto some solid less than the cone ABCDL, in triple proportion of that in which the diameter FH is to the diameter BD: which is proved to be impossible. Wherefore the cone ABCDL is not to any solid greater than the cone EFGHN in triple proportion of that in which the diameter BD is to the diameter FH, and it is also proved that it is not to any less. Wherefore the cone ABCD is to the cone EFGHN in triple proportion of that in which the diameter BD is to the diameter FH.  
But as cone is to cone, so is cylinder to cylinder (by the 15. of the fift) for the cylinder is triple to the cone which is described on the one and self same base, 
Second par● which concerneth Cillinders.  and having one and the self same altitude with the cone. For it is proved (by the 10. of the twelfth) that every cone is the third part of a cylinder, having one and the self same base with it, and one & the self same altitude. Wherefore the cylinder is unto the cylinder in triple proportion of that in which the diameter BD is to the diameter FH. Wherefore like cones & cylinders are in triple proportion of that in which the diameters of their bases are: which was required to be proved.   

The 13. Theorem. The 13. Proposition. If a Cylinder be divided by a plain superficies being a parallel to the two opposite plain superficieces: then as the one Cylinder is to the other Cylinder, so is the axe of the one to the axe of the other.  
SVppose that there be a cylinder AD, whose axe let be EF, and let the opposite bases be the circles AEB, and CFD, and let AD be divided by the superfices GH being a parallel to the two opposite plain superficieces, AB, and CD. Then I say that as the cylinder BG is to the cylinder GD, so is the axe EK to the axe KF. Extend the axe EF on either side to the points L & M. And unto the axe EK put as many axes equal as you will, namely, 
Construction.  EN, & NL, & likewise unto the axe FK put as many axes equal as you will, namely, FX & XM. And by the points L, N, & X, M, extend plain superficieces parallels to the two superficieces AB, & CD (by the corollary of the 15. of the eleventh): & in the plain superficieces thus extended by the points L, N, X, M, imagine to be be drawn these circles, namely, OP, RS, TV, and ZW, having to their centres the points L, N, M, X, and le● them be equal to either of the circles AB, and CD and upon those circles imagine those cylinders PR, RB, DT, TW to be set. 
Demonstration.  Now forasmuch as the axes LN, NE, and EK are equal the one to the other: Therefore the cylinders PR, RB, and BG are (by the 11. of the twelfth) in proportion the one to the other as their bases are. But the bases are equal. Wherefore also the cylinders PR, RB, and BG are equal the one to the other. And forasmuch as the axes LN, NE, and EK are equal the one to the other, and the cylinders PR, RB, and BG are also equal the one to the other, and the multitude of the axes LN, NE, and EK is equal to the multitude of the cylinders PR, RB, and BC: therefore how multiplex the whole axe KL is to the axe EK, so multiplex is the whole cylinder PG to the cylinder BG. And (by the same reason) also how multiplex the whole axe MK is the axe KF, so multiplex is the whole cylinder WG to the cylinder GD. Wherefore if the axe KL be equal the axe KM, the cylinder PG is equal to cilinder GW. And if the axe KL be greater th●n the axe KM, the cylinder PG is greater than the cylinder GW. And if it be less it is less. Now therefore there are four magnitudes, namely, the two axes EK, and KF, and the two cylinders BG, and GD, and unto the axe EK, and to the cylinder BG, namely, to the first and the third are taken equemultiplices, namely, the axe KL, and the cylinder PG. And likewise unto the axe GF, and unto the cylinder GD, namely, the second and the fourth, are taken other equemultiplices, namely, the axe KM, and the cylinder GW. And it is proved that if the axe KL, exceed the axe KM, the cylinder PG exceedeth the cylinder GW and that if it be equal it is equal, and if it be less it is less. Wherefore (by the 6. definition of the fift) as the axe EK is to the axe KF, so is the cylinder BG to the cylinder GD. If therefore a cylinder be divided by a plain superficies being a parallel to the two opposite plain superficieces: then as the one cylinder is to the other cilinder, so is the axe of the one to the axe of the other: which was required to be proved.   

¶ The 14. Theorem. The 14. Proposition. Cones and Cylinders consisting upon equal bases, are in proportion the one to the other as their altitudes.  
SVppose that the cylinders FD, and EB, and the cones AGB, and CKD, do consist upon equal bases, namely, upon the circles AB, and CD. 
Construction.  Then I say that as the cylinder EB is to the cylinder FD, so is the axe GH to the axe KL. Extend the axe KL directly to the point N, and unto the axe GH, put the axe LN equal: and about the axe LN imagine a cylinder CM. 
Demonstration touching Cylinders.  Now forasmuch as the cilinders EB, and CM, are under equal altitudes, therefore (by the 11. of the twelfth) they are in the proportion the one to the other as their bases are. But the bases are equal the one to the other. Wherefore also the cylinders EB, and CM are equal the one to the other. And forasmuch as the whole cylinder FM, is divided by a plain superficies CD being a parallel to either of the opposite plain superficieces: therefore (by the 13. of the twelfth) as the cylinder CM is to the cylinder FD, so is the axe LN to the axe LK. But the cylinder CM is equal to the cylinder EB, and the axe LN to the axe GH. Wherefore as the cylinder EB is to the cylinder FD, so is the axe GH to the axe KL.  
But as the cylinder EB is to the cylinder FD, so (by the 15. of the fift) is the cone ABG to the cone CDK, 
Demonstration touching Cones.  for the cylinders are in triple proportion to the cones (by the 10. of the twelfth). Wherefore (by the 11. of the fift) as the axe GH is to the axe KL, so is the cone ABG to the cone CDK, & the cylinder EB to the cylinder FD. Wherefore cones & cylinders consisting upon equal bases are in proportion the one to the other as their altitudes: which was required to be demonstrated.   

¶ The 15. Theorem. The 15. Proposition. In equal Cones and Cylinders, the bases are reciprocal to their altitudes. And cones and Cylinders whose bases are reciprocal to their altitudes, are equal the one to the other.  
SVppose that these cones ACL, EGN or these cylinders AX, EO, whose bases are the circles ABCD, EFGH, and axes KL, and MN (which axes are also the altitudes of the cones & cylinders) be equal the one to the other. 
First part of the proposition demonstrated touching Cones.  Then I say that the bases of the cylinders XA & EO are reciprokal to their altitudes, that is, that as the base ABCD is to the base EFGH, 
Two cases in this proposition.  so the altitude MN to the altitude KL. For the altitude KL is either equal to the altitude MN or not. First let it be equal. 
The first case.  But the cylinder AX is equal to the cylinder EQ. But cones and cylinders consisting under one and the self same altitude, are in proportion the one to the other as their bases are (by the 11. of the twelfth). Wherefore the base ABCD is equal to the base EFGH. Wherefore also they are reciprokal: as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL.  
But now suppose that the altitude LK be not equal to the altitude M. N, 

Second case.  
Construction.   but let the altitude MN be greater. And (by the 3. of the first) from the altitude MN take away PM equal to the altitude KL, so that let the line PM, be put equal to the line KL. And by the point P let there be extended a plain superficies TVS, which let cut the cylinder EO, and be a parallel to the two opposite plain super●icieces, that is, to the circles EFGH, and RO. 
Demonstration touching cylinders.  And making the base the circle EFGH, & the altitude MP imagine a cylinder ES. And for that the cylinder AX is equal to the cylinder EO, and there is an other cylinder ES, therefore (by the 7. of the fift) as the cylinder AX is to the cylinder ES, so is the cylinder EO to the cylinder ES. But as the cylinder AX is to the cylinder ES, so is the base ABCD to the base EFGH. For the cylinders AX, and ES are under one and the self same altitude. And as the cylinder EO, is to the cylinder ES, so is the altitude MN to the altitude MP. For cylinders consisting upon equal bases are in proportion the one to the other as their altitudes. Wherefore as the base ABCD is ●o the base EFGH, so is the altitude MN to the altitude MP. But the altitude PM is equal to the altitude KL. Wherefore as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL. Wherefore in the equal cylinders AX, and EO the bases are reciprocal to their altitudes. But now suppose that the bases of the cylinders AX, and EO be reciprokal to their altitudes, that is, as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL. 
Second part demonstrated.  Then I say that the cylinder AX is equal to the cylinder EO. For the self same order of construction remaining, for that as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL, but the altitude KL is equal to the altitude PM. Wherefore as the base ABCD is to the base EFGH, so is the altitude MN to the altitude PM. But as the base ABCD is to the base EFGH, so is the cylinder AX to the cylinder ES, for they are under equal altitudes: and as the altitude MN is to the altitude PM, so is the cylinder EO to the cylinder ES (by the 14. of the twelfth). Wherefore also as the cylinder AX is to the cylinder ES, so is the cylinder EO to the cylinder ES. Wherefore the cylinder AX is equal to the cylinder EO (by the 9 of the fift). And so also is it in the cones which ha●● the self same bases and altitudes with the cylinders. Wherefore in equal cones and cylinders, the bases are reciprocal to their altitudes etc. which was required to be demonstrated.  

A corollary added by Campane and Flussas.  
Hitherto hath been showed the passions and proprieties of cones and cylinders whose altitudes fall perpendicularly upon the bases. Now will we declare that cones and cilinders whose altitudes fall obliquely upon their bases have also the self same passions and proprieties which the foresaid cones and cilinders have.  
Forasmuch as in the tenth of this book it was said, that every Cone is the third part of a cilinder having one, and the self same base, & one & the self same altitude with it, which thing was demonstrated by a cilinder given, whose base is cut by a square inscribed in it, and upon the sides of the square are described Isosceles triangles, making a polygonon figure, and again upon the sides of this polygonon figure are infinitely after the same manner described other Isosceles triangles taking away more than the half, as hath oftentimes been declared: therefore it is manifest, that the solids set upon these bases, being under the same altitude that the cilinder inclined is, and being also included in the same cilinder, do take away more than the half of the cilinder, and also more than the half of the residue, as it hath been proved in erected cylinders. For these inclined solids being under equal altitudes and upon equal bases with the erected solids are equal to the erected solids by the corollary of the ●0. of the eleventh. Wherefore they also in like sort as the erected, take away more than the half. If therefore we compare the inclined cilinder, to a cone set upon the self same base, and having his altitude erected, and reason by an argument leading to an impossibility by the demonstration of the tenth of this book, we may prove that the sided solid included in the inclined cylinder is greater than the triple of his pyramid, and it is also equal to the same which is impossible. And this is the first case, wherein it was proved that the cilinder not being equal to the triple of the cone is not greater than the triple of the same. And as touching the second case, we may after the same manner conclude that that ●ided solid contained in the cylinder is greater than the cylinder: which is very absurd Wherefore if the cylinder be neither greater than the triple of the cone, nor less, it must needs be equal to the same. The demonstration of these inclined cylinders most plainly followeth the demonstration of the erected cylinders: for it hath already been proved, that pyramids, and sided solids (which are also called generally prisms) being set upon equal bases and under one and the self same altitude, whether the altitude be erected or inclined, are equal the one to the other, namely, are in proportion as their bases are by the ●. of this book. Wherefore a cylinder inclined shall be triple to every cone (although also the cone be erected) set upon one and the same base with it, and being under the same altitude. But the cilinder erected was the triple of the same cone by the tenth of this book. Wherefore the cilinder inclined is equal to the cilinder erected being both set upon one and the self same base, and having one and the self same altitude. The same also cometh to pass in cones, which are the third parts of equal cilinders, & therefore are equal the one to the other.  
Wherefore according to the eleventh of this book it followeth, that cylinders and cones inclined or erected, being under one and the self same altitude, are in proportion the one to the other as their bases are. For forasmuch as the erected are in proportion as their bases are, and to the erected cilinders the inclined are equal: therefore they also shall be in proportion as their bases are.  
And therefore by the 12. of this book like cones and cylinders being inclined are in triple proportion of that in which the diameters of the bases are. For forasmuch as they are equal to the erected which have the proportion by the 12. of this book, and their bases also are equal with the bases of the erected, therefore they also shall have the same proportion.  
Wherefore it followeth by the 13. of this book, tha● cylinder inclined, being cut by a plain superficies parallel to the opposite plain superficieces thereof shall be cut according to the proportions of the axes. For suppose that upon one and the self same base ●e set an erected cylinder and an inclined cylinder, being both under one and the self same altitude, which 〈…〉 a plain superficies parallel to the opposite bases. Now it is manifest that the sections of the one cylinder are equal to the section of the other cylinder, for they are set upon equal bases, and under one and the self same altitude, namely, between the parallel plain superficieces. And their axes also are by those parallel plain superfici● 〈◊〉 proportionally by the 1ST of ●he ●leuen●h. Wherefore the inclined cylinders (being equal to the erected cylinders) shall have the proportion of thei● axes, a● also have the erected. For in ech● the proportion of the axes is one and the same.  
Wherefore inclined Cones and Cylinders being set upon equal bases, shall by the 14. of this book be in 〈◊〉 as their altitudes 〈…〉 forasmuch a● the inclined are equal to the erected which have the self same bases and altitude, and the erected are i● proportion as their altitudes: therefore the inclined shall be in proportion the one to the other as ●he self same altitudes which are common to each, namely, to the inclined and to the erected.  
And therefore in equal cones and cylinders whether they be inclined or erected, the bases shall be reciprocally proportional with the altitudes, and contrariwise by the 15. of this book. For forasmuch as we have oftentimes showed that the inclined cones and cylinders are equal to the erected, having the self same bases and altitudes with them, and the erected unto whom the inclined are equal, ha●e their bases reciprocal proportionally with their altitudes, therefore it followeth, that the inclined (being equal to the erected) have also their bases and altitudes (which are common to each) reciprocally proportional. Likewise if thei● altitudes & bases be reciprocally proportional, they themselves also shall be equal, for that they are equal to the erected cylinders and cones set upon the same bases and being under the same altitudes which erected cylinders are equal the one to the other by the same 15. of this book. Wherefore we may conclude, that those passions & proprieties which in this twelfth book have been proved to be in cones and cylinders whose altitudes are erected perpendicularly to the 〈…〉 set obliquely vpo● their bases. Howbeit this is to be noted that such inclined cones or cylinders are not perfect round as are the erected: so that if they be cut by a plain superficies passing at right angles with their altitude, this section is a conical section, which is called Ellipsis, and shall not describe in their superficies a circle as it doth in erected cylinders & cones, but a certain figure, whose less diameter is in cylinders equal to the dimetient of the base: that is, is one and the same with it. And the same thing happeneth also in cones inclined, being cut after the same manner.    

The 1. Problem. The 16. Proposition. Two circles having both one and the self same centre being given, to inscribe in the greater circle a polygonon figure, which shall consist of equal and even sides, and shall not touch the superficies of the less circle.  
SVppose that there be two circles ABCD, and EFGH having one & the self same centre, namely, K. It is required in the greater circle which let be ABCD to inscribe a polygonon figure which shallbe of equal and even sides and not touch the circle EFGH. Draw by the centre K a right line BD. 
Construction.  And (by the 11. of the first) from the point G raise up unto the right line BD a perpendicular line AG, and extend it to the point C. Wherefore the line AC toucheth the circle EFGH (by the 15. of the third). Now therefore if (by the 30. of the third) we divide the circumference BAD into two equal parts, and again the half of that into two equal parts, and thus do continually, we shall (by the corollary of the 1. of the tenth) at the length leave a certain circumference less than the circumference AD. Let the circumference left be LD. And from the point L. Draw (by the 12. of the first) unto the line BD a perpendiculare line 
* Note this LM, because of KZ in the next proposition, and here the point M: for the point Z in the next demonstration.  LM, 
I Dee.  and extend it to the point N. And draw these right lines LD and DN. And forasmuch as the angles DML, and DMN are right angles, therefore (by the 3. of the third) the right line BD divideth the right line LN into two equal parts in the point M. Wherefore (by the 4. of the first) the rest of the sides of the triangles DML, and DMN, namely, the lines DL, and DN shallbe equal. And forasmuch as the line AC is a parallel to the LN (by the 28. of the first): But AC toucheth the circle EFGH, wherefore the line LM toucheth not the circle EFGH, and much less do the lines LD, and DN touch the circle EFGH. If therefore there be applied right lines equal to the line LD continually into the circle ABCD (by the 1. of the fourth) there shallbe described in the circle ABCD a polygonon figure which shallbe of equal, and 
* For that the sections were made by the number two: that is, by taking halves, and of the residue the hal●e● and so, to LD, being an half, and a residue which shall be a common measure back again, to make sides of the Poligonon figure.  even sides, and shall not touch the less circle, namely, EFGH: (by the 14. of the third or by the 29.) which was required to be done.  

¶ Corollary.  
Hereby it is manifest that a perpendicular line drawn from the point L to the line BD toucheth not one of the circles.   

¶ An Assumpt added by Flussas.  
If a Sphere be cut of a plain superficies, the common section of the superficieces, shall be the circumference of a circle.  
Suppose that the sphere ABC be cut by the plain superficies AEB, and let the centre of the sphere be the point D. 
Construction.  And from the point D, let there be drawn unto the plain superficies AEB a perpendicular line (by the 11. of the eleventh) which let be the line DE. And from the point E draw in the plain superficies AEB unto the common section of the said superficies and the sphere, lines how many so ever, namely, EA and EB. And draw these lines DA and DB. Now forasmuch as the right angles DEA and DEB are equal (for the line DE is erected perpendicularly to the plain superficies). 
Demonstration.  And the right lines DA & DB which subtend those angles are by the 12. definition of the eleventh equal: which right lines moreover (by the 47. of the first) do contain in power the squares of the lines DE, EA, and DE, EB: if therefore from the squares of the lines DE, EA, and DE, EB ye take away the square of the line DE which is common unto them, the residue namely the squares of the lines EA & EB shall be equal. Wherefore also the lines EA and EB are equal. And by the same reason may we prove, that all the right lines drawn from the point E to the line which is the common section of the superficies of sphere and of the plain superficies, are equal. Wherefore that line shall be the circumference of a circle, by the 15. definition of the first. 
* The circles so made: or so considered in the sphere, are called the greatest circles: All other, not having the centre of the sphere, to be their centre also● are called less circles.  But if it happen the plain superficies which cutteth the sphere, to pass by the centre of the sphere, the right lines drawn from the centre of the sphere to their common section, shal● be equal by the 12. definition of the eleventh. For that common section is in the superficies of the sphere. Wherefore of necessity the plain superficies comprehended under that line of the common section shall be a circle, and his centre shall be one and the same with the centre of the sphere.  

john Dee.  
Euclid hath among the definition of solids omitted certain, which were easy to conceive by a kind of Analogy. As a segment of a sphere, a sector of a sphere, the vertex, or top of the segment of a sphere: with such like. But that (if need be) some farthe● light may be given, in this figure next before, 
Note these descriptions.  understand a segment of the sphere ABC to be that part of the sphere contained between the circle AB, (whose centre is E) and the spherical superficies AFB. To which (being a less segment) add the cone ADB (whose base is the former circle: and top the centre of the sphere) and you have DAFB a sector of a sphere, or solid sector (as I call it). DE extended to F, showeth the top or vertex of the segment, to be the point F and EF is the altitude of the segment spherical. Of segments, some are greater than the half sphere, some are less. As before ABF is less, the remanent, ABC is a segment greater than the half sphere.    

¶ A Corollary added by the same Flussas.  
By the foresaid assumpt it is manifest, that if from the centre of a sphere the lines drawn perpendicularly unto the circles which cut the sphere, be equal: those circles are equal. And the perpendicular lines so drawn fall upon the centres of the same circles.  
For the line which is drawn from the centre of the sphere to the circumference, containeth in power, the power of the perpendicular line, and the power of the line which joineth together the ends of those lines. Wherefore from that square or power of the line from the centre of the sphere to the circumference or common section drawn, which is the semidiameter of the sphere, taking away the power of the perpendicular, which is common to them ●ound about, it followeth, that the residues how many so ever they be, be equal powers, and therefore the lines are equal the one to the other. Wherefore they will describe equal circles, by the first definition of the third. And upon their centres fall the perpendicular lines by the 9 of the third.  

* another Corollary.  And those circles upon which falleth the greater perpendicular lines are the less circles. For the powers of the lines drawn from the centre of the sphere to the circumference being always one and equal, to the powers of the perpendicular lines and also to the powers of the lines drawn from the centres of the circles to their circumference, the greater that the powers of the perpendicular lines taken away from the power containing them both are, the less are the powers and therefore the lines remaining, which are the semidiameters of the circles and therefore the less are the circles which they describe. Wherefore if the circles be equal, the perpendicular lines falling from the centre of the sphere upon them, shall also be equal. For if they should be greater or less, the circles should be unequal as it is before manifest. But we suppose the perpendiculars to be equal. 
* another Corollary.  Also the perpendicular lines falling upon those bases are the lest of all, that are drawn from the centre of the sphere: for the other drawn from the centre of the sphere to the circumference of the circles, are in power equal both to the powers of the perpendiculars and to the powers of the lines joining these perpendiculars and these subtendent lines together: making triangles rectangleround about: as most easily you may conceive of the figure here annexed.  
A the Centre of the Sphere.  
AB the lines from the Centre of the Sphere to the Circumference of the Circles made by the Section.  
BCB the Diameters of Circles made by the Sections.  
AC the perpendiculars from the Centre of the Sphere to the Circles● whose diameters BCB are one both sides or in any situation else.  
CB the Semidiameters of the Circles made by the Sections.  
AO a perpendicular long than AC: and therefore the Semidiameter OB is less.  
ACB, & AOB triangles rectangle.    

¶ The 2. Problem. The 17. Proposition. Two spheres consisting both about one & the self same centre, being given, to inscribe in the greater sphere a solid of many sides (which is called a Polyhedron) which shall not touch the superficies of the less sphere.  
SVppose that there be two spheres about one & the self same centre, namely, about A. It is required in the greater sphere to inscribe a Polyhedron, or a solid of many sides, which shall not with his superficies touch the superficies of the less sphere. Let the spheres be cut by some one plain superficies passing by the centre A. 
Construction.  Then shall their sections be circles. 
* This is also proved in the Assumpt before added out o● Flussas.  [For (by the 12. definition of the eleventh) the Diameter remaining fixed, and the semicircle being turned round about maketh a sphere. Wherefore in what position so ever you imagine the semicircle to be, the plain superficies which passeth by it shall make in the superficies of the sphere a circle. And it is manifest that is also a greater circle, for the diameter of the sphere which is also the diameter of the semicircle, and therefore also of the circle is (by the 15. of the third) greater then all the right lines drawn in the circle or sphere] which circles shall have both one centre being both also in that one plain superficies, 
Note what a greater or greatest circle in a Spear is.  by which the spheres were cut. Suppose that that section or circle in the greater sphere be BCDE, and in the less, be the circle FGH. Draw the diameters of those two circles in such sort that they make right angles, and let those diameters be BD, and CE. 
First part of the Construction.  And let the line AG, being part of the line AB, be the semidiameter of the less sphere and circle, as AB is the semidiameter of the greater sphere and greater circle: both the spheres and circles having one and the same centre. Now two circles that is, BCDE, and FGH consisting both about one, and the self same centre being given, let there be described (by the proposition next going before) in the greater circle BCDE a polygonon figure consisting of equal and even sides, not touching the less circle FGH. And let the sides of that figure in the fourth part of the circle, namely, in BE, be BK, KL, LM, and ME. And draw a right line from the point K to the point A, and extend it to the point N. And (by the 12. of the eleventh) from the A raise up to the superficies of the circle BCDE a perpendicular line AX, and let it light upon the superficies of the greater sphere in the point X. 
Note●  And by the line AX, and by either of these lines BD, and KN extend plain superficieces. Now by that which was before spoken, those plain superficieces shall in the 
* 
You know full well, that in the superficies of the sphere, ●●●ly the circumferences of the circles are: but by th●se circumferences the limitation and assigning of circles is used: and so, the circumference of a circle, usually called a circle, which in this place can not offend.  
This figure is restored by M. Dee his diligence. For in the greek and Latin Euclides, the line GL, the line AG, and the line KZ, (in which three lines the chief pinch of both the demonstrations doth stand), are untruly drawn: as by comparing, the studious may perceive.   superficies of the sphere make two greater circles. Let their semicircles consisting upon the diameters BD, and KN be BXD. and KXN. And forasmuch 

Note.  
You must imagine 〈◊〉 right line AX, to be perpendicular upon the diameters BD and CE: though here AC the semidiater, seem to be part of AX. And so in other points, in this figure, and many other, strengthen your imagination, according to the tenor of constructions: though in the delineation in plain, sense be not satisfied.   as the line XA is erected perpendicularly to the plain superficies of the circle BCDE. Therefore all the plain superficieces which are drawn by the line XA are erected perpendicularly to the superficies of the circle BCDE (by the 18. of the eleventh). Wherefore the semicircles BXD, and KXN are erected perpendicularly to the plain superficies of the circle BCDE. And forasmuch as the semicircles BED, BXD, and KXN are equal, for they consist upon equal diameters ●D, and KN: therefore also the fourth parts or quarters of those circles, namely, BE, BX and KX are equal the one to the other. Wherefore how many sides of a polygonon figure there are in the fourth part or quarter BE, so many also are there in the other fourth parts or quarters BX, and KX, equal to the right lines BK, KL, LM, and ME. Let those sides be described, and let them be BO, OP, PR, RX, KS, ST, TV, and VX: 

Note.  
BO equal to BK, in respect of M. Dee his demonstration following.   and draw these right lines SO, T●, and V●. And from the points O and ●. Draw to the plain superficies of the circle BCDE perpendicular lines, which perpendicular lines will fall upon the common section of the plain superficieces, namely, upon the lines BD, & KN, (by the 38. of the eleventh) for that the plain superficieces of the semicircles BXD, KXN are erected perpendicularly to the plain superficies of the circle BCDE. Let those perpendicular lines be GZ, 
† Note ●his point Z that you may the better understand M. Dee his demonstration.  and SW. And draw a right line from the point Z to the point W. And forasmuch as in the equal semicircles BXD, and KXN the right lines BO, and KS are equal, from the ends whereof are drawn perpendicular lines OZ, and SW, therefore (by the corollary of the 35. of the eleventh) the line OZ is equal to the line SW, & the line BZ is equal to the line KW. [Flussas proveth this an other way thus: Forasmuch as in the triangles SWK, and OZB, the two angles SWK, and OZB are equal, for that by construction they are right angles, and by the 27. of third the angles WKS, and ZBO are equal, for they subtend equal circumferences SXN and OXD, and the side SK is equal to the side OB as it hath before been proved. Wherefore (by the 26. of the first) the other sides & angles are equal, namely, the line OZ to the lines SW, and the line BZ to the line KW,] But the whole line BA is equal to the whole line KA: by the definition of a circle Wherefore the residue ZA is equal to the residue WA. Wherefore the line ZW is a parallel to the line BK (by the 2. of the sixth). And forasmuch as either of these lines OZ, & SW is erected perpendicularly to the plain superficies of the circle BCDE, therefore the line OZ is a parallel to the line SW, (by the 6. of the eleventh) and it is proved that it is also equal unto it. Wherefore the lines WZ and SO are also equal and parallels (by the 7, of the eleventh, and the 33. of the first, and by the 3. of the first) And forasmuch as WZ is a parallel to SO: But ZW is a parallel is to KB: Wherefore SO is also a parallel to KB, (by the 9 of the eleventh). And the lines BO, and KS do knit them together. Wherefore the four sided figure BOKS is in one and the self same plain superficies. For (by the 7. of the eleventh) if there be any two parallel right lines, and if in either of them be taken a point at allaventures, a right line drawn by these points is in one, & the self same plain superficies with the parallels: And by the same reason also every one of the four sided figures SOPT, and TPKV is in one and th● self same plain superficies. And the triangle VRX is also in one and the self same plain superficies (by the ●. of the eleventh). Now if we imagine right lines drawn from the points O, S, P, T, R, V, to the point 〈◊〉, there shallbe described a Polyhedron or a solid figure of many sides, between the circumferences BX and KX composed of pyramids, whose bases are the four sided figures BKOS, SOPT, TPRV, and the triangle VRX, and top the point A. And if in every one of the sides K●L, LM and ME we use the self same construction that we did in BK, and moreover in the other three quadrants or quarters, and also in the other half of the sphere, there shall then be made a Polyhedron or solid figure consisting of many sides described in the sphere, which Polyhedron is made of the pyramids whose bases are the foresaid four sided figure's, and the triangle VRK, and others which are in the self same order with them: and common top to them all in the point A. Now I say that the foresaid polihedron solid of many sides toucheth not the superficies of the less sphere in which is the circle FGH: 
Second part of the construction.  Draw (by the 11. of the eleventh) from the point A to the plain superficies of the four sided figure KBOS a perpendicular line AY, and let it fall upon the plain superficies in the point Y. And draw these right lines BY & YK. And forasmuch as the line AY is erected perpendicularly to the plain superficies of BKOS, therefore the same AY is erected perpendicularly to all the right lines that touch it, and are in the plain superficies of the four sided figure (by the ●. definition of the eleventh). 
Second part of the demonstration.  Wherefore the line AY is erected perpendicularly to either of these lines BY, and YK. And forasmuch as (by the 15. definition of the first) the line A● is equal to the line AK, therefore the square of the line AB is equal to the square of the line AK. And to the square of the line A● are equal the squares of the lines AY, and YB (by the 47. of the first) for the angle ●. YA is a right angle. And to the square of the line AK are equal the squares of the lines AY and YK. Wherefore the squares of the lines AY, and YB are equal to the squares of the lines AY, and TK: take away the square of the line AY which is common to them both. Wherefore the residue, namely, the square of the BY is equal to the residue, namely, to the square of the line YK. Wherefore the line BY is equal to the line YK. In like sort also may we prove that right lines drawn from the point Y to the points O, and S, are equal to either of the lines BY, and YK. Wherefore making the centre the point Y, and the space either the line BY or the line YK describe a circle, and it shall pass by the points O, and S, and the four sided figure KBOS shallbe inscribed in the circle. And forasmuch as the line KB is greater than the line WZ (by the 2. of the sixth, because AK is greater than AW), but the line WZ is equal to the line SO. Wherefore the line BK is greater than the line SO. But the line BK is equal to either of the lines KS, and BO by construction. Wherefore either of the lines KS, and ●O is greater th●n the line SO. And forasmuch ●s in the circle is a four sided figure KBOS, and the sides BK, BO and KS are equal, and the side OS is less than any one of them, and the line BY is drawn from the centre of the circle: therefore the square of the line KB is greater than the double of the square of the line BY (by the 12. of the second) afore that it sub●endeth an angle greater than a right angle contained of the two equal 〈◊〉 B●, and YK, which angle BYK is an obtuse angle. For the 4. angles at the centre Y are equal to 4● right angles: of which three, namely, the angles BYK, KYS, and BYO are equal by the 4. of the first, and the fourth namely, the angle SYO is less than any of those three angles, by the 25. of the first.) Draw (by the 12. of the first) from the point K to the line BZ, a perpendicular line 
✚ Which of necessity shall fall upon Z, as M. Dee proveth it: and his proof is set after at this mark ✚ following.  KZ. And forasmuch as the line BD is less than the double to the line DZ (for the line BD is double to the line DA, which is less than the line DZ): but as the line BD is to the line DZ, so is the parallelogram contained under the lines DB and BZ, to the parallelogram contained under the lines DZ and ZB (by the 1. of the sixth): therefore if ye describe upon the line BZ a square, and making perfect the parallelogram contained under the lines ZD, and ZB, that which is contained under the lines DB & BZ, shall be less than the double to that which is contained under the lines DZ and ZB. And if ye draw a right line from the point K to the point D, that which is contained under the lines DB and BZ, equal to the square of the line BK (by the Corollary of the 8. of the sixth) for the angle BKD is a right angle, by the 3●. of the third, for it is in the semicircle BED): and that which is contained under the lines DZ and ZB, is equal to the square of the line KZ (by the same Corollary). Wherefore the square of the line KB, is less than the double to the square of the line KZ. But the square of the line KB is greater than the double to the square of the line BY, as before hath been proved. Wherefore the square of the line KZ, is greater than the square of the line BY (by the 10. of the fift). And forasmuch as (by the 15. definition of the first) the line BA is equal to the line KA: therefore also the square of the line BA is equal to the square of the line KA. But (by the 47. of the first) unto the square of the line AB, are equal the squares of the lines BY & YA: (for the angle BYA is by construction, a right angle). And (by the same reason) to the square of the line KA, are equal the squares of the lines KZ and ZA: (for the angle KZA is also by construction, a right angle). Wherefore the squares of the lines BY and YA, are equal to the squares of the lines KZ and ●A● Of which the square of the line KZ is greater than the square of the line BY, as hath before been proved. Wherefore the residue, namely, 
I Dee.  the square of the line ZA, is less than the square of the line YA. Wherefore the line YA is greater than the line AZ. 
* But AZ is greater than AG, as in the former proposition, KM was evident to be greater than KG: so may it also be made manifest that KZ doth neither touch nor cut the circle FG●H.  Wherefore the line AY is much greater the● the line AG. But the line AY falleth upon one of the bases of the Polihedron, and the line AG falleth upon the superficies of the less sphere. Wherefore the Polihedron toucheth not the superficies of the less sphere.  
An other and more ready demonstration to prove that the line AY is greater than the line AG. 
An other prove that the line AY is greater than the line AG.  Raise up (by the 11. of the first) from the point G to the line AG a perpendicular line GL. And draw a right line from the point A to the point L. Now then dividing (by the 30. of the third) the circumference E● into halves, & again that half into halves, & thus doing continually, we shall at the length by the corollary of the first of the tenth, leave a certain circumference, which shall be less than the circumference of the circle BCD which is subtended of a line equal to the line GL. Let the circumference left be KB. Wherefore also the right line KB is less than the right line GL. And forasmuch as the four sided figure BKOS is in a circle, and the lines OB, BK, and KS are equal, and the line OS is less, therefore the angle BYK is an obtuse angle. Wherefore the line BK is greater than the line BY. But the line GL is greater than the line KB. Wherefore the line GL is much greater than the line BY. Wherefore also the square of the line GL is greater than square of the line BY. And forasmuch as (by the 15. definition of the first) the line AL is equal to the line AB, therefore the square of the line AL is equal to the square of the line AB. But unto the square of the line AL are equal the squares of the lines AG and GL, and to the square of the line AB are equal the squares of the lines BY and YA. Wherefore the squares of the lines AG and GL are equal to the squares of the lines BY and YA, of which the square of the line BY is less than the square of the line GL. Wherefore the residue, namely, the square of the line YA is greater than the square of the line AG. Wherefore also the line AY is greater than the line AG. Wherefore two spheres consisting both about one and the self same centre, being given, there is inscribed in the greater sphere a polihedron or solid of many sides which toucheth not the superficies of the less sphere● which was requinred to be done.  

¶ Corollary.  
And if in the other sphere, namely, in the less sphere be inscribed a Polihedron or solid of many sides like to the polihedron inscribed in the sphere BCDE, than the polihedron inscribed in the sphere BCDE is to the polihedron inscribed in the other sphere in triple proportion of that in which the diameter of the sphere BCDE is to the diameter of the other sphere. For those solids being divided into pyramids equal in number and equal in order, the pyramids shall be like. But like pyramids are (by the 8. of the twelfth) the one to the other in triple proportion of that in which side of like proportion is to side of like proportion. Wherefore the pyramid whose base is the four sided figure KBOS and top the point A is to that pyramis which is of like order in the other sphere, in triple proportion of that in which side of like proportion is to side of like proportion, that is of that in which the side AB which is drawn from the centre of the sphere which is about the centre A, is to the side which is drawn from the centre of the other sphere. And in like sort also every one of the pyramids which is in the sphere which is about the centre A is to every one of the pyramids of the self same order in the other sphere in triple proportion of that in which the side AB is to the side which is drawn from the centre of the other sphere. But as one of the antecedentes is to one of the consequences, so are all the antecedents to all the consequentes by the 12. of the fifth. Wherefore the whole polihedron solid of many sides which is in the sphere which is about the centre A, is to the whole polihedron or solid of many sides which is in the other sphere, in triple proportion of that in which the side AB is to the side which is drawn from the centre of the other sphere, that is, of that which the diameter BD is to the diameter of the other sphere, by the 15. of the fifth: which was required to be demonstrated.  
M. Dee his devise, to help the imagination to young students in Geometry● and to make his demonstration more evident as concerning the errors by him corrected in Euclides figure, by the ignorant, mislined.   

I D●e.  
This figure is answerable to the first plain: which, cutting the two Spheres by their common centre A, made two concentrical circles (having the same centre with the two Spheres) namely BCDE, and FGH. Upon which, ● you aptly rear perpendicularly, the second figure containing two concentrical circles, (to the first equal) and make the points noted with like letters to agreed, and afterward upon the second figure, set on the third figure being here for the better handling made a semicircle: which upon the first figure must also be erected perpendicularly: And lastly if you take the little quadrangled figure books, and make every point to touch, his like: & then read the construction & weigh the demonstration (twice o● thrice being read over) shall you in this delineation in apt pastborde, or like ma●ter framed, find all things in this problem very evident.  
I need not warn you, that the line AY may easily be imagined, or with a fine thread supplied: or of the right lines imaginable between P and T, and between R and V, I need say nothing, trusting that the great exercise past, by that time you are orderly come to this place, will have made you sufficient perfect to supply any farther thing herein to be considered.  
The little fowercornerd pieces remaining to the semicircle, are to be let through the first ground plain: thereby to stay this semicircle the better in his apt place and situation: which it will the more aptly do, if ye do 〈…〉, the contrary arrases of the slit of it, and of the slit of the second figure, into which it is to be let: abating them alike much: a little will serve. Experience, by advise, will teach sufficiently.   

¶ Master Dee his advise and demonstration, reforming a great error in the designation of the former figure of Euclides second Problem: with two Corollaries (by him inferred) upon his said demonstration.  

A Theorem.  
✚ That a right line drowen from the point K, perpendicularly upon the line BZ, doth fall upon the point Z: we will, thus, make evident.  
BY the premises, it is manifest, that the point Z is that point where a right line from the point O being perpendicularly let ●all to the circle BCDE, doth touch the same circle. Which point Z also is proved to be in the right line BAD, the common section of two circles cutting each other: being one to the other perpendicularly erected. These things, with other, before demonstrated, I here make my suppositions. Consider now the two triangles rectangles OZB and KZB: Of which, the angle OZB i● equal to the angle KZB. For, by construction, they are both right angles: 
* This, as an assumpt, is presently proved.  and the angle Z●O is equal to the angle ZBK. For, if from D to K you imagine a right line: and the like from D to O: you have two triangles in equal semicircles, rectangles, namely, DKB and DOB: which have the diameter BD common: and BK, the Chord, equal to BO the Chord, by construction. Wherefore (by the 47. of the ●irst) the third side, namely, DK, is equal to the third, namely, DO: Wherefore (by the 5. of the sixth) the angle ZBO, which is DBO, is equal to ZBK, which is DBK. (For the line ZB, by construction, is part of DB). And seeing two angles of OZB, are proved equal, to two angles of KZB, of necessity the third, namely, ZOB, is equal to the third, namely, ZKB, by the 32. of the first. Wherefore the two triangles rectangles OZB and KZB, are proved equiangled. By the fourth, therefore, of the sixth, their sides are proportional: therefore by the premises, proved, as BO is to BK, so is OZ to KZ, and the third line, which subtendeth the angle ZOB, to the third line which subtendeth the angle ZKB. But, by construction, BO is equal to BK: therefore OZ is equal to KZ: And the third also is equal to the third: Wherefore the point Z, in respect of the two triangles rectangles, OZB and KZB determineth one and the same magnitude, i● the line BZ. Which can not be: if any other point, in the line BZ, were assigned, nearer, or farther of, from the point B. One only point therefore, is that, at which the two perpendiculars KZ and OZ fall: But, by construction, OZ falleth at Z the point and therefore at the same Z, doth the perpendicular, drawn from K, fall likewise: Which was required to be demonstrated.  
Although a brief monition, might herein have served for the pregnant or the humble learner, yet for them that are well pleased to have things made plain, with many words, and for the stiff-necked busy body, it was necessary, with my controlment of other, to annex the cause & reason thereof, both invincible and also evident.   

A Corollary. 1.  
Hereby it is manifest, that two equal circles cutting one the other by the whole diameter, if from one and the same end of their common diameter, equal portions of their circumferences be taken: and from the points ending those equal portions, two perpendiculars be let down to their common diameter, those perpendiculars shall fall upon one and the same point of their common diameter.   

2.  
Secondly it followeth that those perpendiculars are equal.   

¶ Note.  
From circles in our first supposition each to other perpendicularly erected, we proceed and infer now these Corollaries, whether they be perpendicularly erected or no: by reasou the demonstration hath a like force, upon our suppositions here used.     

¶ The 16. Theorem. The 18. Proposition. Spheres are in triple proportion the one to the other of that in which their diameters are.  
SVppose that there be two spheres ABC and DEF, and let their diameters be BC and EF. Then I say that the sphere ABC is to the sphere DEF in triple proportion of that in which the diameter BC is to the diameter EF. 
Two cases in this proposition.  For if not, than the sphere ABC is in triple proportion of that in which BC is to EF, either to some sphere less than the sphere DEF, or to some sphere greater. First let it be unto a less, namely, to GHK. 
The first case.  And imagine that the spheres DEF and GHK be both about one and the self same centre. 
Demonstration leading to an impossibility.  And (by the proposition next going before) describe in the greater sphere DEF a polihedron or a solid of many sides not touching the superficies of the less sphere GHK. And suppose also that in the sphere ABC be inscribed a polihedron like to the polihedron which is in the sphere DEF. Wherefore (by the corollary of the same) the polihedron which is in the sphere ABC, is to the polihedron which is in the sphere DEF in triple proportion of that in which the diameter BC is to the diameter EF. But by supposition the sphere ABC is to the sphere GHK in triple proportion of that in which the diameter BC is to the diameter EF. Wherefore as the sphere ABC is to the sphere GHK, so is the polihedron which is described in the sphere ABC to the polihedron which is described in the sphere DEF by the 11. of the fift. Wherefore alternately (by the 16. of the fift) as the sphere ABC is to the polihedron which is described in it, so is the sphere GHK to the polihedron which is in the sphere DEF. But the sphere ABC is greater than the polihedron which is described in it. Wherefore also the sphere GHK is greater than the polihedron which is in the sphere DEF by the 14. of the fift. But it is also less, for it is contained in it, which impossible. Wherefore the sphere ABC is not in triple proportion of that in which the diameter BC, is to the diameter EF, to any sphere less than the sphere DEF. In like sort also may we prove that the sphere DEF is not in triple proportion of that in which the diameter EF is to the diameter BC, to any sphere less than the sphere ABC. 
Second case.  Now I say that the sphere ABC is not in triple proportion of that in which the diameter BC is to the diameter EF to any sphere greater than the sphere DEF. For if it be possible, let it be to a greater, namely, to LMN. Wherefore by conversion the sphere LMN is to the sphere ABC in triple proportion of that in which the diameter EF is to the diameter BC. But as the sphere LMN is to the sphere ABC, so is the sphere DEF to some sphere less than the sphere ABC, 
* As it is ●asi● to gather by the assumpt put after the seco●● of this boo●●.  as it hath before been proved, for the sphere LMN is greater than the sphere DEF. Wherefore the sphere DEF is in triple proportion of that in which the diameter EF is to the diameter BC to some sphere less than the sphere ABC, which is proved to be impossible. Wherefore the sphere ABC is not in triple proportion of that in which BE is to EF to any sphere greater than the sphere DEF. And it is also proved that it is not to any less. Wherefore the sphere ABC is to the sphere DEF in triple proportion of that in which the diameter BC is to the diameter EF: which was required to be demonstrated.  

A corollary added by Flussas.  
Hereby it is manifest, that spheres are the one to the other, as like Polihedrons' and in like sort described in them are: namely, each are in triple proportion of that in which the diameters.  

A Corollary added by M● Dee.  
It is then evident, how to give two right lines, having that proportion between them, which, any two spheres given, have the one to the other.  
For, if to their diameters, as to the first and second lines (of four in continual proportion) you adjoin a third and a fourth line in continual proportion (as I have taught before): The first and fourth lines, shall answer the Problem. 
Note: a general rule.  How general this rule is, in any two like solids, with their correspondent (or Omologall) lines, I need not, with more words, declare.    

¶ Certain Theorems and Problems (whose use is manifold, in Spheres, Cones, Cylinders, and other solids) added by joh. Dee.  

A Theorem. 1.  
The whole superficies of any Sphere, is quadrupla, to the greatest circle, in the same sphere contained.  
It is needless to bring Archimedes demonstration hereof, into this place: seeing his book of the Sphere and Cylinder, with other his wo●kes, are every where to be had, and the demonstration thereof, easy.   

A Theorem. 2.  
Every sphere, is quadrupl●, to that Cone, whose base is the greatest circle, & height, the semidiameter of the same sphere.  
This is the 32. Proposition of Archimedes fi●st book of the Sphere and Cylinder.  

A Problem. 1.  
A Sphere being given, to make an upright Cone, equal to the same: or in any other proportion, given, between two right lines.  
And as concerning the other part of this Problem, 
The second part of the Problem two ways executed.  it is now easy to execute, and that two ways: I mean to A the sphere given, to make an upright cone in any proportion given between two right lines. For, let the proportion given, be that which is between X and Y. By the order of my additions, upon the 2. of this twelfth book: to the circle EKG make an other circle in that proportion that X is to Y: which let be Z. Upon the centre of Z, rear a line perpendicular and equal to FL. I say that the cone, whose base is Z, and the height equal to FL, is to A, in the proportion of X to Y. For the cone upon Z, by construction, hath height equal to the height of the cone LEKG: and Z, by construction, is to EKG● as X is to Y: Wherefore, by the 11. of this twelfth, the cone upon Z, is to the cone LEKG, as X is to Y. But the cone LEKG is proved equal to the sphere A. Wherefore the cone upon Z, is to A, as X is to Y, by the 7. of the fift. To a Sphere given therefore, we have made a cone in any proportion given, between two right lines. Secondly, as X is to Y, so to FL, let there be a fourth line, by the 12. of the sixth: and suppose it to be W. I say that a cone, whose base is equal to EKG, and height the line W, is to A, as X is to Y. For, by the 14. of this twelfth, cones being set on equal bases, are one to the other, as their heights are: But, by construction, the height W, is to the height FL, a● X is to Y. Wherefore the cone which hath his base equal to EKG, and height the line W, is to the cone LEKG, as X is to Y. And it is proved, that to the cone LEKG, the Sphere A is equal: Wherefore, by the 7. of the fift, the cone, whose base is equal to EKG, and height the line W, is to A, as X is to Y. Therefore a Sphere being given, we have made an upright cone, 
An upright Cone.  in any proportion given between two right lines. And before, we made an upright cone, equal to the Sphere given. Wherefore a Sphere being given, we have made an upright cone, equal to the same, or in any other proportion, given between two right lines. I call that an upright cone, whose axe is perpendicular to his base.    

¶ A Corollary.  
Of the first part of the demonstration, it is evident: A Sphere being propounded, that a Cone, whose base hath his semidiameter, equal to the diameter thereof, and height equal to the semidiameter of the same Sphere, is equal to that sphere propounded.  

¶ A Problem. 2.  
A Sphere being given, and a circle, to rear an upright Cone, upon that circle (as a base) equal to the Sphere given: or in any proportion between two right lines assigned.  
And the second part of this Problem is thus performed. Suppose the proportion given to be that which is between X & Y. 
The second part of the Problem.  Then, as X is to Y, so let an other right line found, be to the h●ight of F: which line let be G. For this G, the found height (by construction) being to the height of F, as X is to Y, doth cause this cone (which let be M) upon C, the circle given (or an other to it equal) duly reared, to be unto the cone F, as X is to Y, by the 14. of this twelfth. But F is proved equal to the Sphere given: Wherefore M, is to the Sphere given, as X is to Y. And M, is ●eared upon the circle given: or his equal. Wherefore, a Sphere being given, & a circle, we have reared an vp●ight cone, upon that given circle (as a base) equal to the Sphere given: or in any proportion, between two right lines assigned: which was required to be done.   

¶ A Problem. 3.  
A Sphere being given, and a right line, to make an upright cone, equal to the Sphere given, or in any other proportion given between two right lines: which made cone, shall have his height equal to the right line given.  

The second ●a●● o● the problem.  For the second part: find a circle, which shall have to the base of L, any proportion appointed in ●ight lines: as the proportion of X to Y: which, by my additions, upon the second of this book, ye have learned to do. Then, with the height, equal to the height of L reared upon this last found circle, which l●t be T, as a base, you shall satisfy the Problem. L●t that Cone be V For this last cone V, is to L, as his base is to the base of L, by the 11. of this twelfth. But L is proved equal to the Sphere given: Wherefore by the 7. of the fift, this l●●t cone V, hath to R, the Sphere given, that proportion which is between X and Y assigned: and forasmuch as the height of this cone V, is equal to the height of L: and the height of L, equal to S, the right line given (by construction): it is evident, that a Sphere being given, & a right line, we have made an upright cone, equal to the Sphere given, or in any other proportion given between two right lines: which made cones, have their height equal to the right line given: which aught to be done.  
Vuwilling I am to use thus many words, in matters so plain● and ease. 
☜  But this (I think) can not hinder them, that by nature are not so quick of invention, as to lead every thing, generally spoken, to a particular execution.    

¶ A Theorem. 3.  
Every Cylinder which hath his base, the greatest Circle in a Sp●er●, & heith equal to the diameter of that Sphere, is Sesquialtera to that Sphere. Also the superficies of that Cylinder, with his two bases is Sesquilatera to the superficies of the Sphere: and without his two bases, is equal to the superficies of that Spher●  
Suppose, a sphere to be signified by A, and an upright cylinder having to his base a circle equal to the greatest circle in A contained, and his heith equal to the diameter of A, let be signified by FG. I say that FG, is sesquialter to A: Secondly I say that the crooked cylindrical superficies of FG, together with the superfici●ces of his two opposite bases, is sesquialtera to the whole superficies spherical of A. Thirdly I say that the cylindrical superficies of FG, omitting his two opposite bases, is equal to the superficies of the spear A. Let the base of FG, be the circle FLB: whose centre, suppose M, and diameter FB. And the axe of the same FG, let be, MH. Which is his heith (for we suppose the cylinder to be upright): and suppose H, to be his top or vertex. Forasmuch as, by supposition MH is equal to the diameter of A. Let MH be divided into two equal parts in the point N, by a plain superficies passing by the point N, and being parallel to the opposite bases of FG. By the thirteenth of this twelfth book, it than followeth, that the cylinder FG, is also divided into two equal parts: being cylinders: which two equal cylinders let be IG, and FK: the axe of IG suppose to be HN: and of FK the axe to be NM. And for that, FG, is an upright cylinder, and at the point N, cut by a plain Superficies parallel to his opposite bases, the common section of that plain superficies and the cylinder FG, must be a 
* This may easily be demonstrated, as in th● 17. proposition the section of a sphere was proved to be a circle.  circle, equal, to his base FLB, and have his centre, the point N. Which circle, let be JOK: And seeing that FLB is, by supposition, equal to the greatest circle in A, JOK, also, shall be equal to the greatest circle, in A, contained: Also, by reason MH, is by supposition, equal to the diameter of A: and NH, by construction, half of MH, it is manifest that NH is equal to the semidiameter of A. If, therefore, you suppose a cone to have the circle JOK to hi● base: and NH to his heith, the sphere A, shall be to that Cone, quadrupla, by the 2. Theorem. Let that cone be HIOK. Wherefore A, is quadrupla to HIOK. And the Cylinder IG having the same base, with HIOK (the circle JOK) and the same heith, (the right line NH) is triple to the cone HIOK by the 10. of this twelfth book. But to IG, the whole cylinder FG, is double, as is proved: Wherefore FG, is triple and triple, to the cone HIOK, that is, sextuple. And A is proved quadrupla to the same HIOK. Wherefore FG is to HIOK, as 6. to 1: and A, is to HIOK, as 4. to 1: 
* For taking away all doubt, this, a● a Lemma, afterward is demonstrated.  Therefore FG is to A, as 6, to 4: which in the lest terms, is, as 3 to 2. but 3 to 2, is the terms of sesquialtera proportion. Wherefore the cylinder FG, is to A sesquialtera in proportion. Secondly, forasmuch as the superficies of a cylinder (his two opposite bases excepted) is equal to that circle whose semidiameter is middle proportional between the side of the cylinder, and the diameter of his base: (as unto the 10. of this book, I have added.) But of FG, the side BG, being parallel and equal to the axe MH, must also be equal to the diameter of A. And the base FLB, being (by supposition) equal to the greatest circle in A contained, must have his diameter (FB) equal to the said diameter of A. The middle proportional therefore between BG and FB, being equal each to other, shal● be a line, equal to either of them. [As i● 〈◊〉 set BG and FB together, as one line, and upon that line composed, as a diameter make a semicircle and from the centre, to the circumference draw a lin● perpendicular to the said diameter: 
A Lemma (as it were) presently demonstrated.  by the 〈◊〉 of the sixth, that perpendicular, is middle proportional between FB and BG, the semidiameters● and he himself also a semidiameter: and therefore by the definition of a circle, equal to FB, and likewise, to BG.] And a circle, having his semidiameter, equal to the diameter FB, is quadruple to the circle FLB. [For the square of every whole line, is quadruple to the squ●re of his half line, as may be proved by the 4. of the second: and by the second of this twelfth, circles are one to the other, as the squares of their diameters, are.] Wherefore the superficies cylindricall of FG, alone, is quadrupla to his base FLB. But if a certain quantity be dupla to one thing, and an other, quadrupla to the same one thing, those two quantities together are sextupla to the same one thing. Therefore seeing the base, opposite to FLB, (being equal to to FLB) added to FLB, maketh that compound, double to FLB: that double added to the cylindrical superficies of FG, doth make a superficies sextupla to FLB. And the superficies of A, is quadrupla to the same FLB, by the first Theorem. Therefore the cylindrical superficies of FG, with the superficieces of his two bases is to the superficies FLB, as 6 to 1, and the superficies of A to FLB, is as 4 to 1. Wherefore the cylindrical superficies of FG, & his two bases, together, are to the superficies of A, as 6 to 4: that is, in the smallest terms, as 3 to 2. Which is proper to ses●uialtera proportion. Thirdly, it is already made evident that the superficies cylindrical, of FG (only by itself) is quadrupla to FLB. And also it is proved, that the superficies of the sphere A, is quadrupla to the same FLB, Wherefore by the 7. of the fifth, the cylindrical superficies of FG, is equal to the superficies of A. Therefore, every cylinder, which hath his base the greatest circle in a sphere, and heith equal to the diameter of that sphere, is sesquialtera to that spear: Also the superficies of that cylinder with his two bases, is sesquialtera, to the superficies of the sphere: and without his two bases is equal to the superficies of the sphere: which was to be demonstrated.   

The Lemma.  
If A be to C, as 6, to 1: and B, to C, as 4 to 1: A, is to B, as 6, to 4.  
For, seeing, B is to C, as 4 to 1, by supposition: therefore backward, by the 4. of the fifth, C is to B, as 1, to 4. Imagine now two orders of qnantities, the first, A, C, and B the second, 6, 1, and 4. Forasmuch as, A, is to C, as 6, to 1, by supposition: and C is to B, as 1, to 4, as we have proved: wherefore, A is to B, as 6 to 4, by the 22 of the fift. Therefore, if A be to C as 6 to 1, and B to C, as 4 to 1: A is to B, as 6, to 4. which was to be proved.   

Note.  
Slight things (some times) lacking evident proof, breed doubt or ignorance. And, I need not warn● you, how gen●rall, this demonstration is: for if you put in the place of 6 and 4, any other numbers, the like manner of conclusion will follow. So likewise, in place of 1. any other one number may be, as, if A be to C as ● to 5: and B unto C, be as 7 to 5: A, shall be to B, as 6 to 7. etc.  

A Problem. 4.  
To a Sphere given, to make a cylinder equal, or in any proportion given between two right lines.  
Suppose the given Sphere to be A: and the proportion given to be that between X and Y. I say that a cylinder is to be made, equal to A● or else in the same proportion to A, that is between X to Y. Let a cylinder be made (such one as the Theorem next before supposed) that shall have his base equal to the greatest circle in A, 
Construction.  and height equal to the diameter of A: Let that cylinder b● the upright cylinder BC. Le● the one side of BC, be the right line QC. Divide QC into three equal part● of which, let QE contain two, and let the third part be CE. By the point E suppose a plain (parallel to the bases of BC) to pass through the cylinder BC, cutting the same by the circle DE. I say that the cylinder BE is equal to the Sphere A. For seeing BC, being an upright cylinder, is cu● by a plain, parallel to his bases, by construction: therefore as the cylinder DC, is to the cylinder BE, so is the axe of DC, to the axe of BE, by the 13. of this twelfth. 
Demonstration.  Wherefore as the axe is to the axe, so is cylinder to cylinder. But axe is to axe, as side to side, namely, CE to QE, because the axe is parallel to any side of an upright cylinder: by the definition of a cylinder. And the circle of the section, is parallel to the bases, by construction. Wherefore in the parallelogram (made of the axe, and of two semidiameters, on one side parallels, one to the other, being coupled together by a line drawn between their ends in their circumferences, which line is the side QC) it is evident, that the axe of BC is cut in like proportion, that the side QC is cut. Wherefore the cylinder DC, is to the cylinder BE, as EC is to QE. Wherefore, by composition, the cylinders DC and BE, that is, whole BC, are to the cylinder BE, as CE and QE (the whole right line QC) are to QE. But by construction, QC is of 3. such parts, as QE containeth 2. Wherefore the cylinder BC, is of 3. such parts, as BE containeth 2. Wherefore BC the cylinder, is to BE, as 3. to 2: which is sesquialtera proportion. But by the former Theorem, BC is sesquialtera to the Sphere A: Wherefore, by the 7. of the fift, BE is equal to A. Therefore to a Sphere given, we have made a cylinder equal.    

Or thus more briefly omitting all cutting of the Cylinder.  
Forasmuch as BC is an upright Cylinder: his sides are equal to his axe or heith: therefore the two cylinders, whereof one hath the heith QC and the other the heith QE, having both their bases, the greatest circle in the Sphere A, are one to the other as QC is to QE, by the 14. of this twelfth, but QC is to QE, as 3. to 2, by construction: and 3. to 2. is in Sesquialtera proportion: therefore the cylinder BC having his heith QC, & his base the greatest circle in A contained, is Sesquialtera to the cylinder which hath his base the greatest circle in A contained, and heith the line QE. But by the former Theorem, BC, is also Sesquialtera to A: wherefore the cylinder having the base BQ (which by supposition, 
The second part of the Problem.  is equal to the greatest circle in A contained) and heith, QE, is equal to the sphere A, by the 7. of the fift. And now it can not be hard, to give a cylinder, to A, in that proportion, which is between X and Y. For let the side QE, be to QP, as Y is to X, by the 12. of the sixth. Therefore backward, QP is to QE as X to Y. Wherefore the cylinder having the base the greatest circle in A and heith the line QP, 
*  is to the cylinder having the same base, and heith the line QE, as X is to Y, by the 14. of this twelfth: but the cylinder having the heith QE, & his base the greatest circle in A, contained, is proved equal to the Sphere A. Wherefore by the 7. of the fift, the cylinder whose heith is QP and base the greatest circle in A, contained, is to the sphere A, as X to Y. Therefore to a sphere given: we have made a cylinder, in any proportion given between two right lines, and also, before we have to a sphere given, made a cylinder equal. Therefore to a sphere given, we have made a cylinder equal, or in any proportion given between two right lines.  

A Problem. 5.  
A Sphere being given, and a circle, upon that circle as a base, to rear a cylinder, equal to the sphere given: or in any proportion, given between two right lines.   

A Problem. 6.  
A Sphere being given, and a right line, to make a cylinder, equal to the sphere given, or in any other proportion, between two right lines, given.  
In this 5. and 6. problem, first make a cylinder equal to the sphere given, by the 4. problem: and then by the order of the 2. and 3. problems, in cones, execute these accordingly in cylinders.   

A Problem. 7.  
Two unlike Cones or Cylinders, being given, to find two right lines, which have the same proportion one to the other, that the two given cones or cylinders have one to the other.  
Upon one of their bases rear a cone (if cones be compared) or a cylinder (if cylinders be compared) equal to the other: by the order of the second and third problems: and the heith of the cone or cylinder, on whose base you reared an equal cone or cylinder, with the new heith found, have that proportion, which the cones or cylinders have, one to the other, by the 14. of this twelfth book.   

A Problem. 8.  
An upright Cone, and Cylinder, being given: to find two right lines, having that proportion, the one to the other, which the Cone and Cylinder have, one to the other.  
Suppose, QEK an upright cone and AB an upright cylinder given. I say two right lines are to be given which shall have that proportion one to the other, 
Construction.  which QEK and AB have one to the other. Upon the base BH, erect a Cone, equal to QEK: by the order of the second problem: which let be OBH, and his heith let be OC, 
Demonstration.  and let the heith of AB, (the cylinder), be CS, produce CS to P: so that CP, be tripla to CS, & make perfect the cone PBH. I say that FC and OC have that proportion, which AB hath to QEK. For, by constr●ction, OBH is equal to QEK: and PBH is equal to AB, as we will prove. (Assumpt wise). And PBH, and OBH are upon one base, namely BH: wherefore by the 14. of this twelfth, ●BH and OBH are one to the other as their heithes PC, and OC, are one to the other) wherefore the cylinder and cone equal to PBH and OBH are as PC is to OC: by the 7. of the fifth. But AB the cylinder QEK the cone, are equal to PBH and OBH: by construction: wherefore AB the cylinder, is to QEK the cone, as PC, is to OC. Wherefore we have found two right lines having that proportion that a cone and a cylinder given, have one to the other. Which thing we may execute upon the base of the cone given, as we did upon the base of the cylinder given, on this manner. Upon the base of the cone QEK, 
another way of executing this problem.  which base, let be EK, erect a cylinder, equal to AB, by the order of my second problem. Which cylinder let be ED, and GT his heith, and let the heith of the cone QEK, be QG. Take the line GR, the ●hird part o● QG, (by the 9 of the sixth): and with a plain passing by R, parallel to EK, cut of the cylinder E●: which ●hall be equal to the cone QEK, by the assumpt following I say now, that AB, the cylinder, is to QEK the cone, as GT, is to GR. For the cylinder ED is to the cylinder EF, as GT is to GR, by the 14. of the twelfth: and to ED is the cylinder AB equal: by construction: and to EF, we have proved the cone QEK, equal, wherefore by the 7. of the fifth, AB is to QEK, as GT i● to GR. Wherefore an upright cone, & a cylinder being given, we have found two right lines having the same proportion between them, which the cone and the cylinder, have one to the other: which was requisite to be done.    

An assumpt.  
If a co●e, and a cylinder, being both on one base, are equal one to the other: the heith of the cone ●s tripla to the heith of the cylinder. And if a cone and a cylinder being both on one base, the heith of the cone be tripla to the heith of the cylinder, the cone and the cylinder are equal.  
We will use the cylinder AB & the cone PBH in the ●o●mer problem: with their base & heithes so noted as before. I say if PBH be equal to AB, that CP the heith of PBH, is tripla to CS the heith of the cylinder AB. Suppose upon the base BH, a cone to be reared of the heith of CS, which let be SBH: it is manifest that AB is tripla to that cone SBH, by the 10. of this twelfth. Wherefore a cone equal to AB the cylinder is tripla to SBH the cone, by the 7. of the fifth, but PBH is supposed equal to AB. Therefore PBH is tripla to SBH, therefore the heith of PBH shall be tripla to the heith of SBH by the 14. of the twelfth. But the heith of PBH, is CP: and of SBH, the heith is CS: wherefore CP is tripla to CS. And CS is the heith of the cylinder AB by supposition. Therefore a cone and a cylinder, being both on one base, and equal, the heith of the cone is tripla to the heith of the cylinder. And the second part, as easily may be confirmed. For if AB a cylinder, and PBH a cone, have one base, as the circle about BH: and the heith of PBH be tripla to the heith of AB, I say that PBH, and AB are equal. 
The converse of the assumpt.  The heith of AB let be (as afore) CS: and of PBH, the heith, let be CP: of the heith CS, imagine a Cone upon the same base BH: by the 10. of this twelfth, AB shall be triple to that cone. And the cone PBH having heith CP, (by supposition) tripla to CS, shall also be tripla to that cone SBH: by the 14. of this twel●th. Wherefore by the 7. of the fifth AB and PBH are equal. Therefore, if a cone and a cylinder being both on one base, the heith of the cone be tripla to the heith of the cylinder, the cone and the cylinder are equal. So have we demonstrated both parts: as was required.   

¶ A Theorem. 4.  
The superficies of the segment or protion of any sphere, is equal to the circle, whose semidiameter, is equal to that right line which is drawn from the top of that segment to the circumference of the circle, which is the base of that portion or segment.  
This hath Archimedes demonstrated in this first book of the Sphere and Cylinder, in his 40. and 41 propositions: and I remit them thither, that will herein demonstratively be certified: I would wish all mathematicians, as well of verities easy, as of verities rare and obscure, to seek the causes demonstrative, the final fruit thereof, is perfection in this art.  

Note.  
Besides all other uses and commodities, that are of the Crooked superficieces of the Cone, Cylinder, and Sphere, so easily and certainly, of us to be dealt with all: this is not the lest, that a notable Error, which among Sophistical brablers, and ungeometrical Masters and Doctors, hath a long time been upholden: may most evidently, hereby be confuted, and utterly rooted out of all men's fantasies for ever. The Error is this, Curui, ad rectum, nulla est proportio, that is: 
A great error commonly maintained.  Between crooked and strait, is no proportioned This error, in line●, sup●●ficieces, and solids, may with more true demonstrations be overthrown, than the favourers of that fond fantasy are able, with argument, either probable or Sophistical to make show or pretence to the contrary. In lines, I omit (as now) Archimedes two ways, for the finding of the proportion of the circles circumference to a strait line. I mean, by the inscription and circumscription of like polygonon figures, and that other, by spiral lines. And I omit likewise (as now) in solids, of a parallelipipedon, equal to a Sphere, Cone, or Cylinder● or any segment or sector of the said solids. And only, here require you to consider in this twelfth book, the ways brought to your knowledge, how to the crooked superficies of a cone and cylinder, and of a sphere, 
Between strait and crooked all manner of proportion may be given.  (the whole, any segment or sector thereof) a plain and strait superficies may be given equal: Namely, a Circle to be given equal, to any of the said crooked superficieces assigned, and given. And then farther by my Additions upon the second proposition, you have means to proceed in all proportions, that any man can in right lines given or assign. Therefore, Curui ad rectum, proportio omnimoda potest dari. One thing it is, to demonstrate, that between a crooked line and a strait or a crooked superficies and a plain or strait superficies, etc. there is proportion. And an other thing it is, to demonstrate a particular and special kind of proportion, being between a crooked superficies and a strait or plain superficies. For this also confi●meth the first. This short warning will 'cause you to avoid the said error, and make you also able to cure them, which are infected therewith.    

A Theorem. 5.  
Any two Spheres being given, as the Spherical Superficies of the one, is to the spherical Superficies of the other: So is the greatest Circle contained in that one, to the greatest Circle contained in that other. And as greatest Circle, is to greatest Circle, so is Spherical superficies, to Spherical superficies.  
For the superficies of every sphere is quadrupla to his greatest circle, by my first Theorem: wherefore, of two given spheres, as the spherical superficies of the one, is to his greatest circle, so is the spherical superficies of the other, to his greatest circle: therefore by alternate proportion, as spherical superficies is to spherical superficies, so is greatest circle to greatest circle. And therefore also as greatest circle, is to greatest circle, so is spherical superficies to spherical superficies: which was to be demonstrated.  

A Problem. 9  
A Sphere being given, to give an other Sphere, to whose Spherical superficies, the superficies Spherical of the Sphere given shall have any proportion, between two right lines given.  
Suppose A, to be a sphere given, and the proportion given, to be that, which is between the right lines X and Y. I say that a sphere is to be given to whose spherical superficies, the superficies spherical of A, shall have that proportion which X hath to Y. Let the greatest circle, contained in A the sphere be the circle BCD. 
Construction.  And by the problem of my additions, upon the second proposition of this book, as X is to Y, so let the circle BCD be to an other circle found, let that other circle be EFG: and his diameter EG. I say that the spherical superficies of the sphere A, hath to the spherical superficies of the sphere, whose greatest circle is EFG, (or his equal) that proportion, which X hath to Y. 
Demonstration.  For (by construction) BCD is to EFG, as X is to Y: and by the theorem next before● as BCD is to ●FG, so is the spherical superficies of A (whose greatest circle is BCD, by supposition) to the spherical superficies, of the sphere, whose greatest circle is EFG: wherefore, by the 11. of the fifth as X is to Y: So is the spherical superficies of A, to the spherical superficies of the sphere, whose greatest circle is EFG: wherefore, the sphere whose diameter is EG, (the diameter also of EFG) is the sphere, to whose spherical superficies, the spherical superficies of the sphere A, hath that proportion which X hath to Y. A sphere being given therefore, we have given an other sphere, to whose spherical superficies, the superficies spherical of the sphere geum hath any proportion given, between two right lines: which aught to be done.   

A Problem. 10.  
A sphere being given, and a Circle less than the greatest Circle, in the same Sphere contained. to coped in the Sphere given, a Circle equal to the Circle given.  
Suppose A, to be the sphere given: and the circle given less than the greatest circle in A contained, to be FKG. 
The definition of a circled ●●ap●●d in a sp●er●.  I say, that in the Sphere A, a circle, equal to the circle FKG, is to be coapted. First understand, what we mean here, by coapting of a circle in a Sphere. We say, that circle to be coapted in a Sphere, whose whole circumference is in the superficies of the same Sphere. Let the greatest circle in the Sphere A contained, be the circle BCD. Whose diameter suppose to be BD, and of the circle FKG, let FG be the diameter. 
Construction.  By the 1. of the fourth, let a line equal to FG, be coapted in the circle BCD. Which line coapted, let be BE. And by the line BE, suppose a plain to pass, cutting the Sphere A, and to be perpendicularly erected to the superficies of BCD. 
Demonstration.  Seeing that the portion of the plain remaining in the sphere, is called their common section: the said section shall be a circle, as before is proved. And the common section of the said plain, and the greatest circle BCD, (which is BE by supposition) shall be the diameter of the same circle, as we will prove. For, let that circle be BLEM. Let the centre of the sphere A, be the point H: which H, is also the centre of the circle BCD, because BCD is the greatest circle in A contained. From H, the centre of the sphere A, let a line perpendicularly be let fall to the circle BLEM. Let that line be HO: and it is evident that HO shall fall upon the common section BE, by the 38. of the eleventh. And it divideth BE, into two equal parts, by the second part of the third proposition of the third book: by which point O, all other lines drawn in the circle BLEM, are, at the same point O, divided into two equal parts. As if from the point M, by the point O, a right line be drawn one the other side coming to the circumference, at the point N: it is manifest that NOM is divided into two equal parts at the point O: 
This is manifest: if you consider the two triangles rectangles, HOM and HON: and then with all, use the 47. of the first of Euclid.  by reason, if from the centre H, to the points N and M, right lines be drawn, HN and HM, the squares of HM, and HN are equal: for that all the semidiameters of the sphere are equal: and therefore their squares are equal one to the other: and the square of the perpendicular HO, is common: wherefore the square of the third line MO is equal to the square of the third line NO: and therefore the line MO to the line NO. So therefore is NM equally divided at the point O. And so may be proved of all other right lines, drawn in the circle BLEM, passing by the point O, to the circumference one both sides. Wherefore O is the centre of the circle BLEM: and therefore BE passing by the point O is the diameter of the circle BLEM. Which circle (I say) is equal to FKG: for by construction BE is equal to FG: and BE is proved the diameter of BLEM, and FG is by supposition the diameter of the circle FKG: wherefore BLEM is equal to FKG the circle given: and BLEM is in A the sphere given. Wherefore we have in a sphere given coapted a circle equal to a circle given: which was to be done.  

A Corollary.  
Besides our principal purpose, in this Problem, evidently demonstrated, this is also made manifest: that if the greatest circle in a Sphere, be cut by an other circle, erected upon him at right angles, that the other circle is cut by the centre, and that their common section is the diameter of that other circle: and therefore that other circle divided is into two equal parts.    

A Problem. 11.  
A Sphere being given, and a circle, less than double the greatest circle in the same Sphere contained, to cut of, a segment of the same Sphere, whose Spherical superficies, shall be equal to the circle given.  
Suppose K to be a Sphere given, whose greatest circle let be ABC: and the circle given, suppose to be DEF. 
Construction.  I say, that a segment of the Sphere K, is to be cut of, so great, that his Spherical superficies, shall be equal to the circle DEF. Let the diameter of the circle ABC, be the line AB. At the point A, in the circle ABC, coped a right line equal to the semidiameter of the circle DEF (by the first of the fourth). Which line suppose to be AH. From the point H, to the diameter AB, let a perpendicular line be drawn: which suppose to be high. Produce high to the other side of the circumference, and let it come to the circumference at the point L. By the right line HIL (perpendicular to AB) suppose a plain superficies to pass, perpendicularly erected upon the circle ABC: and by this plain superficies, the Sphere to be cut into two segments: one less than the half Sphere, namely, HALI: and the other greater than the half Sphere, namely, HBLI. I say, that the Spherical superficies of the segment of the Sphere K, in which the segment (of the greatest circle) HALI, is contained, (whose base is the circle passing by HIL, and top the point A) is equal to the circle DEF. 
Demonstration  For the circle, whose semidiameter is equal to the line AH, is equal to the Spherical superficies of the segment HAL, by the 4. Theorem here added. And (by construction) AH is equal to the semidiameter of the circle DEF: therefore the Spherical superficies of the segment of the Sphere K (cut of by the circle passing by HIL) whose top is the point A, is equal to the circle DEF. Wherefore, a Sphere being given, and a circle less than double the greatest circle in the same Sphere, we have cut of, a segment of the same Sphere, whose Spherical superficies, is equal to the circle given: which was requisite to be done.  

¶ An advise.  
In noting or signifying of Spheres, sometimes we use by one and the same circle, in plain designed, to represent a Sphere and also the greatest circle in the same contained: and likewise, by a segment of that circle, signify a segment of the same Sphere, as by a strait line, we often signify the circle, which is the base of a segment of a Sphere, Cone, or Cylinder: and so in such like. Wherein, consider our suppositions: and take heed when we shifted from one signification to an other, in one and the same designation: and withal remember the principal intent of our drift: and such light things, can not either trouble or offend thee. compendiousness and artificial custom, procureth such means: sufficient, to stir up imagination Mathematical: or to inform the practiser Mechanical.    

A Problem. 12.  
To cut a Sphere given, into two such segments, that the Spherical superficieces of the segment●● shall have one to the other, any proportion given between two right lines.  
Suppose F to be a Sphere given: and the proportion given, to be that, which is between GH and high. I say, that the Sphere F, is to be cut into two such segments, that the Spherical superficies of those segments, shall have that proportion, one to the other, which, the right line GH, hath to the right line high. 
Construction.  Suppose ABCE to be a greatest circle, in the Sphere F, contained: and his diameter, to be AB. Divide ABinto two such parts, as GI' is divided into, in the point H (by the 10. of the sixth) Let those parts be AD, and DB. So that, as GH is to high, so is AD to DB. By the point D, let a plain superficies pass, cutting the Sphere F, and the diameter AB: So, that unto that cutting plain, the diameter AB, be perpendicular: and the Sphere also thereby by divided into two segments, whose common base suppose to be the circle CE, having the centre, the point D: and the top of the one to be the point A, and the top of the other to be the point B: and the segments themselves, to be noted by EAC, and EBC: Draw from the two tops, A and B, to C (a point in the circumference of their common base) two right lines AC and BC. I say now, that the Spherical superficies of the segment EAC, hath to the Spherical superficies of the segment EBC, the same proportion, 
Demonstration  which GH hath to high. For, forasmuch as circles have that proportion, one to the other, that the squares of their diameters have one to the other (by the 2. of this twelfth). 
This in manner of a Lemm● is presently proved.  And the squares of their semidi●meters, have the same proportion one to the other, which the squares of their diameters have. [For like parts have that proportion one to the other, that the whole magnitudes● whose like parts they are, have the one to the other: by the 15. of the fift. But the square of every diameter is quadruple to the the square of his semidiameter: as hath often before● been proved: therefore, circle haue● one to an other, that proportion, that the squares of their semidiameters have one to the other]. Wherefore, seeing AC and BC are semidiameters of two circles, whereof each is equal to the Spherical superficies of the segments, between whose tops and circumference of their base, they are drawn: by the 4. Theorem of these additions: it followeth that both those circles whose semidiameters they are: and also those Spherical superficieces, which are equal to those circles, have the one to the other, the same proportion, which the square of AC hath to the square of BC. But AC is drawn between the circumference of the base, and top of the segment Spherical, EAC, by construction: and likewise BC is drawn between the 〈◊〉 and the 〈…〉 of the base Spherical segments ABC, by construction: Wherefore the Spherical superficies of the segment EAC, is to the Spherical superficies of the segment EBC, as the square of AC is to the square of BC. But the square of AC is to the square of BC, as AD is to DB: by the Corollary of the Problem of my additions upon the second of this twelfth: And AD is to DB, as GH is to high: by construction. Wherefore the Spherical superficies of the segment EAC, is to the Spherical superficies of the segment EBC, as GH is to high. We have therefore, cut the Sphere given, into two such segments, that the Spherical superficieces of the segments, have one to the other any proportion given between two right lines: which was to be done.  

¶ A Corollary. 1.  
Here it appears demonstrated, that, circles are one to the other, as the squares of their semidiameter are, one to the other.  
Whereby (as occasion shall serve) you may, by force of the former argument, use other like parts of the diameter, as well as halves.   

¶ A Corollary. 2.  
It is also evident, that the Spherical superficieces of the two segment●s of any Sphere, to whose common base, the diameter (passing to their two tops) is perpendicular, have that proportion the o●e to the other, that the portions of the said diameter● have the one to the other: that superficies and that portion of the diameter on the one side of the common base, being compared to that superficies, and that portion of the diameter● on the other side of the common base.   

¶ A Corollary. 3.  
It likewise evidently followeth, that the two Spherical superficieces of two segments of a Sphere: which two segments are equal to the Sphere, are in that proportion the one to the other, that their axes (perpendicularly erected to their bases) are in, one to the other: where soever in the Sphere those segments be taken.  
I say that the Spherical superficies of the segment CAE, and the Spherical superficies of the segment FGH, having their axes AD and GI' (perpendicular to their bases): are in proportion one to the other, as AD is to GI': if the segment of the Sphere containing CAE with (the segment of the same Sphere) FGH be equal to the whole Sphere. For seeing the diameter (of axe) AD, extended to the other pole or top, opposite to A (which opposite top, let be Q) doth make with the segment CAE, the compliment of the whole Sphere: 
Note here of Axe, base, & solidity, more than I need to bring any farther proof for.  and by supposition, the segment FGH, whith the segment CAE, are equal to the whole Sphere: Wherefore from equal, taking CAE (the segment common) remaineth the segment CQE, equal to the segment FGH. And thereby, Axe, Base, Solitie, and superficies Spherical of the segment FGH, must (of ncessitie) be equal to the Axe, Base, Solidity, and superficies Spherical of the segment CQE: Wherefore, by the second Corollary here, and the 7. of the fift, our conclusion is inferred, the superficies Spherical, of the segment CAE, to be, to the superficies Spherical of the segment FGH, as AD is to GI'.     

A Theorem. 6.  
To any solid sector of a Sphere, that upright C●●e is equal, whose base is equal to the connex Spherical superficies of that sector, and heith equal to the semidiameter of the same Sphere.  
Hereof the demonstration in respect of the premises: and the common argument of inscription and circumscription of figures is easy: and nevertheless, if your own writ will not help you sufficiently: you may take help at Archimedes hand, in his first book & last proposition of the sphere and cylinder. Whether if ye have recourse, you shal● perceive how your Theorem here amendeth the common translation there: and also our delineation giveth more s●u●y show of the chief circumstances necessary to the construction, than there you shall find. Of the sphere here imagined to be A, we note a solid sector by the letter● PQRO. So that PQR doth signify the spherical superficies, to that solid sector belonging: 
Note.  (which is also common to the segment of the same sphere PRQ) and therefore a line drawn from the top of that segment● (which top suppose to be Q) is the semidiameter of the circle, which is equal to the spherical superficies of the said solid sector, or segment● as before is taught. Let that line be QP. By Q draw a line contingent: which let be SQT. At the point Q from the line QS, cut a line equal to PQ which let be SQ. And unto SQ, make QT equal, then draw the right lines OSO● and OQ. About which OQ (as an axe fa●●ened) if you imagine the triangle OST, to make an 
* I say half a circular revolution for that sufficeth in the whole diameter ST, to describe a circle by: i● it be moved ●●out his centre Q etc.  half circular revolution, you shall have the upright cone OST: (whose heith is OQ, the semidiameter of the sphere, and base the circle, whose diameter is ST,) equal to the solid sector PQRO.   

A Theorem. 7.  
To any segment, or portion of a Sphere, that cone i● equal, which hath that circle to his base, which is the b●se of the segment, and heith, a right line, which unto the heith of the segment hath that proportion, which the semidiameter of the Sphere together with the heith of the other segment remay●●●g hath to the heith of the same other segment remayn●ng.  
This is well demonstrated by Archi●●des & therefore needeth no invention of mine, 
Lib 2 prop 2. de Sphe●a & Cylindr●.  to confirm the same: and for that the said demonstration is over long here to be added, I will refer you thither for the demonstration: and here supply that which to Archimedes demonstration shall give light, and to your farther speculation and practice, shall be a great aid and direction. Suppose K to be a sphere: & the greatest circle K in contained, let be ABCE, and his diameter BE, & centre D. Let the sphere K, be cut by a plain superficies, perpendicularly erected upon the said greatest circle ABCE: let the section be the circle about AC: And let the segments of the sphere be the one that wherein is ABC, whose ●oppe is B● and the other let be that wherein is AEC and his top, let be E: I say that a cone which hath his base the circle about AC, & held a line which to BF (the heith of the segment, whose top is B,) hath that proportion that a line compo●ed of DE, the semidiameter of the sphere, and EF (the heith of the other remaining segment, whose top is E) hath to EF, (the heith of that other segment remaining), is equal to the segment of the sphere K, whose top is B. To make this cone, take my easy order thus. Frame your work for the finding of the fourth proportional line● by making EF the first and a line composed of DE and EF, the second● and the third, let be BF: then by the 12. of the six●h, let the fourth proportional line be found: which let be FG● upon F the centre of the base of the segment, whose top is B, erect a line perpendicular equal to FG found and draw the lines GA and GC: and so make perfect the cone GAC. I say, that the cone GAC, is equal to the segment (of the sphere K) whose top is B. In like manner, for the other segment whose top is E, to find the heith due for a cone equal to it: by the order of the Theorem you must thus frame your lines: let the first be BF: the second DB and BF, composed in one right line, and the third must be EF: where by the 12. of the sixth, finding the fourth, it shall be the heith, to rear upon the base, (the circle about AC,) to make an upright cone, equal to the segment, whose top is E.   

¶ Logistically. ¶  
The logistical finding hereof is most easy: the diameter of the sphere being given, and the portions of the diameter in the segments contained (or axes of the segments) being known. Then order your numbers in the rule of proportion, as I here have made most plain, in ordering of the lines: for the ●ought heith will be the producte.  

A Corollary. 1.  
Hereby, and other the premises it is evident that to any segment of a Sphere, 
Note.  whose whole diameter is known and the Axe of the segment given, An upright cone may be made equal: or in any proportion, between two right lines assigned● and therefore also a cylinder may to the said segment of the Sphere, be made equal ●r in any proportion given, between two right lines.   

A Corollary. 2.  
Manifestly also, of the former theorem, it may be inferred that a Sphere, and his diameter being divided, by one and the same plain superficies, to which the said diameter is perpendicular● the two segments of the Sphere, are one to the other in that proportion, in which a rectangle parallelipipedon having for his base the square of the greater part of the diameter, and his heith a line composed of the less portion of the diameter, and the semidiameter: to the rectangle parallelip●pedon, having for his base the square of the less portion of the diameter, & his heith a line composed of the semidiameter & the greater part of the diameter.    

A Theorem. 8.  
Every Sphere, to the cube, made of his diameter, is (in manner) as 11. to 21.  
As upon the first and second propositions of this book, I began my additions with the circle (being the chief among plain figures) and therein brought manifold considerations, about circles: as of the proportion between their circumferences and their diameters of the content or Area of circles: of the proportion of circles to the squares described of their diameters: & of circles to be given in all pro●portions, to other circles: with diverse other most necessary problems (whose use is partly there specified): So have I in the end of this book, added some such Problems & Theoremes● about the sphere (being among solids the chief) as of the same, either in itself considered, or to cone and cylinder, compared (by reason of superficies, or solidity, in the hole, or in part●) such certain knowledge demonstrative may arise, and such mechanical exercise thereby be devised, that (sure I am) to the sincere & true student great light, aid, and comfortable courage (farther to wade) will enter into his heart; and to the Mechanical, witty, and industrous deviser, new manner of inventions, & executions in his works will (with small travail for feet application) come to his perceiverance and understanding. Therefore, even a, manifold speculations & practices may be had with the circle, his quantity being not known in any kind of smallest certain measure: So likewise of the sphere many Problems may be executed and his precise quantity, in certain measure, not determined, or known: yet, because, both one of the first (human) occasions of inventing and stablishing this Art, was measuring of the earth (and therefore called Geometria, that is, Earthmeasuring), and also the chief and general end (in deed) is measure: and measure requireth a determination of quantity in a certain measure by number expressed: It was needful for Mechanical earthmeasures, not to be ignorant of the measure and contents of the circle, neither of the sphere his measure and quantity, as near as sense can imagine or wish. And (in very deed) the quantity and measure of the circle, being known, maketh not only, the cone and cylinder, but also the sphere his quantity to be as precisely known, and certain. Therefore seeing in respect of the circles quantity (by Archimedes specified) this Theorem is noted unto you: I will, by order, upon that (as a supposition) infer the conclusion of this our Theorems.  

Note. 1.  
Wherefore if you divide the one side (as TQ) of the cube TX into 21. equal parts, and where 11. parts do end, reckoning from T, suppose the point P: and by that point P, imagine a plain (passing parallel to the opposite bases) to cut the cube TX: and thereby, the cube TX, to be divided into two rectangle parallelipipedons, namely, TN, and PX: It is manifest, 
* A rectangle parallelipipedon given, equal to a Sphere given.  TN, to be equal to the Sphere A, by construction: and the 7. of the fift.   

Note. 2.  
Secondly, the whole quantity, of the Sphere A, 

To a Sphere, or to any part of a Sphere assigned: as a third, fourth, fifth etc. to give a parallelipipedon equal.  
Sided Columes Pyramids, and prisms to be given equal to a Sphere, or to any certain part thereof.  
To a Sphere or any segment, or sector of the same, to give a cone or cylinder equal or in any proportion assigned.   being contained in the rectangle parallelipipedon TN, you may easily transform the same quantity, into other parallelipipedons rectangles, of what height, and of what parallelogram base you list: by my first and second Problems upon the 34. of this book. And the like may you do, to any assigned part of the Sphere A: by the like means dividing the parallelipipedon TN: as the part assigned doth require. As if a third, fourth, fifth, or sixth, part of the Sphere A, were to be had in a parallelipipedon, of any parallelogra●●e base assigned, or of any heith assigned: then dividing TP, into so many parts (as into 4. if a fourth part be, to be transformed: or into five, if a fifth part, be to be transformed etc.) and then proceed, ●s you did with cutting of TN, from TX. And that I say of parallelipipedons, may in like sort (by my ●●yd two problems, added to the 34. of this book) be done in any sided columns, pyramids, and prisme●: so th●● in pyramids and some prisms you use the cautions necessary, in respect of their quan 〈…〉 odyes' having parallel, equal, and opposite bases: whose parts 〈…〉re in their propositions, is by Euclid demonstrated. And finally, 〈…〉 additions, you have the ways and orders how to give to a Sphere, or any segment o● the same, Cones, or Cylinders equal, or in any proportion between two right lines, given: with many other most necessary speculations and practices about the Sphere. I trust that I have sufficiently ●raughted your imagination, for your honest and profitable study herein, and also given you rea●●, ●●tter, whe●● with to s●●p the mouths of the malicious, ignorant, and arrogant, despisers of the most excellent discourses, travails, and inventions mathematical. Sting aswell the heavenly spheres, & stars their spherical solidity, 
Farther use of Spherical Geometry.  with their convene spherical superficies, to the earth at all times respecting, and their distances from the earth, as also the whole earthly Sphere and globe itself, and infinite other cases, concerning Spheres or globes, may hereby with as much ease and certainty be determined of, as of the quantity of any bowl, ball, or bullet, which we may gripe in our hands (reason, and experience, being our witnesses): and without these aids, such things of importance never able of us, certainly to be known, or attained unto.  
Here end M. john Dee his additions upon the last proposition of the twelfth book.     

A proposition added by Flussas.  
If a Sphere touch a plain superficies● a right line drawn from the centre to the touch, shall be erected perpendicularly to the plain superficies.  
Suppose that there be a Sphere BCDL: whose centre let be the point A. And let the plain superficies GCI touch the Spear in the point C, and extend a right line from the centre A to the point C. Then I say that the line AC is erected perpendicularly to t●e plain GIC. Let the sphere be cut by plain superficieces passing by the right line LAC: which plains let be ABCDL and ACEL, which let cut the plain GCI by the right lines GCH and KCI. Now it is manifest (by the assumpt put before the 17. of this book) that the two sections of the sphere shall be circles, having to their diameter the line LAC, which is also the diameter of the sphere. Wherefore the right lines GCH and KCI which are drawn in the plain GCI, do at the point C fall without the circles BCDL and ECL. Wherefore they touch the circles in the point C, by the second definition of the third. Wherefore the right line LAC maketh right angles with the lines GCH and KCI by the 16. of the third. Wherefore by the 4. of the eleventh the right line AC is erected perpendicularly to to the plain superficies GCI wherein are drawn the lines GCH and KCI. If therefore a Sphere touch a plain superficies, a right line drawn from the centre to the touch, shall be erected perpendicularly to the plain superficies: which was required to be proved.    
The end of the twelfth book of Euclides Elements.   

¶ The thirteenth book of Euclides Elements.  
IN THIS thirteenth BOOK are set forth certain most wonderful and excellent passions of a line divided by an extreme and mean proportion: 
The argument of the thirteenth book.  a matter undoubtedly of great and infinite use in Geometry, as ye shall both in this book, and in the other books following most evidently perceive. It teacheth moreover the composition of the five regular solids, and how to inscribe them in a Sphere given, and also setteth forth certain comparisons of the said bodies both the one to the other, and also to the Sphere, wherein they are described.  

The 1. Theorem. The 1. Proposition. If a right line be divided by an extreme and mean proportion, and to the greater segment, be added the half of the whole line: the square made of those two lines added together shallbe quintuple to the square made of the half of the whole line.  
SVppose that the right line AB be divided by an extreme and mean proportion in the point C. And let the greater segment thereof, be AC. And unto AC, add directly a right line AD, and let AD be equal to the half of the line AB. 
Construction.  Then I say that the square of the line CD is quintuple to the square of the line DA. Describe (by the 46. of the first) upon the lines AB and DC squares, namely, AE & DF. And in the square DF, describe and make complete the figure. And extend the line FC, to the point G. 
Demonstration.  And forasmuch as the line AB is divided by an extreme and mean proportion in the point C, therefore that which is contained under the lines AB and BC is equal to the square of the line AC. But that which is contained under the lines AB and BC, is the parallelogram CE, and the square of the line AC is the square HF. Wherefore the parallelogram CE is equal to the square HF. And forasmuch as the line BA, is double to the line AD, by construction: 〈◊〉 the line BA is equal to the line KA, and the line AD, to the line AH: therefore also, the line KA, is double to the line AH. But as the line KA is to the line AH, so is the parallelogram CK to the parallelogram CH: Wherefore the parallelogram CK is double to the parallelogram CH. And the parallelograms LH and CH are double to the parallelogram CH (for supplements of parallelograms are b● the 4●. of the first equal the one to the other). Wherefore the parallelogram CK is equal to the parallelograms LH & CH. And it is proved that the parallelogram CE is equal to the square FH. Wherefore the whole square AE is equal to the gnomon MXN. And forasmuch as the line BA, i● double to the line AD, therefore the square of the line BA is, by the 20. of the sixth, quadruple to the square of the line DA, that is, the square AE to the square DH. But the square AE is equal to the gnomon MXN, wherefore the gnomon MXN, is also quadruple to the square DH. Wherefore the whole square DF is quintuple to the square DH. But the square DF, i● the square of the line CD, and the square DH is the square of the line DA. Wherefore the square of the line CD, is quintuple to the square of the line DA. If therefore a right line be divided by an extreme and mean proportion, and to the greater segment, be added the half of the whole line: the square made of those two lines added together shallbe quintuple to the square made of the half of the whole line: Which was required to be demonstrated.  
This proposition is an other way demonstrated after the fifth proposition of this book.   

The 2. Theorem. The ●. Proposition. If a right line, be in power quintuple to a segment of the same line: the double of the said segment is divided by an extreme and mean proportion, and the greater segment thereof is the other part of the line given at the beginning.  

* The Assumpt proved.  Now, that the double of the line AD (that is AB) is greater than the line AC may thus be proved. For if not, then if if it be possible let the line AC be double to the line AD, wherefore the square of the line AC is quadruple to the square of the line AD. Wherefore the squares of the lines AC and AD are quintuple to the squares of the line AD. And it is supposed that the square of the line DC is quintuple to the square of the line AD, wherefore the square of the line DC is equal to the square of the lines AC and AD: which is impossible (by the 4. of the second). Wherefore the line AC is not double to the line AD. In like sort also may we prove that the double of the line AD is not less than the line AC, for this is much more absurd: wherefore the double of the line AD is greater than the line AC● which was required to be proved.  
This proposition also is an other way demonstrated after the fifth proposition of this book.  

Two Theorems, (in Euclides Method necessary) added by M. Dee.  

A Theorem. 1.  
A right line can be divided by an extreme and mean proportion, but in one only point.  
Suppose a line divided by extreme and mean proportion, to be AB. And let the greater segment be AC. I say, that AB can not be divided by the said proportion, in any other point then in the point C. If an adversary would contend that it may, in like sort, be divided in an other point: let his other point, be supposed to be D: making AD; the greater segment of his imagined division. Which AD, also, let be less than our AC: for the first discourse. Now, forasmuch as by our adversaries opinion, AD, is the greater segment, of his divided line● the parallelogram contained under AB, and DB, is equal to the square of AD, by the third definition and 17. proposition of the sixth Book. And by the same definition and proposition, the parallelogram under AB, and CB, contained, is equal to the square of our greater segment AC. Wherefore, as the parallelogram, under AB, and D●, is to the square of AD: so i● 〈◊〉 parallelogram, under AB, and CB, to the square of AC. For proportion of equality, is concluded in them both. But, forasmuch as D●● i● (by● 
* Because AC is supposed greater than AD: therefore his residue is less, than the residue of AD, by the common sentence. Wherefore, by the supposition: DB is greater than ●C.  supposition) greater than CB, the parallelogram under AB, and DB, is greater than the parallelogram under AC, and CB: by the first of the sixth (for AB is their equal heith.) Wherefore, the square of AD, shallbe greater than the square of AC: by the 14. of the fifth. But the line AD, is less than the line AC, by supposition: wherefore the square of AD is less than the square of AC. And it is concluded also to be greater than the square of AC: Wherefore the square of AD, is both greater, than the square of AC● and also less. Which is a thing impossible. The square therefore of AD, is not equal to the parallelogram under AB, and DB. And therefore by the third definition of the sixth, AB is not divided by an extreme and mean proportion, in the point D: as our adversary imagined. And (Secondly) in like sort will the inconueniency fall out: if we assign AD, our adversaries greater segment, to be greater than our AC. Therefore seeing neither on the one side of our point C: neither on the other side of the same point C, any point can be had, at which the line AB can be divided by an extreme and mean proportion, it followeth of necessity, that AB can be divided by an extreme and mean proportion in the point C, only. Therefore, a right line can be divided by an extreme and mean proportion, but in one, only point: which was requisite to be demonstrated.   

A Theorem. 2.  
What right line so ever, being divided into two parts, hath those his two parts, proportional, to the two segments of a line divided by extreme and mean proportion: is also itself divided by an extreme and mean proportion: and those his two parts, are his two segments, of the said proportion.  
Suppose, AB, to be a line divided by an extreme and mean proportion in the point C, and AC to be the greater segment. Suppose also the right line DE, to be divided into two parts, in the point F: and that the part DF, is to FE, as the segment AC, is to CB: or DF, to be, to AC, as FE is to CB. For so these parts are proportional, to the said segments. I say now, that DE is also divided by an extreme and mean proportion in the point F. And that DF, FE, are his segments of the said proportion. For, seeing, as AC, is to CB: so is DF, to FE: (by supposition). Therefore, as AC, and CB (which is AB) are to CB: so is DF, and FE, (which is DE) to FE: by the 18. of the fifth. Wherefore (alternately) as AB is to DE: so is CB, to FE. And therefore, the residue AC, is to the residue DF, as AB is to DE, by the fifth of the fift. And then alternately, AC is to AB, as DE, is to DF. Now therefore backward, AB is to AC, as DE is to DF. But as AB is to AC, so is AC to CB: by the third definition of the sixth book. Wherefore DE is to DF, as AC is to CB: by the 11. of the fifth. And by supposition, as AC is to CB, so is DF to FE: wherefore by the 11. of the fifth, as DE is to DF: so is DF to FE. Wherefore by the 3. definition of the sixth, DE is divided by an extreme and mean proportion, in the point F. Wherefore DF, and FE are the segments of the said proportion. Therefore, what right line so ever, being divided into two parts, hath those his two parts, proportional to the two segments of a line divided by extreme and mean proportioned is also itself divided by an extreme and mean proportion, and those his two parts are his two segments, of the said proportioned which was requisite to be demonstrated.  

Note.  
Many ways, these two Theorems, may be demonstrated: which I leave to the exercise of young students. But utterly to want these two Theorems, and their demonstrations: in so principal a line, or rather the chief pillar of Euclides Geometrical palace, 
The chie●e line in all Euclides Geometry.  was hitherto, (and so would remain) a great disgrace. Also I think it good to note unto you, what we meant, by one only point. We m●●●●, that the quantities of the two segments, can not be altered, the whole line being once given. And though, from either end of the whole line, the greater segment may begin: 
What is meant here by, A section in one only poi●t.  And so as it were the point of section may seem to be altered: yet with us, that is no alteration: forasmuch as the quantities of the segments, remain all one. I mean, the quantity of the greater segment, is all one: at which end so ever it be taken: And therefore, likewise the quantity of the less segment is all one, etc. The like consideration may be had in Euclides tenth book, in the binomial lines. etc.  
Io●n Dee. 1569. Decemb. 18.      

The 3. Theorem. The 3. Proposition. If a right line be divided by an extreme and mean proportion, and to the less segment be added the half of the gerater segment: the square made of those two lines added together, is quintuple to the square made of the half line of the greater segment.  
SVppose that the right line AB be divided by an extreme and mean proportion in the point C. And let the greater segment thereof be AC. And divide AC into two equal parts in the point D. Then I say that the square of the line BD, it quintuple to the square of the line DC. 
Construction.  Describe (by the 46. of the first) upon the line AB a square AE. And describe and make perfect the figure (that is divide the line AT like unto the division of the line AB, by the 10. of the sixth, in the points R, H, by which points draw (by the 31. of the first) unto the line AB parallel lines RM and HN. So likewise draw by the points D, C, unto the line BE these parallel lines DL and CS, & draw the diameter BT). 
Demonstration.  And forasmuch as the line AC is double to the line DC, therefore the square of AC, is quadruple to the square of DC, by the 20. of the sixth, that is, the square RS to the square FG. And forasmuch as that which is contained under the lines AB and BC is equal to the square of the line AC, and that which is contained under the lines AB and BC is equal to the parallelogram CE, & the square of the line AC is the square RS: wherefore the parallelogram CE is equal to the square RS. But the square RS is quadruple to the square FG: wherefore the parallelogram CE also is quadruple to the square FG. Again forasmuch as the line AD is equal to the line DC, therefore the line HK is equal the line KF, wherefore also the square GF is equal to the square HL: wherefore the line GK is equal to the line KL, that is, the line MN to the line NE: wherefore the parallelogram MF is equal to the parallelogram FE. But the parallelogram MF is equal to the parallelogram CG, wherefore the parallelogram CG is also equal to the parallelogram FE. Put the parallelogram CN, common to them both: 
* Note, how CE and the gnonom XOP, are proved equal, for it serveth in the converse demonstrated by M. Dee, here next after.  Wherefore the gnomon XOP is equal to the parallelogram CK. But the parallelogram CE is proved to be quadruple to G● the square, wherefore the gnomon XOP is quadruple to the square GF. Wherefore the square DN is quintuple to the square FG. And DN is the square of the line DB, and GF the square of the line DC. Wherefore the square of the line DB is quintuple to the square of the line DC. If therefore a right line be divided by an extreme and mean proportion, and to the less segment be added the half of the greater segment: the square made of those two lines added together, is quintuple to the square made of the half line of the greater segment. Which was required to be demonstrated.  
You shall find this proposition an other way demonstrated after the fifth proposition of this book.  

Here followeth M. Dee, his additions.  

¶ A Theorem. 1.  
If a right line, given, be quintuple in power, to the power of a segment of himself: the double of that segment, 
This proposition, ●the converse of the former.  and the other part remaining, of the first given line, make a line, divided by extreme and mean proportion: and that double of the segment, is the greater part thereof.  
Forasmuch as, this, is the converse of Euclides third proposition: we will use the same suppositions and constructions there specified: so far, as they shall serve our purpose. Beginning therefore at the conclusion, we must infer the part of the proposition, before granted. It was concluded, that the square of the line DB, is quintuple, to the square of the line DC, his own segment. Therefore DN (the square of DB) is quintuple, to GF, (the square of DC). But the squa●e of AC (the double of DC) which is RS, is quadruple to GF, (by the second Corollary of the 20. of the sixth): and therefore RS, with GF, are quintuple to GF: and so it is evident, that the square DN, is equal to the square RS, together with the square GF. Wherefore, from those two equals, taking, the square GF, (common to them both): remaineth the square RS, equal to the Gnomon XOP. But to the Gnomon XOP, the parallelogram 
* As we ha●e noted the place of the peculiar pro●e there ●in the demonstration of the 3.  CE is equal: Wherefore the square of the line AC, which is RS, is equal to the parallelogram C●. Which parallelogamme is contained under BE, (equal to AB:) and CB, the part remaining of the first line given: which was DB. And the line AB, is made of the double of the segment DC, and of CB● the other part of the line DB, first given. Wherefore the double of the segment DC, with CB, the part remaining (which altogether, is the whole line AB) is to AC, (the double of the segment DC) as that same, AC, is to CB: by the second part of the 16. of the sixth. Therefore by the 3. definition of the sixth book, the whole line AB, is divided by an extreme and mean proportion, & AC, (the double of the segment DC) being middle proportional, is the greater part thereof. Wherefore, if a right line, be quintuple in power, etc. (as in the proposition) which was to be demonstrated.  

Or, thus it may be demonstrated.  
Forasmuch as the square, DN is quintuple to the square GF, (I mean the square of DB the line given, to the square o● DC the segment). And the same square DN, is equal to the parallelogram under AB, CB, with the square made of the line DC: by the sixth of the second: (for unto the line AC, equally divided: the line, CB, is, as it were adjoined). Wherefore the parallelogram under AB, CB, together with the square of DC, which is GF, is quintuple to the square GF, made o● th● line DC. Taking then, that square GF, from the parallelogram under AB, CB: that parallelogram (under AB, CB) remaining alone, is but quadruple to the said square of the line DC. But, (by the 4. of the second, or the second Corollary of the 20. of the sixth) RS, ●he square of the line AC, is quadrupla to the same square GF● Wherefore by the 7. of the fifth, the square of the line AC, is equal to the parallelogram under AB, CB, and so, by the second part of the 16. of the sixth: AB, AC, and CB, are three lines in continual proportion. And seeing AB is greater than AC, the same AC, the double of the line DC, shall be greater than the part BC, remaining: Wherefore by the 3. definition of the sixth, AB, (composed or made of the double of DC, and the other part of DB remaining) is divided by an extreme and middle proportion: and also his greater segment is AC the double of the segment DC. Wherefore, If a right line be quintuple in power etc. as in the propositions which was to be demonstrated.    

A Theorem. 2.  
If a right line, divided by an extreme and mean proportion, be given, and to the great segment ●herof, he directly adjoined a line equal to the whole line given, that adjoined line, and the said greater segment, do make a line divided by extreme and mean proportion, whose greater segment is the line adjoined.  
Suppose the line given, divided by extreme and mean proportion, to be AB divided in the point C, and his greater segment, let be AC: unto AC directly adjoin a line equal to AB: let that be AD: I say, that AD, together with AC, (that is DC) is a divided by extreme and middle proportion, whose greater segment is AD, the line adjoined. Divide AD, equally in the point E. Now, forasmuch as AE, is the half of AD, (by construction,) it is also, the half of AB (equal, to AD, by construction): Wherefore by the 1. of the thirteenth, the square of the line composed of AC and AE (which ●ne is EC) is quintuple to the square of the line AE. Wherefore the double of AE, and the line AC composed, (as in one right line) is a line divided by extreme and mean proportion, by the converse of this third (by me demonstrated): and the double of AE, is the greater segment. But DC is the line composed of the double of AE, & the line AC: and with all, AD is the double of AE. Wherefore, DC, is a line divided by extreme and mean proportion, and AD, i● hi● greater segment. If a right line, therefore, divided by extreme and mean proportion, be given, and to the greater segment thereof, be directly adjoined a line equal to the whole line given, that adjoined line, and the said greater segment, do make a line divided by extreme and mean proportion, whose greater segment, is the line adjoined: Which was required to be demonstrated.  

Two other brief demonstrations of the same.  
Forasmuch as, AD is to AC: as AB, is to AC (because AD is equal to AB, by construction): but as AB is to AC, so is AC to CB: by supposition. Therefore by the 11. of the fifth, as AC, is to CB, so is AD to AC. 
* Therefore, by my second Theorem added upon the second proposition, DC is divided by extreme and mean proportion in the point A. And because AC is bigger than CB: therefore DA is greater than AC: wherefore if a right line etc. as in the proposition. Which was to be demonstrated.  Wherefore, as AC, and CB, (which is AB) is to CB: so is AD, and AC (which is DC) to AC. Therefore, eversedly, as AB, is to AC: so is DC to AD. And it is proved, AD, to be to AC: as AC is to CB. Wherefore as AB is to AC, and AC, to CB: so is DC, to AD, and AD, to AC. But AB, AC, and CB are in continual proportion, by supposition: Wherefore DC, AD, and AC, are in continual proportion. Wherefore, by the 3. definition of the sixth book, DC, is divided by extreme and middle proportion, and his greatest segment, is AD. Which was to be demonstrated. Note from the mark 
* Therefore, by my second Theorem added upon the second proposition, DC is divided by extreme and mean proportion in the point A. And because AC is bigger than CB: therefore DA is greater than AC: wherefore if a right line etc. as in the proposition. Which was to be demonstrated. , how this hath two demonstrations. One I have set in the margin by.   

¶ A Corollary. 1.  
Upon Euclides third proposition demonstrated, it is made evident: that, of a line divided by extreme and mean proportion, if you produce the less segment, equally to the length of the greater: the line thereby adjoined, together with the said less segment, make a new line divided by extreme and middle proportion: Whose less segment, is the line adjoined.  
For, if AB, be divided by extreme and middle proportion in the point C, AC, being the greater segment, and CB be produced, from the point B, making a line, with CB, equal to AC, which let be CQ: and the line thereby adjoined, let be BQ: I say that CQ, is a line also divided by an extreme and mean proportion, in the point B: and that BQ (the line adjoined) is the less segment. For by the third, it is proved, that half AC, (which, let be, CD) with CB, as one line, composed, hath his power or square, quintuple to the power of the segment CD: Wherefore, by the second of this book, the double of C D, is divided by extreme and middle proportioned and the greater segment thereof, shallbe CB. But, by construction, CQ, is the double of CD, for it is equal to AC. Wherefore CQ is divided by extreme and middle proportion, in the point B: and the greater segment thereof shallbe, CB. Wherefore BQ, is the less segment, which is the line adjoined. Therefore, a line being divided, by extreme and middle proportion, if the less segment, be produced equally to the length of the greater segment, the line thereby adjoined together with the said less segment, make a new line divided, by extreme & mean proportion, who●e less segment, is the line adjoined. Which was to be demonstrated.   

¶ A Corollary. 2.  
If● from the greater segment, of a line divided, by extreme and middle proportion, a line, equal to the less segment be cut of: the greater segment, thereby, is also divided by extreme and mean proportion, whose greater segment● shall be 〈◊〉 that part of it, which is cut of.  
For, taking from AC, a line equal to CB: let AR remain. I say, that AC, is divided by an extreme and mean proportion in the point R: and that CR, the line cut of, is the greater segment. For it is proved in the former Corollary that CQ is divided by extreme and mean proportion in the point B. But AC, is equal to CQ, by construction: and CR is equal to CB by construction: Wherefore the residue, AR is equal to BQ the residue. Seeing therefore the whole AC is equal to the whole CQ: and the greater part of AC, which is CR is equal to CB the greater part of CQ and the less segment also equal to the less: and withal seeing CQ is proved to be divided by extreme & mean proportion in the point B, it followeth of necessity, that, AC, is divided by extreme and mean proportion in the point R. And seeing CB, is the greater segment of CQ: CR shall be the greater segment of AC. Which was to be demonstrated.   

A Corollary. 3.  
It is evident thereby, a line being divided by extreme and mean proportion, that the line whereby the greater segment exceedeth the less, together with the less segment, do make a line divided by extreme and mean proportion: whose less segment, is the said line of exceesse, or difference between the segments.  

john Dee.  
¶ Two new ways, to divide any right line given by an extreme and mean proportion: demonstrated and added by M. Dee.    

A Problem.  
To divide by an extreme and mean proportion, any right line given, in length and position.  
Suppose a line given in length and positions to be AB. I say that AB is to be divided by an extreme, and mean proportion. 
Construction.  Divide AB into two equal parts as in the point C. Produce AB directly, from the point B, to the point D: making BD, equal to BC. To the line AD, and at the point D, let a line be drawn 
* Though I say, perpendicular: yes you may perceive how infinite other positions will serve: so that DIEGO and AD make an angle for a triangle to have his sides proportionally cut. etc.  perpendicular: by the 11. of the first, which let be DF: (of what length you will). From DF and at the point D, cut of the sixth part of DF: by the 9 of the sixth. And let that sixth part, be the line DG. Upon DF, as a diameter, describe a semicircle: which let be DHF. From the point G, rear a line perpendicular to DF, which suppose to be GH: and let it come to the circumference of DHF, in the point H. Draw right lines, HD, and HF. Produce DH, from the point H, so long, till a line adjoined with DH, be equal to HF, which let be DI, equal to HF. From the point H, to the point B, (the one end of our line given) let a right line be drawn: as HB. From the point I, let a line be drawn, to the line AB: so that it be also parallel to the line HB. Which parallel line suppose to be IK: cutting the line AB, at the point K. I say that AB, is divided by an extreme & mean proportion, in the point K. For the triangle DKI, having HB, parallel to IK, hath his sides DK and DIEGO, 
Demonstration.  cut proportionally, by the 2. of the sixth. Wherefore as IH is to HD: so is KB, to BD. And therefore compoundingly, (by the 18. of the fifth) as DIEGO, is to DH: so is DK to DB. But by construction DIEGO is equal to HF: wherefore by the 7. of the fifth, DI is to DH, as HF is to DH. Wherefore by the 11. of the fifth, DK is to DB, as HF is to DH. Wherefore the square of DK is to the square of DB, as the square of HF, is to the square of DH: by the 22. of the sixth. But the square of HF, is to the square of DH: as the line GF is to the line GD● by my corollary upon the 5, problem of my additions to the second proposition of the twelfth. Wherefore by the 11. of the fifth, the square of DK is to the square of DB, as the line GF is to the line GD. But by construction, GF is quintuple to GD. Wherefore the square of DK is quintuple to the square of DB: and therefore, the double of DB, is divided by an extreme and mean proportion, and BK is the greater segment thereof, by the 2. of this thirteenth. Wherefore seeing AB is the double of DB by construction: the line AB is divided by an extreme and mean proportion: and his greater segment, is the line BK. Wherefore AB is divided by an extreme and mean proportion, in the point K. We have therefore divided by extreme and mean proportion any line given in length and position. Which was requisite to be done.  

The second way to execute this problem.  
Suppose the line given to be AB. Divide A● into two equal parts: as suppose it to be done in the point C. Produce AB from the point B: adjoining a line equal to BC, which let be BD. To the right line AD, and at the point D, erect a perpendicular line equal to BD, let that be DE. Produce ED from the point D to the point F: making DF to contain five such equal parts, as DE is one. Now upon EF as a diameter, describe a semicircle which let 〈◊〉 EKF, and let the point where the circumference of EKF, doth cut the line AB, be the point K. I say that AB, is divided in the point K, by an extreme and mean proportion. For by the 13. of the sixth ED, DK, & DF, are three lines in continual proportion, (DK being the middle proportional) ● Wherefore by the corollary of the 20. of the sixth, as ED is to DF, so is the square of ED, to the square of DK, but by construction, ED, is subquintuple to DF. Wherefore the square of ED, is subquintuple to the square of DK. And therefore the square of DK, is quintuple to the square of ED. And ED is equal to ED, by construction, therefore the square of DK, is quintuple to the square of E D. Wherefore the double of BD, is divided by an extreme and mean proportion: whose greater segment is BK ● by the second of this thirte●th. But by construction, AB, is the double of ●D ● Wherefore AB, is divided by extreme and mean proportion, and his greater segment, is BK: and thereby, K ● the point of the division. We have therefore divided by extreme and mean proportion, any right line given, in length and position. Which was to be done.    

Note●  
Each of these ways, may well be executed: But in the first, you have this advantage: that the diameter is taken at pleasure. Which ●n the second way, is ever just thrice so long, as the line given to be divided.  
john Dee.      

¶ The 4. Theorem. The 4. Proposition. If a right line be divided by an extreme and mean proportion: the squares made of the whole line and of the less segment, are triple to the square made of the greater segment.  
SVppose that the right line AB, be divided by an extreme & mean proportion, in the point C. And let the greater segment thereof be AC. Then I say, that the squares made of the lines AB, and BC, are triple to the square of the line AC. Describe (by the 46. of the first) upon the line AB, a square ADEB. And make perfect the figure. Now forasmuch as the line AB, 
Demonstration.  is divided by an extreme and mean proportion, in the point C: and the greater segment thereof, is the line AC, therefore that which is contained under the lines AB and BC is equal to the square of the line AC. But that which is contained under the lines AB and CB is the parallelogram AK, and the square of the line AC is the square FD. Wherefore the parallelogram AK is equal to the square FD. And the parallelogram AF is equal to the parallelogram FE, put the square CK common to them both wherefore the whole parallelogram AK is equal to the whole parallelogram CE. Wherefore the parallelograms CE and AK are double to the parallelogram AK. B●t the parallelograms AK and CE, are the gnomon LMN, and the square CK. Wherefore the gnomon LMN and the square CK, are double to the parallelogram AK. But it is proved that the parallelogram AK is equal to the square DF. Wherefore the gnomon LMN and the square CK are double to the square DF. Wherefore the gnomon LMN and the squares CK and DF, are triple to the square DF. But the gnomon LMN and the squares CK and DF, are the whole square AE together with the square CK, which are the squares of the lines AB and BC. And DF is the square of the line AC. Wherefore the squares of the lines AB and BC, are triple to the square of the line AC. If therefore a right line be divided by an extreme and mean proportion, the squares made of the whole line and of the less segment, are triple to the square made of the greater segment: which was required to be proved.  
Look for an other demonstration of this proposition after the fifth proposition of this book.   

¶ The 5. Theorem. The 5. Proposition. If a right line be divided by an extreme and mean proportion, and unto it be added a right ●ine, 
I Dee. This is most evident of my second Theorem, added to the third proposition. For to add to a whole line, a line equal to the greater segment: & to add to the greater segment a line equal to the whole line, is all one thing, in the line produced. By the whole line, I mean the line divided by extreme and mean proportion.  equal to the greater segment, the whole right line is divided by an extreme and mean proportion, and the greater segment thereof, is the right line given at the beginning.  
SVppose that the right line AB be divided by an extreme and mean proportion in the point C, and let the greater segment thereof be AC. And unto the line AB, add the line AD equal to the line AC. Then I say that the line D●, is divided by an extreme and mean proportion in the point A: and the greater segment thereof, is the right line p●t at the beginning, namely, AB. Describe (by the 46. of the first) upon on the line AB a square AE, and make perfect the figure. And forasmuch as the line AB, is divided by an extreme and mean proportion in the point C, therefore that which is contained under the lines AB and BC is equal to the square of the line AC. But that which is contained under the lines AB and BC is the parallelogram CE● and the square of the ●●ne AC is the square CH. Wherefore the parallelogram CE is equal to the square CH. But unto the square CH is equal the square DH, by the first of the sixth: and unto the parallelogram CE, is equal the parallelogram HERALD Wherefore the parallelogram DH is equal to the parallelogram HERALD Add the parallelogram HB, common to them both. Wherefore the whole parallelogram DK is equal to the whole square AE. And the parallelogram DK, is that which is contained under the lines BD and DA, for the line AD is equal to the line DL, & the square AE is the square of the line AB. Wherefore that which is contained under the lines AD and DB is equal to the square of the line AB. Wherefore as the line DB is to the line BA, so is the line BA to the line AD, by the 17. of the sixth. But the line DB is greater than the line BA. Wherefore the line BA is greater than the line AD. Wherefore the line BD is divided by an extreme and mean proportion in the point A, and his greater segment is the line AB. If therefore a right line be divided by an extreme and mean proportion, and unto ●t be added a right line, equal to the greater segment: the whole right line is divided by an extreme and mean proportion, and the greater segment thereof, is the right line given at the beginning: which was required to be demonstrated.  
This proposition is again afterward demonstrated.  

A Corollary added by Campane.  
Hereby it is 〈…〉 from the grea●●● 〈◊〉 of a line divided by an extreme & mean proportion, 
This is before demonstrated most evidently and briefly by M. Dee, after the 3. proposition.  be 〈◊〉 away 〈◊〉 segment: the said great a segment shall be divided by an extreme and mean proportion, and the greater segment thereof shall be the line taken away.  
As let the line ●●, be divided by an extreme and mean● proportion, in the point C. And le● the 〈…〉 line 〈…〉 A. D. I say that AC is also divided by an extreme and mean● proportion in the point D, and that his greater portion is DC. For, by the definitino (of a line so divided) AB, is to AC, as AC is to CB. But as AC is to CB, so is AC to DC, by the 7. of the 〈…〉 is equal to CB) wherefore, by the 11. of the fifth, as AB is to AC, so is AC to CD: and therefore by the 19 of the fifth, as AB is to AC, so is the residue CB, to the residue AD. But CB is to AD, as DC is to AD (by the 7. of the fifth) for DC is by construction equal to C●. Wherefore, 〈…〉 so by the definition of A line divided by an extreme and mean 〈…〉 the point D, to be divided, by an extreme and mean proportion: which was to be proved.  
Two Corollaries (added by M. Dee) following chiefly upon the verity, and demonstration of his Additions, unto the 3. porposition annexed, and partly upon this fifth, by Euclid demonstrated.   

A Corollary. 1.  
As any line being divided by an extreme and middle proportion, doth give us three right lines, 〈…〉 〈…〉 〈…〉 〈…〉 these four ways, we have two line● divided by an extreme and mean proportion; (it is to w●●t●, 〈◊〉 given, and the other made) and with all, in every way, we have four 〈◊〉 continual proportion. 
Note.   
Of these two lines, (by extreme and mean proportion divided,) their demonstrations● are after Euclides 3, proposition added: and here in this fifth by Euclid proved. But of the four lines in continual proportion, seeing, the demonstration is most easy for any man to frame. I will here, but note the lines v●to you: as in every of the four places, t●e constructions have them lettered and specified. As in my first way added after the third proposition, 
Note 4. Proportional lines.  DC, AD, AC, and CB● are four lines in continual proportion. And in the second way, AB, CQ (equal to AC) CB, and BQ: are four lines in continual proportion. And in the third way, AB, AC, C●, (equal to CB) and AR, are the four lines in continual proportion. And in the fourth way (by Euclid declared) DB, AB, AC (equal to AD) and CB● are four lines in continual proportion. 
Note two middle proportionals.  So, that in the first way you have AD, and AC, middle proportionals between DC, and CB. In the second way, you have CQ, and CB, between AB, and BQ. In the third way, you have AC, CR, between AB and AR: and in the fourth way, you have AB, and AC, between DB, and CB.   

A Corollary. 2.  
It is also manifest, that you may by any of the four ways, here specified, proceed infinitely in the proportion of a line divided by extreme & middle proportion: 
Note 4. ways of progression, in the proportion of a line divided by extreme and middle proportion.  And in the first, and fourth ways, increasing continually the quantities of the lin●s made: but in the second and third ways, diminishing continually the quantities of the said whole lines, made (and thereby their segments). And ye●, nevertheless retaining in every line made (by ●●y of the ways) and in his segments all, and the same properties, which the first line, and his segment ●● have. After which ●are of Progression, as the terms in continual proportion do increase, and are m●e in number: So, likewise, do the middle proportionalls, (accordingly) become more: But ever four in number, by two, than the terms of the Progression are.  

¶ What Resolution is.  
Resolution, is the assumption or taking of the thing which is to be proved, as granted, and by things which necessarily follow it, 
What resolution and composition is, hath before been taught in the beginning of the first book.  to pass unto s●me truth granted.   

¶ What Composition is.  
Composition, is an assumption or taking of a thing granted, and by things which of necessity follow it, to pass unto the finding out of the thing sought or to be proved.   

Resolution of the first Theorem.  
Suppose that a certain right line AB, be divided by an extreme & mean proportion in the point C, & let the greater segment thereof be AC, unto which add a line equal to the half of the line AB, and let that line be AD. Then I say that the square of the line CD is quintuple to the square of AD. For forasmuch as the square of the line CD is quintuple to the square of AD: but the square of the line CD is (by the 4. of the second) equal to that which is composed of the squares of the lines CA, & AD, together with that which is contained under the lines CA, and AD twice. Wherefore that which is composed of the squares of the lines CA, & AD, together with that which is contained under the lines CA, & AD twice is quintuple to the square of the line AD. Wherefore, that which is composed of the square of the line CA together with that, which is contained under the lines CA, and AD twice is quadruple to the square of the line AD. But unto that, which is contained under the lines CA, and AD, twice, is equal that which is contained under the lines CA, and AB, for the line AB is double to the line AD. And unto the square of the line AC, is equal that which is contained under the lines AB & BC, for the line AB, is by supposition divided by an extreme and mean proportion, in the point C. Wherefore, that, which is contained under the lines AB, and AC, together with that which is contained under the lines AB, and BC, is quadruple to the square of the line AD. But that, which is composed of that which is contained under the lines AB, and AC, together with that which is contained under the lines AB, and BC, is the square of the line AB (by the 2. of the second.) Wherefore the square of the line AB is quadruple to the square of the line AD. And so is it in deed: for the line AB is double to the line AD: as was at the first supposed.   

Composition of the first Theorem.  
Now forasmuch as the square of the line AB is quadruple to the square of the line AD, but the square of the line AB, is that which is contained under the lines AB, and AC, together with that, which is contained under the lines BA, and BC. Wherefore that which is contained under the lines BA, and AC, together with that which is contained under the lines BA, and BC is quadruple to the square of the line AD. But that which is contained under the lines BA, and AC, is equal to that which is contained under the lines DA, and AC twice (by the 1. of the sixth), and that which is contained under the lines AB, and ●C is equal to the square of the line AC, by the definition of a line divided by extreme and mean proportion. Wherefore the square of the line AC, together with that, which is contained under the lines DA, and AC twice, is quaduple to the square of the line DA. Wherefore that which is composed of the squares of the lines DA, and AC, together with that which is contained under the lines DA, and AC, twice, is quintuple to the square of the line DA. But that which is composed of the squares of the lines DA, and AC, together with that which is contained under the lines DA, and AC twice, is equal to the square of the line CD (by the 4. of the second). Wherefore the square of the line CD is quintuple to the square of the line AD: which was required to be demonstrated.   

Resolution of the 2. Theorem.  
Suppose that a certain right line, CD, be quintuple to a segment of the same line, namely, to DA: and let the double of the line DA, be AB. Then I say that the line AB is divided by an extreme and mean proportion in the point C● and the greater segment thereof is AC, which is the rest of the right line put at the beginning. For forasmuch as the line AB is divided by an extreme and mean proportion in the point C, and the greater segment thereof is the line AC, therefore that which is contained under the lines A●, and BC, is equal to the square of the line AC. But that which is contained under the lines BA, and AC, is equal to that which is contained under the lines DA, and AC twise● for the line BA, is double to the line AD. Wherefore that which is contained under the lines AB, and BC together with that which is contained under the lines BA, and AC, which is the square of the line AB (by the 2. of the second) is equal to that which is contained under the lines DA, & AC, twice together with the square of the line AC. But the square of the line AB, is quadruple to the square of the line DA. Wherefore that which is contained under the lines DA, and AC, twice, together with the square of the line AC, is quadruple to the square of the line AD. Wherefore the squares of the lines DA, and AC, together with that which is contained under the lines DA, and AC, twice, which is the square of the line DC, are quintuple to the square of the line DA. And so are they in deed by supposition.   

Composition of the 2. Theorem.  
Now forasmuch as the square of the line CD, is quintuple to the square of the line DA ● But the square of the line CD, is that which is composed of the squares of the lines DA, & AC, together with that which is contained under the lines DA, & AC, twice: Wherefore the squares of the line DA, & AC together with that which is contained under the lines DA, & AC, twice, are quintuple to the square of the line DA. Wherefore, by division, that which is contained under the lines DA, and AC, twice together with the square of the line CA, is quadruple to the square of the line AD. And the square of the line AB, is quadruple to the square of the line AD. Wherefore that which is contained under the lines DA, and AC twice, which is that, which is contained under the lines BA, and AC once, together with the square of the line AC, is equal to the square of the line AB. But the square of BA, is that which is contained under the lines BA, and AC, together with that which is contained under the lines BA, and BC. Wherefore that which is contained under the lines BA, & AC, together with that which is contained under the lines AB, & BC, is equal to that which is contained under the lines BA, and AC together with the square of the line AC. Now then taking away that which is common to them both, namely, that which is contained vnde● the lines BA, and AC, the residue, namely, that which is contained under the lines AB, BC is equal to the square of the line AC. Wherefore as the line BA, is to the line AC, so is the line AC to the line CB. But the line BA is greater than the line AC, wherefore the line AC also is greater than the line CB. Wherefore the line AB is divided by an extreme, and mean proportion in the point C, and the greater segment thereof is the line AC, which was required to be demonstrated.   

Resolution of the 3. Theorem.  
Suppose that a certain right line AB, be divided by an extreme, and mean proportion in the point C ● and let the greater segment thereof be the line AC, and let the half of the line AC, be the line CD. Then I say that the square of the BD is quintuple to the square of the line CD. For forasmuch as the square of the line BD, is quintuple to the square of the line CD. But the square of the line DB, is that which is contained under the lines AB, and BC, together with the square of the line DC (by the 6. of the second). Wherefore that which is contained under the lines AB, and BC, together with the square of the line DC, i● quintuple to the square of the line DC. Wherefore, that which is contained under the lines AB, and BC, is quadruple to the square of the line DC. But unto that which is contained under the lines AB, and BC, is equal the square of the line AC: for the line AB, is divided by an extreme and mean proportion in the point C. Wherefore the square of the line AC is quadruple to the square of the line DC: and so is it in deed, for the line AC is double to the line DC.   

Composition of the 3. Theorem.  
Forasmuch as the line AC is double to the line DC, therefore the square of the line AC is quadruple to the square of the line DC (by the 20. of the sixth). But unto the square of the line AC, is equal that which is contained under the lines AB, and BC, by supposition: wherefore that which is contained under the lines AB, and BC, is quadruple to the square of the line CD. Wherefore, that which is contained under AB, and BC, other with the square the line DC, which is the square of the line DB (by the 6. of the second) is quintuple to the square of the line DC: which was required to be demonstrated.   

Resolution of the 4. Theorem.  
Suppose that a certain right line AB, be divided by an extreme and mean proportion in the point C. And let the greater segment thereof be AC. Then I say that the squares of the lines AB, and BC, are triple to the square of the line AC. For forasmuch as the squares of the lines AB, and BC, are triple to the square of the line AC, but the squares of the lines AB, and BC, are that which is contained under AB, and BC, twice together with the square of the line AC (by the 7. of the second). Wherefore that which is contained under the lines AB, and BC, twice, together with the square of the line AC, is triple to the square of the line AC. Wherefore, that which is contained under the lines AB, & BC, twice, is double to the square of the line AC. Wherefore that which is contained under the lines AB, and BC, once, is equal to the square of the line AC. And so it is in deed. For the line AB is divided by an extreme, and mean proportion in the point C.   

Composition of the 4. Theorem.  
Forasmuch therefore as the line AB, is divided by an extreme and mean proportion in the point C, and the greater segment thereof is the line AC● therefore that which is contained under the lines AB, and BC is equal to the square of the line AC. Wherefore that which is contained under the lines AB, and BC twice is double to the square of AC. Wherefore that which is contained under the lines AB, and BC, twice, together with the square of the line AC, is triple to the square of the line AC. But that which is contained under the lines AB, and BC, twice, together with the square of the line AC, is the squares of the lines AB, and BC (by the 7. of the second). Wherefore the squares of the lines AB, and BC, are triple to the square of the line AC: which was required to be demonstrated.   

Resolution of the 5. Theorem.  
Suppose that a certain right line AB, be divided by an extreme and mean proportion in the point C. And let the greater segment thereof be the line AC. And unto the line AB, add a line equal to the line AC, and let the same be AD. The● I say that the line DB, is divided by an extreme and mean proportion in the point A. And the greater segment thereof is the line AB. For forasmuch as the line DB is divided by an extreme & mean proportion in the point A, and the greater segment thereof is the line AB, therefore as the line DB, is to the line BA, so is the line BA, to the line AD: but the line AD, is equal to the line AC: wherefore as the line DB, is to the line BA, so is the line BA to the line AC. Wherefore by conversion as the line BD is to the line DA, so is the line AB to the line BC (by the corollary of the 19 of the fifth): wherefore by division, by the 17. of the fifth, as the line BA, is to the line AD, ●o is the line AC, to the line CB. But the line AD is equal to the line AC. Wherefore as the line BA, is to the line AC, so is the line AC to the line CB. And so it is indeed, for the line AB is, by supposition, divided by an extreme and mean proportion in the point C.   

Composition of the 5. Theorem.  
Now forasmuch as the line AB, is divided by an extreme and mean proportion in the point C: therefore as the line BA is to the line AC, so is the line AC to the line CB: but the line AC is equal to the line AD. Wherefore as the line BA is to the line AD, so is the line AC to the line CB. Wherefore by composition (by the 18. of the fifth) as the line BD is to the line DA, so is the line AB to the line BC. Wherefore by conversion (by the corollary of the 19 of the fifth) as the line DB is to the line BA, so is the line BA to the line AC: but the line AC is equal to the line AD. Wherefore as the line DB is to the line BA, so is the line BA to the line AC. Wherefore the line DB, is divided by an extreme and mean proportion in the point A: and his greater segment is the line AB: which was required to be demonstrated.   

An advise, by john Dee, added.  
Sing, it is doubtless, that this parcel of Resolution and Composition, is not of Euclides doing: it can not ●ustly be imputed to Euclid, that he hath, thereby, either superfluity or any part disproportioned in his whole Composition Elemental. And though, for one thing, one good demonstration well sufficeth: for stablishing of the verity: yet, o● one thing diversly demonstrated: to the diligent examiner of the diverse means, by which, that variety ariseth, doth grow good occasions of inventing demonstrations, where matter is more strange, hard, and barren. Also, though resolution were not in all Euclid before used: yet thanks are to be given to the Greek Scholic writer, who did leave both the definition, and also, so short and easy examples of a Method, so ancient, and so profitable. The antiquity of it, is above 2000 years: it is to we●e, ever since Plato his time, and the profit, thereof so great, that thus I find in the Greek recorded. 
* Proclus in the Greek: in the 58. page.  〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉. Proclus having spoken of some by nature, excellent in inventing demonstrations, pithy and brief sayeth: Yet are there Methods given [for that purpose]. And in deed, that, the best, which, by Resolution, reduceth the thing inquired of, to an undoubted principle. Which Method, Plato, taught Leodamas (as i● reported) And he is registered, thereby, to have been the inventor of many things in Geometry.  
And, verily, in Problems, it is the chief aid for winning and ordering a demonstration: first by Supposition, of the thing inquired of, to be done: by due and orderly Resolution to bring it to a stay, at an undoubted verity. In which point of Art, great abundance of examples, are to be seen, in that excellent and mighty Mathematician, Archimedes: & in his expositor, Eutocius, in Menaechmus likewise: and in Diocles' book, de Pyti●s: and in many other. And now, for as much as, our Euclid in the last six Propositions of this thirteenth book propoundeth, and concludeth those Problems, which were the end, Scope, and principal purpose, to which all the premises of the 12. books, and the rest of this thirteenth, are directed and ordered: It shall be artificially done, and to a great commodity, by Resolution, backward, from these 6. Problems, to return to the first definition of the first book: I mean, to the definition of a point. Which, is nothing hard to do. And I do counsel all such, as desire to attain, to the profound knowledge of Geometry, Arithmetic, or any branch of the sciences Mathematical, so by Resolution, (discretely and advisedly) to resolve, unlose, unjoint and disseaver every part of any work Mathematical, that, therby● aswell, the due placing of every verity, and his proof: as also, what is either superfluous, or wanting, may evidently appear. For so to invent, & there with to order their writings, was the custom of them, who in the old time, were most excellent. And I (for my part) in writing any Mathematical conclusion, which requireth great discourse, at length have found, (by experience) the commodity of it, such: that to do other ways, were to me a confusion, and an unmethodical heaping of matter together: besides the difficulty of inventing the matter to be disposed and ordered. I have occasion, thus to give you friendly advise, for your be●ofe● because some, of late, have inveighed against Euclid, or Theon in this place, otherwise than I would wish they had.     

The 6. Theorem. The 6. Propositions If a rational right line be divided by an extreme and mean proportion: either of the segments, is an irrational line of that kind, which is called a residual line.  
SVppose that AB, being a rational line be deuidedly 〈◊〉 extreme and mean proportion in the point C, and let the greater segment thereof be AC. Then I say that either of the lines AC, and CB, is an irrational line of that kind, which is called a residual. Extend the line AB, to the point D: 
Construction.  and let the line AD, be equal to half of the line AB. Now forasmuch as the right line AB, 
Demonstration.  is divided by an extreme & mean proportion in the point C, and unto the greater segment AC is added a line AD, equal to the half of the right line AB: therefore (by the 1. of the thirteenth) the square of the line CD, is quintuple to the square of the line AD. Wherefore the square of the line CD, hath to the square of the line AD, that proportion that number hath to number. Wherefore the square of the line CD is commensurable to the square of the line AD. But the square of the line DA is rational, for the line DA is rational; forasmuch as it is the half of the rational line AB. Wherefore the square of the line CD, is rational. Wherefore also the lin● CD, is rational. And forasmuch as the square of the line CD; hath not to the square of the line AD, that proportion that a square number hath to a square number, therefore (by the 9 of the tenth) the line CD, is incommensurable in length, to the line AD. Wherefore the lines CD, and DA are rational commensurable in power only. Wherefore the line AC is a residual line, by the 73. of the tenth. Again, forasmuch as the line AB, is divided by an extreme and mean proportion, and the greater segment thereof is AC, therefore that which is contained under the lines AB, &. BC, is equal to the square of the line AC. Wherefore the square of the line AC, applied to the rational line AB, maketh the breadth BC. But the square of a residual line, applied to a rational line maketh the breadth a first residual line (by the 97. of the tenth). Wherefore the line CB, is a first residual line. And it is proved that the line AC, is also a residual line. If therefore a rational right line be divided by an extreme and mean● proportion, either of the segments, is an irrational line of that kind, which is called a residual line, which was required to be demonstrated.  

¶ A Corollary added by Campane.  
Hereby it is manifest, that if the greater segment be a rational line: the less segment shallbe a residual line.  
For if the greater segment AC, of the right line ACB be divided into two equal parts in the point D, the square of the line DB, shallbe quint●pl● to the square of the line DC, (by the 3. of this book.) And forasmuch as the line CD, (being the ●alfe of the rational line supposed AC) is rational by the 6. definition of the tenth: And unto the square of the line DC, the square of the line DB is commensurable, (for it is quintuple unto it) wherefore the square of the line DB, is rational. Wherefore also the line DB is rational. And forasmuch as the squares of the lines DB & DC, are not in proportion, as a square number is to a square number: therefore the lines DB and DC are incommensurable in length (by the 9 of the tenth.) Wherefore they are commensurable in power only: Wherefore by the 73. of the tenth, the line BC, which is the less segment, is a residual line.    

The 7. Theorem. The 7. Proposition. If an equilater Pentagon have three of his angles, whether they follow in order, or not in order, equal the one to the other: that Pentagon shallbe equiangle.  
SVppose that ABCDE, be an equilater pentagon. And let the angles of the said Pentagon, 
Two cases in this proposition.  namely, first, three angles following in order, which are at the points A, B, C, be equal the one to the other. Then I say that the Pentagon ABCDE is equiangle. 
Construction.  Draw these right lines AC, BE, and FD. Now forasmuch as these two lines CB, 
Th● first case.  and BA, are equal to these two lines BA, and AE, the one to the other, and the angle CBA is equal to the angle BAE: 
Demonstration.  therefore (by the 4. of the first) the base AC is equal to the base BE, and the triangle ABC is equal to the triangle ABE, and the rest of the angles are equal to the rest of the angles, under which are subtended equal sides. Wherefore the angle BCA is equal to the angle BEA, and the angle ABE to the angle CAB. Wherefore also the side AF, is equal to the side BF (by the 6. of the first). And it was proved that the whole line AC is equal to the whole line BE. Wherefore the residue CF is equal to the residue F●. And the line CD, is equal to the line DE. Wherefore these two lines FC, and CD are equal to these two lines FE, & ED, and the base FD, is common to them both. Wherefore the angle FCD, is equal to the angle FED (by the 8. of the first). And it is proved that the angle BCA, is equal to the angle AEB. Wherefore the whole angle BCD is equal to the whole angle AED. But the angle BCD, is supposed to be equal to the angles A, and B. Wherefore the angle AED, is equal to the angles A and B. In like sort also may we prove that the angle CDE, is equal to the angles A, and B. Wherefore the Pentagon ABCDE is equiangle.  

The second case.  But now suppose that three angles, which follow not in order, be equal the one to the other, namely, let the angles A, C, D, be equal. Then I say that in this case also the Pentagon ABCDE is equiangle. Draw a right line from the point B, to the point D. Now forasmuch as these two lines BA, and A●, are equal to these two lines BC, and CD, and they comprehend equal angles, therefore (by the 4. of the first) the base BE, is equal to the base BD. And the triangle ABE, is equal to the triangle BDC, and the rest of the angles are equal to the rest of the angles, under which are subtended equal sides: wherefore the angle AEB, is equal to the angle CDB. And the angle BED, is equal to the angle BDE, (by the 5. of the first) for the side BE, is equal to the side BD. Wherefore the whole angle AED, is equal to the whole angle CDE. But the angle CDE, is supposed to be equal to the angles A, and C. Wherefore the AED, is equal to the angles A, and C. And by the same reason also the angle ABC, is equal to the angles A; C, and D. Wherefore the Pentagon ABCDE is equiangle. If therefore an equilater Pentagon have three of his angles, whither they follow in order, or not in order, equal the one to the other: that Pentagon shallbe equiangle: which was required to be proved.   

The 8. Problem. The 8. Proposition. If in an equilater & equiangle Pentagon two right lines do subtend two of the angles following in order: those lines do divide the one the other by an extreme and mean proportion: and the greater segments of those lines are each equal to the side of the Pentagon.  
SVppose that ABCDE be an equilater and equiangle Pentagon. And let two right lines AC, and BE, subtend the two angles A, and B, which follow in order. And let them cut the one the other in the point H. Then I say that either of those lines is divided by an extreme & mean proportion in the point H: And that each of the greater segments of those lines are equal to the side of the Pentagon Circumscribe (by the 14. of the fourth) about the Pentagon. ABCDE, 
Construction.  a circle ABCDE. And forasmuch as these two right lines EA, and AB, are equal to these two right lines AB, and BC, 
Demonstration.  and they contain equal angles: therefore (by the 4. of the first) the base BE, is equal to the base AC: and the triangle ABE, is equal to the triangle ABC, and the angles remaining, are equal to the angles remaining, the one to the other, under which are subtended equal sides. Wherefore the angle BAC, is equal to the angle ABE. Wherefore the angle AHE is double to the angle BAH, (by the 32. of the first) for it is an outward angle of the triangle ABH. And the angle EAC is double to the angle● BAC (by the last of the sixth). For the circumference EDC is double to the circumference CB. Wherefore the angle HAE is equal to the angle AHE. Wherefore also the right line ●E, is (by the 6. of the first) equal to the right line EA, that is to the line AB. And forasmuch as the right line BA is equal to the right line AE, therefore the angle ABE is equal to the angle AEB. But it is proved that the angle ABE is equal to the angle BAH: wherefore also the angle BEA is equal to the angle BAH. And in the two triangles ABE, and ABH, the angle ABE is common to them both, wherefore the angle remaining, namely, BAE is equal to the angle remaining, namely, to AHB (by the corollary of the 32. of the first). Wherefore the triangle ABE, is equiangle to the triangle ABH. Wherefore proportionally as the line EB, is to the line BA, so is the line AB to the line BH (by the 4. of the sixth). But the line BA is equal to the line EH. Wherefore as the line B● is to the line EH, so is the line EH to the line H●. But the line BE is greater the● the line BARNES wherefore the line EH, also is greater than the line HB. Wherefore the line BE● is divided by an extreme and mean proportion in the point H (by the 3. definition of the sixth) and his greater segment EH is equal to the side of the Pentagon. In like sort also may we prove that the line AC is divided by an extreme and mean proportion in the point H, and that his greater segment CH, is equal to the side of the Pentagon. (For the whole line AC is equal to the whole line BE, and it hath been proved that the parts taken away BH, and AH are equal wherefore the residue CH is equal to the residue EH (by the 10. of the fifth). If therefore in an equilater and equiangle Pentagon two right lines 〈◊〉 subtend two of the angles following in order: those lines do divide the one the other by an extreme and mean proportion: and the greater segments of those lines are each equal to the side of the Pentagon: which was required to be demonstrated.   

¶ The 9 Theorem. The 9 Proposition. If the side of an equilater hexagon, and the side of an equilater decagon or ●●u●gled figure, which both are inscribed in one & the self same circle, be added together: the whole right line made of them is a line divided by a● extreme and mean proportion, and the greater segment of the same is the side of the hexagon.  
SVppose that there be a circle ABC. And let the side of a decagon or tenangled figure inscribed in the circle ABC, be BC, and let the side of an hexagon or six angled figure inscribed in the same circle, be CD. And let the lines BC and CD be so joined together directly that they both make one right line, namely, BD. Then I say that the line BD is divided by an extreme and mean proportion in the point C: and that the greater segment thereof is the line CD. 
Construction.  Take (by the 1. of the third) the centre of the circle. And let it be the point E: and draw these right lines EB, EC, and ED. And extend the line BE to the point A. Now forasmuch as BC is the side of an equilater decagon, 
Demonstration.  therefore the circumference or semicircle ACB is quintuple to the circumference CB. Wherefore the circumference AC is quadruple to the circumference CB. But as the circumference AC is to the circumference CB, so is the angle AEC to the angle CEB, by the last of the sixth. Wherefore the angle AEC is quadruple to the angle CEB. And forasmuch as the angle EBC is equal to the angle ECB (by the 5. of the first,) for the line EB is equal to the line EC, by the definition of a circle, therefore the angle AEC is double to the angle ECB, by the 32. of the first. And forasmuch as the right line EC is equal to the right line CD, by the corollary of the 15. of the fourth (for either of them is equal to the side of the hexagon inscribed in the circle ABC) therefore the angle CED is equal to the angle CDE wherefore the angle ECB is double to the angle EDC, by the 32. of the first. But it is proved that the angle AEC is double to the angle ECB, wherefore the angle AEC is quadruple to the angle EDC. And it is proved that the angle AEC is quadruple to the angle BEC. Wherefore the angle EDC is equal to the angle BEC. And the angle EBD is common to the two triangles BEC and BED: wherefore the angle remaining BED is equal to the angle remaining ECB, by the corollary of the 32. of the first. Wherefore the triangle EBD is equiangle to the triangle EBC. Wherefore, by the 4. of the sixth, proportionally, as the line BD is to the line BE, so is the line BE to the line BC. But the line EB is equal to the line CD. Wherefore as the line BD is to the DC, so is the line DC to the line CB. But the line BD is greater than the line DC: wherefore also the line DC is greater than the line CB. Wherefore the right line BD is divided by an extreme and mean proportion in the point C: and his greater segment is DC. If therefore the side of an equilater hexagon, and the side of an equilater decagon or tenangled figure, which both are inscribed in one and the self same circle, be added togethers the whole right line made of them, is a line divided by an extreme and mean proportion, and the greater segment of the same is the side of the hexagon: which was required to be proved.  

A Corollary added by Flussas.  

This Corollary is the 3. proposition of the ●4. book after Campane.  Hereby it is manifest, that the side of an exagon inscribed in a circle being cut by an extreme and mean proportion, the greater segment thereof is the side of the decagon inscribed in the same circle. For if from the right line DC be cut of a right line equal to the line CB, we may thus reason, as the whole DB is to the whole DC, so is the part taken away DC to the part taken away CB: wherefore by the 19 of the fifth, the residue is to the residue as the whole is to the whole. Wherefore the line DC is cut like unto the line DB: and therefore is cut by an extreme and mean proportion.   

Campane putteth the converse of this proposition after this manner.  
If a line be divided by an 〈◊〉 and ●●ane proportion, of 〈◊〉 circle the greater segment is the side of an equilater Hexagon, of the same shall the less segment be the side of an equilater Decagon. And of what circle the less segment is the side of an equilater Decagon, of the same is the greater segment the side of an equilater Hexagon.  
For the former figure remaining, suppose that the line BD be divided by an extreme and mean proportion in the point C: and let the greater segment thereof be DC. Then I say that of what circle the line DC is the side of an equilater Hexagon, of the same circle is the line CB the side of an equilater decagon: and of what circle, the line BC is the side of an equilater Decagon, of the same is the line DC the side of an equilater Hexagon. 
Demonstration of the first part.  For if the line DC be the side of an Hexagon inscribed in the circle, then by the corollary of the 15. of the fourth, the line DC is equal to the line BE. And forasmuch as the proportion of the line BD to the line DC is as the proportion of the line DC to the line CB, by supposition: therefore (by the 7. of the fifth) the proportion of the line BD to the line BE, is as the proportion of the line BE to the line BC. Wherefore (by the 6. of the sixth) the two triangles DEB, EBC are equiangle (for the angle B is common to each triangle). Wherefore the angle D is equal to the angle CEB: for they are subtended of sides of like proportion. And forasmuch as the angle AEC is quadruple to the angle D (by the 32. of the first twice taken, and by the 5. of the same) therefore the same angle AEC is quadruple to the angle CEB. Wherefore (by the last of the sixth) the circumference AC is quadruple to to the circumference CB. Wherefore the line BC is the side of a decagon inscribed in the circle ACB.  
But now if the line BC be the side of a decagon inscribed in the circle ABC, the line CD shallbe the side of an Hexagon inscribed in the same circle. 
Demonstration of the second part.  For let DC be the side of an Hexagon inscribed in the circle H. Now by the first part of this proposition the line BC shallbe the side of a decagon inscribed in the same circle. Suppose that in the two circles ACB and H be inscribed equilater decagons, all whose sides shallbe equal to the line CB. And forasmuch as every equilater figure inscribed in a circle is also equiangle, therefore both the decagons are equiangle. And forasmuch as all the angles of the one taken together are equal to all the angles of the other taken together, as it is easy to be proved by that which is added after the 32. of the first, therefore one of these decagons is equiangle to the other: and therefore the one is like to the other by the definition of like superficieces. And for that if there be two like rectiline figures inscribed in two circles, the proportion of the sides of like proportion of those figures, shallbe as the proportion of the Diameters of those circles, as it is easy to prove by the corollary of the 20. of the sixth, and first of this book●: but the sides of the like deca●ons inscribed in the two circles ABC and H are equal: therefore their Diameters also are equal. Wherefore also their semidiameters are equal. But the s●midi●meters and the side of the Hexagon are equal, by the corollary of the 1●. of the fourth. Wherefore the line DC is the side of an hexagon in the circle ABC, as also it is the side of an hexagon inscribed in the circle F, which is equal to the circled ABC: which was required to be proved.    

¶ The 10. Theorem. The 10. Proposition. If in a circle be described an equilater Pentagon, the side of the Pentagon containeth in power both the side of an hexagon and the side of a decagon, being all described in one and the self same circle.  
SVppose that ABCDE be a circle. And in the circle ABCDE describe (by the 11. of the fourth) a pentagon figure ABCDE. Then I say, that the side of the pentagon ●igure ABCDE containeth in power both the side of an hexagon figure, and of a decagon figure, being described in the circle ABCDE. 
Construction.  Take (by the 1. of the third) the centre of the circle, and let the same be F. And drawing a right line from the point A to the point F, extend it to the point G. And draw a right line from the point F to the point B. And from the point F draw (by the 12. of the first) unto the line AB a perpendicular line FH: and extend it to the point K. And draw the right lines AK and KB. And again, from the point F draw (by the same) unto the line AK, a perpendicular line FN: and extend FN to the point M, which line let cut the line AB in the point L. And draw a right line from the point K to the point L. Now forasmuch as the circumference ABCG is equal to the circumference AEDG, 
Demonstration.  of which the circumference ABC is equal to the circumference AED, therefore the rest of the circumference, namely, CG, is equal to the rest of the circumference, namely, to DG. But the circumference CD is subtended of the side of a pentagon: Wherefore the circumference CG is subtended of the side of a decagon ●igure. And forasmuch as the line BH is equal to the line HA (by the 3. of the third) and the line FH is common to them both, & the angles at the point H are right angles. Wherefore (by the 4. of the first) the angle AFK is equal to the angle KFB. Wherefore also (by the 26. of the third) the circumference AK, is equal to the circumference KB. Wherefore the circumference AB is double to the circumference BK. Wherefore the right line AK is the side of a decagon figure. And by the same reason also the circumference AK is double to the circumference KM. And forasmuch as the circumference AB is double to the circumference BK, but the circumference CD is equal to the circumference AB: wherefore the circumference CD is double to the circumference BK. But the circumference CD is double to the circumference CG: wherefore the circumference CG is equal to the circumference BK. But the circumference BK is double to the circumference KM (for that the circumference KA is double thereunto) ● wherefore also the circumference CG is double to the circumference KM. But the circumference CB is also double to the circumference BK (for the circumference CB is equal to the circumference BA). Wherefore the whole circumference GB is double to the whole circumference BM, by the 12. of the fift. Wherefore also the angle GFB is double to the angle BFM, by the last of the sixth. But the angle GFB is double to the angle FAB (by the 32. of the first, or 20. of the third). For the angle FAB is equal to the angle ABF. Wherefore the angle BFL is equal ●o the angle FAB, by the 15. of the fift. And the angle ABF is common to the two triangles ABF and BFL. Wherefore (by the Corollary of the ●●. of the first) the angle remaining AFB is equal to the angle remaining BLF. Wherefore the triangle ABF is equiangle to the triangle BFL. Wherefore (by the 4. of the sixth) proportionally as the right line AB is to the right line BF, so is the same right line FB to the right line BL: wherefore that which is contained under the lines AB and BL is equal to the square of the line BF, by the 17. of the sixth. Again forasmuch as the line AL is equal to the line LK [for by the last of the sixth, the angle KFL is equal to the angle AFL, which equal angles are contained under the lines FK, FL, and FA, FL, & the line FK is equal to the line FA, and the line FL is common to them both. Wherefore, by the 4. of the first, the line AL is equal to the line LK] and the line LN is common to them both, & maketh right angles at the point N, and (by the 3. of the third) the base AN is equal to the base KN. Wherefore also the angle LKN is equal to the angle LAN. But the angle LAN is equal to the angle KBL (by the 5. of the first). Wherefore the angle LKN is equal to the angle KBL. And the angle KAL is common to both the triangles AKB and AKL. Wherefore the angle remaining AKB is equal to the angle remaining ALK (by the Corollary of the 32. of the first). Wherefore the triangle KBA is equiangle to the triangle KLA. Wherefore (by the 4. of the sixth) proportionally as the right line BA is to the right line AK, so is the same right line KA to the right line AL. Wherefore that which is contained under the lines BA and AL is equal to the square of the line AK (by the 17. of the sixth). And it is proved that that which is contained under the lines AB and BL, is equal to the square of the line BF. Wherefore that which is contained under AB & BL together with that which is con●●ined under BA and AL (which by the 2. of the second, is the square of the line BA) is equal to the squares of the lines AF and AK. But the line BA is the side of the pentagon figure, and AF the side of the hexagon figure (by the Corollary of the 15. of the fourth), and AK the side of the decagon figure. Wherefore the side of a pentagon figure, containeth in power both the side of an hexagon figure, and of a decagon figure, being described all in one and the self same circle: which was required to be demonstrated.  

¶ A Corollary added by Flussas.  
A perpendicular line from any angle drawn to the base of a pentagon, passeth by the centre.  
For if we draw a right line from the point A to the point C, and an other from the point A to the point D: those right lines sh●●l be equal, by the 4. of the first: and therefore in the triangle ACD the angles at the points C and D are, by the 5. of the first● equal. But the angles made at the point where the line AG cutteth the line CD, are by supposition right angles: wherefore, by the 26. of the first, the line CD is by the line AG divided into two equal parts, and it is also divided perpendicularly: wherefore by the corollary of the first of the third in the line AG is the centre of the circle: and therefore the line AG passeth by the centre.    

¶ The 11. Theorem. The 11. Proposition. If in a circle having a rational line to his diameter be inscribed an equilater pentagon: the side of the pentagon is an irrational line, and is of that kind which is called a less line.  
SVppose that in the circle ABCDE having a rational line to his diameter be inscribed a pentagon figure ABCDE. Then I say that the side of the pentagon figure ABCDE, namely, the side AB, is an irrational line of that kind which is called a less line. Take (by the 1. of the third) the centre of the circle, and let the same be the point F, and draw a right line from the point A to the point F, 
Construstion.  and an other from the point F to the point B, and extend those lines to the points G and H. And draw a right line from the point A to the point C. And from the semidiameter FH take the fourth part (by the 9 of the sixth) and let the same be FK: But the line FH is rational (for that it is the half of the diameter which is supposed to be rational), wherefore also the line FK is rational. And the line or semidiameter BF is rational. Wherefore the whole line BK is rational. 
Demonstration.  And forasmuch as the circumference ACG is equal to the circumference ADG, of which the circumference ABC is equal to the circumference AED, wherefore the residue CG is equal to the residue GD. Now if we draw a right line from the point A to the point D, it is manifest that the angles ALC and ALD are right angles. For forasmuch as the circumference CG is equal to the circumference GD, therefore (by the last of the sixth) the angle CAG is equal to the angle DAG. And the line AC is equal to the line AD, for that the circumferences which they subtend are equal, and the line AL is c●mmon to them both, therefore there are two lines AC and AL equal to two lines AD and AL, and the angle CAL is equal to the angle DAL. Wherefore (by the 4. of the first) the base CL is equal to the base LD, and the rest of the angles to the rest of the angles, and the line CD is double to the line CL. And by the same reason may it be proved, that the angles at the point M are right angles, and that the line AC is double to the line CM. Now forasmuch as the angle ALC is equal to the angle AMF, for that they are both right angles, and the angle LAC is common to both the triangles ALC and AMF: wherefore the angle remaining, namely, ACL, is equal to the angle remaining AFM, by the corollary of the 32. of the first. Wherefore the triangle ACL is equiangle to the triangle AMF. Wherefore proportionally, by the 4. of the sixth, as the line LC is to the line CA, so is the line MF to the line FA. And in the same proportion also are the doubles of the antecedents LC and MF (by the 15. of the fifth). Wherefore as the double of the line LC is to the line CA, so is the double of the line MF to the line FA. But as the double of the line MF is to the line FA, so is the line MF to the half of the line FA, by the 15. of the fifth, wherefore as the double of the line LC is to the line CA, so is the line MF to the half of the line FA, by the 11. of the fifth. And in the same proportion, by the 15. of the fifth, are the halves of the consequents, namely, of CA and of the halue of the line AF. Wherefore as the double of the line LC is to the half of the line AC, so is the line MF to the fourth part of the line FA. But the double of the line LC is the line DC, and the half of the line CA is the line CM, as hath before been proved, and the fourth part of the line FA is the line FK (for the line FK is the fourth part of the line FH by construction). Wherefore as the line DC is to the line CM, so is the line MF to the line FK. Wherefore by composition (by the 18. of the fifth) as both the lines DC and CM are to the line CM, so is the whole line MK to the line FK. Wherefore also (by the 22. of the sixth) as the squares of the lines DC and CM are to the square of the line CM, so is the square of the line MK to the square of the line FK. And forasmuch as (by the 8. of the thirteenth) a line which is subtended under two sides of a pentagon figure, as is the line AC, being divided by an extreme & mean proportion, the greater segment is equal to the side of the pentagon figure, that is, unto the line DC: and (by the 1. of the thirteenth) the greater segment having added unto it the half of the whole, is in power quintuple to the square made of the half of the whole: and the half of the whole line AC is the line CM. Wherefore the square that is made of the lines DC and CM, that is, of the greater segment and of the half of the whole, as of one line, is quintuple to the square of the line CM, that is, of the half of the whole. But as the square made of the lines DC and CM, as of one line, is to the square of the line CM, so is it proved, that the square of the line MK is to the square of the line FK. Wherefore the square of the line MK is quintuple to the square of the line FK. But the square of the line KF is rational, as hath before been proved, wherefore also the square of the line MK is rational, by the 9 definition of the tenth, for the square of the line MK hath to the square of the line KF that proportion that number, hath to number, namely, that 5. hath to 1. and therefore the said squares are commensurable, by the 6. of the tenth. Wherefore also the line MK is rational. And forasmuch as the line BF is quadruple to the line FK (for the semidiameter BF is equal to the semidiameter FH), therefore the line BK is quintuple to the line FK. Wherefore the square of the line BK is 25. times so much as the square of the line KF, by the corollary of the 20 of the sixth. But the square of the line MK, is quintuple to the square of the FK, as is proved. Wherefore the square of the line BK is quintuple to the square of the line KM. Wherefore the square of the line BK, hath not to the square of the line KM, that proportion that a square number hath to a square number, by the corollary of the 25. of the eight. Wherefore (by the 9 of the tenth) the line BK is incommensurable in length to the line KM, and either of the lines is rational. Wherefore the lines BK and KM are rational commensurable in power only. But if from a rational line be taken away a rational line being commensurable in power only to the whole, that which remaineth is irrational, and is (by the 73. of the tenth) called a residual line. Wherefore the line MB is a residual line. And the line conveniently joined unto it, is the line MK. Now I say that the line BM is a fourth residual line. Unto the excess of the square of the line BK above the square of the line KM, let the square of the line N be equal (which excess how to find out, is taught in the assumpt put after the 13. proposition of the tenth). Wherefore the line BK is in power more than the line KM by the square of the line N. And forasmuch as the line KF is commensurable in length to the line FB, for it is the fourth part thereof, therefore (by the 16. of the tenth) the whole line KB is commensurable in length to the line FB. But the line FB is commensurable in length to the line BH, namely, the semidiameter● to the diameter: wherefore the line BK is commensurable in length to the line BH, by the 12. of the tenth. And forasmuch as the square of the line BK is quintuple to the square of the line KM, therefore the square of the line BK hath to the square of the line KM that proportion that five hath to one. Wherefore by conversion of proportion (by the corollary of the 19 of the fifth) the square of BK hath to the square of the line N, that proportion that five hath to four: & therefore it hath not that proportion that a square number hath to a square number, by the corollary of the 25. of the eight. Wherefore, the line BK is incommensurable in length to the line N (by the 9 of the tenth). Wherefore the line BK is in power more than the line KM, by the square of a line incommensurable in length to the line BK. Now then forasmuch as the whole line BK is in power more than the line conveniently joined, namely, then KM, by the square of a line incommensurable in length to the line BK, and the whole line BK is commensurable in length to the rational line given BH: therefore the line MB is a fourth residual line, by the definition of a fourth residual line. But a rectangle parallelogram contained under a rational line and a fourth residual line, is irrational, and the line which containeth in power the same parallelogram is also irrational, and is called a less line (by the 94. of the tenth). But the line AB containeth in power the parallelogram contained under the lines HB and BM (for if we draw a right line from the point A to the point H, the triangle ABH shall be like to the triangle ABM, by the 8. of the sixth. For from the right angle BAH is drawn to the base BH a perpendicular line. And therefore as the line BH is to the line BA, so is the line AB to the line BM. this followeth also of the corollary of the said 8. of the sixth. Wherefore the line AB which is the side of the pentagon figure, is an irrational line of that kind which is called a less line. If therefore in a circle having a rational line to his diameter be inscribed an equilater pentagon, the side of the pentagon is an irrational line, and is of that kind which is called a less line: which was required to be demonstrated.   

¶ The 12. Theorem. The 12. Proposition. If in a circle be described an equilater triangle: the square made of the side of the triangle, is triple to the square made of the line, which is drawn from the centre of the circle to the circumference.  
SVppose that ABC be a circle, and in it describe an equilater triangle ABC. Then I say that the square made of the side of the triangle ABC is triple to the square made of the line drawn from the centre of the circle ABC to the circumference. 
Constr●yction.  Take (by the 1. of the third) the centre of the circle, and let the same be D. A●d draw a right line from the point A to the point D, and extend it to the point E. And draw a right line from the point B to the point E. 
Demonstration.  Now forasmuch as the triangle ABC is equilater, therefore each of these three circumferences AB, AC, & BEC is the third part of the whole circumference of the circle ABC wherefore the circumference BE is the sixth part of the circumference of the circle [for the circumference of the semicircle ABE is equal to the circumference of the semicircle ACE, from which taking away equal circumferences AB and AC, the circumference remaining BE shallbe equal to the circumference remaining EC]: wherefore the right line BE is the side of an equilater hexagon figure described in the circle. Wherefore it is equal to the line drawn from the centre of the circle to the circumference, that is unto the line DE (by the corollary of the 15. of the sixth). And forasmuch as the line AE is double to the line DE, therefore the square of the line AE is quadruple to the square of the line DE (by the 4. of the second): that is, to the square of the line BE. But the square of the line AE is equal to the squares of the lines AB, and BE (by the 47. of the first) for the angle ABE is (by the 31. of the third) a right angle. Wherefore the squares of the line AB & BE are quadruple to the square of the line BE. Wherefore taking away the square of the line BE, the squa●e of the line AB sh●lbe treble to the square of BE: but the line BE is equal to the line DE. Wh●r●fore the square of the line AB is triple to the square of the line DE. Wherefore the square made of the side of the triangle, is triple to the square made of the line drawn from the centre of the circle to the circumference: which was required to be proved.  

A Corollary added by Campane.  
Hereby it is manifest, that the line BC, which is the side of the equilater triangle, divideth the semidiameter DE into two equal parts. For let the point of the division be F. And suppose a line to be drawn from the point D to the B, and an other from the point D to the point C. Now it is manifest (by the 4. of the first) that the line BF is equal to the line FC, and therefore (by the 3. of the third) all the angles at the point F are ●ight angles. Wherefore (by the 47. of the first) the square of the line BD is equal to the squares o● the line●●F and FD, and by the same the square of the line BE is equal to the squares of the lines BF, and FE: b●t the line BD is equal to the line BE (as hath before been proved). Wherefore by the common sentence the two squares of the two lines BF and FD are equal to the two squares of the line● BF, and FE. Wherefore taking away the square of the line BF which is common to them both: the residue, namely, the square of the line DF shallbe equal to the residue, namely, to the square of the line FE. Wherefore also the line FD is equal to the line FE. Wherefore hereby it is manifest that a perpendicular line drawn from the centre of a circle to the side of an equilater triangle inscribed in it, is equal to half of the line drawn from the centre of the same circle, to the circumference thereof.   

A Corollary added by Flussas.  
The side of an equilater triangle is in power sesquitertia to the perpendicular line which is drawn from one of the angles to the opposite side. For of what parts the line AB containeth in power 12. of such parts the line BF which is the half of AB contained in power 3. 
This Corollary is the 11. proposition of the 14. book after Campane.  Wherefore the residue, namely, the perpendicular line AF containeth in power of such parts 9 (for the squares of the lines AF, and BF are by the 47. of the first equal to the square of the line AB). Now 1●. to 9 is sesquitertia● wherefore the power of the line AB is to the power of the line AF in sesquitertia proportion.  
Moreover the side of the triangle is the mean proportional between the diameter and the perpendicular line: For (by the Corollary of the 8. of the sixth) the line AE is to the line AB as the line AB is to the line AF.  
Farther the perpendicular line drawn from the angle divideth the base into two equal parts and passeth by the centre. 
This Corollary is the 3. Corollary after the 17. proposition of the 14 book after Campane.  For if there should be drawn any other right line from the point A to the point F, then that which is drawn by the point D, two right lines should include a superficies, which is impossible. Wherefore the contrary followeth, namely, that the line, which being drawn from the angle passeth by the centre, is a perpendicular line to the base (by the 3. of the third).    

The 1. Problem. The 13. Proposition. To make a 
* By the name o● a Pyramid both here & i● this book following, understand a Tetrahedron.  Pyramid, and to comprehend it in a sphere given: and to prove that the diameter of the sphere is in power sesquialtera to the side of the Pyramid.  
[In the semicircle ADB of the former figure draw the line DN. And divide the line KG into two equal parts in the point M. 
another construction and demonstration of the second part after F●ussas.  And draw a line from M to G. And forasmuch as by construction the line KH is equal to the line AC, and the line HL to the line CB: therefore the whole line AB is equal to the whole line KL. Wherefore also the half of the line KL, namely, the line LM, is equal to the semidiameter BN: wherefore taking away from those equal lines, equal parts BC and LH, the residues NC, and MH shallbe equal. Wherefore in the two triangles MHG and NCD, the two sides about the equal right angles DCN and GHM, namely, the sides DC, CN, and GH, HM, are equal: wherefore the bases MG and ND are equal (by the 4. of the first.) And by the same reason may it be proved that right lines drawn from the point M to the points E and F are equal to the line ND. But the right line ND is equal to the line AN, which is drawn from the centre to the circumference: wherefore the line MG is equal to the line MK, & also to the lines ME, MF and ML. Wherefore making the centre the point M, and the space MK or MG describe a semicircle KGL: and the diameter KL abiding fixed let the said semicircle KGL be moved round about until it return to the same place from whence it began to be moved: and there shallbe described a sphere about the centre M (by the 12. definition of the eleventh) touching every one of the angles of the Pyramid which are at the points K, E, F, G: for those angles are equally distant from the centre of the sphere, namely, by the semidiameter of the said sphere, as hath before been proved. Wherefore in the sphere given whose diameter is the line KL, or the line AB, is inscribed a Tetrahedron EFGK.]  
Now I say, that the diameter of the sphere is in power sesquialtera to the side of the Pyramid. 
Third part of the demonstration.  For forasmuch as the line AC is double to the line CB (by construction) therefore the line AB is triple to the line BC. Wherefore by conversion) by the corollary of the 19 of the fifth) the line AB is sesquialtera to the line AC. But as the line BA is to the line AC, so is the square of the line BA, to the square of the line AD. For if we draw a right line from the point B, to the point D, as the line BD is to the line AD, so is the same AD to the line AC, by reason of the likeness of the ttriangles DAB, & DAC (by the 8. of the sixth): & by reason also that as the first is to th' third so is the square of the first to the square of the second (by the corollary of the 20. of the sixth). Wherefore the square of the line BA, is sesquialter to the square of the line AD. But the line BA is equal to the diameter of the sphere given, namely, to the line KL, as hath been proved, & the line AD is equal to the side of the pyramid inscribed in the sphere. Wherefore the diameter of the sphere is in power sesquialter to the side of the pyramid. Wherefore there is made a pyramid comprehended in a sphere given, and the diameter of the sphere is sesquialtera to the side of the pyramid: which was required to be done and proved.  

¶ An other demonstration to prove that as the line AB is to the line BC, so is the square of the line AD to the square of the line DC.  
Let the description of the semicircle ADB be as in the first description. And upon the line AC describe (by the 46. of the first) a square EC, and make perfect the parallelogram FB. Now forasmuch as the triangle DAB is equiangle to the triangle DAC (by the 32. of the sixth: therefore as the line BA is to the line AD, so is the line DA to the line AC, by the 4. of the sixth. Wherefore that which is contained under the lines BA and AC, is equal to the square of the line AD, by the 17. of the sixth. And for that as the line AB is to the line BC, so is the parallelogram EB to the parallelogram FB, by the 1. of the sixth: and the parallelogram EB is that which is contained under the lines BA and AC (for the line EA is equal to the line AC): and the parallelogram BF is that which is contained under the lines AC and BC. Wherefore as the line AB is to the line BC, so is that which is contained under the lines BA and AC, to that which is contained under the lines AC and CB. But that which is contained under the lines BA and AC, is equal to the square of the line AD, by the Corollary of the 8. of the sixth, and that which is contained under the lines AC and CB, is equal to the square of the line CD, for the perpendicular line DC is the mean proportional between the segments of the base, namely, AC and CB, by the former Corollary of the 8. of the sixth, for that the angle ADB is a right angle. Wherefore as the line AB is to the line BC, so is the square of the line AD to the square of the line DC, by the 11. of the fift: which was required to be proved.   

¶ Two Assumptes added by Campane.  

First Assumpt.  
Suppose that upon the line AB be erected perpendicularly the line DC, which line DC let be the mean proportional between the parts of the line AB, namely, AC & CB: so that as the line AC is to the line CD, so let the line CD be to the line CB. And upon the line AB describe a semicircle. Then I say, that the circumference of that semicircle shall pass by the point D, which is the end of the perpendicular line. But if not, than it shall either cut the line CD, or it shall pass above it, and include it not touching it. First let it cut it in the point E. And draw these right lines EB and EA. Wherefore by the 31. of the third, the whole angle AEB is a right angle. Wherefore by the first part of the Corollary of the 8. of the sixth, the line AC is to the line ●C, as the line EC is to the line CB. But by the 8. of the fift, the proportion of the line AC to the line EC, is greater than the proportion of the same line AC to the line CD (for the line CE is less than the line CD). Now for that the line CE is to th● line CB, as the line AC is to the line CE, and the line CD is to the line CB, as the line AC is to the line CD, therefore by the 13. of the fift, the proportion of the line EC to the line CB, is greater than the proportion of the line CD to the line CB. Wherefore by the 10. of the fift, the line EC is greater than the line DC, namely, the part greater than the whole: which is impossible. Wherefore the circumference shall not cut the line CD. Now I say, that it shall not pass above the line CD, and not touch it in the point D. For if it be possible, let it pass above it, and extend the line CD to the circumference, and let it cut it in the point ●. And draw the lines FB and FA, and it shall follow as before that the line CD is greater than the line CF: which is impossible. Wherefore that is manifest which was required to be proved.   

¶ Second Assumpt.  
If there be a right angle unto which a base is subtended, and if upon the same be described a semicircle: the circumference thereof shall pass by the point of the right angle.  
The converse of this was added after the demonstration of the 31. of the third, out of Pelitarius. And these two Assumptes of Campane are necessary, for the better understanding of the demonstration of the seco●d part of this 13. Proposition, wherein is proved that the pyramid is contained in the Sphere given.    

¶ Certain Corollaryes added by Flussas.  

First Corollary.  
The diameter of the Sphere is in power quadruple sesquialtera to the line which is drawn from the centre to the circumference of the circle which containeth the base of the pyramid.  
For forasmuch as it hath been proved, that the diameter KL is in power sesquialter to the side EF: and it is proved also, by the 12. of this book, that the side EF is in power triple to the line EH (which is drawn from the centre of the circle containing the triangle EFG). But the proportion of the extremes, namely, of the diameter to the line EH, consisteth of the proportions of the means, namely, of the proportion of the diameter to the line EF, and of the proportion of the line EF to the line EH, by the 5. definition of the sixth: which proportions, namely, triple, and sesquialter, added together, make quadruple sesquialter (as it is easy to prove by that which was taught in the declaration of the 5. definition of the sixth book). Wherefore the Corollary is manifest.   

¶ Second Corollary.  
Only the line which is drawn from the angle of the pyramid to the base opposite unto it, 
This Corollary is the 15. proposition of the 14. book after Campane.  & passing by the centre of the Sphere, is perpendicular to the base, and falleth upon the centre of the circle which containeth the base.  
For if any other line (than the line KMH which is drawn by the centre of the Sphere to the centre of the circle) should fall perpendicularly upon the plain of the base, then, from one and the self same point should be drawn to one and the self same plain two perpendicular lines, contrary to the 13. of the eleventh: which is impossible. Farther if from the top K should be drawn to the centre of the base, namely, to the point H, any other right line not passing by the centre M, two right lives should include a superficies contrary to the last common sentence: which were absurd. Wherefore only the line which is drawn by the centre of the Sphere to the centre of the base, is perpendicular to the said base. And the line which is drawn from the angle perpendicularly to the base, shall pass by the centre of the Sphere.   

Third Corollary.  
The perpendicular line which is drawn from the centre of the Sphere to the base of the pyramid, 
This Corollary Campane putteth as a Corollary after the 17. proposition of the 14. book.  is equal to the sixth part of the diameter of the Sphere.  
For it is before proved, that the line MH (which is drawn from the centre of the Sphere to the centre of the base) is equal to the line NC: which line NC is the sixth part of the diameter AB, and therefore the line MH is the sixth part of the diameter of the Sphere. For the diameter AB is equal to the diameter of the Sphere, as hath also before been proved.     

¶ The 2. Problem. The 14. Proposition. To make an octohedron, and to comprehend it in the sphere given, namely, that wherein the pyramid was comprehended: and to prove that the diameter of the sphere is in power double to the side of the octohedron.  

Construction.  TAke the diameter of the former sphere given, which let be the line AB: and divide it (by the 10. of the first into two equal parts in the point C. And describe upon the line AB a semicircle ADB. And (by the 11. of the first) from the point C raise up unto the line AB a perpendicular line CD. And draw a right line from the point D to the point B. And describe a square EFGH having every one of his sides equal to the line BD. and draw the diagonal lines FH & EG, cutting the one the other in the point K. And (by the 12. of the eleventh) from the point K (namely, the point where the lines FH and EG cut the one the other) raise up to the plain superficies, wherein the square EFGH is, a perpendicular line KL, and extend the line KL on the other side of the plain superficies to the point M. And let each of the lines KL and KM be put equal to one of these lines KE, KF, KH or KG. And draw these right lines LE, LF, LG, LH, ME, MF, MG, and MH. 
●rist part of the demonstration.  Now forasmuch as the line KE is (by the corollary of the 34. of the first) equal to the line KH, and the angle 
* For the 4. angles at the point K are equal to four right angles by the Corollary of t●e 15. of the first: and those 4. angle, are equal the one to the other by the ●. of the ●irst: and therefore each is a right angle.  EKH is a right angle, therefore the square of HE is double to the square of EK, by the 47. of the first: Again forasmuch as the line LK is equal to the line KE by position, and the angle LKE is, by the second definition of the eleventh, a right angle: therefore the square of the line EL is double to the square of the line EK: and it is proved that the square of the line HE is double to the square of the line EK. Wherefore the square of the line LE is equal to the square of the line EH. Wherefore also the line LE is equal to the line EH. And by the same reason the line LH is also equal to the line HERALD Wherefore the triangle LHE is equilater. In like sort may we prove that every one of the rest of the triangles whose bases are the sides of the square EFGH, and tops the points L and M, are equilater. The said eight triangles also are equal the one to the other, for every side of each is equal to the side of the square EFGH. Wherefore there is made an octohedron contained under eight triangles whose sides are equal. Now it is required to comprehend it in the sphere given, and to prove that the diameter of the sphere is in power double to the side of the octohedron.  

Second part of the demonstration.  Forasmuch as these three lines LK, KM, and KE are equal the one to the other, therefore a semicircle described upon the line LM shall pass also by the point E. And by the same reason, if the semicircle be turned round about, until it return unto the self same place from whence first it began to be moved, it shall pass by the points F, G, H, and the octohedron shall be comprehended in a sphere. I say also that it is comprehended in the sphere given. For forasmuch as the line LK is equal to the line KM, by position, and the line KE is common to them both and they contain right angles by the 3. definition of the eleventh, therefore (by the 4. of the first) the base LE is equal to the base EM. And forasmuch as the angle LEM is a right angle, by the 31. of the third, for it is in a semicircle, as hath been proved, therefore the square of the line LM is double to the square of the line LE by the 47. of the first. Again forasmuch as the line AC is equal to the line BC, therefore the line AB is double to the line BC, by the definition of a circle. But as the line AB is to the line BC, so is the square of the line AB to the square of the line BD, by the corollaries of the 8. and ●0. of the sixth. Wherefore the square of the line AB is double to the square of the line BD. And it is proved that the square of the line LM is double to the square of the line LE. Wherefore the square of the line BD is equal to the square of the line LE. For the line EH which is equal to the line LF, is put to be equal to the line DB. Wherefore the square of the line AB is equal to the square of the line LM. Wherefore the line AB is equal to the line LM. And the line AB is the diameter of the sphere given, wherefore the line LM is equal to the diameter of the sphere given. Wherefore the octoedron is contained in the sphere given: 
* For the square of the line AB, which is proved equal to the square of the line LM, is double to the square of the line BD which is also equal to the square of the line LE.  and it is also proved that the diameter of the sphere is in power double to the side of the octohedron. Wherefore there is made an octohedron, and it is comprehended in the sphere given, wherein was comprehended the Pyramid: and it is proved that the diameter of the sphere is in power double to the side of the octohedrn: which was required to be done, and to be proved.  

Certain Corollaries added by Flussas.  

First Corollary.  
The side of a Pyramid is in power sesquitertia to the side of an octahedron inscribed in the same Sphere.  
For forasmch as the diameter is in power double to the side of the octohedron, therefore of what parts the diameter containeth in power 6. of the same, the side of the octohedron containeth in power 3. but of what parts the diameter containeth 6. of the same, the side of the pyramid containeth 4. by the 13. of this book. Wherefore of what parts the side of the pyramid containeth 4. of the same the side of the octohedron containeth 3.   

Second Corollary.  
An octohedron is divided into two equal and like Pyramids.  
The common bases of these Pyramids are set upon every square contained of the sides of the octohedron, upon which square are set the ●● triangles of the octohedron: 
This Corollary is the 16. proposition of the 14. book after Campane.  which pyramids are by the ●. definition of the eleventh equal and like. And the foresaid square common to those Pyramids, is the half of the square of the diameter of the sphere, for it is the square of the side of the octohedron.   

Third Corollary.  
The three diameters of the octohedron, do cut the one the other perpendicularly into two equal parts, in the centre of the sphere which containeth the said octohedron.  
As it is manifest by the three diameters EG, FH, and LM which cut the one the other in the centre K equally and perpendicularly.     

¶ The 3. Problem. The 15. Proposition. To make a solid called a cube, and to comprehend it in the sphere given, namely, that Sphere wherein the former two solids were comprehend●d● and to prove that the diameter of the sphere, is in power triple to the side of the cube.  
TAke the diameter of the sphere given, namely, AB, and divide it in the point C● So that let the line AC be double to the line BC by the 9 of the sixth. And upon the line AB describe a semicircle ADB. And (by the 11. of the first) from the p●ynt C r●yse up unto the line AB a perpendicular line CD. And draw a right lin● DB. And describe a square EFGH, having every one of his sides equal to the line DB, And from the points E, F, G, H, raise up (by the 12. of the eleventh) unto the plain superficies of the square EFGH perpendicular lines EK, FL, GM, and HN: and let every one of the lines EK, FL, GM, and HN be put equal to one of the lines EF, FG, GH, or HE, which are the sides of the square, and draw these right lines KL, LM, MN, and NK. 
First part of the demonstration.  Wherefore there is made a cube namely FN which is contained under six equal squares. 
Second part of the Construction.  Now it is required to comprehend the same cube in the sphere given, and to prove that the diameter of the sphere is in power ble to the side of the cube. 
Second part of the demonstration.  Draw these right lines KG and EG. And forasmuch as the angle KEG is a right angle, for that the line KE is erected perpendicularly to the plain superficies E●, and therefore also to the right line EG, by the 2. definition of the eleventh, wherefore a semicircle described upon the line KG shall 
* By the 2. Assumpt of the 13. of this book.  pass by the point E. Again forasmuch as the line FG is erected perpendicularly to either of these lines FL and FE, by the definition of a square, & by the 2. definition of the eleventh, therefore the line FG is erected perpendicularly to the plain superficies FK, by the 4. of the eleventh. Wherefore if we draw a right line from the point F to the point K, the line GF shall be erected perpendicularly to the line KF, by the 2. definition of the eleventh. And by the same reason again a semicircle described upon the line GK shall pass also by the point F. And likewise shall it pass by the rest of the points of the angles of that cube. If now the diameter KG abiding fixed the semicircle be turned round about until it return into the self same place from whence it began first to be moved, the cube shallbe comprehended in a sphere. I say also that it is comprehended in the sphere given.  

Third part of the demonstration.  For forasmuch as the line GF is equal to the lin●●E and the angle F is a right angle, therefore the square of the line EG is by the 47. of the first double to the square of the line ●F. But the line EF is equal to the line EK. Wherefore the square of the line EG is double to the square of the line EK. Wherefore the squares of EG and EK that is the square of the line GK, by the 47. of the first, are triple to the square of the line EK. And forasmuch as the line AB is triple to the line BC, but as the line AB is to the line BC, so is the square of the line AB to the square of the line BD, by the corollaries of the 8. and 20. of the sixth. Wherefore the square of the line AB is triple to the square of the line BD. And it is proved that the square of the line GK is triple to the square of the line KE, and the line KE is put equal to the line BD. Wherefore the line KG is also equal to the line AB. And the line AB is the diameter of the sphere given. Wherefore the line KG is equal to the diameter of the sphere given. Wherefore the cube is comprehended in the sphere given: and it is also proved that the diameter of the Sphere is in power triple to the side of the cube: which was required t●●e done, and to be proved.  

another demonstration after Flussas.  
Suppose that the diameter of the Sphere given in the former Propositions, be the line A●. And let the centre be the point C, upon which describe a semicircle ADB. And from the diameter AB cut of a third part BG, by the 9 of the sixth. And from the point G raise up unto the line AB a perpendicular line DG, by the 11. of the first. And draw these right lines DA, DC, and DB. And unto the right line DB put an equal right line ZI: and upon the line ZI describe a square EZIT. And from the points E, Z, I, T, erect unto the superficies EZIT perpendicular lines EK, ZH, IM, TN (by the 12. of the eleventh): and put every one of those perpendicular lines equal to the line ZI. And draw these right lines KH, HM, MN, and NK, each of which shall be equal and parallels to the line ZI, and to the rest of the lines of the square, by the 33. of the first. And moreover they shall contain equal angles (by the 10. of the eleventh): and therefore the angles are right angles, for that EZIT is a square: wherefore the rest of the bases shall be squares. Wherefore the solid EZITKHMN being contained under 6. equal squares, is a cube, by the 21. definition of the eleventh. Extend by the opposite sides KE and MI of the cube, a plain KEIM: and again by the other opposite sides NT and HZ, extend an other plain HZTN. Now forasmuch as each of these plains divide the solid into two equal parts, namely, into two prisms equal and like (by the 8. definition of the eleventh): therefore those plains shall cut the cube by the centre, by the Corollary of the 39 of the eleventh. Wherefore the common section of those plains shall pass by the centre. Let that common section be the line LF. And forasmuch as the sides HN and KM of the superficieces KEIM and HZTN do divide the one the other into two equal parts, by the Corollary of the 34. of the first, and so likewise do the sides ZT and EI: therefore the common section LF is drawn by these sections, and divideth the plains KEIM and HZTN into two equal parts, by the first of the sixth: for their bases are equal, and the altitude is one and the ●ame, namely, the altitude of the cube. Wherefore the line LF shall divide into two equal parts the diameters of his plains, namely, the right lines KI, EM, ZN, and NT, which are the diameters of the cube. Wherefore those diameters shall concur and cut one the other in one and the self same point, let the same be O. Wherefore the right lines OK, OE, HEY, OM, OH, OZ, OT, and ON, shall be equal the on● to the other, for that they are the halves of the diameters of equal and like rectangle parallelograms. Wherefore making the centre the point O, and the space any of these lines OE, or OK. etc. a Sphere described, shall pass by every one of the angles of the cube, namely, which are at the points E, Z, I, T, K, H, M, N, by the 12. definition of the eleventh, for that all the lines drawn from the point O to the angles of the cube are equal. But the right line EI containeth in power the two equal right lines EZ, and ZI, by the 47. of the first. Wherefore the square of the line EI is double to the square of the line ZI. And forasmuch as the right line KI subtendeth the right angle KEI (for that the right line KE● is erected perpendicularly to the plai●e superficies of the right lines EZ and ZT (by the 4. of the eleventh) ● therefore the square of the line KI is equal to the squares of the lines EI and EK, but the square of the line EI is double to the square of the line EK (for it is double to the square of the line ZI, as hath been proved, and the bases of the cube are equal squares). Wherefore the square of the line KI is triple to the square of the line KE, that is, to the square of the line ZI. But the right line ZI is equal to th● right line DB, by position, unto whose square the square of the diameter AB is triple, by that which was demonstrated in the 13. Proposition of this book. Wherefore the diameters KI & DB are equal. Wherefore there is described a cube KI, and it is comprehended in the Sphere given wherein the other solids were contained, the diameter of which Sphere is the line AB. And the diameter KI or AB of the same Sphere, is proved to be in power triple to the side of the cube, namely, to the line DB, or ZI.   

¶ Corollaryes added by Flussas.  

First Corollary.  
Hereby it is manifest, that the diameter of a Sphere containeth in power the sides both of a pyramid and of a cube inscribed in it.  
For the power of the side of the pyramid is two thirds of the power of the diameter (by the 13. of this book). And the power of the side of the cube is, by this Proposition, one third of the power of the said diameter. Wherefore the diameter of the Sphere containeth in power the sides of the pyramid and of the cube ..   

¶ Second Corollary.  
All the diameters of a cube cut the one the other into two equal parts in the centre of the sphere which containeth the cube. And moreover those diameters do in the self same point cut into two equal parts the right lines which join together the centres of the opposite bases.  
As it is manifest to see by the right line LOF. For the angles LKO, and FIO, are equal, by the 29. of the first: and it is proved, that they are contained under equal sides: Wherefore (by the 4. of the first) the bases LO and FO are equal. In like sort may be proved, that the rest of the right lines which join together the centres of the opposite bases do cut the one the other into two equal parts in the centre O.     

¶ The 4. Problem. The 16. Proposition. To make an Icosahedron, and to comprehend it in the Sphere given, wherein were contained the former solids, and to prove that the side of the Icosahedron is an irrational line of that kind which is called a less line.  

Second part of the demonstration.  Now forasmuch as the lines QP, QV, QT, QS, and QR do each subtend right angles contained under the sides of an equilater hexagon & of an equilater decagon inscribed in the circle PRSTV or in the circle EFGHK (which two circles are equal) therefore the said lines are each equal to the side of the pentagon inscribed in the foresaid circle by the 10. of this book, and are equal the one to the other, by the 4. of the first, (for all the angles at the point W which they subtend are right angles). Wherefore the five triangles QPV, QPR, QRS, QST, and QTV, which are contained under the said lines QV, QP, QR, QS, QT, and under the sides of the pentagon VPRST, are equilater, and equal to the ten former triangles. And by the same reason the five triangles opposite unto them, namely, the triangles YML, YMN, YNX, YXO, and YOL, are equilater and equal to the said ten triangles. For the lines YL, YM, YN, YX, and YO do subtend right angles contained under the sides. of an equilater hexagon and of an equilater decagon inscribed in the circle EFGHK, which is equal to the circle PRSTV. Wherefore there is described a solid contained under 20. equilater triangles. Wherefore by the last definition of the eleventh there is described an Icosahedron.  
Now it is required to comprehend it in the sphere given, and to prove that the side of the Icosahedron is an irrational line of that kind which is called a less line. Forasmuch as the line ZW is the side of an hexagon, & the line WQ is the side of a decagon, therefore the line ZQ is divided by an extreme and mean proportion in the point W, and his greater segment is ZW (by the 9 of the thirteenth). Wherefore as the line QZ is to the line ZW, so is the line ZW to the line WQ. But the ZW is equal to the line ZL by construction, and the line WQ to the line ZY by construction also: Wherefore as the line QZ is to the line ZL, so is the line ZL to the line ZY, and the angles QZL● and LZY are right angles (by the 2. definition of the eleventh): If therefore we draw a right line from the point L to the point Q, the angle YLQ shallbe a right angle, by reason of the likeness of the triangles YLQ and ZLQ (by the 8. of the sixth). Wherefore a semicircle described upon the line QY, shall pass also by the point L (by the assumpts added by Campane after the 13. of this book). And by the same reason also, for that as the line QZ is the line ZW, so is the line ZW to the line WQ, 

For the line QW, is equal to the line IZ, & the line ZW, is common to them both.  
This part is again afterward demonstrated by Flussas.   but the line ZQ is equal to the line YW, and the line ZW to the line PW: wherefore as the line YW is to the line WP, so is the line PW to the line WQ. And therefore again if we draw a right line from the point P to the point Y, the angle YPQ shallbe a right angle. Wherefore a semicircle described upon the line QY shall pass also by the point P, by the former assumpts: & if the diameter QY abiding fixed the semicircle be turned round about, until it come to the self same place from whence it began first to be moved, it shall pass both by the point P, and also by the rest of the points of the angles of the Icosahedron, and the Icosahedron shallbe comprehended in a sphere. I say also that it is contained in the sphere given.  
Divide (by the 10. of the first) the line ZW into two equal parts in the point a. And forasmuch as the right line ZQ is divided by an extreme and mean proportion in the point W, and his less segment is QW, therefore the segment QW having added unto it the half of the greater segment, namely, the line Wa, is (by the 3. of this book) in power quintuple to the square made of the half of the greater segment: wherefore the square of the line Qa is quintuple to the square of the line ●W. But unto the square of the Qa, the square of the line QY is quadruple (by the corollary of the 20. of the sixth) for the line QY is double to the line Qa: and by the same reason unto the square of the WA the square of the line ZW is quadruple: Wherefore the square of the line QY is quintuple to the square of the line ZW (by the 15. of the fifth). And forasmuch as the line AC, is quadruple to the line CB, therefore the line AB is quintuple to the line CB. But as the line AB is to the line BC, so is the square of the line AB to the square of the line BD (by the 8 of the sixth, and corollary of the 20. of the same). Wherefore the square of the line AB is quintuple to the square of the line BD. And it is is proved that the square of the line QY is quintuple to the square of the line ZW. And the line BD is equal to the line ZW, for either of them is by position equal to the line which is drawn from the centre of the circle EFGHK to the circumference. Wherefore the line AB is equal to the YQ. But the line AB is the diameter of the sphere given: Wherefore the line YQ, which is proved to be the diameter of the sphere containing the Icosahedron, is equal to the diameter of the sphere given. Wherefore the Icosahedron is contained in the sphere given. Now I say that the side of the Icosahedron is an irrational line of that kind which is called a less line. For forasmuch as the diameter of the sphere is rational, and is in power quintuple to the square of the line drawn from the centre of the circle OLMNX: wherefore also the line which is drawn from the centre of the circle OLMNX is rational: wherefore the diameter also being commensurable to the same line (by the 6. of the tenth) is rational. But if in a circle having a rational line to his diameter be described an equilater pentagon, the side of the pentagon is (by the 11. of this book) an irrational line, of that kind which is called a less line. But the side of the pentagon OLMNX is also the side of the Icosahedron described, as hath before been proved. Wherefore the side of the Icosahedron is an irrational line of that kind which is called a less line. Wherefore there is described an Icosahedron and it is contained in the sphere given, and it is proved that the side of the, Icosahedron is an irrational line of that kind which is called a less line. Which was required to be done, and to be proved.  

A Corollary.  
Hereby it is manifest that the diameter of the sphere, is in power quintuple to the line which is drawn from the centre of the circle to the circumference, on which the Icosahedron is described. And that the diameter of the sphere: is composed of the side of an hexagon, and of two sides of a decagon described in one and the self same circle.  
Flussas' proveth the Icosahedron described, to be contained in a sphere, by drawing right lines from the point, a, to the points P and G after this manner.  
Forasmuch as the lines ZW, WP are put equal to the line drawn from the centre to the circumference● and the line drawn from the centre to the circumference is double to the line awe, by construction: therefore the line WP is also double to the same line awe. Wherefore the square of the line WP is quadruple to the square of the line awe (by the corollary of the 20. of the sixth). And those lines PW and Wa containing a right angle PWa (as hath before been proved) are subtended of the right line aP. Wherefore (by the 47. of the first) the line aP containeth in power the lines PW, and Wa. Wherefore the right line aP is in power quintuple to the line Wa. Wherefore the right lines aP, and aQ being quintuple to one and the same line Wa, are (by the 9 of the flueth) equal. In like sort also may we prove that unto those lines aP and aQ, are equal the rest of the lines drawn from the point a to the rest of the angles R, S, T, V For they subtend right angles contained of the line Wa, and of the lines drawn from the centre to the circumference. And forasmuch as unto the line Wa is equal the line Valerio, which is likewise erected perpendicularly unto the other plain superficies OLMNX: therefore lines drawn from the point a to the angles O, L, M, N, X, and subtending right angles at the point Z contained under lines drawn from the centre to the circumference, and under the line aZ, are equal not only the one to the other, but also to the lines, drawn from the said point a, to the former angles at the points, P, R, S, T, V●. For the lines drawn from the centre to the circumference of each circle are equal, & the line awe is equal to the line aZ. But the line aP is proved equal to the line aQ, which is the half of the whole QY. Therefore the residue aye is equal to the foresaid lines aP, aQ etc. Wherefore making the centre the point a, and the space one of those lines aQ, aP, etc. extend the superficies of a sphere, & it shall touch the 12. angles of the Icosahedron, which are at the points O, L, M. N, X, P, K, S, T, V, Q, Y: which sphere is described, if upon the diameter QY, be drawn a semicircle, and the said semicircle be moved about, till it return unto the same place from whence it began first to be moved.   

¶ A Corollary added by Flussas.  
The opposite sides of an Icosahedron are parallels. For the diameters of the sphere do fall upon the opposite angles of the Icosahedron: as it was manifest by the right line QY. If therefore there be imagined to be drawn the two diameters PN, and OM they shall concur in the point F: wherefore the right lines which join them together, PV, and LN, are in one and the self same plain superficies, by the 2. of the eleventh. And forasmuch as the alternate angles at the ends of the diameters are equal (by the 8. of the first): for the triangles contained under equal semidiamete●s and the side of the Icosahedron are equiangle: therefore (by the 28. of the first) the lines PV and LN are paralles.    

¶ The 5. Problem. The 17. Proposition. To make a Dodecahedron, and to comprehend it in the sphere given, wherein were comprehended the foresaid solids: and to prove that the side of the dodecahedron is an irrational line of that kind which is called a residual line.  

The pentagon VBWCR, proved to be in one and the self same plain superficies.  Now I say that it is in one and the self same plain superficies. Forasmuch as the line ZV is a parallel to the line SR (as was before proved) but unto the same line SR, is the line CB a parallel (by the 28. of the first). Wherefore (by the 9 of the eleventh) the line VZ is a parallel to the line CB. Wherefore, by the seventh of the eleventh, the right lines which join them together are in the self same plain wherein are the parallel lines. Wherefore the Trapesium BVZC is in one plain. And the triangle BWC is in one plain (by the 2. of the eleventh). Now to prove that the Trapesium BVZC & the triangle BWC are in one and the self same plain, we must prove that the right lines YH, and HW are made directly one right line: which thing is thus proved. Forasmuch as the line HP is divided by an extreme and mean proportion in the point T, and his greater segment is the line PT, therefore as the line HP is to the line PT, so is the line PT to the line TH. But the line HP is equal to the line HO, and the line PT to either of these lines TW and OY. Wherefore as the line HO is to the line OY, so is the line WT to the line TH. But the lines HO and TW being sides of like proportion are parallels (by the 6. of the eleventh): (For either of them is erected perpendicularly to the plain superficies BD) ● and the lines TH and OY are parallels, which are also sides of like proportion, by the same 6. of the eleventh, (For either of them is also erected perpendicularly to the plain superficies BF.) But when there are two triangles, having two sides proportional to two sides, so set upon one angle, that their sides of like proportion are also parallels (as the triangles YOH and HTW are) ● whose two sides, OH & HT, being in the two bases of the cube making an angle at the point H, the sides remaining of those triangles shall (by the 32. of the sixth) be in one right line. Wherefore the lines YH & HW make both one right line. But every right line is (by the 3. of the eleventh) in one & the self same plain superficies. Wherefore if ye draw a right line from B to Y, there shallbe made a triangle BWY, which shallbe in one and the self same plain (by the 2. of the eleventh). And therefore the whole pentagon figure VBWCZ is in one and the self same plain superficies.  
Now also I say that it is equiangle. 
The pentagon VBWCZ, it proved equiangle.  For forasmuch as the right line NO is divided by an extreme and mean proportion in the point R, and his greater segment is OR, therefore as both the lines NO and OR added together is to the line ON, so (by the 5. of this book) is the line ON to the line OR. But the line OR is equal to the line OS. Wherefore as the line SN is to the line NO, so is the line NO to the line OS. Wherefore the line SN is divided by an extreme and mean proportion in the point O, and his greater segment is the line NO. Wherefore the squares of the lines NS and SO are triple to the square of the line NO (by the 4. of this book). But the line NO is equal to the NB, and the line SO to the line SZ: wherefore the squares of the lines NS and ZS are triple to the square of the line NB: wherefore the squares of the lines ZS, SN and NB, are quadruple to the square of the line NB. But unto the squares of the lines SN & NB (by the 47. of the first) is equal the square of the line SB: wherefore the squares of the lines BS and SZ, that is, the square of the line BZ, by the 47. of the first, (for the angle ZSB is a right angle by position) is quadruple to the square of the line NB. Wherefore the line BZ is double to the line BN (by the Corollary of the 20. of the sixth). But the line BC is also double to the line BN. Wherefore the line BZ is equal to the line BC. Now forasmuch as these two lines BV and VZ are equal to these two lines BW and WC, and the base BZ is equal to the base BC, therefore (by the 8. of the first) the angle BVZ is equal to the angle BWC. And in like sort (by the 8. of the first) may we prove that the angle VZC is equal to the angle BWC (proving first that the lines CB and CV are equal: which are proved equal by this, that the line NS is equal to the line XR, and therefore the line CR is equal to the line BS, by the 47. of the first: wherefore also by the same the line CV is equal to the line BZ, that is, to the line BC (for the lines BC & BZ are proved equal.) Wherefore the three angles BWC, BVZ, and VZC are equal the one to the other. But if in an equilater pentagon figure there be three angles equal the one to the other, the pentagon is (by the 7. of the thirteenth) equiangle: wherefore the pentagon BVZCW is equiangle. And it is also proved that it is equilater. Wherefore the pentagon BVZCW is both equilater & equiangle. And it is made upon one of the sides of the cube, namely, upon BC. 
* Look for a farther construction after Flussas at the end of the demonstration.  If therefore upon every one of the twelve sides of the cube be used the like construction, there shall then be made a dodecahedron contained under twelve pentagons' equilater and equiangle.  
Now I say, that the side of the dodecahedron is an irrational line of that kind which is called a residual line. 
That the side of the dodecahedron is a residual line.  For forasmuch as the line NO is divided by an extreme and mean proportion in the point R, and his greater segment is the line OR, and the line OX is also divided by an extreme and mean proportion in the point S, and his greater segment is the line OS. Wherefore the whole line NX is divided by an extreme and mean proportion, and his greater segment is the line RS. (For for that as the line ON is to the line OR, so is the line OR to the line NR, and in the same proportion also are their doubles (for the parts of equemultiplices have one and the self same proportion with the whole, by the 15. of the fifth). Wherefore as the line NX is to the line RS, so is the line RS to both the lines NR and SX added together. But the line NX is greater than the line RS, by both the lines NR and SX added together. Wherefore the line NX is divided by an extreme and mean proportion, and his greater segment is the line RS. But the line RS is equal to the line VZ, as hath before been proved. Wherefore the line NX is divided by an extreme and mean proportion, and his greater segment is the line VZ. And forasmuch as the diameter of the Sphere is rational, and is in power triple to the side of the cube, by the 15. of this book, therefore the line NX, being the side of the cube, is rational. But if a rational line be divided by an extreme and mean proportion, either of the segments is (by the 6. of this book) an irrational line of that kind which is called a residual line. Wherefore the line VZ being the side of the dodecahedron, is an irrational line of that kind which is called a residual line. Wherefore there is made a dodecahedron, and it is comprehended in the Sphere given, wherein the other solids were contained, and it is proved that the side of the dodecahedron is a residual line: which was required to be done, and also to be proved.  

¶ Corollary.  
Hereby it is manifest, that the side of a cube being divided by an extreme and mean proportion, the greater segment thereof is the side of the dodecahedron. As it was manifest by the line VZ which was proved to be the greater segment of the right line NX, namely, of the side of the cube.  

A further construction of the dodecahedron after Flussas.  
Forasmuch as it hath been proved that the pentagon BVZCW is equilater and equiangle and toucheth one of the sides of the cube. 
Draw in the former figure these lines, ctA, ctL, c●D.  Let us show also by what means upon each of the 12. sides of the cube may in like sort be applied pentagons joining one to the other, and composing the 12. bases of the dodecahedron. Draw in the former figure these right lines AI, ID, IL, ctK. Now forasmuch as the line PL was in the point ct divided like unto the lines PH, ON. or OX, and upon the points T, P, ct, were erected perpendicular lines equal unto the line OY, and the rest: namely, unto the greater segment: and the lines T W and ct I were proved parallels, therefore the lines WI and Tct are parallels, by the 7. of the eleventh, and 33. of the first. Wherefore also, by the 9 of the eleventh, the lines WI and DC are parallels. Wherefore by the 7. of the eleventh CWID is a plain superficies. And the triangle AID is a plain superficies, by the 2. of the eleventh. Now it is manifest that the right lines ID, & IA are equal to the right line WC. For the right lines AL & ●ct (which are equal to the right lines BH, & HT) do make the subtended lines A ct and BT equal by the 4. of the first. And again forasmuch as the lines BT and TW contain a right angle BTW, as also do the right lines Act and ctI contain the right angle ActI (for the right lines WT, and Ict are erected perpendicularly unto one and the self same plain ABCD by supposition). And the squares of the lines BT and TW are equal to the squares of the lines Act, and ctI (for it is proved that the line BT is equal to the line Act, and the line TW to the line ctI). And unto the squares of the lines BT and TW is equal the square of the line BW, by the 47. of the first: likewise by the same unto the squares of the lines Act and ctI is equal the square of the line AI. Wherefore the square of the line BW is equal to the square of the line AI, wherefore also the line BT is equal to the line AI. And by the same reason are the lines ID and WC equal to the same lines. Now forasmuch as the lines AI and ID, and the lines AL and LD are equal, and the base IL is common to them both, the angles ALI and DLI shallbe equal, by the 8. of the first: and therefore they are right angles, by the 10. definition of the first. And by the same reason are the angles WHB, and WHC right angles. And forasmuch as the two lines HT and TW are equal to the two lines Lct and ctI, and they contain equal angles, that is, right angles by supposition, therefore the angles WHT, and ILct, are equal by the 4. of the first. Wherefore the plain superficies AID is in like sort inclined to the plain superficies ABCD, as the plain superficies BWC is inclined to the same plain ABCD, by the 4. definition of the eleventh. In like sort may we prove that the plain WCDI is in like sort inclined to the plain ABCD, as the plain BVZC is to the plain EBCF. For that in the triangles YOH and ctPK which consist of equal sides (each to his correspondent side), the angles YHO, and ctKP, which are the angles of the inclination, are equal. And now if the right line ctK be extended to the point a, and the pentagon CWIDa be made perfect, we may, by the same reason, prove that that plain is equiangle and equilater, that we proved the pentagon BVZCW to be equaliter and equiangle. And likewise if the other plains BWIA and AID be made perfect, they may be proved to be equal and like pentagons and in like sort situate, and they are set upon these common right lines BW, WC, WI, AI, and ID. And observing this method, there shall upon every one of the 12. ●id●s of the cube be set every one of the 12. pentagons which compose the dodecahedron.    

¶ Certain Corollaryes added by Flussas.  

First Corollary.  
The side of a cube, is equal to the right line which subtendeth the angle of the pentagon of a dodecahedron contained in one and the self same sphere with the cube.  
For the angles BWC and AID, are subtended of the lines BC and AD. Which are sides of the Cube●   

¶ Second Corollary.  
In a dodecahedron there are six sides every two of which are parallels and opposite, whose sections into two equal parts, are coupled by three right lines, which in the centre of the sphere which containeth the dodecahedron, divide into two equal parts and perpendicularly both themselves and also the sides.  
For upon the six bases of the cube are set six sides of the dodecahedron, as it hath been proved (by the lines ZV, WI etc.) which are cut into two equal parts by right lines, which join together the centres of the bases of the cube, as the line YO produced, and the other like. Which lines coupling together the centres of the bases are three in number, cutting the one the other perpendicularly (for they are parallels to the sides of the cube) and they cut the one the other into two equal parts in the centre of the sphere which containeth the cube (by that which was demonstrated in the 15. of this book). And unto these equal lines, joining together the centres of the bases of the cube, are without the bases added equal parts OY, P ct, and the other like, which by supposition are equal to half of the side of the dodecahedron. Wherefore the whole lines, which join together the sectoins of the opposite sides of the dodecahedron, are equal, and they cut those sides into two equal parts and perpendicularly.   

Third Corollary.  
A right line joining together the points of the sections of the opposite sides of the dodecahedron into two equal parts, being divided by an extreme and mean proportion: the greater segment thereof shallbe the side of the cube, and the less segment the side of the dodecahedron contained in the self same sphere.  
For it was proved that the right line YQ is divided by an extreme and mean proportion in the point O, and that his greater segment OQ is half the side of the cube, and his less segment OY is half of the side VZ (which is the side of the dodecahedron). Wherefore it followeth (by the 15. of the fifth) that their doubles are in the same proportion. Wherefore the double of the line YQ which joineth the point opposite unto the line Y, is the whole: and the greater segment is the double of the line OQ which is the side of the cube: & the less segment is the double of the line YO, which is equal to the side of the dodecahedron, namely, to the side VZ.     

¶ The 6. Problem. The 18. Proposition. To find out the sides of the foresaid five bodies, and to compare them together.  
TAke the diameter of the Sphere given, and let the same be AB, and divide it in the point C, so that let the line AC be equal to CB, by the 10. of the first: and in the point D, so that let AD be double to DB, by the 9 of the sixth. And upon the line AB describe a semicircle AEB. And from the points C and D, raise up (by the 11. of the first) unto the line AB perpendicular lines CE and DF. And draw these right lines AF, FB, and BE. Now forasmuch as the line AD is double to the line DB, therefore the line AB is triple to the line DB. Wherefore the line BA is sesquialter to the line AD (for it is as 3. to 2.). But as the line BA is to the line AD, so is the square of the line BA to the square of the line AF (by the 6. of the sixth, or by the Corollary of the same, and by the Corollary of the 20. of the same): for the triangle AFB is equiangle to the triangle AFD. Wherefore the square of the line BA is sesquialter to the square of the line AF. But the diameter of a sphere is in power sesquialter to the sid● of the pyramid, 
The side of a pyramid.  by the 13. of this book, and the line AB is the diameter of the sphere. Wherefore the line AF is equal to the side of the pyramid.  
Again forasmuch as the line AB is triple to the line BD: but as the line AB is to the line BD, so is the square of the line AB to the square of the line FB, by the Corollaries of the 8. and 20. of the sixth. Wherefore the square of the line AB is triple to the square of the line FB. But the diameter o● a sphere is in power triple to the side of the cube (by the 15. of this book) and the diameter of the sphere is the line AB. Wherefore the line BF is the side of the cube. 
The side of a cube.   
And forasmuch as the line FB is the side of a cube, let it be divided by an extreme and mean proportion in the point N, and let the greater segment thereof be NB. Wherefore the line NB is the side of a Dodecahedron, 
The sides of a dodecahedron.  by the Corollary of the 17. of this book.  

Comparison of the five sides of the foresaid bodies.  And forasmuch as it hath been proved, by the 13. of this book, that the diameter of the sphere is in power sesquialter to AF the side of the pyramid, and is in power double to BE the side of the octohedron, by the 14. of the same, and is in power triple to FB the side of the cube, by the 15. of the same. Wherefore it followeth, that of what parts the diameter of the sphere containeth six, of such parts the side of the pyramid containeth four: and the side of the octohedron three: and the side of the cube two. Wherefore the side of the pyramid is in power to the side of the octohedron in sesquitertia proportion: and is in power to the side of the cube in double proportion. And the side of the octohedron is in power to the side of the cube in sesquialtera proportion. Wherefore the foresaid sides of the three figures, that is, of the pyramid, of the octohedron, and o● the cube, are the one to the other in rational proportions. Wherefore they are rational. But the other two sides, namely, the sides of the Icosahedron and of the Dodecahedron, are neither the one to the other, nor also to the foresaid sides, in rational proportions: ●or they are irrational lines, namely, a less line, and a residual line.  
another way to prove that the line MB is greater than the line NB. 
another demommonstration to prove that the side of the Icosahedron is greater than the side of the dodecahedron.  Forasmuch as the line AD is double to the line DB, therefore the line AB is triple to the line DB. But as AB is to BD, so is the square of the line AB to the square of the line BF, by the 8. of the sixth (for the triangle FAB is equiangle to the triangle FDB). Wherefore the square of the line AB is triple to the square of the line BF. And it is before proved, that the square of the line AB is quintuple to the square of the line KL. Wherefore five squares made of the line KL, are equal to three squares made of the line FB. But three squares made of the line FB, are greater than six squares made of the line NB, as is strait way proved. Wherefore five squares made of the line KL, are greater than six squares made of the line NB. Wherefore also one square made of the line KL, is greater than one square made of the line NB. Wherefore the line KL is greater than the line NB. But the line KL is equal to the line LM. Wherefore the line LM is greater than the line NB. Wherefore the line MB is much greater than the line NB: which was required to be proved.  
But now let us prove that three squares made of the line FB, 
That 3. squares of the line FB are greater than 6. squares of the line NB.  are greater than six squares made of the line NB. Forasmuch as the line BN is greater than the line NF, for it is the greater segment of the line BF divided by an extreme and mean proportion, therefore that which is contained under the lines BF, and BN, is greater than that which is contained under the lines BF and FN, by the 1. of the sixth. Wherefore that which is contained under the lines BF and BN, together with that which is contained under the lines BF and FN, is greater than that which is contained under the lines BF and FN twice. But that which is contained under the lines BF and FN; together with that which is contained under the lines BF and BN, is the square of the line BF, by the 2. of the second, and that which is contained under the lines BF and FN once, is equal to the square of NB. For the line FB is divided by an extreme and mean proportion in the point N: (and (by the 17. of the sixth) that which is contained under the extremes, is equal to the square made of the middle line). Wherefore the square of the line FB, is greater than the double of the square of the line BN. Wherefore one of the squares made of the line BF, is greater than two squares made of the line BN. Wherefore also three squares made of the line FB, are greater than six squares made of the line BN: which was required to be proved.  

A Corollary.  
Now also I say that besides the five foresaid solids there can not be described any other solid comprehended under figures equilater & equiangle the one to the other. 
That there can be no other solid besides these five, contained under equilater and equiangle bases.  For of two triangles, or of any two other plain superficieces can not be made a solid angle (for, that is contrary to the definition of a solid angle). Under three triangles is contained the solid angle of a pyramid: under four, the solid angle of an octohedron: under five, the solid angle of an Icosahedron: of six, equilater & equiangle triangles set to one point can not be made a solid angle. For forasmuch as the angle of an equilater triangle is two third parts of a right angle, the six angles of the solid shallbe equal to four right angles, which is impossible. For every solid angle is (by the 21. of the eleventh) contained under plain angles less than four right angles. And by the same reason can not be made a solid angle contained under more than six plain superficial angles of equilater triangles. Under three squares is contained the angle of a cube. Under four squares it is impossible that a solid angle should be contained: for then again it should be contained under four right angles. Wherefore much less can any solid angle be contained under more squares than four. Under three equilater and equiangle pentagons is contained the solid angle of a dodecahedron. But under four it is impossible. For forasmuch as the angle of a pentagon is a right angle and the fift part more of a right angle, the four angles shallbe greater than four right angles: which is impossible. And therefore much less can a solid angle be composed of more pentagons than four. Neither can a solid angle be contained under any other equilater and equiangle figures of many angles, for that that also should be absurd. For the more the sides increase, the greater are the angles which they contain, and therefore the farther of are the superficial angles contained of those sides from composing of a solid angle. Wherefore besides the foresaid five figures there can not be made any solid figure contained under equal sides and equal angles: which was required to be proved.   

An Assumpt.  
But now that the angle of an equilater and equiangle pentagon is a right angle and a fi●th par● more of a right angle, may thus be proved. Suppose that ABCDE be an equilater and equiangle pentagon. 
That the angle of an equilater and equiangle Pentagon is one right angle and a 〈◊〉 part 〈◊〉 which thing was also before proved in the 〈◊〉 of the 32. of the ●irst.  And (by the 14. of the fourth) describe about it a circle ABCDE. And take (by the 1. of the third) the centre thereof, and let the same be F. And draw these right lines FA, FB, FC, FD, FE. Wherefore those lines do divide the angles of the pentagon into two equal parts in the points A, B, C, D, E, by the 4. of the first. And forasmuch as the five angles that are at the point F a●e equal to four right angles, by the corollary of the 15. of the first, and they are equal the one to the other by the 8. of the first: therefore one of those angles, as ●or example sake, the angle AFB is a fi●th part less than a right angle. Wherefore the angles remaining, namely, FAB, & ABF, are one right angle and a fifth part over. But the angle FAB is equal to the angle FBC. Wherefore the whole angle ABC being one of the angles of the pentagon is a right angle and a fifth part more than a right angle: which was required to be proved.   

¶ A Corollary added by Flussas.  
Now, let us teach, how those five solids, have each like inclinations of their bases.  
●i●st let us take a Pyramid, and divide one of the sides thereof into two equal parts: and from the two angles opposite unto that side, d●aw perpendiculars, which shall fall upon the section, by the corollary of the 12. of the thirteenth, 
The sides of the angle of the incl●●●tion of the 〈◊〉 of the 〈◊〉 are 〈◊〉 rational.  and at the said point of division (as may easily be proved). Wherefore they shall contain the ang●e of the inclination of the plains, by the 4. definition of the eleventh, which angle is subtended of the opposite side of the pyramid. Now forasmuch as the rest of the angles of the inclination of the plains of the Pyramid, are contained under two perpendicular lines of the triangles, and are subtended of the side of the Pyramid, it followeth, by the 8. of the fir●t, that those angles are equal. Wher●fo●e (by the 5. definition of the eleventh) the superficieces are in like sort inclined the one to the other.  

The sides of the angle of the inclination of the 〈◊〉 ●f t●e 〈…〉.  One of the sides of a Cube being divided into two equal parts, if from the said section be drawn in two of the bases thereof, two perpendicular lines, they shallbe parallels and equal to the sides of the square which contain a right angle. And forasmuch as all the angles of the bases of the Cube are right angles: therefore those perpendiculars falling upon the section of the side common to the two bases, shall cont●yne a right angle (by the 10. of the eleventh): which self angle is the angle of inclination (by the 4. definition of the eleventh) and is subtended of the diameter of the base of the Cube. And by the same reason may we prove that the rest of the angles of the inclination of the bases of the cube are right angles. Wherefore the inclinations of the superficieces of the cube the one to the other are equal (by the 5. definition of the eleventh.  
In an Octohedron take the diameter which coupleth the two opposite angles. 
That the plains of an octohedron are in li●e sort inclined.  And from those opposite angles draw to one and the sel●e same side of the Octohedron, in two bases thereof, two perpendicular lines, which shall divide that side into two equal parts and perpendicularly (by the Corollary of the 12. of the thirteenth). Wherefore those perpendiculars shall contain the angle of the inclination of the bases (by the 4. definition of the eleventh): and the same angle is subtended of the diameter of the Octohedron. Wherefore the rest of the angles after the same manner described in the rest of the bases, being comprehended and subtended of equal sides, shall (by the 8. of the first) be equal the one to the other. And therefore the inclinations of the plains in the Octohedron, shall (by the 5. definition of the eleventh) be equal.  
In an Icosahedron let there be drawn from the angles of two of the bases, 
That the plains of an Icosahedron are in like sort inclined.  to one side common to both the said bases perpendiculars, which shall contain the angle of the inclination of the bases (by the 4. definition of the eleventh): which angle is subtended of the right line which subtendeth the angle of the pentagon which containeth five sides of the Icosahedron, by the 16. of this book: for it coupleth the two opposite angles of the triangles which are joined together. Wherefore the rest of the angles of the inclination of the bases being after the same manner found out, they shallbe contained under equal sides, and subtended of equal bases, and therefore (by the 8. of the fi●st) those angles shallbe equal. Wherefore also all the inclinations of the bases of the Icosahedron the one to the other shalb● equal, by the 5. definition of the eleventh.  
In a Dodecahedron, 
That the plains of a D●●●●●hedron are 〈◊〉 like sort inclined.  from the two opposite angles of two next pentagons draw to their common side perpendicular lines, passing by the centres of the said pentagons, which shall, where they fall, divide the side into two equal parts by the 3. of the third. (For the bases of a Dodecahedron are contained in a circle) And the angle contayn●d under those perpendicular lines is the inclination of those bases (by the 4. definition of the eleventh). And the foresaid opposite angles are coupled by a right line equal to the right line which coupleth the opposite sections into two equal parts of the sides of the dodecahedron (by the 33. of the first). For they couple together the half sides of the dodecahedron, which halves are parallels and equal, by the 3. corollary of the 17. of this book: which coupling lines also are equal, by the same corollary. Wherefore the angles being contained of equal perpendicular lines, and subtended of equal coupling lines, shall (by the 8. of the first) be equal. And they are the angles of the inclinations. Wherefore the bases of the dodecahedron are in like sort inclined the one to the other (by the 5. definition of the eleventh).  
Flussas after this teacheth how to know the rationality or irrationality of the sides of the triangles, which contain the angles of the inclinations of the superficieces of the foresaid bodies.  
In a Pyramid the angle of the inclination is contained under two perp●dicular lines of the triangles, 
The sides of the angle of the inclination of the supe●ficieces of the Tetrahedron are proved rational.  and is subtended of the side of the Pyramid Now the side of the pyramid is in power sesquitertia to the perpendicular line, by the corollary of the 12. of this book: and therefore the triangle contained of those perpendicular lines and the side of pyramid, hath his sides rational & commensurable in power the one to the other.  
Forasmuch as the two sides of a Cube (or right lines equal to them) subtended under the diameter of one of the bases, 
The sides of the angle of the inclination of the superficieces of the cube proved rational.  do make the angle of the inclination: and the diameter of the cube is in power sesquialter to the diameter of the base, which diameter of the base is in power double to the side (by the 47. of the first): therefore those lines are rational and commensurable in power.  
In an Octohedron, 
The sides of th● angle etc. of the octohedron proved rational.  whose two perpendiculars of the bases contain the angle of the inclination of the Octohedron, which angle also is subtended of the diameter of the Octohedron, the diameter is in power double to the side of the Octohedron, & the side is in power sequitertia to the perpendiclar line, by the 12. of this book: wherefore the diameter thereof is in power duple superbipartiens tertias to the perpendicular line. Wherefore also the diameter and the perpendicular line are rational and commensurable (by the 6. of the tenth.)  
As touching an Icosahedron, it was proved in the 16. of this book, that the side thereof is a less line, when the diameter of the sphere is rational. And forasmuch as the angle of the inclination of the bases thereof, is contained of the perpendicular lines of the triangles, and subtended of the right line which subtendeth the angle of the Pentagon which containeth five sides of the Icosahedron: and unto the perpendicular lines the side is commensurable (namely, is in power sesquitertia unto them, by the Corollary of the 12. of this book): therefore the perpendicular lines which contain the angles are irrational lines, namely, less lines (by the 105. of the tenth book.) And forasmuch as the diameter containeth in power both the side of the Icosahedron, and the line which subtendeth the foresaid angle, if from the power of the diameter which is rational, be taken away the power of the side of the Icosahedron which is irrational, it is manifest that the residue which is the power of the subtending line shallbe irrational. For if it should be rational, the number which measureth the whole power of the diameter, and the part taken away of the subtending line, should also, by the 4. common sentence of the seventh measure the residue, namely, the power of the side: 
The sides of the angle etc. of the Icosahedron proved irrational.  which is irrational for that it is a less line, which were absurd. Wherefore it is manifest that the right lines which compose the angle of the inclination of the bases of the Icosahedron are Irrational lines. For the subtending line hath to the line containing, a greater proportion, than the whole hath to the greater segment.  
The angle of the inclination of the bases of a dodecahedron, is contained under two perpendiculars of the bases of the dodecahedron, and is subtended of that right line, whose greater segment is the side of a Cube inscribed in the dodecahedron, which right line is equal to the line which coupleth the sections, into two equal parts, of the opposite sides of the dodecahedron. And this coupling line we say is an irrational line, for that the diameter of the sphere containeth in power both the coupling line, and the side of the dodecahedron: but the side of the dodecahedron is an irrational line, namely, a residual line (by the 17. of this book). Wherefore the residue namely, the coupling line is an irrational line, as it is ●asy to prove by the 4. common sentence of the seventh. And that the perpendicular lines which contain the angle of the inclination are irrational, is thus proved.  
By the proportion of the subtending line (of the foresaid angles of inclination) to the lines which contain the angle, is found out the obliquity of the angle. 
How to know whether the angle of the inclination be a right angle, an acute angle, or an obliqne angle.  For if the subtending line be in power double to the line which containeth the angle, then is the angle a right angle (by the 48. of the first,) But if it be in power less than the double it is an acute angle (by the 23. of the second). But if it be in power more than the double, or have a greater proportion than the whole hath to the greater segment● the angle shallbe an obtuse angle (by the 12. of the second and 4. of the thirteenth). By which may be proved that the square of the whole is greater than the double of the square of the greater segment.  
This is to be noted that that which Flussas hath here taught touching the inclinations of the bases of the ●iue regular bodies, Hypsicles teacheth after the 5 proposition of the 15. book. Where he confesseth, that he received it of one Isidorus, and seeking to make the matter more clear, he endeavoured himself to declare, that the angles of the inclination of the solids are given, and that they are either acute or obtuse, according to the nature of the solid: although euclid in all his 15. books hath not yet showed, what a thing given is. Wherefore Flussas framing his demonstration upon an other ground proceedeth after an other manner, which seemeth more plain, and more aptly hereto be placed then there. Albeit the reader in that place shall not be frustrate of his also.    
The end of the thirteenth Book of Euclides Elements.   

¶ The fourteenth book of Euclides Elements.  
IN this book, which is commonly accounted the 14. book of Euclid is more at large entreated of our principal purpose: 
The argument of the fourteenth book.  namely, of the comparison and proportion of the five regular bodies (customably called the 5. figures or forms of Pythagoras) the one to the other, and also of their sides together, each to other: which things are of most secret use, and inestimable pleasure, and commodity to such as diligently search for them, and attain unto them. Which things also undoubtedly for the worthiness and hardness thereof (for things of most price are most hardest) were first searched, and found out of Philosophers, not of the inferior or mean sort, but of the deepest and most grounded Philosophers, and best exercised in Geometry. And albeit this book with the book following, namely, the 15. book, hath been hitherto of all men for the most part, and is also at this day numbered and accounted amongst Euclides books, and supposed to be two of his, namely, the 14. and 15. in order: as all exemplars (not only new and lately set abroad, but also old monuments written by hand) do manifestly witness: yet it is thought by the best learned in these days, that these two books are none of Euclides, but of some other author, no less worthy, nor of less estimation and authority, notwithstanding, than Euclid. Apollonius a man of deep knowledge a great Philosopher and in Geometry marvelous (whose wonderful books written of the sections of cones, which exercise & occupy thewittes of the wisest and best learned, are yet remaining) is thought, and that not without just cause, to be the author of them, or as some think Hypsicles himself. For what can be more plainly, then that which he himself witnesseth in the preface of this book. Basilides of Tire (saith Hypsicles) and my father together, scanning, and peysing a writing or books of Apollonius, which was of the comparison of a dodecahedron to an Icosahedron inscribed in one and the self same sphere, and what proportion these figures had the one to the other, found that Apollonius had failed in this matter. But afterward (saith he) I found an other copy or book of Apollonius, wherein the demonstration of that matter was full and perfect, and showed it unto them, whereat they much rejoiced. By which words it seemeth to be manifest that Apollonius was the first author of this book, which was afterward set forth by Hypsicles. For so his own words after in the same preface seem to import.  

The Preface of Hypsicles before the fourteenth book.  
Friend Protarchus, when that Basilides of Tire came into Alexandria, having familiar friendship with my father by reason of his knowledge in the mathematical sciences, he remained with him a long time, yea even all the time of the pestilence. And sometime reasoning between themselves of that which Apollonius had written touching the comparison of a dodecahedron and of an Icosahedron inscribed in one and the self same sphere, what proportion such bodies have the one to the other, they judged that Apollonius had somewhat erred therein. Wherefore they (as my father declared unto me) diligently weighing it, wrote it perfectly. Howbeit afterward I happened to find an other book written of Apollonius, which contained in it the right demonstration of that which they sought for: which when they saw, they much rejoiced. As for that which Apollonius wrote, may be seen of all men, for that it is in ●uery man's hand. And that which was of us more diligently afterward written again, I thought good to sand and dedicated unto you, as to one whom I thought worthy commendation, both for that deep knowledge which I know you have in all kinds of learning, and chiefly in Geometry, so that you are able readily to judge of those things which are spoken, and also for the great love and good will which you bear towards my father and me. Wherefore vouchsafe gently to accept this, which I sand unto you. But now is it time to end our preface, and to begin the matter.  

¶ The 1. Theorem. The 1. Proposition. 
First proposition after Flussas.  A perpendicular line drawn from the centre of a circle to the side of a Pentagon described in the same circle: is the half of these two lines, namely, of the side of an hexagon figure, and of the side of a decagon figure being both described in the self same circle.  
SVppose that the circle be ABC. 
Construction.  And let the side of an equilater Pentagon described in the circle ABC, be BC. And (by the 1. of the third) take the centre of the circle, and let the same be D. And (by the 12. of the first) from the point D draw unto the line BC a perpendicular line DE. And extend the right line DE directly to the point F. Then I say, that the line DE (which is drawn from the centre to BC the side of the pentagon) is the half of the sides of an hexagon and of a decagon taken together and described in the same circle. Draw these right lines DC and CF. And unto the line EF put an equal line GOE And draw a right line from the point G to the point C. 
Demonstration.  Now forasmuch as the circumference of the whole circle is quintuple to the circumference BFC (which is subtended of the side of the pentagon) and the circumference ACF is the half of the circumference of the whole circle, and the circumference CF (which is subtended of the side of the decagon) is the half of the circumference BCF: therefore the circumference ACF is quintuple to the circumference CF (by the 15. of the ●i●t). Wherefore the circumference AC is qradruple to the circumference FC. But as the circumference AC is to the circumference FC, so is the angle ADC to the angle FDC, by the last of the sixth. Wherefore the angle ADC is quadruple to the angle FDC. But the angle ADC is double to the angle EFC, by the 20. of the third: Wherefore the angle EFC is double to the angle GDC. But the angle EFC is equal to the angle EGC, by the 4. of the first. Wherefore the angle EGC is double to the angle EDC. Wherefore the line DG is equal to the line GC (by the 32. and 6. of the first). But the line GC is equal to the line CF, by the 4. of the first. Wherefore the line DG is equal to the line CF. And the line GE is equal to the line EF (by construction). Wherefore the line DE is equal to the lines EF and FC added together. Unto the lines EF and FC add the line DE. Wherefore the lines DF and FC added together, are double to the line DE. But the line DF is equal to the side of the hexagon: and FC to the side of the decagon. Wherefore the line DE is the half of the side of the hexagon, and of the side of the decagon being both added together and described in one and the self same circle.  
It is manifest 
* This is manifest by the 12. proposition of the thirtenh book as Campane well gathereth in a Corollary of the same.  by the Propositions of the thirteenth book, that a perpendicular line drawn from the centre of a circle to the side of an equilater triangle described in the same circle; is half of the semidiameter of the circle. Wherefore by this Proposition, a perpendicular drawn from the c●ntre of a circle to the side of a Pentagon, is equal to the perpendicular drawn from the centre to the side of the triangle, ●nd to half of the side of the decagon described in the same circle.   

¶ The 2. Theorem. The 2. Proposition. One and the self same circle comprehendeth both the Pentagon of a Dodecahedron, and the triangle of an Icosahedron, 
The 4. p●pos●tiō after Flussas.  described in one and the self same Sphere.  
THis Theorem is described of Aristeus in that book whose title is, The comparison of the five figures, and is described of Apollonius in his second edition of the comparison of a Dodecahedron to an Icosahedron, which is, 
* This is afterward proved in the 4. proposition.  that as the superficies of a Dodecahedron is to the superficies of an Icosahedron, so is the Dodecahedron to an Icosahedron, for that a perpendicular line drawn from the centre of a sphere to the pentagon of a dodecahedron and to the triangle of an Icosahedron is one and the self same. Now must we also prove that one and the self same circle comprehendeth both the pentagon of a Dodecahedron, and also the triangle of an Icosahedron described in one and the self same sphere, first this being proved: 
This Assumpt is the 3. proposition after Flussas.  If in a circle be described an equilater pentagon, the squares which are made of the side of the pentagon, and of that right line which is subtended under two sides of the pentagon, are quintuple to the square of the semidiameter o● the circle. Suppose that ABC be a circle. 
Construction of the Assumpt.  And let the side of a pentagon in the circle ABC, be AC. And take (by the 1. of the third) the centre of the circle and let the same be D. And (by the 12. of the first) from the point D draw unto the line AC a perpendicular line DF. And extend the line DF on either side to the points B and E. And draw a right line from the point A to the point B. Now I say, that the squares of the lines BA and AC are quintuple to the square of the line DE. Draw a right line from the point A to the point E. Wherefore the line AE is the side of a decagon figure. And forasmuch as the line BE is double to th● line DE: 
Demonstration of the Assumpt.  therefore the square of the line BE is quadruple to the square of DE (by the 20. of the sixth). But unto the square of the line BE, are equal the squares of the lines BA and AE (by the 47. of the first, for the angle BAE is a right angle, by the 31. of the third). Wherefore the squares of the lines BA and AE, are quadruple to the square of the line DE. Wherefore the squares of the lines AB, AE, and DE, are quintuple to the square of the line DE. But the squares of the lines DE and AE, are equal to the square of the line AC (by the 10. of the thirteenth). Wherefore the squares of the lines BA and AC, are quintuple to the square of the line DE.  
This being thus proved, now is to be demonstrated that one and the self same circle comprehendeth both the pentagon of a dodecahedron, & the triangle of an Icosahedron described in one & the self same circled. 
Construction of the proposition.  Take the diameter of the sphere, & ● let the same be AB. And in the same sphere describe a dodecahedron, & also an Icosahedron. And let one of the pentagons of the dodecahedron be CDEFG, & let one of the triangles of the Icosahedron be KLH. Now I say that the semidiameters of the circles which are described about them are equal, that is, that one and the self same circle containeth both the pentagon CDEFG, and the triangle KLH. Draw a right line from the point D to the point G. Wherefore the line DG is the side of a cube (by the corollary of the 17. of the thirteenth). Take a certain right line MN. And let the square of the line AB be quintuple to the square of the line MN, by the assumpt put after the 6. proposition of the tenth. But the diameter of a sphere is in power quintuple to the square of the semidiameter of the circle, on which is described the Icosahedron (by the corollary of the 16. of the thirteenth). Wherefore the line MN is the semidiameter of the circle on which is described the Icosahedron. Divide (by the 30. of the sixth) the line MN by an extreme and mean proportion in the point X. And let the greater segment thereof be MX. Wherefore the line MX is the side of a decagon described in the same circle (by the corollary of the 9 of the thirteenth). 
Demonstration of the proposition.  And forasmuch as the square of the line AB is quintuple to the square of the line MN: But the square of the line BA is triple to the square of the line DG (by the corollary of the 15, of the thirteenth). Wherefore three squares of the line DG are equal to five squares of the line MN. 
*  But as three squares of the line DG are to ●iue squares of the line MN, so are three squares of the line CG to five squares of the line MX. Wherefore three squares of the line CG are equll to five squares of the line MX. But five squares of the line CG are equal to ●iue squares of the line MN & to five squares of the MX. For (by the 10. of the thirteenth) one square of the line CG is equal to one square of the line MN & to one square of the line MX. Wherefore five squares of the line CG are equal to three squares of the line DG and to three squares of the line CG (as it is not hard to prove, marking what hath before been proved). But three squares of the line DG, together with three squares of the line CG, are equal to fifteen squares of the semidiameter of the circle described about the pentagon CDEFG (for it was before proved in the assumpt put in this proposition) that the square● of DG and G C taken once, are quintuple to the square of the semidiameter of the circle described about the pentagon CDEFG). And five squares of the line KL are equal to fifteen squares of the semidiameter of the circle described about the triangle KLH. (For by the 12. of the thirteenth, one square of the line LK is triple to one square of the line drawn from the centre to the circumference). Wherefore fifteen squares of the line drawn from the centre to the circumference (of the circle which containeth the pentagon CDEFG) are equal to fifteen squares of the line drawn from the centre to the circumference of the circle which containeth the triangle KLH): wherefore one of the squares which is drawn from the centre to the circumference of the one circle, is equal to one of the squares which is drawn from the centre to the circumference of the other circle. Wherefore the diameter is equal to the diameter, wherefore one and the self same circle comprehendeth both the pentagon of a dodecahedron and the triangle of an Icosahedron described in one and the self same circle: which was required to be proved.   

¶ The 3. Theorem. The .3 Proposition. 
Th● 5. proposition a●t●r 〈◊〉.  If there be an equilater and equiangle pentagon, and about it be described a circle, and from the centre to one of the sides be drawn a perpendicular line, that which is contained under one of the sides and the perpendicular line thirty times, is equal to the superficies of the dodecahedron.  
SVppose that ABCD be an equilater and equiangle pentagon. And about the same pentagon, describe (by the 14. of the fourth) a circle. 
Construction.  And let the centre thereof be the point F. And from the point F draw (by the 12. of the first) unto the line CD a perpendicular line FG. Now I say that that which is contained under the lines CD and GF thirty times, is equal to 12. pentagons of the same quantity that the pentagon ABCD is. Draw these right lines CF and FD. 
Demonstration.  Now forasmuch as that which is contained under the lines CD and FG is double to the triangle CDF (by the 41. of the first) therefore that which is contained under the lines CD and FG five times is equal to ten of those triangles. But ten of those triangles are two pentagons, and six times ten of those triangles are all the pentagons. Wherefore that which is contained under the lines CD and FG thirty times is equal to 12. pentagons But 12. pentagons are the superficies of dodecahedron. Wherefore that which is contained under the lines CD and FG thirty times is equal to the superficies of the dodecahedron.  
In like sort also may we prove that if there be an equilater triangle, as for example, 
The 5. proposition a●ter F●ussas.  the triangle ABC, and about it be described a circle, and the centre of the circle be the point D, and the perpendicular line be the line DE: that which is contained under the lines BC and DE thirty times, is equal to the superficies of the Icosahedron. 
Demonstration.  For again forasmuch as that which is contained under the lines DE and BC is double to the triangle DBC (by the 41. of the first): therefore two triangles are equal to that which is contained under the lines DE and BC, and three of those triangles contain the whole triangle. Wherefore six such triangles as DBC is, are equal to that which is contained under the lines DE and BC thrice. But six s●ch triangles as DBC is, are equal to two such triangles as ABC is. Wherefore that which is contained under the lines DE and BC thrice, is equal to two such triangles as ABC is. But two of those triangles taken ten times containeth the whole Icosahedron. Wherefore that which is contained under the lines DE & BC thirty times, is equal to twenty such triangles as the triangle ABC is, that is, to the whole superficies of the Icosahedron. 
* This is the reason of the Corollary following.  Wherefore as the superficies of the dodecahedron is to the superficies of the Icosahedron, so is that which is contained under the lines CD and FG to that which is contained under the lines BC and DE.  

¶ Corollary.  
By this it is manifest, that as the superficies of the Dodecahedron is to the superficies of the Icosahedron, 
A Corollary which also Flussas putteth as a Corollary after the 5. proposition in his order.  so is that which is contained under the side of the Pentagon, and the perpendicular line which is drawn from the centre of the circle described about the Pentagon to the same side, to that which is contained under the side of the Icosahedron and the perpendicular line which is drawn from the centre of the circle described about the triangle to the same side: so that the Icosahedron and Dodecahedron be both described in one and the self same Sphere.    

¶ The 4. Theorem. The 4. Proposition. 
The 6. p●●positiō●●ter Flussas.  This being done, now is to be proved, that as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so is the side of the cube to the side of the Icosahedron.  

Construction.  TAke (by the 2. Theorem of this book) a circle containing both the pentagon of a Dodecahedron, and the triangle of an Icosahedron, being both described in one and the self same sphere, and let the same circle be DBC. And in the circle DBC describe the side of an equilater triangle, namely, CD, and the side of an equilater pentagon, namely, AC. And take (by the 1. of the third) the centre of the circle, and let the same be E. And from the point E draw unto the lines DC and AC, perpendicular lines EF and EG. And extend the line EG directly to the point B. And draw a right line from the point B to the point C. And let the side of the cube be the line H. Now I say, that as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so is the line H to the line CD. 
Demonstration.  Forasmuch as the line made of the lines EB and BC added together (namely, of the side of the hexagon, and of the side of a decagon) is (by the 9 of the thirteenth) divided by an extreme and mean proportion, and his greater segment is the line BE: and the line EG is also (by the 1. of the fo●retenth) the half of the same line, and the line EF is the half of the line BE (by the Corollary of the 12. of the thirteenth). Wherefore the line EG being divided by an extreme and mean proportion, 
* This is not hard to prove by the 15. 16. and 19 of the ●●●eth.  his greater segment shall be the line EF. And the line H also being divided by an extreme & mean proportion, his greater segment is the line CA, as it was proved 
† In the Corollary of the 17. of the t●irtēth.  in the Dodecahedron. 
* 〈◊〉 again is required the Assumpt which is afterward proved in this 4 proposition.  Wherefore as the line H is to the line CA, so is the line EG to the line EF. Wherefore (by the 16. of the sixth) that which is contained under the lines H and EF, is equal to that which is contained under the lines CA and EG. And for that as the line H is to the line CD, so is that which is contained under the lines H and EF, to that which is contained under the lines CD and EF (by the 1. of the sixth). But unto that which is contained under the lines H and EF, is equal that which is contained under the lines CA and EG. Wherefore (by the 11. of the fift) as the line H is to the line CD, so is that which is contained under the lines CA and EG, to that which is contained under the lines CD and EF, that is (by the Corollary next going before) as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so is the line H to the line CD.  

another demonstration to prove that as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so is the side of the cube to the side of the Icosahedron. 
† But first the Assumpt following, the construction whereof here begins, is to be proved.   
LEt there be a circle ABC. And in it describe two sides of an equilater pentagon (by the 11. of the fift) namely, AB and AC: and draw a right line from the point B to the point C. And (by the 1. of the third) take the centre of the circle, and let the same be D. And draw a right line from the point A to the point D, and extend it directly to the point E, and let it cut the line BC in the point G. And let the line DF be half to the line DA, and let the line GC be triple to the line HC, by the 9 of the sixth. 
The Assumpt, which also Flussas putteth as an Assumpt a●ter the 6. proposition.  Now I say, that that which is contained under the lines AF and BH, is equal to the pentagon inscribed in the circle ABC. Draw a right line from the point B to the point D. Now forasmuch as the line AD is double to the line DF, therefore the line AF is sesquialter to the line AD. Again, 
Demonstration of the Assumpt.  forasmuch as the line GC is triple to the line CH, therefore the line GH is double to the line CH. Wherefore the line GC is sesquialter to the line HG. Wherefore as the line FA is to the line AD, so is the line GC to the line GH. Wherefore (by the 16. of the sixth) that which is contained under the lines AF & HG, is equal to that which is contained under the lines DA and GC. But the line GC is equal to the line BG (by the 3. of the third). Wherefore that which is contained under the lines AD and BG, is equal to that which is contained under the lines AF and GH. But that which is contained under the lines AD and BG, is equal to two such triangles as the triangle ABD is (by the 41. of the first). Wherefore that which is contained under the lines AF and GH, is equal to two such triangles as the triangle ABD is. Wherefore that which is contained under the lines AF and GH ●iue times, is equal to ten triangles. But ten triangles are two pentagons. Wherefore that which is contained under the lines AF and GH five times, is equal to two pentagons. And forasmuch as the line GH is double to the line HC, therefore that which is contained under the lines AF and GH, is double to that which is contained under the lines AF and HC (by the 1. of the sixth). Wherefore that which is contained under the lines AF and CH twice, is equal to that which is contained under the lines AF and GH once. Take each of those parallelograms five times. Wherefore that which is contained under the lines AF and HC ten times, is equal to that which is contained under the lines AF & GH five times, that is, to two pentagons. Wherefore that which is contained under the lines AF and HC five times, is equal to one pentagon. But that which is contained under the lines AF and HC five times, is equal (by the 1. of the sixth) to that which is contained under the lines AF and HB, for the line HB is quintuple to the line HC (as it is easy to see by the construction) and they are both under one & the self same altitude, namely, under AF. Wherefore that which is contained under the lines AF and BH, is equal to one pentagon.   

This being proved, now let there be drawn a Circle comprehending both the Pentagon of a Dodecahedron, and the triangle of an Icosahedron, being both described in one and the self same Sphere.  

Construction pertaining to the second demonstration of the 4. proposition.  LEt the circle be ABC. And in it describe as before, two sides of an equilater pentagon, namely BA and AC● and draw a right line from the point B to the point C: and take the centre of the circle and let the same be E. And from the point A to the point E draw a right line AE: and extend the line AE to the point F. And let it cut the line BC in the point K. And let the line AE be double to the line EG, & let the line CK be triple to the line CH, by the .9. of the sixth. And from the point G raise up (by the .11. of the first) unto the line AF a perpendicular line GM: and extend the line GM directly to the point D. Wherefore the line MD is the side of an equiliter triangle, by the corollary of the. 1●. of the thirteenth: draw these right lines AD and AM. Wherefore ADM is an equilater triangle. 
Second demonstration o● the 4. proposition.  And for as much as that which is contained under the lines AG and BH is equal to the pentagon (by the former assumpt) and that which is contained under the lines AG and GD is equal to the triangle ADM: therefore as that which is contained under the lines AG and HB is to that which is contained under the lines AG and GD, so is the pentagon to the triangle. But as that which is contained under the lines BH & AG is to that which is contained under the lines AG and GD, so is the line BH to the line DG (by the .1. of the sixth) wherefore (by the .15. of the fifth) as 12. such lines as BH is, are to .20 such lines as DG is, so are 12. pentagons to 20. triangles, that is the superficies of the Dodecahedron, to the superficies of the Icosahedron. And 12. such lines as BH is, are equal to ten such lines as BC is (for the line HB is quintuple to the line HC): and the line BC is sextuple to the line CH● Wherefore six such lines as BH is, are equal to five such lines as BC are: and in the same proportion are their doubles: and 20. such lines as the line DG is, are equal to .10. such lines as the line DM is: for the line DM is double to the line DG. Wherefore as 10. such lines as BC is, are to 10. such lines as DM is, that is, as the line BC is to the line DM, so is the superficies of the Dodecahedron to the superficies of the Icosahedron. But the line BC is the side of the cube, and the line DM the side of the Icosahedron: wherefore (by the 11. of the fifth) as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so is the line BC to the line DM, that is, the side of the cube to the side of the Icosahedron.   


The 7● proposition after Flussas.  Now will we prove that a right line being divided by an extreme and mean proportion, what proportion the line containing in power the squares of the whole line and of the greater segment, hath to the line containing in power the squares of the whole line and of the less segment, the same proportion hath the side of the cube to the side of the Icosahedron, being both described in one and the self same sphere.  

Construction.  SVppose that AB be a circle containing both the pentagon of a Dodecahedron & the triangle of an Icosahedron described both in one and the self same sphere. Take the centre of the circle, and let the same be C. And from the point C extend to the circumference a right line at all adventures, and let the same BC. And (by the 30. of the sixth) divide the line BC by an extreme and mean proportion in the point D, and let the greater segment thereof be CD. Wherefore the line CD is the side of a Decagon described in the same circle (by the corollary of the 9 of the thirteenth). Take the side of an Icosahedron, and let the same be the line E, and the side of a Dodecahedron, and let the same be the line F, and the side of a cube & let the same be the line G. 
Demonstration.  Wherefore the line E is the side of an equilater triangle, and F of an equaliter pentagon described in one and the self same circle. And the line G being divided by an extreme and mean proportion, his greater segment is the line F, by the corollary of the 17. of the thirteenth. Now forasmuch as the line E is the side of an equilater triangle, but (by the 12. of the thirteenth) the side of an equilater triangle is in power triple to the line BC, (which is drawn from the centre to the circumference) therefore the square of the line E is triple to the square of the line BC: but the squares of the line BC and BD are (by the 4. of the thirteenth) triple to the square of the line CD. Wherefore as the square of the line E is to the square of the line CB, so are the squares of the lines CB and BD to the square of the line CD. Wherefore alternately (by the 16. of the fifth) as the square of the line E is to the squares of the lines CB and BD, so is the square of the line CB to the square of the line CD. 
* Here again is required the Assumpt afterward proved in this 4. proposition.  But as the square of the line BC is to the square of the line CD, so is the square of the line G (the side of the cube) to the square of the line F, the side of Dodecahedron. For the line F is the greater segment of the line G (as was before proved.) Wherefore (by the .11. of the fift) as the square of the line E is to the squares CB and BD, so is the square of the line G, to the square of the line F. Wherefore alternately (by the 16. of the fifth) & also by conversion (by the corollary of the 4. of the fift) as the square of the line G, is to the square of the line E, so is the square of the line F, to the squares of the lines CB & BD. But unto the square of the line F are equal the squares of the lines BC & CD, for the side of a pentagon containeth in power both the side of a six angled figure, and the side of a ten angled figure (by the 10. of the thirteenth.) Wherefore as the square of the line G, is to the square of the line E, so are the squares of the lines BC and CD to the squares of the lines CB and BD. But as the squares of the lines CB and CD are to the squares of the lines CB & BD, 
† As may by the Assumpt afterward in this proposition be plainly proved.  so (any right line what so ever it be, being divided by an extreme and mean proportion) is the line containing in power the squares of the whole line, and of the greater segment, to the line containing in power the squares of the whole line, and of the less segment: wherefore (by the 11. of the fifth) as the square of the line G (the side of the cube) is to the square of the line E, so (any right line being divided by an extreme and mean proportion) is the line containing in power the squares made of the whole line, and of the greater segment, to the line containing in power the squares made of the whole line, and of the less segment: but the line G is the side of the Cube, and the line E of the Icosahedron (by supposition.) If therefore a right line be divided by an extreme and mean proportion, as the line containing in power the squares of the whole line, and of the greater segment, is to the line containing in power the squares of the whole line and of the less segment: so is the side of the cube to the side of the Icosahedron, being both described in one and the self same sphere.   

Now will we prove that as the side of the Cube is to the side of the Icosahedron, 
The 8. prodition a●ter Flussas.  so is the solid of the Dodecahedron to the solid of the Icosahedron.  
Forasmuch as equal circles comprehend both the pentagon of a Dodecahedron, and the triangle of an Icosahedron, being both described in one and the self same sphere, by the 2. of this book: but in a sphere equal circles are equally distant from the centre (for the perpendicular lines drawn from the centre of the sphere to the plain superficieces of the circles are equal, and do fall upon the centres of the circles. 
† By the Corollary added by Flussas after has Assumpt put after the 17. proposition of the 12. book.  Wherefore perpendicular lines drawn from the centre of the sphere, to the centre of the circle, comprehending both the triangle of an Icosahedron, and the pentagon of a Dodecahedron are equal: wherefore the pyramids, whose bases are the pentagons of the Dodecahedron, are of equal altitude with the pyramids whose bases are the triangles of the Icosahedron. But pyramids of equal altitude, are in that proportion the one to the other, that their bases are (by the 5. of the twelfth) wherefore as the pentagon is to the triangle, so is the pyramid whose base is the pentagon of the Dodecahedron and top the centre of the sphere, to the pyramid whose base is the triangle and top the centre also of the sphere. Wherefore (by the 15. of the fifth) as 12. pentagons are to 20. triangles, so are 12. pyramids having pentagons to their bases to 20. pyramids having triangles to their bases. But 12. pentagons are the superficies of the D●decahedron, and 20. triangles are the superficies of the Icosahedron. Wherefore as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so are 12. pyramids having pentagons to their bases to 20. pyramids having triangles to their bases. But 12. pyramids having pentagons to their bases, are the solid of the Dodecahedron, and 20. pyramids having triangles to their bases are the solid o● the Icosahedron. 
Corollary of the 8. after Flussas.  Wherefore (by the 11. of the fifth) as the superficies of the Dodecahedron is to the superficies of the Icosahedron ●o is the solid of the Dodecahedron to the solid of the Icosahedron. But as the superficies of the Dodecahedron, is to the superficies of the Icosahedron, so have we proved that the side o● the cube is to the side of the Icosahedron. Wherefore, by the 11. of the fifth, as the side of the cube is to the side of the Icosahedron, so is the solid of the Dodecahedron to the solid of the Icosahedron.   


This Assumpt is the 3. proposition a●ter ●lussas, and is it which 〈◊〉 times hath been taken a● granted in this book, and o●ce also in the last proposition of the 13. book: as we have be●ore noted.  If two right lines be divided by an extreme and mean proportion, they shall every way be in like proportion: which thing is thus demonstrated.  
LEt the line AB be (by the 30. of the sixth) divided by an extreme and mean proportion in the point C, and let the greater segment thereof be the line CA And likewise also let the line DE be divided by an extreme and mean proportion in the point F, and let the greater segment thereof be the line DF. Then I say that as the whole line AB is to the greater segment thereof AC, so is the whole line DE to the greater segment thereof DF. For forasmuch as that which is contained under the lines AB and BC is equal to the square of the line AC (by the definition of a line divided be an extreme and mean proportion): 
Demonstration.  and that which is contained under the lines DE and EF is also equal to the square of the line DF (by the same definition): therefore as that which is contained under the lines AB and BC is to the square of the line AC, so is that which is contained under the lines DE and EF to the square of the line DF. For in each is the proportion of equality. Wherefore as that which is contained under the lines AB and BC four times, is to the square of the line AC, so is that which is contained under the lines DE and EF four times to the square of the line DF (by the 15. of the fifth). Wherefore by composition (by the 18. of the ●i●th) as that which is contained under the lines AB and BC four times, together with the square of the line AC, is to the square of the line AC, so is that which is contained under the lines DE and EF four times, together with the square of the line DF, to the square of the line DF. Wherefore as the square which is made of the lines AB and BC added together and made one line (which square by the 8. of the second is equal to that which is contained under the lines AB and BC four times together with square of the line AC) is to the square of the line AC, so is the square made of the lines DE & EF added together and made one line (which square is also, by the same, equal to that which is contained under the lines DE and EF four times together with the square of the line DF) to the square of the line DF. Wherefore also as the lines AB & BC added together are to the line AC, so are the lines DE & EF added together to the line DF (by the 22. of the sixth). Wherefore by composition (by the 18. of the fifth) as both the lines AB & BC added the one to the other, together with the line AC, that is, as two such lines as AB is, are to the line AC, so are both the lines DE and EF added the one to the other together with the line DF, that is two such lines as DE is to the line DF. And in the same proportion are the halves of the antecedents by the 15. of the fifth. Wherefore as the line AB is to the line AC, so is the line DE to the line DF. (And therefore by the 19 of the fifth, as the line AB is to the line BC, so is the line DF to the line FE. Wherefore also by division by the 17. of the fifth, as the line AC is to the line CB, so is the line DF to the line DE).  
Now that we have proved, 
* In the 4. section ●f this proposition.  that, any right line whatsoever being divided by an extreme and mean proportion, what proportion the line containing in power the squares made of the whole line and of the greater segment added together, hath to the line containing in power the squares made of the whole line and of the less segment added together, the same proportion hath the side of the cube to the side of the Icosahedron: Now also that we have proved, 
† In the 1. and 3 section of the same proposition.  that as the side of the cube is to the side of the Icosahedron, so is the superficies of the Dodecahedron to the superficies of the Icosahedron, being both described in one and the self same sphere: and moreover seeing that we have proved, 
† In the 5. section of the same proposition.  that as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so is the Dodecahedron to the Icosahedron, for that both the pentagon of the Dodecahedron, and the triangle of the Icosahedron are comprehended in one and the self same circle: 
A Corollary.  All these things I say being proved, it is manifest, that if in one and the self same sphere be described a Dodecahedron, and an Icosahedron, they shall be in proportion the one to the other, as, a right line whatsoever being divided by an extreme and mean proportion, the line containing in power the squares of the whole line and of the greater segment added together, is to the line containing in power the squares of the whole line and of the less segment added together. For for that as the Dodecahedron is to the Icosahedron, so is the superficies of the Dodecahedron to the superficies of the Icosahedron, that is, the side of the cube to the side of the Icosahedron: but as the side of the cube is to the side of the Icosahedron, so, any right line what so ever being divided by an extreme and mean proportion, is the line containing in power the squares of the whole line and of the greater segment added together, to the line containing in power the squares of the whole line and of the less segment added together. Wherefore as a Dodecahedron is to an Icosahedron described in one and the self same sphere, so, any right line what so ever being divided by an extreme and mean proportion, is the line containing in power the squares of the whole line & of the greater segment added together, to the line containing in power the squares of the whole line and of the less segment added together.     
The end of the fourteenth Book of Euclides Elements after Hypsicles.   

¶ The fourteenth book of Euclides Elements after Flussas.  
FOr that the fouretenth Book, as it is set forth by Flussas, containeth in it more Propositions than are found in Hypsicles, & also some of those Propositions which Hypsicles hath, are by him somewhat otherwise demonstrated, I thought my labour well bestowed for the reader's sake to turn it also all whole, notwithstanding my travail before taken in turning the same book after Hypsicles. Where note ye, that here in this 14. book after Flussas, and in the other books following, namely, the 15. and 16. I have in alleging of the Propositions of the same 14. book, followed the order and number of the Propositions, as Flussas hath placed them.  

¶ The first Proposition. A perpendicular line drawn from the centre of a circle, to the side of a Pentagon inscribed in the same circle: 
The first proposition after Campane.  is the half of these two lines taken together, namely, of the side of the hexagon, and of the side of the decagon inscribed in the same circle.  
TAke a circle ABC, and inscribe in it the side of a pentagon, which let be BC, and take the centre of the circle, which let be the point D: 
Construction.  and from it draw unto the side BC a perpendicular line DE: which produce to the point ●. And unto the line E F put the line EG equal. And draw these right lines CG, CD, and CF. Then I say, that the right line DE (which is drawn from the centre to BC the side of the pentagon) is the half of ●he side● of the decagon and hexagon, taken together. 
Demonstration.  Forasmuch as the line DE is a perpendicular ●nto the line BC: therefore the sections BE and EC shall be equal (by the 3. of the third): and the line EF is common unto them both; and the angles FEC and FEB, are right angles, by supposition. Wherefore the bases BF and FC are equal (by the 4. of the first). But the line BC is the side of a pentagon, by construction. Wherefore FC which subtendeth the half of the side of the pentagon, is the side of the decagon inscribed in the circle ABC. But unto the line FC is, by the 4. of the first, equal the line CG, for they subtend right angles CEG, and CEF, which are contained under equal sides. Wherefore also the angles CGE, and CFE, of the triangle CFG, are equal, by the 5. of the first. And forasmuch as the ark FC is subtended of the side of a decagon, the ark CA shall be quadruple to the ark CF: Wherefore also the angle CDA shall be quadruple to the angle CDF (by the last of the six●). And forasmuch as the same angle CDA, which is set at the centre, is double to the angle CFA, which is set at the circumference, by the 20. of the third: therefore the angle CFA, or CFD, is double to the angle CD●, namely, the half of quadruple. But unto the angle CFD or CFG, is proved equal the angle CGF: Wherefore the outward angle CGF, is double to the angle CDF. Wherefore the angles CDG and DCG, shall be equal. For unto those two angles the angle CGF is equal, by the 32. of the first. Wherefore the sides GC and GD, are equal, by the 6. of the first. Wherefore also the line GD is equal to the line FC, which is the side of the decagon. But unto the right line FE is equal the line EG, by construction. Wherefore the whole line DE is equal to the two lines C● and FE. Wherefore those lines taken together (namely, the lines DF and FC) shall be double to the line DE. Wherefore the line DE (which is drawn from the centre perpendicularly to the side of the pentagon) shall be the half of both these lines taken together, namely, of DF the side of the hexagon, and CF the side of the decagon. For the line DF which is drawn from the centre, is equal to the side of the hexagon, by the Corollary of the 15. of the fourth. Wherefore a perpendicular line drawn from the centre of a circle, to the side of a pentagon inscribed in the same circle: is the half of these two lines taken together, namely, of the side of the hexagon, and of the side of the decagon inscribed in the same circle: which was required to be proved.  

A Corollary.  
If a right line drawn perpendicularly from the centre of a circle to the side of a pentagon, be divided by an extreme and mean proportion: the greater segment shall be the line which is drawn from the same c●●tre to the side of an equilater triangle inscribed in the same circle. For, that li●● (drawn to the side of the triangle) is (by the Corollary of the 12. of the thirteenth) the half of the line drawn from the centre to the circumference, that is, of the side of the hexagon: Wherefore the residue shall be the half of the side of the decagon. For the whole line is the half of the two sides, namely, of the side of the hexagon, and of the side of the decagon. But of the side of a decagon and of an hexagon taken together, the greater segment is the side of the hexagon (by the 9 of the thirteenth). Wherefore the greater segment of their halves shall be the half of the hexagon, by the 15. of the fift: which half is the perpendicular line drawn from the centre to the side of the triangle, by the Corollary of the 12. of the thirteenth.    

¶ The second Proposition. If two right lines be divided by an extreme and mean proportion: 
The 2. proposition after Campane.  they shall be divided into the self same proportions.  
SVppose that these two right lines AB and DE be each cut by an extreme and mean proportion in the points F and Z. Then I say, that these two lines are divided into the self same proportions, that is, that the line AB is in the point F divided in like sort as the line DE is in the point Z. For if they be not in like sort cut, let one of them, namely, DE, be cut like unto the line AB in the point C. 
Demonstration leading to an impossibility.  So that let the line DE be to DC the greater part, as the greater part DC is to CE the less part, by the 3. definition of the sixth. But (by suppositition) the line DE is to the line DZ, as the line DZ is to the line ZE. Wherefore the right line DE is divided by an extreme and mean proportion in two points C and Z. But the proportion of DE to DC the less line, is greater than the proportion of the same DE to DZ the greater line, by the 2. part of the 8. of the fift. But as DE is to DC, so is DC to CE: Wherefore the proportion of DC to CE, is greater than the proportion of DZ to ZE. And forasmuch as DZ is greater than DC, the proportion of DZ to CE shall be greater than the proportion of DC to CE, by the 8. of the fift. Wherefore the proportion of DZ to CE, is much greater than the proportion of DZ to ZE. Wherefore one and the self same magnitude, namely, DZ, hath to CE the greater line, a greater proportion than it hath to ZE the less line, contrary to the second part of the 8. of the fift: which is impossible. Wherefore the right lines AB & DE, are not cut unlike. Wherefore they are cut like, and into the self same proportions. And the same demonstration also will serve, if the point C fall in any other place. For always some one of them shall be the greater. If therefore two right lines be cut by an extreme and mean proportion: they shall be cut into the self same proportions: which was required to be proved.   

¶ The third Proposition. If in a circle be described an equilater Pentagon: 
The 4. proposition after Campane.  the squares made of the side of the Pentagon and of the line which subtendeth two sides of the Pentagon, these two squares (I say) taken together, are quintuple to the square of the line drawn from the centre of the circle to the circumference.  
SVppose that in the circle BCG the side of a Pentagon be BG: and let the line BC subtend two sides thereof. 
Construction.  And let the line BG be divided into two equal parts by a right line drawn from the centre D: namely, by the diameter CDE produced to the point Z. And draw the right line BZ. Then I say, that the right lines BC and BG, are in power quintuple to the right line DZ, which is drawn from the centre to the circumference. For forasmuch as (by the 47. of the first) the squares of the lines CB and BZ, 
Demonstration.  are equal to the square of the diameter CZ: therefore they are quadruple to the square of the line DZ, by the 20. of the sixth (for the line CZ is double to the line DZ). Wherefore the right lines CB, BZ, and ZD, are in power quintuple to the line ZD. But the right line BG containeth in power the two lines BZ and ZD, by the 10. of the thirteenth. For DZ is the side of an hexagon, & BZ the side of a decagon. Wherefore the lines BC and BG (whose powers are equal to the powers of the lines CB, BZ, ZD) are in power quintuple to the line DZ. If therefore in a circle be described an equilater Pentagon: the squares made of the side of the Pentagon and of the line which subtendeth two sides of the Pentagon, th●se two squares (I say) taken together, are quintuple to the square of the line drawn from the centre of the circle to the circumference.  

¶ A Corollary.  
If a Cube and a Doderahedron be contained in one and the self same Sphere: 
This Corollary Campane also ●utteth after the 4. proposition in his order.  the side of the Cube, and the side of the Dodecahedron, are in power quintuple to the line which is drawn from the centre of the circle which containeth the Pentagon of the Dodecahedron. For it was proved in the 17. of the thirteenth, that the side of the Cube subtendeth two sides of the Pentagon of the Dodecahedron, where the said solids are contained in one and the self same Sphere. Wherefore the side of the Cube subtending two sides of the Pentagon, and the side of the same Pentagon, are contained in one and the self same circle. Wherefore, by this Proposition, they are in power quintuple to the line which is drawn from the centre of the same circle which containeth the Pentagon of the Dodecahedron.    

The 4. Proposition. One and the self same circle containeth both the Pentagon of a Dodecahedron, 
The 5. proposition after Campane.  and the triangle of an Icosahedron described in one and the self same sphere.  
LEt the diameter of the sphere given be AB, and let the bases of the Icosahedron and Dodecahedron described in it, 
Construction.  be the triangle MNR, and the pentagon FKH, and about them let there be described circles. by the 5. and 14. of the fourth. And let the lines drawn from the centres of those circles to the circumferences be LN and OK. Then I say that the lines LN and OK are equal, and therefore one and the self same circle containeth both those figures. Let the right line AB, be in power quintuple to some one right line, as to the line CG (by the Corollary of the 6. of the tenth.) And making the centre the point C, & the space CG, describe a circle DZG. And let the side of a pentagon inscribed in that circle (by the 11. of the fourth) be the line ZG. And let EG (subtending half of the ark ZG) be the side of a Decagon inscribed in that circle. And by the 30. of the sixth, divide the line CG by an extreme & mean proportion in the point I 
Demonstration.  Now forasmuch as in the 16. of the thirteenth, it was proved, that this line CG (unto whom the diameter AB of the sphere is in power quintuple) is the line which is drawn from the centre of the circle, which containeth five angles of the Icosahedron, and the side of the pentagon described in that circle DZG, namely the line ZG is side of the Icosahedron described in the Sphere, whose diameter is the line AB: therefore the right line ZG, is equal to the line MN, which was put to be the side of the Icosahedron, or of his triangular base. Moreover, by the 17. of the thirteenth, it was manifest that the right line ●H (which subtendeth the angle of the pentagon of the Dodecahedron inscribed in the foresaid sphere) is the side of the Cube, inscribed in the self same sphere. (For upon the angles of the cube, were made the angles of the Dodecahedron.) Wherefore the diameter AB is in power triple to FH, the side of the Cube (by the 15. of the thirteenth). But the same line AB is (by supposition) in power quintuple to the line CG. Wherefore five squares of the line CG, are equal to three squares of the line FH: (for each is equal to one and the self same square of the line AB). And forasmuch as EG the side of the Decagon, cutteth the right line CG by an extreme and mean proportion (by the corollary of the 9 of the thirteenth): Likewise the line HK, cutteth the line FH, the side of the Cube by an extreme and mean proportion (by the Corollary of the 17. of the thirteenth): therefore the lines CG and FH, are divided into the self same proportions, by the second of this book: and the right lines CI and EG, which are the greater segments of one and the self same line CG, are equal: And forasmuch as five squares of the line CG are equal to three squares of the lines FH: therefore five squares of the line GE, are equal to three squares of the line HK (for the lines GE and HK are the greater segments of the lines CG and FH). Wherefore five squ●re● of the line● CG & GE are equal to the squares of the 〈◊〉 ●H & HK, by the 1● of the ●ift. But unto the squares of the lines CG and GE● is ●qual the squ●re of thetine ZG, by the 10. of the thirteenth: and unto the line ZG the line MN was equal: wherefore five squares of the line MN, are equal to three squares of the lines FH, HK. But the squares of the lines ●● and HK, 〈◊〉 quintuple to the square of the line OK (which is drawn from the centre) by the third of this book. Wherefore three squares of the lines FH and HK make 15. squares of the line OK. And forasmuch as the square of the line MN is triple to the square of the line LN (which is drawn from the centre) by the 12. of the thirteenth, therefore five squares of the line MN are equal to 15. squares of the line LN. But five squares of the line MN are equal unto three squares of the lines FH and HK. Wherefore one square of the line LN is equal to one square of the line OK (being each the fifteenth part of equal magnitudes) by the 15. of the fif●●. Wherefore the lines LN and OK, which are drawn from the centres, are equal. Wherefore also the circles NRM, and FKH which are described of those lines, are equal. And those circles contain (by supposition) the bases of the Dodecahedron and of the Icosahedron described in one and the self same sphere. Wherefore one and the self same circle, etc. a● in th● proposition: which was required to be proved.   

The 5. Proposition. If in a circle be inscribed the pentagon of a Dodecahedron, and the triangle of an Icosahedron, 
This is the 6. and 7. propositions after Campane.  and from the centre to one of their sides, be drawn a perpendicular line: That which is contained 30. times under the side, & the perpendicular line falling upon it, is equal to the superficies of that solid, upon whose side the perpendicular line falleth.  
SVppose that in the circle AGE, be described the pentagon of a Dodecahedron, which let be ABGDE, and the triangle of an Icosahedron described in the same sphere, which let be AFH. 
Construction.  And let the centre be the point C. ●●on which draw perpendicularly the line CI to the side of the Pentagon, and the line CL to the side of the triangle. Then I say that the rectangle figure contained under the lines CI and GD 30. times, is equal to the superficies of the Dodecahedron: and that that which is contained under the lines CL & AF 30. times is equal to the superficies of the Icosahedron described in the same sphere. Draw these right lines CA, CF, CG and CD. 
Demonstration.  Now forasmuch as that which is contained under the base GD & the altitude IC, is double to the triangle GCD, by the 41. of the first: And five triangles like and equal to the triangle GCD do make the pentagon ABGDE of the Dodecahedron: wherefore that which is contained under the lines GD and IC five times is equal to two pentagons. Wherefore that which is contained under the lines GD and IC ●0. times is equal to the 12. pentagons, which contain the superficies of the Dodecahedron.  
Again that which is contained under the lines CL and AF, is double to the triangle ACF: wherefore that which is contained under the lines CL and AF three times is equal to two such triangles as AFH is, which is one of the bases of the Icosahedron (for the triangle ACF, is the third part of the triangle AFH, as it is easy to prove, by the 8. & 4. of the first.) Wherefore that which is contained under the lines CL and AF. 30 times times, is equal to 10. such triangles as AFH i●, which contain the superficies of the Icosahedron. And forasmuch as one and the self same spher● containeth the Dodecahedron of this pentagon, and the Icosahedron of this triangle (by the 4. of this book ●) and the line CL falleth perpendicularly upon the side of the Icosahedron, and the line CI upon the side of the Dodecahedron: that which is 30. times contained under the side, and the perpendicular line falling upon it, is equal to the superficies of that solid, upon whose side the perpendicular falleth. If therefore in a circle etc. as in the proposition: which was required to be demonstrated.  

A Corollary.  

This Corollary Campane also addeth after the 7. proposition i● his order.  The superficieces of a Dodecahedron and of an Icosahedron described in one and the self same sphere are the one to the other, as that which is contained under the side of the one and the perpendicular line drawn unto it from the centre of his base, to that which is contained under the side of the other, and the perpendicular line drawn to it from the centre of his base. For a● thirty● tim●s is to thirty times, so is once to once by the 15. of th● fifth.    

The 6. Proposition. The superficies of a Dodecahedron, is to the superficies of an Icosahedron described in one and the self same sphere, 
The 5. proposition a●ter Campane.  in that proportion, that the side of the Cube is to the side of the Icosahedron contained in the self same sphere.  

Construction.  SVppose that there be a circle ABG, & in it (by the 4. of this book) let there be inscribed the side● of a Dodecahedron and of an Icosahedron contained in on● and the self same sphere. And let the side o● the Dodecahedron be AG, and the side of the Icosahedron be DG. And let the centre be the point E: from which draw unto those s●des, perpendicular lines EI and EZ. And produce the line EI to the point B, and draw the lin● BG. And let the side of the cube contained in the self same sphere be GC. Then I say that the superficies of the Dodecahedron i● to the superficies of the Icosahedron, as the line ●G, i● to the li●● GD. For forasmuch as the line EI bein● divided by an extreme and mean proportion, the greater segment thereof shall be the lin● EZ, by the corollary of the first of this book: 
Demonstration.  and the line CG being divided by an extreme and mean proportion, his greater segment is the line AG, by the corollary of the 17. of the thirteenth: Wherefore the right lines EI and CG ●r● cut proportionally by the second of this b●oke. Wh●r●fore as the line CG, is to the line AG, so is the line EI to the line EZ. Wher●fore that which it contained under the extremes CG and EZ, is squall to that which i● contayn●d under the means AG and EI. (by the 16. of the sixth.) But as that which i● contained under the lin●● CG and ●Z is to that which is contained under the lines DG and EZ, so (by the first of the sixth) i● the lin● CG to the line DG, for both those parallelograms have o●● and the self same altitude, namely the line EZ. Wherefore as that which is contained under the lines EI and AG (which i● proved equal to that which is contained under the line● CG and EZ) is to that which is contained under the lines DG and EZ, so is the line CG to the li●● DG. But as that which is contained under the lines EI and AG is to that which is contained under the lines DG and EZ, so (by the corollary of the former proposition) is the superficies of the Dodecahedron, to the superficies of the Icosahedron. Wherefore as the superficies ●● the Dodecahedron is to the superficies of the Icosahedron, so is CG the side of the cube, to GD the side of the Icosahedron. The superficies therefore of a Dodecahedron is to the superficies● etc. as in the proposition, which was required to be proved.  

An Assumpt.  
The Pentagon of a Dodecahedron, 
This Assumpt Campane also hath after the 8. proposition, in his order.  is equal to that which is contained under the perpendicular line which falleth upon the base of the triangle of the Icosahedron, and five sixth parts of the side of the cube, the said three solids being described in one and the self same sphere.  
Suppose that in the circle ABEG, the pentagon of a Dodecahedron be A●CIG, and let two sides thereof AB and AG be subtended of the right line BG. And let the triangle of the Icosahedron inscribed in the self same sphere, 
Construction.  by the 4. of this book, be AFH. And let the centre of the circle be the point D, and let the diameter be ADE, cutting FH, the side of the triangle in the point Z, and cutting the line BG in the point K. And draw the right line BD. And from the right line KG cut of a third part TG, by the 9 of the sixth. Now than the line BG subtending two sides of the Dodecahedron, shallbe the side of the cube inscribed in the same sphere, 
Demonstration.  by the 17. of the thirteenth: and the triangle of the Icosahedron of the same sphere shallbe A●H by the 4. of this book. And the line AZ which passeth by the centre D shall fall perpendicularly upon the side of the triangle. For forasmuch as the angles GAE & BAE are equal (by the 27. of the third● for they are see upon equal circumferences): therefore the ●ases BK and KG are (by the ●. of the first) equal. Wherefore the line BT containeth 5. sixth parts of the line BG. Then I say that that which is contained under the lines AZ and BT, is equal to the pentagon A●C●G. For forasmuch as the line ●Z is sesquialter to the line AD (for the line D● is divided into two equal parts in the point Z, by the corollary of the ●2● of the thirteenth). Likewise by construction, the line KG is sesquialter to the line KT: therefore as the line AZ is to the line AD, so is the line KG to the 〈◊〉 ●T. Wherefore that which is contained vnde● the 〈◊〉 AZ and KT, is equal to that which is contained under the means AD and KG, by the 16. of the sixth. But unto the line KG is the line ●K ●roued equal. Wherefore that which is contained under the lines AZ and KT is equal to that which is contained under the lines AD and BK. But that which is contained under the lines AD and BK is (by the 41. of the first) double to the triangle ABD. Wherefore that which is contained under the lines AZ and KT is double to the same triangle ABD. And forasmuch as the pentagon ABCIG contayneth● 〈…〉 equal ●o the triangle ABD, and that which is contained under the lines AZ and KT containeth two such triangles: therefore the pentagon ABCIG is duple sesquialter to the rectangle parallelogram contained under the lines AZ and KT. And 〈…〉 1. of the sixth, that which is contained under the lines AZ and BT is to that which is contained under the lines AZ and KT, as the base BT is to the base ●●T● therefore that which is contained under the lines AZ and BT is duple sesquialter to that which is contained vn●●r the line AZ & KT. But unto that which is contained under the lines AZ and KT the pentagon ABCIG is proved duple sesquialter. Wherefore the pentagon ABCIG of the Dodecahedron is equal to that which is contained under the perpendicular line AZ, and under the line BT which is five six parts of the line BG.    

¶ The 7. Proposition. 
The 9 proposition after Campane.  A right line divided by an extreme and mean proportion: what proportion the line containing in power the whole line and the greater segment, hath to the line containing in power the whole and the less segment: the same hath the side of the cube to the side of the Icosahedron contained in one and the same sphere.  

Construction.  TAke a circle ABE: and in it (by the 11. of the fourth) inscribe an equilater pentagon BZECH: and (by the second of the same) an equilater triangle ABI. And let the centre thereof be the point G. And draw a line from G to B. And divide the line GB by an extreme and mean proportion in the point D (by the 30. of the sixth). And let the line ML contain in power both the whole line GB and his less segment BD (by the corollary of the 13. of the tenth). And draw the right line B● subtending the angle of the pentagon, which shall be the side of the cube (by the corollary of the 17. of the thirteenth) ● and the line BY shall be the side of the Icosahedron, and the line ●Z the side of the Dodecahedron by the 4. of this book. Then I say that BE the side of the cube is to BY the side of the Icosahedron, as the line containing in power the lines BG & GD is to the line containing in power the lines GB and BD. 
Demonstration.  For forasmuch as (by the 12. of the thirteenth) the line BY is in power triple to the line BG, and (by the 4. of the same) the squares of the line GB & BD are triple to the square of the line GD. Wherefore (by the 15. of the fifth) the square of the line BY is to the squares of the lines GB & BD (namely, triple to triple), as the square of the line B● is to the square of the line GD (namely, as one is to one). But as the square of the line BG is to the square of the GD, so is the square of the line BE to the square of the line BZ. (For the lines BG, GD, and BE, BZ are in one and the same proportion, by the second of this book. For BZ is the greater segment of the line BE, by the corollary of the 17. of the thirteenth). Wherefore the square of the line BE is to the square of the line BZ, as the square of the line BY is to the squares of the lines BG and BD. Wherefore alternately the square of the line BE is to the square of the line BY, as the square of the line BZ is to the squares of the lines GB and BD. But the square of the line BZ is equal to the squares of the lines BG and GD (by the 10. of the thirteenth). For the line BG is equal to the side of the hexagon, and the line GD to the side of the decagon, by the corollary of the 9 of the same. Wherefore the squares of the lines BG and GD, are to the squares of the lines G● and BD, as the square of the line BE is to the square of the line BY. But the line ZB containeth in power the lines BG and GD, and the line ML containeth in power the lines GB and BD by construction. Wherefore as the line ZB (which containeth in power the whole line BG and the greater segment GD) is to the line ML (which containeth in in power the whole line GB and the less segment BD) so is BE the side of the cube to BY the side of the Icosahedron, by the 22. of the sixth. Wherefore a right line divided by an extreme and mean proportion: what proportion the line containing in power the whole line and the greater segment, hath to the line containing in power the whole line and the less segment: the same hath the side of the cube to the side of the Icosahedron contained in one and the same sphere which was required to be proved.   

¶ The 8. Proposition. 
This Campane putteth a● a Corollary in the 9 proposition after his order.  The solid of a Dodecahedron is to the solid of an Icosahedron: as the side of a Cube is to the side of an Icosahedron, all those solids being described in one and the self same Sphere.  
FOrasmuch as in the 4. of this book, it hath been proved, that one and the self same circle containeth both the triangle of an Icosahedron, and the pentagon of a Dodecahedron described in one and the self same Sphere: Wherefore the circles, which contain those bases, being equal, the perpendiculars also which are drawn from the centre of the Sphere to those circles, shall be equal (by the Corollary of the Assumpt of the 16 of the twelfth). And therefore the Pyramids set upon the bases of those solids have one and the self same altitude: For the altitudes of those Pyramids concurr● in the centre. Wherefore they are in proportion as their bases are, by the 5. and 6. of the twelfth. And therefore the pyramids which compose the Dodecahedron, ar● to the pyramids which compose the Icosahedron, as the bases are, which bases are the superficieces of those solids. Wherefore their solids are the one to the other, as their superficieces are. But the superficies of the Dodecahedron is to the superficies of the Icosahedron, as the side of the cube is to the side of the Icosahedron, by the 6. of this book. Wherefore by the 11. of the fifth, as the solid of the Dodecahedron is to the solid of the Icosahedron, so is the side of the cube to the side of the Icosahedron, all the said solids being inscribed in one and the self same Sphere. Wherefore the solid of a Dodecahedron is to the solid of an Icosahedron: as the side of a cube is to the side of an Icosahedron, all those solids being described in one and the self same Sphere: which was required to be proved.  

A Corollary.  
The solid of a Dodecahedron is to the solid of an Icosahedron, 
This Corollary is the 9 proposition after Campane.  as the superficieces of the one are to the superficieces of the other, being described in one and the self same Sphere: Namely, as the side of the cube is to the side of the Icosahedron, as was before manifest: for they are resolved into pyramids of one and the self same altitude.    

¶ The 9 Proposition. If the side of an equilater triangle be rational: the superficies shall be irrational, 
The 12. proposition after Campane.  of that kind which is called medial.  
SVppose that ABG be an equilater triangle: and from the point A draw unto the side BG a perpendicular line AD: and let the line AB be rational. Then I say, that the superficies ABG is medial. 
Construction.  Forasmuch as the line AB is in power sesquitertia to the line AD (by the Corollary of the 12. of the thirteenth): of what parts the line AB containeth in power 12, of the same parts the line AD containeth in power 9: 
Demonstration.  wherefore the residue BD containeth in power of the same parts 3. (●or the line AB containeth in power the lines AD and BD, by the 47. of the first). Wherefore the lines AD and DB are rational and commensurable to the rational line set AB, by the 6. of the tenth. But forasmuch as the power of the line AD is to the power of the line DB in that proportion that 9 a square number is to 3. a number not square: therefore they are not in the proportion of square numbers, by the Corollary of the 25. of the eight. And therefore they are not commensurable in length, by the 9 of the tenth. Wherefore that which is contained under the lines AD and DB, which are rational lines commensurable in power only, is medial, by the 22. of the tenth. But that which is contained under the lines AD and DB, is double to the triangle ABD, by the 41. of the first. Wherefore that which is contained under the lines AD and DB, is equal to the whole triangle ABG (which is double to the triangle ABD, by the 1. of the sixth). Wherefore the triangle ABG is medial. If therefore the side of an equilater triangle be rational: the superficies shall be irrational, of that kind which is called medial: which was required to be proved.  

A Corollary.  
If an Octohedron and a Tetrahedron be inscribed in a Sphere whose diameter is rational: 
The 13. proposition after Campane.  their superficieces shall be medial. For those superficieces consist of equilater triangles, whose sides are commensurable to the diameter which is rational, by the 13. and 14. of the thirteenth, and therefore they are rational. But they are commensurable in power only to the perpendicular line, and therefore they contain a medial triangle, as it was before manifest.    

¶ The 10. Proposition. 
The 14. proposition after Campane.  If a Tetrahedron and an Octohedron be inscribed in one and the self same Sphere: the base of the Tetrahedron shall be sesquitertia to the base of the Octohedron, and the supersicieces of the Octohedron shall be sesquialtera to the superficieces of the Tetrahedron.  

Demonstration of the first part.  FOrasmuch as the diameter of the Sphere is in power sesquialtera to the side of the Tetrahedron (by the 13. of the thirteenth) and the same diameter is in power duple to the side of the Octohedron (by the 14. of the same book): therefore of what parts the diameter containeth in power six, of the same the side of the Tetrahedron containeth in power 4, and of the same the side of the Octohedron containeth in power 3. Wherefore the power of the side of the Tetrahedron, is to the power of the side of the Octohedron in the same proportion that 4. is to 3: which is sesquitertia. And like triangles (which are the bases of the solids) described of those sides, shall have the one to the other the same proportion that the squares made of those sides have. For both the triangles are the one to the other, and also the squares are the one to the other, in double proportion of that in which the sides are, by the 20. of the sixth. 
Demonstration of the second part.  Wherefore of what parts one base of the Tetrahedron was 4: of the same are four bases of the Tetrahedron 16: likewise of what parts of the same one base of the Octohedron was 3: of the same are 8. bases of the Octohedron 24. Wherefore the bases of the Octohedron are to the bases of the Tetrahedron, in that proportion that 24. is to 16: which is sesquialtera. If therefore a Tetrahedron and an Octohedron be inscribed in one and the self same Sphere: the base of the Tetrahedron shall be sesquitertia to the base of the Octohedron, and the superficieces of the Octohedron shall be sesquialtera to the superficieces of the Tetrahedron: which was required to be proved.   

¶ The 11. Proposition. A Tetrahedron is to an Octohedron inscribed in one and the self same Sphere, 
The 17. proposition after Campane.  in proportion, as the rectangle parallelogram contained under the line, which containeth in power 27. sixty four parts of the side of the Tetrahedron, & under the line which is subsesquiocta●a to the same side of the Tetrahedron, is to the square of the diameter of the sphere.  

Fir●t part of the construction.  LEt us suppose a Sphere, whose diameter let be the line AB, and let the centre be the point H. And in it let there be inscribed a Tetrahedron ADC, and an Octohedron AEKBG. And let the line NL contain in power ●7/64 of AC the side of the Tetrahedron. And let the line ML be in length subsesquioctava to the same side. Then I say, that the Tetrahedron ACD is to the Octohedron AEB, as the rectangle parallelogram contained under the lines NL and LM, 
First part of the Demonstration.  is to the square of the line AB. Forasmuch as the line drawn from the angle A by the centre H perpendicularly upon the base of the Tetrahedron, falleth upon the centre T of the circle which containeth that base, and maketh the right line HT the sixth part of the diameter AB (by the Corollary of the 13. of the thirteenth): therefore the line HA (which is drawn from the centre to the circumference) is triple to the line HT: and therefore the whole line AT is to the line AH, 〈…〉. Let the Tetrahedron ADC be cut by a plain GHK passing by the centr● H, and being parallel unto the base DTC, by the Corollary of the 11. of the eleventh. Now than the triangle ADC of the Tetrahedron, shall be cut by the right line KG, which is parallel to the line DC, by the 16. of the eleventh. Wherefore as the line AT is to the line AH, so is the line AC to the line AG (by the 2. of the sixth). Wherefore the line AC is to the line AG sesquitertia, that is, as 4. to 3. And forasmuch as the triangles ADC, AKG, and the rest which are cut by the plain KHG, are like the one to the other, by the 5● of the sixth: the pyramids ADC and AKG, shall be like the one to the other, by the 7. definition of the ●leuenth. Wherefore they are in triple proportion of that in which the sides AC and AG are, by the 8. of the twelfth. But the proportion of the sides AC to AG is, as the proportion of 4. to 3. Now then, if, by the 2. of the eight, ye find out 4. of the jest numbers in continual proportion, and in that proportion that 4. is to 3: which shall be 64.48.36. and 27: it is manifest, by the 15 definition of the fifth, that the extremes ●4. to 27. are in triple proportion of that in which the proportion given 4. to 3. is: Or the quantity of the proportion of 4. to 3. (which is 1. and 1/●) being twice multiplied into itself, there shall be produced the proportion of 64. to 27. Wherefore the Pyramid or Tetrahedron ADC is to the pyramid AKG, as 64. is to ●7: which is triple to the proportion of 4. to 3. And forasmuch as the line AC is unto the line AG in length sesquitertia: of what parts the line AC containeth in power 64: of the same parts doth the line AG contain in power 36. For (by the 2. of the sixth) the proportion of the powers or squares, is duple to the porportion of the sides which are as 64. is to 48.  
Now then upon the line RS which let be equal to the line AG, let there be an equilater triangle QRS described (by the first of the first). 
Second par● of the construction.  And from the angle Q, draw to the base RS a perpendicular line QT. And extend the line RS to the point X. And as 27. is to 64. (so by the corollary of the 6. of the tenth) let the line RS be to the line RX. And divide ●he line RX into two equal parts in the point V, and draw the line QU. And forasmuch as the line RS is equal to the line AG, of what parts the line AC containeth in power 64. of the same part the line RS containeth in power 36. for it is proved that the line AG containeth in power 36. of those parts: 
Second part of the Demonstration.  And of what parts the line RS containeth in power 36 of the same parts the 〈◊〉 QT containeth in power ●7. by the corollary of the 12. of the thi●tenth. Wherefore of what parts the line AZ containeth in power 64. of the same parts the line QT containeth in power 27. Wherefore the right line QT shall be equal to the right line LN by supposition. Again forasmuch as the line RS is put equal to the line AG: and of what parts the line RS containeth in length 27. of the same parts is the line RX put to contain in length 64. and of what parts the line RX containeth in length 64. of the same the line AC (which is in length sesquitertia to the line AG or RS) containeth 36. Wherefore the line RV (which is the half of the line RX) containeth in length of the same parts 32. of which the line AC contained in length 36. Wherefore the line RV is to the line AC subsesquioctava: and therefore the line RV is equal to the line LM which is also subsesquioctava to the same line AC. And forasmuch as the line NL is equal to the line QT, and the line LM to the line RV (as before hath been proved) the rectangle parallelogram contained under the lines QT and RV, shall be equal to the rectangle parallelogram, contained under the line NL which is in power ●7/64 to the side AC, and under the line LM, which is in length subsesquioctava to the same side AC. But that which is contained under the lines QT and RV is double to the triangle QVR by the 41. of the first: and to the same triangle QVR is the triangle QXR duple by the first of the sixth. Wherefore the whole triangle QXR is equal to that which is contained under the lines QT and RV, and therefore is equal to the parallelogram MN. And forasmuch as the line RX by supposition containeth in length 64. of those parts of which the line RS containeth 27: and the triangles QRX, and QRS are, by the first of the sixth, in the proportion of their bases, that is, as 64. is to 27: but as 64. is to 27. so is the pyramid or tetrahedron ADC to the pyramid AKG: wherefore as the parallelogram NM or the triangle QRX, is to the triangle QRS, so i● the pyramid ADC to the pyramid AKG. And forasmuch as the semidiameter AH is the altitude of the pyramid AKG, and also of the two equal and like pyramids of the octohedron which have their common base in the square of the octohedron (by the corollary of the 14. of the thirteenth): therefore as the base of the pyramid AKG (which is the triangle QRS) is to two squares of the octoh●dron, that is, to the square of the diameter AB, which is equal to those squares (by the 47. of the first), so is the pyramid AKG to the octohedron AEB, by the 6. of the twelfth. And forasmuch as the parallelogram MN is to the base QRS, as the pyramid ADC is to the pyramid AKG, and the base QRS is to the square of the line BE, as the pyramid AKG is to the octohedron AEB: therefore by proportion of equality ta●ing away th● meane● (by the 22. of the fifth) as the parallelogram NM is to the square of the line BE, so is the pyramid ADC to the octohedron AEB inscribed in one and the self same sphere. But the parallelogram NM is contained under the line NL which by supposition is in power ●7/●● to AC the side of the tetrahedron ADC, and under the line LM which is also by supposition in length subsesquioctava to the same line AC. Wherefore a tetrahedron & an octohedron inscribed in one and the self same sphere, are in proportion, as the rectangle parallelogram contained under the line, which containeth in power 27. sixty four parts of the side of the Tetrahedron, and under the line which is subsesquioctava to the same side of the Tetrahedron, is to the square of the diameter of the sphere: which was required to be proved.   

¶ The 12. Proposition. 
The 18. proposition after Campane.  If a cube be contained in a sphere: the square of the diameter doubled, is equal to all the superficieces of the cube taken together. And a perpendicular line drawn from the centre of the sphere to any base of the cube, is equal to half the side of the cube.  

Demonstration of the first part.  FOr forasmuch as (by the 15. of the thirteenth) the diameter of the sphere is in power triple to the side of th● cube: therefore the square of the diameter doubled is sextuple to the base of the same cube. But the sextuple of the power of one of the sides containeth the whole superficies of the cube 〈◊〉 or the cube is composed of six square superficieces (by the 2●. definition of th● eleventh) whose sides therefore are equal: wherefore the square of the diameter doubled is equal to the whole superficies of the cube. And forasmuch as the diameter of the cube, and the line which falleth perpendicularly upon the opposite bases of the cube, 
Demonstration of the second part.  do cut the one the other into two equal parts in the centre of the sphere which containeth the cube (by the 2. corollary of the 15. of the thirteenth) and the whole right line which coupleth the centres of the opposite bases, is equal to the side of the cube by the 33. of the first, for it coupleth the equal and parallel semidiameters of the bases: therefore the half thereof shall be equal to the half of the side of the cube by the 15. of the fifth. If therefore a cube be contained in a sphere: the square of the diameter doubled is equal to all the superficieces of the cube taken together. And a perpendicular line drawn from the centre of the sphere to any base of the cube, is equal to half the side of the cube: which was required to be prou●d.  

¶ A Corollary.  
If two thirds of the power of the diameter of the sphere be multiplied into the perpendicular line equal to half the side of the cube, 
The Corollary of the 8. proposition after Campane.  there shall be produced a solid equal to the solid of the cube. For it is before manifest that two third parts of the power of the diameter of the sphere are equal to two bases of the cube. If therefore unto each of those two thirds be applied half the altitude of the cube, they shall make each of those solids equal to half of the cube, by the 31. of the eleventh: for they have equal bases. Wherefore two of those solids are equal to the whole cube.  
You shall understand (gentle reader) that Campane in his 14. book of Euclides Elements hath 18. propositions with divers corollaries following of them. Some of which propositions and corollaries I have before in the twelfth and thirteenth books added out of Flussas as corollaries (which thing also I have noted on the side of those corollaries, namely, with what proposition or corollary of Campanes 14. book they do agreed). The rest of his 18. propositions and corollaries are contained in the twelve former propositions and corollaries of this 14. book after Flussas: where ye may see on the side of each proposition and corollary with what proposition and corollary of Campanes they agreed. But the eight propositions following together with their corollaries, Flussas hath added of himself, as he himself affirmeth.    

The 13. Proposition. One and the self same circle containeth both the square of a cube, and the triangle of an Octohedron described in one and the self same sphere.  
SVppose that there be a cube ABG, and an Octohedron DEF described in one and the self same sphere, whose diameter let be AB, or DH. And let the lines drawn from the centres (that is the semidiameters of the circles which ctonaine the bases of those solids) ● be CA and ID. Then I say that the lines CA and ID are equal. Forasmuch as AB the diameter of the sphere which containeth the cube, 
Demonstration.  is in power triple to BG the side of the cube (by the 15. of the thirteenth) unto which side, AG the diameter of the base of the cube, is in power double (by the 47. of the first): which line AG is also the diameter of the circle, which containeth the base (by the 9 of the fourth:) therefore AB the diameter of the sphere is in power sesquialter to the line AG: namely, of what parts the line AB, containeth in power 12. of the same the line AG, shall contain in power 8. And therefore the right line AC which is drawn from the centre of the circle to the circumference, containeth in power of the same parts 2. Wherefore the diameter of the sphere is in power sextuple to the line which is drawn from the centre to the circumference of the circle which containeth the square of the cube But the Diameter of the self same Sphere which containeth the Octohedron, is one and the self same with the diameter of the cube, namely, DH, is equal to AB: and the same diameter is also the diameter of the square which is made of the sides of the Octohedron: wherefore the said diameter is in power double to the side of the same Octohedron, by the 14. of the thirteenth. But the side DF is in power triple to the line drawn from the centre to the circumference of the circle which containeth the triangle of the octohedron (namely to the line ID) by the 12. of the thirteenth. Wherefore the self same diameter AB or DH, which was in power sextuple to the line drawn from the centre to the circumference of the circle which containeth the square of the cube, is also sextuple to the line ID drawn from the centre to the circumference of the circle, which containeth the triangle of the Octohedron: Wherefore the lines drawn from the centres of the circles to the circumferences which contain the bases of the cube and of the octohedron are equal. And therefore the circles are equal, by the first definition of the third. Wherefore one and the self same circle containeth etc. as in the proposition: which was required to be proved.  

A Corollary.  
Hereby it is manifest, that perpendiculars coupling together in a sphere, the centres of the circles which contain the opposite bases of the cube and of the Octohedron, are equal. For the circles are equal, by the second corollary of the assumpt of the 16. of the twelfth: and the lines which passing by the centre of the sphere, couple together the centres of the bases, are also equal, by the first corollary of the same. Wherefore the perpendicular which coupleth together the opposite bases of the Octohedron, is equal to the side of the cube. For either of them is the altitude erected.    

The 14. Proposition. An Octohedron is to the triple of a Tetrahedron contained in one and the self same sphere, in that proportion that their sides are.  
SVppose that there be an octohedron ABCD, and a Tetrahedron EFGH: upon whose base FGH erect a Prism, 
Construction.  which is done by erecting from the angles of the base perpendicular lines equal to the altitude of the Tetrahedron: which prism shallbe triple to the Tetrahedron EFGH, by the first corollary of the 7. of the twelfth. Then I say that the octohedron ABCD is to the prism which is triple to the Tetrahedron, EFGH, as the side BC is to the side FG. For forasmuch as the sides of the opposite bases of the octohedron, 
Demonstration.  are right lines touching the one the other, and are parellels to other right lines touching the one the other, for the sides of the squares which are composed of the sides of the octohedron, are opposite: Wherefore the opposite plain triangles, namely, ABC & KID, shallbe parallels, and so the rest by the 15. of the eleventh. Let the diameter of the Octohedron, be the line AD. Now then the whole Octohedron is cut into four equal and like pyramids set upon the bases of the octohedron, and having the same altitude with it, & being about the Diameter AD: namely the pyramid set upon the base BID, and having his top the point A, and also the pyramid set upon the base BCD, having his top the same point A. Likewise the pyramid set upon the base IKD, & having his top the same point A: and moreover the pyramid set upon the base CKD, and having his top the former point A: which pyramids shallbe equal by the 8. definition of the eleventh (for they each consist of two bases of the octohedron, and of two triangles contained under the diameter AD and two sides of the octohedron). Wherefore the prism which is set upon the base of the Octohedron, and having the same altitude with it, namely, the altitude of the parallel bases, as it is manifest by the former, is equal to three of those pyramids of the Octohedron, by the first corollary of the seventh of the twelft. Wherefore that prism shall have to the other prism under the same altitude, composed of the 4. pyramids of the whole octohedron, the proportion of the triangular bases, by the 3. corollary of the same. And forasmuch as 4. pyramids are unto 3. pyramids in sesquitercia proportion, therefore the trianguler base of the prism which containeth 4. pyramids, is in sesquitertia proportion to the base of the prism which containeth three pyramids of the same octohedron, and are set upon the base of the Octohedron and under the altitude thereof: that is, in sesquitercia proportion to the base of the Octohedron. But the base of the same octohedron is in sesquitertia proportion to the base of the pyramid, by the ●enth of this book: Wherefore the triangular bases, namely, of the prism which containeth four pyramids of the octohedron, and is under the altitude thereof, are equal to the triangular bases of the prism, which containeth three pyramids under the altitude of the pyramid EFGH. But the prism of the octohedron is equal to the octohedron: and the prism of the pyramid EFGH is proved triple to the same pyramid EFGH. Now than the prisms set upon equal bases, are the one to the other as their altitudes are (by the corollary of the 25. of the eleventh) namely, as are the parallelipidedons their doubles, by the corollary of the 31. of the eleventh. But the altitude of the Octohedron is equal to the side of the cube contained in the same sphere, by the corollary of the 13. of this book. And the side of the cube is in power to the altitude of the Tetrahedon in that proportion that 12. is to 16, by the 18. of the thirteenth: And the side of the octohedron is to the side of the pyramid in that proportion that 18. is to 24. (by the same 18. of the thirteenth) which proportion is one & the self same with the proportion of 12. to 16. Wherefore that prism which is equal to the Octohedron, is to the prism which is triple to the Tetrahedron, in that proportion that the altitudes, or that the sides are. Wherefore an octohedron is to the triple of a Tetrahedron contained in one and the self same sphere, in that proportion that their sides are: which was required to be demonstrated.  

A Corollary.  
The sides of a Tetrahedron & of an Octohedron are proportional with their altitudes. For the sides & altitudes were in power sesquitercia. Moreover the diameter of the sphere is to the side of the Tetrahedron, as the side of the Octohedron is to the ●●de of the cube● namely, the powers of each is in sesquialter proportion, by the 18. of the thirteenth.    

The 15. Proposition. If a rational line containing in power two lines, make the whole and the greater segment, and again containing in power two lines, make the whole and the less segment: the greater segment shallbe the side of the Icosahedron, and the less segment shallbe the side of the Dodecahedron contained in one and the self same sphere.  
SVppose that AG be the diameter of the sphere which containeth the Icosahedron ABGC. And let BG subtend the sides of the pentagon described of the sides of the Icosahedron (by the 16. of the thirteenth.) Moreover upon the same diameter AG, or DF equal unto it, 
Construction.  let there be described a dodecahedron DEFH, by the 1●. of the thirteenth, whose opposites sides ED and FH let be cut into two equal parts in the points I and K, and draw a line from I to K. And let the line EF couple two of the opposite angles of the bases which are joined together. Then I say that AB the side of the Icosahedron is the greater segment which the diameter AG containeth in power together with the whole line, and line ED is the less segment, which the same diameter AG or DF containeth in power together with the whole. 
Demonstration.  For forasmuch as the opposite sides AB and GC of the Icosahedron being coupled by the diameters AG and BC, are equal & parallels, by the 2. corollary of the 16. of the thirteenth: the right lines BG & AC which couple them together are equal & parallels by the 33. of the first. Moreover the angles BAC & ABG being subtended of equal diam●ters, shall by the 8. of the first be equal, & by the 29 of the 〈◊〉, they shall be right angles. Wherefore the right line AG 〈◊〉 in power the too lines AB and BG, by the 47. of 〈…〉. And forasmuch as the line BG subtendeth the angle of the pentagon composed of the sides of the Icosahedron, the greater segment of the right line BG, shallbe the right line AB, by the ●. of the thirteenth: which line AB, together with the whole line BG, the line AG containeth in power. And forasmuch a● the line IK coupling the opposite and parallel sides ED and FH of the Dodecahedron, maketh at those points right angles, by the 3. corollary of the 17. of t●e thirteenth: the right line EF which coupleth together equal and parallel lines EI & FK, shallbe equal to the same line IK, by the 33. of the first. Wherefore the angle DEF shallbe ● right angle by the 29. of the first. Wherefore the diameter DF containeth in power the two lines ED and EF. But the less segment of the line IK is ED the side of the Dodecahedron, by the 4. corollary of the 17● of the thirteenth. Wherefore the same line ED is also the less segment of the line EF (which is equal unto the line IK): wherefore the diameter DF containing in power the two lines ED and EF (by the 47. of the first) containeth in pow●r● ED the side of the dodecahedron, the less segment, together with the whole. If therefore a rational line AG or DF containing in power two lines AB and BG, do make the whole line and the greater segment, and again containing in power two lines EF and ED, do make the whole line and the less segment: the greater segment AB, shall be the side of the Icosahedron, and the less segment ED shall be the side of the Dodecahedron contained in one and the self same sphere.   

The 16. Proposition. If the power of the side of an Octohedron be expressed by two right line● joined together by an extreme and me●ne proportion: the side of the Icosahedron contained in the same sphere, shallbe duple to the less segment.  
LEt AB the side of the Octohedron ABG contain in power the two lines C and H, which let have that proportion that the whole hath to the greater segment (by the corollary of the first proposition added by Flussas after the last proposition of the sixth book). 
Construction.  And let the Icosahedron contained in the same sphere be DEF, whose side let be DE, and let the right line subtending the angle of the pentagon made of the sides of the Icosahedron be the line EF. Then I say that the side ED is in power double to the line H the less of those segments. Forasmuch as by that which was demonstrated in the 15. of this book, 
Demonstration.  it was manifest that ED the side of the Icosahedron is the greater segment of the line EF● and that the diameter DF containeth in power the two lines ED and EF, namely, the whole and the greater segment: but by supposition the side AB containeth in power the two lines C & H joined together in the self same proportion. Wherefore the line EF is to the line ED, as the line C, is to the line H, by the ●. o● this boke● And alternally by the 16. of the fifth, the line EF is to the line C, as the line ED, is to the line H. And forasmuch as the line DF containeth in power the two lines ED and EF, and the line AB containeth in power the two lines C and H: therefore the squares of the lines EF and ED are to the square of the line DF, as the squares of the lines C and H to the square AB. And alternately, the squares of the lines EF and ●D, are to the squares of the lines C and H, as the square of the line DF is to the square of the line AB● But DF the diameter is (by the 14. of the thirten●h) i● power double to AB the side of the octohedron inscribed (by supposition) in the same sphere. Wherefore the squares of the lines EF and ED, are double to the squares of the lines C and H. And therefore one square of the line ED is double to one square of the line H by the 12. of the fifth. Wherefore ED the side of the Icosahedron is in power duple to the line H, which is the less segment. If therefore the powe● of the side of an octohedron be expressed by two right lines joined together by an extreme and mean proportion: the side of the Icosahedron contained in the same sphere, shallbe duple to the less segment.   

The 17. Proposition. If the side of a dodecahedron, and the right line, of whom the said side is the less segment, be so set that they make a right angle: the right line which containeth in power half the line subtending the angle, is the side of an Octohedron contained in the self same sphere.  
SVppose that AB be the side of a Dodecahedron, and let the right line of which that side is the less segment be AG, namely which coupleth the opposite sides of the Dodecahedron, by the 4. corollary of the 17. of the thirteenth: 
Construction.  and let those lines be so set that they make a right angle at the point A. And draw the right line BG. And let the line D contain in power half the line BG (by the first proposition added by Flussas after the last of the sixth). Then I say that the line D is the side of an Octohedron contained in the same sphere. Forasmuch as the line AG maketh the greater segment GC the side of the cube contained in the same sphere (by the same 4. corollary of the 17. of the thirteenth): 
Demonstration.  and the squares of the whole line AG. and of the less segment AB are triple to the square of the greater segment GC, by the 4. of the thirteenth: Moreover the diameter of the sphere, is in power triple to the same line GC the side of the cube (by the 15. of the thirteenth: Wherefore the line BG is equal to the 〈◊〉. For it containeth in power the two lines AB and AG (by the 47. of the first,) and therefore it containeth in power the triple of the line GC. But the side of the Octohedron contained in the same sphere, is in power triple to half the diameter of the sphere by the 14. of the thirteenth. And by supposition the line D contai●●●● in pow●● the half of the line BG. Wherefore the line D (containing in power the half of the same diameter) is the side of an octohedron. If therefore the side of a Dodecahedron and the right line of whom the said side is the less segment, be so set that they make a right angle: the right line which containeth in power half the line subtending the angle, is the side of an Oc●●●edron contained in the self same sphere: Which was required to be proved.  

A Corollary.  
Unto what right line the side of the Octo●edron is in power sesquialter: unto the same line the side of the Dodecahedron inscribed in the same sphere, is the greater segment. For the side of the Dodecahedron is the greater segment of the segment CG, unto which D the side of the Octohedron is in power sesquialter, that is, is half of the power of the line BG, which was triple unto the line CG.    

¶ The 18. Proposition. If the side of a Tetrahedron contain in power two right lines joined together by an extreme and mean proportion: the side of an Icosahedron described in the self same Sphere, is in power sesquialter to the less right line.  
SVppose that ABC be a Tetrahedron, and let his side be AB, 
Construction.  whose power let be divided into the lines AG and GB, joined together by an extreme and mean proportion: namely, let it be divided into AG the whole line, and GB the greater se●ment (by the Corollary of the first Proposition added by Flussas after the last of the sixth). And let ED be the side of the Icosahedron EDF contained in the self same Sphere. And let the line which subtendeth the angle of the Pentagon described of the sides of the Icosahedron be EF. Then I say, that ED the side of the Icosahedron is in power sesquialter to the less line GB. 
Demonstration.  Forasmuch as (by that which was demonstrated in the 15. of this book) the side ED is the gre●ter segment of the line EF which subtendeth the angle of the Pentagon. But as the whole line EF is to the greater segment ED, so is the same gr●●ter segment to the less (by the 30. of the sixth): and by supposition, AG, was the whole line, and G● the greater segment: Wherefore as EF is to ED, so is AG to G●, by the second of the fourteenth. And alternately, the line EF is to the line AG, as the line ED is to the line GB. And forasmuch as (by supposition) the line AB containeth in power the two lines AG and GB: therefore (by the 4●. of the first) the angle AGB is a right angle. But the angle DEF is a right angle, by that which was demonstrated in the 15. of this book. Wherefore the triangles AG● and FED, are equiangle, by the ●. of the sixth. Wherefore their sides are proportional: namely, as the line ED is to the line GB, so is the line FD to the line AB, by the 4. of the sixth. But by that which hath before been demonstrated, FD is the diameter of the Sphere which containeth the Icosahedron: which diameter is in power sesquialter to AB the side of the Tetrahedron inscribed in the● same Sphere, by the 13. of the thirteenth. Wherefore the line ED the side of the Icosahedron, is in power sesquialter to G● the greater segment or less line. If therefore the side of a Tetrahedron contain in power two right lines joined together an extreme and mean proportion: the side of an Icosahedron described in the self same Sphere, is in power sesquialter to the less right line.   

¶ The 19 Proposition. The superficies of a Cube is to the superficies of an Octohedron inscribed in one and the self same Sphere, in that proportion that the solids are.  

Construction.  SVppose that ABCDE be a Cube, whose four diameters let be the lines AC, BC, DC, and EC produced on each side. Let also the Octohedron inscribed in the self same Sphere be FGHK: whose three diameters let be FH, GK, and ON. Then I say, that the cube ABD is to the Octohedron FGH, as the superficies of the cube is to the superficies of the Octohedron. Draw from the centre of the cube to the base ABED, a perpendicular line CR. And from the centre of the Octohedron draw to the base GNH, a perpendicular line ●L. 
Demonstration.  And forasmuch as the three diameters of the cube do pass by the 〈◊〉 C, therefore, by the 2. Corollary of the 15. of the thirteenth, ●here shall be made of the cube six pyramids, as this pyramid ABDEC, equal to the whole cube. For there are in the cube ●ixe bases, upon which fall equal perpendiculars from the cen●●● by the Corollary of the Assumpt of the 16. of the twelfth, for the bases are contained in equal circled of the Sphere. But in the Octohedron the three diameters do make upon the 8. bases, 8. pyramids, having their tops in the centre, by the 3. Corollary of the 14● of the thirteenth. Now the bases of the cube and of the Octohedron are contained in equal circles of the Sphere, by the 13. of this book. Wherefore they shall be equally distant from the centre, and the perpendicular lines CR and ●, shall be equal, by the Corollary of the Assumpt of the 16. of the twelfth. Wherefore the pyramids of the cube shall be under one and the self same altitude with the pyramids of the Octohedron, namely, under the perpendicular line drawn from the centre to the bases. Wherefore six pyramids of the cube, are to 8. pyramids of the Octohedron being under one and the same altitude, in that proportion that their bases are, by the 6. of the twelfth: that is, one pyramid set upon six bases of the cube, and having to his altitude the perpendicular line, which pyramid is equal to the six pyramids, by the same 6. of the twelfth, is to one pyramid set upon the 8. bases of the Octohedron, being equal to the Octohedron, and also under on● and the self same altitude, in that proportion that six bases of the cube, which contain the whole superficies of the cube, are to 8. bases of the Octohedron● which contain the whole superficies of the Octohedron. For the solids of those pyramids are in proportion the one to the other, as their bases are, by the self same 6. of the twelfth. Wherefore ●he superficies of the cube is to the superficies of the Octohedron inscribed in one and the self same Sphere, in that proportion, that the solids are: which was required to be proved.   

¶ The 20. Proposition. If a Cube and an Octohedron be contained in one & the self same Sphere: they shall be in proportion the one to the other, as the side of the Cube is to the semidiameter of the Sphere.  
SVppose that the Octohedron AECDB be inscribed in the Sphere ABCD: and let the cube inscribed in the same Sphere be FGHIM: whose diameter let be high, 
Construction.  which is equal to the diameter AC, by the 15. of the thirteenth: let the half of the diameter be AE. Then I say, that the cube FGHIM is to the Octohedron AECDB, as the side MG is to the semidiameter AE. Forasmuch as the diameter AC is in power double to BK the side of the Octohedron (by the 14. of the thirteenth) and is in power triple to MG the side of the cube (by the 15. of the same): therefore the square BKDL shall be sesquialter to FM the square of the cube. From the line AE cut of a third part AN, and from the line MG cut of likewise a third part GO, by the 9 of the sixth. 
Demonstration.  Now than the line EN shall be two third parts of the line AE, and so also shall the line MO be of the line MG. Wherefore the parallelipipedon set upon the base BKDL, and having his altitude the line EA, is triple to the parallelipipedon set upon the same base, and having his altitude the line AN, by the Corollary of the 31. of the eleventh: but it is also triple to the pyramid ABKDL which is set upon the same base, and is under the same altitude (by the second Corollary of the 7. of the twelfth). Wherefore the pyramid ABKDL is equal to the parallelipipedon, which is set upon the base BKDL, and hath to his altitude the line AN. But unto that parallelipipedom, is double the parallelipipedon which is set upon the same base BKDL, and hath to his altitude a line double to the line EN, by the Corollary of the 31. of the first and unto the pyramid is double the Octohedron ABKLDC, by the 2. Corollary of the 14. of the thirteenth. Wherefore the Octohedron ABKDLC is equal to the parallelipipedon set upon the base BKLD, & having his altitude the line EN (by the 15. of the fifth). But the parallelipipedon set upon the base BKDL, which is sesquialter to the base FM, and having to his altitude the line MO, which is two third parts of the side of the cube MG, is equal to the cube FG: by the 2. part of the 34. of the eleventh. (For it was before proved that the base BKDL is sesquialter to the base FM). Now than these two parallelipipedons, namely, the parallelipipedon which is set upon the base BKDL (which is sesquialter to the base of the cube) and hath to his altitude the line MO (which is two third parts of MG the side of the cube) which parallelipipedon is proved equal to the cube, and the parallelipipedon set upon the same base BKDL, and having his altitude the line EN (which parallelipipedon is proved equal to the Octohedron): these two parallelipipedons (I say) are the one to the other, as the altitude MO, is to the altitude EN (by the Corollary of the 31. of the eleventh). Wherefore also as the altitude MO, is to the altitude EN, so is the cube FGHIM, to the Octohedron ABKDLC, by the 7. of the fifth. But as the line MO is to the line EN, so is the whole line MG to the whole line EA, by the 18. of the fifth. Wherefore as MG the side of the cube, is to EA the semidiameter, so is the line FGHIM to the Octohedron ABKDLC inscribed in one & the self same Sphere. If therefore a cube and an Octohedron be contained in one and the self same Sphere. they shall be in proportion the one to the other, as the side of the cube is to the semidiameter of the Sphere: which was required to be demonstrated.  

A Corollary.  
Distinctly to notify the powers of the sides of the five solids by the power of the diameter of the sphere.  
The sides of the tetrahedron and of the cube do cut the power of the diameter of the sphere into two squares which are in proportion double the one to the other. The octohedron cutteth the power of the diameter into two equal squares. The Icosahedron into two squares, whose proportion is duple to the proportion of a line divided by an extreme and mean proportion, whose less segment is the side of the Icosahedron. And the dodecahedron into two squares, whose proportion is quadruple to the proportion of a line divided by an extreme and mean proportion, whose less segment is the side of the dodecahedron. For AD the diameter of the sphere, containeth in power AB the side of the tetrahedron, and BD the side of the cube, which BD is in power half of the side AB. The diameter also of the sphere containeth in power AC and CD two equal sides of the octohedron. But the diameter containeth in power the whole line AE and the greater segment thereof ED, which is the side of the Icosahedron, by the 15. of this book. Wherefore their powers being in duple proportion of that in which the sides are, by the first corollary of the 20. of the sixth, have their proportion duple to the proportion of an extreme & mean proportion. Farther the diameter containeth in power the whole line AF, and his less segment FD, which is the side of the dodecahedron, by the same 15. of this book. Wherefore the whole having to the less, ● double proportion of that which the extreme hath to the mean, namely, of the whole to the greater segment, by the 10. definition of the fifth, it followeth that the proportion of the power is double to the doubled proportion of the sides, by the same first corollary of the 20. of the sixth: that is, is quadruple to the proportion of the extreme and of the mean, by the definition of the sixth.  

An advertisement added by Flussas.  
By this means therefore, the diameter of a sphere being given, there shall be given the side of every one of the bodies inscribed. And forasmuch as three of those bodies have their sides commensurable in power only, and not in length, unto the diameter given (for their powers are in the proportion of a square number to a number not square: wherefore they have not the proportion of a square number to a square number, by the corollary of the 25. of the eight: wherefore also their sides are incommensurabe in length by the 9 of the tenth): therefore it is sufficient to compare the powers and not the lengths of those sides the one to the other● which powers are contained in the power of the diameter: namely, from the power of the diameter, let there ble taken away the power of the cube, and there shall remain the power of the Tetrahedron: and taking away the power of the Tetrahedron, there remaineth the power of the cube: and taking away from the power of the diameter half the power thereof, there shall be left the power of the side of the octohedron. But forasmuch as the sides of the dodecahedron and of the Icosahedron are proved to be irrational (for the side of the Icosahedron is a less line, by the 16. of the thirteenth: and the side of the dedocahedron is a residual line, by the 17. of the same) therefore those sides are unto the diameter which is a rational line set, incommensurable both in length and in power. Wherefore their comparison can not be defined or described by any proportion expressed by numbers, by the 8. of the tenth: neither can they be compared the one to the other: for irrational lines of divers kinds are incommensurable the one to the other: for if they should be commensurable, they should be of one and the self same kind, by the 103. and 105. of the tenth, which is impossible. Wherefore we seeking to compare them to the power of the diameter, thought they could not be more aptly expressed, then by such proportions, which cut that rational power of the diameter according to their sides: namely, dividing the power of the diameter by lines which have that proportion, that the greater segment hath to the less, to put the less segment to be the side of the Icosahedron: & dividing the said power of the diameter by lines having the proportion of the whole to the less segment, to express the side of the dodecahedron by the less segment: which thing may well be done between magnitudes incommensurable.     
The end of the fourteenth Book of Euclides Elements after Flussas.   

¶ The fifteenth book of Euclides Elements.  
THis finetenth and last book of Euclid, or rather the second book of Appollonius or Hypsicles, 
The argument of the 15. book.  teacheth the inscription and circumscription of the five regular bodies one within and about an other: a thing undoubtedly pleasant and delectable in mind to contemplate, and also profitable and necessary in act to practise. For without practice in act, it is very hard to see and conceive the constructions and demonstrations of the propositions of this book, unless a man have a very deep, sharp, & fine imagination. Wherefore I would wish the diligent student in this book, (to make the study thereof more pleasant unto him) to have presently before his eyes, the bodies formed & framed of pasted paper (as I taught after the definitions of the eleventh book.) And then to draw and describe the lines and divisions, and superficieces, according to the constructions of the propositions. In which descriptions if he be wary and diligent, he shall find all things in these solid matters, as clear and as manifest unto the eye, as were things before taught only in plain or superficial figures. And although I have before in the twelfth book admonished the reader hereof, yet because in this book chief that thing is required, I thought it should not be irksome unto him, again to be put in mind thereof.  
Farther this is to be noted, that in the Greek exemplars are found in this 15. book only 5. propositions, which 5. are also only touched and set forth by Hypsicies: unto which Campane addeth 8. and so maketh up the number of 13. Campane undoubtedly although he were very well learned, and that generally in all kinds of learning, yet assuredly being brought up in a time of rudeness, when all good letters were darkened, & barberousnes had overthrown and overwhelmed the whole world, he was utterly rude and ignorant in the Greek tongue, so that certainly he never red Euclid in the Greek, nor (of like) translated out of the Greek: but had it translated out of the Arabike tongue. The Arabians were men of great study, and industry, and commonly great Philosophers, notable Physicians, and in mathematical Arts most expert, so that all kinds of good learning flourished and reigned amongst them in a manner only. These men turned whatsoever good author was in the Greek tongue (of what Art and knowledge so ever it were) into the Arabike tongue. And from thence were many of them turned into the Latin, and by that means many Greek authors came to the hands of the Latins, and not from the first fountain the Greek tongue, wherein they were first written. As appeareth by many words of the Arabike tongue yet remaining in such books: as are Zenith, nadir, helmuayn, helmuariphe, and infinite such other. Which Arabians also in translating such Greek works, were accustomed to add, as they thought good, & for the fuller understanding of the author, many things: as is to be seen in divers authors, as, namely, in Theodosius de Sphera, where you see in the old translation (which was undoubteldy out of the Arabike) many propositions, almost every third or fourth leaf. Some such copy of Euclid, most likely, did Campanus follow, wherein he found those propositions, which he hath more & above those which are found in the Greek set out by Hypsicles: and that not only in this 15. book, but also in the 14. book, wherein also ye find many propositions more than are found in the Greek, set out also by Hypsicles. Likewise in the books before, ye shall find many propositions added, and many inverted, and set out of order far otherwise, than they are placed in the Greek examplars. Flussas' also a diligent restorer of Euclid, a man also which hath well deserved of the whole Art of Geometry, hath added moreover in this book (as also in the former 14. book he added 8. propositions) 9 propositions of his own, touching the inscription, and circumscription 〈…〉 bodies, very singular ●ndoubtedly and witty. All which, for that nothing should want to the desirous lover of knowledge, I have faithfully with no small pains turned. And whereas Fl●ss●● in the beginning of the eleventh book, namely, in the end of the definitions there ●e●, putteth two definitions, of the inscription and circumscription of solids or corporal figures, within or about the one the other, which certainly are not to be rejected: yet for that until this present 15. book, there is no mention made of the inscription or circumscription of these bodies, I thought it not so convenient th●r● to place them, but to refer them to the beginning of this 15. book: where they are in manner of necessity required to the elucidation of the Propositions and d●monstration● of the same. The definitions are these.  

Definition. 1.  
A solid figure, is then ●aid to be inscribed in a solid figure, when the angles of the figure inscribed touch together at one time, either the angles of the figure circumscribed, or the superficieces, or the sides.   

Definition. 2.  
A solid figure is then said to be circumscribed about a solid figure: when together at one time either the angles, or the superficieces, or the sides of the figure circumscribed, ●ouch the angles of the figure inscribed.  
IN the four●● book in the definitions of the inscription or circumscription of plain rectiline figures one with in or about an other, was required that all the angles of the figu●● inscribed, should at one time touch all the sides of the figure circumscribed: but in the five regular solids (●o whom chiefly these two definitions pertain) for that the number of their angles, superficieces, & sides are not equal, one compared to an other: it is not of necessity, that all the angles of the solid inscribed should together at one time touch either all the angles, or all the superficieces, or all the sides of the solid circumscribed: but it is sufficient, that those angles of the inscribed solid which touch, do at one time together each touch some one angle of the figure circumscribed, or some one base, or some one side: so that if the angles of the inscribed figure do at one time touch the angles of the figure circumscribed, none of them may at the same time touch either the bases or the sides of the same circumscribed figure: and so if they touch the bases, they may touch neither angles nor sides: and likewise if they touch the sides, they may touch neither angles nor bases. And although sometimes all the angles of the figure inscribed can not touch either the angles, or the bases, or the sides of the figure circumscribed, by reason the number of the angles, bases or sides of the said figure circumscribed, wanteth of the number of the angles of the ●igure inscribed yet shall those angles of the inscribed figure which touch, so touch, that the voided places left between the inscribed and circumscribed figures shall on every side be equal and like. As ye may afterward in this fifteenth book most plainly perceive.  

¶ The 1. Proposition. The 1. Problem. In a Cube given to describe 
† In this proposition as also in all the other following, by the name of a pyramid understand a tetrahedron: as I have before admonished.  a trilater equilater Pyramid.  
SVppose that the cube given be ABCDEFGH. In the same cube it is required to inscribe a Tetrahedron. Draw these right lines AC, 
Construction.  CE, AE, AH, EH, HC. 
Demonstration.  Now it is manifest, that the triangles AEC, AHE, AHC, and CHE, are equilater, for their sides are the diameters of equal squares. Wherefore AECH is a trilater equilater pyramid, or Tetrahedron, & it is inscribed in the cube given (by the first definition of this book): which was required to be done.   

¶ The 2. Proposition. The 2. Problem. In a trilater equilater Pyramid given to describe an Octohedron.  
SVppose that the trilater equilater pyramid given be ABCD, whose sides let be divided into two equal parts in the points E, Z, I, K, L, T. 
Construction.  And draw these 12. right lines EZ, ZI, IE, KL, LT, TK, EK, KZ, ZL, LI, IT, and TE. Which 12. right lines are, 
Demonstration.  by the 4. of the first, equal. For they subtend equal plain angles of the bases of the pyramid, and those equal angles are contained under equal sides, namely, under the halves of the sides of the pyramid. Wherefore the triangles TKL, TLI, TIE, TEK, ZKL, ZLI, ZIB, ZEK, are equilater: and they limitate, and contain the solid TKLEZI. Wherefore the solid TKLEZI is an Octohedron: by the 23. definition of the eleventh. And the angles of the same Octohedron do touch the sides of the pyramid ABCD in the points E, Z, I, T, K, L. Wherefore the Octohedron is inscribed in the pyramid (by the 1. definition of this book). Wherefore in the trilater equilater pyramid given, is inscribed an Octohedron: which was required to be done.  

A Corollary added by Flussas.  
Hereby it is manifest, that a pyramid is cut into two equal parts, by every one of the three equal squares, which divide the Octohedron into two equal parts and perpendicularly. For the three diameters of those squares do in the centre cut the one the other into two equal parts and perpendicularly, by the third Corollary of the 1●. of the thirteenth, which squares, as for example, the square EKLI, do divide in sunder the pyramids and the prisms, namely, the pyramid KLTD and the prism KLTEIA from the pyramid EKZB, and the prism EKZILG, which pyramids are equal the one to the other, and so also are the prisms equal the one to the other: by the 3. of the twelfth. And in like sort do the rest of the squares, namely, KZIT and ZLTE: which squares, by the second Corollary of the 14. of the thirteenth, do divide the Octohedron into two equal parts.    

¶ The 3. Proposition. The 3. Problem. In a cube given, to describe an Octohedron.  
TAke a Cube, namely, ABCDEFGH. And divide every one of the sides thereof into two equal parts. 
Construction.  And draw right lines coupling together the sections, as for example, these right lines, PQ and RS, which shall be equal unto the side of the cube (by the 33. of the first) and shall divide the one the other into two equal parts in the midst of the diameter AG in the point I (by the Corollary of the 34. of the first). Wherefore the point I is the centre of the base of the cube. And by the same reason may be found out the centres of the rest of the bases, which let be the points K, L, O, N, M. And draw these right lines LI, IM, MO, OL, KI, KL, KM, KO, NI, NL, NM, & NO. 
Demonstration.  And now forasmuch as the angle IPL is a right angle (by the 10. of the eleventh, for the lines IP and PL are parallels to the lines RA and AB). And the right line IL subtendeth the right angle IPL, namely, it subtendeth the half sides of the cube which contain the right angle IPL, and likewise the right line IM subtendeth the angle IQM which is equal to the same angle IPL, and is contained under right lines equal to the right lines which contain the angle IPL. Wherefore the right line IM is equal to the right line IL (by the 4. of the first). And by the same reason may we prove, that every one of the right lines MO, OL, KI, KL, KM, KO, NI, NL, NM, and NO, which subtend angles equal to the self same angle IPL, and are contained under sides equal to the sides which contain the angle IPL, are equal to the right line IL. Wherefore the triangles KLI, KLO, KMI, KMO, and NLI, NLO, NMI, NMO, are equilater and equal: and they contain the solid IKLONM. Wherefore IKLONM is an Octohedron, by the 23. definition of the eleventh. And forasmuch as the angles thereof do altogether in the points I, K, L, O, N, M, touch the bases of the cube which containeth it, it followeth that the Octohedron is inscribed in the cube (by the first definition of this book.) Wherefore in the cube given, is described an Octohedron: which was required to be done.  

¶ A Corollary a●ded by ●luss●●.  
Hereby it is manifest, that right lines joining together the centres of the opposite bases of the cube, do cut the one the other into two equal parts, and perpendicularly, in the centre of the cube, or in the centre of the Sphere which containeth the cube.  
For forasmuch as the right lines LM and IO which kn●● together the centres of the opposite bases of the cube, do also knit together the opposite angel's of the Octahedron inscribed in the cube, it followeth (by the 3. Corollary of the 14. of the thirteenth) that those lines LM and IO, do cut the one the other into two equal parts in a point. But the diameters of the cube do also cut the one the other into two equal parts, by the 39 of the eleventh. Wherefore that point shall be the centre of the sphere which containeth the c●●●. For making that point the centre, and the space some one of the semidiameters, describe a sphere, and it shall pass by the angles of the cube: and likewise making the same point the centre, and the space half of the line LM, describe a sphere, and it shall also pass by the angles of the Octohedron.    

¶ The 4. Proposition. The 4. Problem. In an Octohedron given, to describe a Cube.  
SVppose that the Octohedron given be ABGDEZ. And let the two pyramids thereof be ABGDE, and ZBGDE. 
Construction.  And take the centres of the triangles of the pyramid ABGDE, that is, take the centres of the circles which contain those triangles: and let those centres be the point●s T, I, K, L. And by these centres let there be drawn parallel lines ●o the sides of the square BGDE: which parallel ●ig●● lin●● let be MTN, NLX, XKO, & OIM. 
Demonstration.  And forasmuch as th●se parallel right lines do (by the 2. of the sixth) cut the equal right lines AB, AG, AD, and AE, proportionally, therefore they concur in the points M, N, X, O. Wherefore the right lines MN, NX, XO, and OM, which subtend equal angles set at the point A, & contained under squall right lines, are equal (by the 4. of the first). And moreover, seeing that they are parallels unto the lines BG, GD, DE, E●, which make a square, therefore MNXO is also a square, by the 10. of the eleventh. Wherefore also, by the 15. of the ●ame, the square MNXO is parallel to the square BGDE. For all t●e right lines touch the one the other in the points of their sections. From the centres T, I, K, L, draw these right lines TI, IK, KL, LT● And draw the right line AIC. And forasmuch as I is the centre of the equilater triangle ABE, therefore the right line AI being extended, cutteth the right line BE into two equal parts (by the Corollary of the 12. of the thirteenth). And forasmuch as MO is a parallel to BE, therefore the triangle AIO is like to the whole triangle ACE (by the Corollary of the 2. of the sixth). And the right line MO is divided into two equal parts in the point I (by the 4. of the sixth). And by the same reason may we prove, that the right lines MN, NX, XO, are divided into two equal parts in the points T, L, K. Wherefore also again, the bases TI, IK, KL, LT, which subtend the angles set at the points M, O, X, N, which angles are right angles, and are contained under equal sides, those bases, I say, are equal. And forasmuch as TIM is an Isosceles triangle, therefore the angles set at the base, namely, the angles MTI and MIT, are equal (by the ●● of the first). But the angle M is a right angle: wherefore each of the angles MIT and MTI, is the half of a right angle. And by the same reason the angles OIK & OKI, are equal. Wherefore the angle remaining, namely, TIK, is a right angle (by the 13. of the first). For the right lines TI and IK are set upon the line MO. And by the same reason may the rest of the angles, namely, IKL, KLT, LTI, be proved right angles, and they are in one and the self same plain superficies, namely, MNXO (by the 7. of the eleventh). Wherefore the right lines which join together the centres of the plain superficial triangles which make the solid angle A, do make the square ITKL. And by the same reason may be proved, that the plain superficial triangles of the rest of the five solid angles of the Octohedron set at the points B, G, Z, D, E, do in the centres of their bases receive squares, So that there are in number six squares, for every Octohedron hath six solid angles: and those squares are equal: for their sides do contain equal angles of inclinations contained under equal sides, namely, under those sides which are drawn from the centre to the side of the equal triangles (by the 2. Corollary of the 18. of the thirteenth). Wherefore ITKLRPUS is a cube (by the 21. definition of the eleventh) and hath his angles in the centres of the bases of the Octohedron, and therefore is inscribed in it (by the first definition of this book). Wherefore in an Octohedron given is described a cube: which was required to be done.   

The 5. Proposition. The 5. Problem. In an Icosahedron given, to describe a Dodecahedron.  
TAke an Icosahedron, one of whose solid angles let be Z. 
Construction.  Now forasmuch as (by those things which have been proved in the 16. of the thirteenth) the bases of the triangles which contain the angle of the Icosahedron do make a pentagon inscribed in a circle, let that pentagon be ABGDE, which is made of the five bases of the triangles, whose plain superficial angles remaining make the solid angle given, namely, Z. And take the centres of the circles which contain the foresaid triangles, which centres let be the points I, T, K, M, L: and draw these right lines IT, TK, KM, ML, LIVELY 
Demonstration.  Now then a perpendicular line drawn from the point Z to the plain superficies of the pentagon ABGDE, shall fall upon the centre of the circle which containeth the pentagon ABGDE (by those things which have been proved in the self same 16. of the thirteenth). Moreover perpendicular lines drawn from the centre to the sides of the pentagon ABGDE shall in the points C, N, O where they fall cut the right lines AB, BG, GD into two equal parts (by the 3. of the third). Draw these right lines CN and NO. And forasmuch as the angles CBN and NGO are equal, and are contained under equal sides, therefore the base CN is equal to the base NO (by the 4. of the first). Moreover perpendicular lines dr●●●e from the point Z to the b●s●● of the pentagon ABGDE, shall likewise cut the bases into two equal parts (by th●● of the third). For the perpendiculars pass by the centre (by the corollary of the 12. of the thirteenth): Wherefore th●se perpendicular lines shall fall upon the points C, N, O. And now forasmuch as the right lines ZI, IG are equal to the right lines ZT, TN, & also to the right lines ZK, KO (by reason of the likeness of the equal triangles): therefore the line IT is a parallel to the line CN, and so also is the line TK to the line NO (by the 2. of the sixth). Wherefore the angles ITK, and CNO are equal (by the 11. of the eleventh). Again forasmuch as the triangles CBN, and NGO are Isoscels triangles, therefore the angles BCN and BNC are equal (by the 5. of the first). And by the same reason the angles GNO, and GON are equal. And moreover the angles BCN and BNC are equal to the angles GNO, and GON, for that the triangles CBN and NGO are equal and like. B●● the three angles BNC, CNO, ONG, are equal to two right angles (by the 13. of the first): for that upon the right line B● are set the right lines CN & ON. And the three angles of the triangle CBN, namely, the angles BNC, BCN or GNO (for the angle GNO is equal to the angle BCN as it hath been proved) and NBC are also equal to two right angles (by the 32. of the first). Wherefore taking away the angles BNC & GNO, the angle remaining, namely, CNO is equal to the angle remaining, namely, to CBN. Wherefore also the angle ITK (which is proved to be equal to the angle CNO) is equal to the angle CBN. Wherefore ITK is the angle of a pentagon. And by the same reason may be pro●ed that the rest of the angles, namely the angles TILL, ILM, LMK, MKT, are equal to the rest of the angles, namely to BAE, AED, edge, DGB. Wherefore ITKML is an equilater and equiangle pentagon (by the 4. of the first) For the equal bases of the pentagon ITKML do subtend equal angles set at the point Z, and comprehended under equal sides. Moreover it is manifest that the pentagon I TKML is in one and the self same plain superficies. For forasmuch as the angles ONC and NCP are in one and the self s●me plain superficies, namely in the superficies ABGDE: But unto the same plain superficies the plain superficieces of the angles KTI and TILL are parallels (by the 15. of the eleventh). And the triangles KTI and TILL concur: wherefore they are in one and the self same plain superficies (by the corollary of the 16. of the eleventh). And by the same reason so may we prove that the triangles ILM, LMK, MKT are in the self same plain superficies wherein are the triangles KTI and TILL. Wherefore the pentagon ITKML is in one and the self same plain superficies. Wherefore the solid angle of the Icosahedron, namely the solid angle at the point Z subtendeth an equilater and equiangle pentagon plain superficies, which pentagon hath his plain superficial angles in the centres of the triangles which make the solid angle Z. And in like sort may we prove that the other eleven solid angles of the Icosahedron, each of which eleven solid angles are equal and like to the solid angle Z (by the 16. of the thirteenth) are subtended unto pentagons equal, and like, and in like sort set to the pentagon ITKML. And forasmuch as in those pentagons the right lines, which join together the centres of the bases, are common sides, it followeth that those 12. pentagons include a solid which solid is therefore a dodecahedron (by the 24. definition of the eleventh): and is, by the first definition of this book, described in the Icosahedron, five sides whereof 〈◊〉 set upon the pentagon ABGDE. Wherefore in an Icosahedron given i● inscribed a dodecahedron: which was required to be done.  

An annotation of Hypsi●les.  
This is to be noted, that if a man should demand 〈◊〉 many sides an Icosahedron hath, we may thus answer: It is manifest that an Icosah●r●n is contained under 20. triangles, and that every triangle consisteth of three right lin●s. Now then multiply the 20. triangles into the sides of one of the triangles, and so shall there be produced 6●. ●he half of which is 30. And so many sides hath an Icosahedron. And in like sort in a dodecahedron, forasmuch as 12. pentagons make a dodecahedron, and every pentagon containeth ●. right lines● multiply ●● into 12. and there shall be produced 60. the half of which is 30. And so many are the sides of a dodecahedron. And the reason why we take the half, i●, for that every side whether it be of a triangle or of a pentagon, or of a square as in a cube, ●s taken twice. And by the same reason may you find out how many sides are in a cube, and in a pyramid, and in an octohedron.  
But now again if ye will find out the number of the angles of every one of the solid figures, when ye have done the same multiplication that ye did before, divide the same sides, by the number of the plain superficieces which comprehend one of the angles of the solids As for example, forasmuch as 5. triangles contain the solid angle of an Icosahedron, divide 60. by 5. and there will come forth 12. and so many solid angles hath an Icosahedron. In a dodecahedron, forasmuch as three pentagons comprehend an angle, divide 60. by 3. and there will come forth 20: and so many are the angles of a dodecahedron. And by the same reason may you find out how many angles are in each of the rest of the solid figures.  

That which here followeth concerning the inclination of the plains of the five solids, was before taught (●hough not altogether after the same manner) out of Flussas in the latter ●nde of the 13 book.  If it be required to be known, how one of the plains of any of the five solids being given, there may be found out the inclination of the said plains the one to the other, which contain each of the solids. This (as saith Isidorus our great master) is fo●●d out after this manner. It is manifest that in a cube, the plains which contain i●, do● 〈◊〉 the one the other by a right angle. But in a Tetrahedron, one of the triangles being given, let the ends of one of the sides of the said triangle be the centres, and let the space be the perpendicular line drawn from the top of the triangle to the base, and describe circumfer●nces of a circle, which shall cut the one the other: and from the intersection to the centres draw right lines, which shall contain the inclination of the plains containing the Tetrahedron. In an Octo●edron, take one of the sides of the triangle thereof, and upon it describe a square, and draw the diagonal line, and making the centres, the ends of the diagonal line, and the space likewise the perpendicular line drawn from the top of the triangle to the base, describe circumferences: and again from the common section to the centres draw right lines, and they shall contain the inclination sought for. In an Icosahedron, upon the side of one of the triangles thereof, describe a pentagon, and draw the line which subtendeth one of the angles of the said pentagon, and making the centres the ends of that line, and the space the perpendicular line of the triangle, describe circumferences: and draw from the common intersection of the circumferences, unto the centres right lines: and they shall contain likewise the inclination of the plains of the icosahedron. In a dodecahedron, take one of the pentagons, and draw likewise the line which subtendeth one of the angles of the pentagon, and making the centres the ends of that line, and the space, the perpendicular line drawn from the section into two equal parts of that line to the side of the pentagon, which is parallel unto it, describe circumferences: and from the point of the intersection of the circumferences draw unto the centres right lines: and they shall also contain the inclination of the plains of the dodecahedron. Thus did this most singular learned man reason, thinking the demonstration in every one of them to be plain and clear. But to make the demonstration of them manifest, I think it good to declare and make open his wordes● and first in a T●trahedron●    
The end of the fifteenth Book of Euclides Elements after 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉   

¶ The 6. Proposition. The 6. Problem. In an Octohedron given, to inscribe a trilater equilater Pyramid.  
SVppose tha● the Octohedron where●● the Tetrahedron is required to be ins●ri●●●, be ABGDEI. Take 〈…〉 bases of the Octo●●dron, that is, 
Construction.  〈…〉 close in the lowe●● triangle BGD, namely, AE●, HED, IGD: and let the fourth be AIB, which is opposite to the lowest triangle before put, namely, to EGD. And take the centres of those four bases, which let be the points H, C, N, ●. And upon the triangle HCN erect a pyramid HCNL. Now forasmuch as these two bases of the Octohedron, namely, AGE and ABI are set upon the right lines EG and BY which are opposite the one to the other● in the square GEBI of the Octohedron, from the poin● A dra●e by the centres of the bases, namely, by the centres H, L, perpendicular lines AHF, ALK, cutting the lines EG and BY 〈◊〉 two equal parts in the points F, K (by the Corollary of the 1● of the thirteenth). Wherefore a right line drawn from the point F to the point K, 
Demonstration.  shall be a parallel and equal to the sides of the Octohedron, namely, to ●● and GI' (by the 33. of the first). And the right line HL which cutteth the 〈…〉 AF, AK, proportionally (for AH and AL are drawn from the centres of equal circles to the circumferences) is a parallel to the right line FK (by the 2. of the sixth) and also to the sides of the Octohedron, namely, to E● and IG (by the 9 of the eleventh). Wherefore as the line AF is to the line AH, so is the line FK to the line HL (by the 4. of the sixth): For the triangles AFK and AHL are like (by th● Corollary of the 2. of the sixth). But the line AF is in sesquialter proportion to the line AH: (for the side EG maketh HF the half of the right line AH, by the Corollary of the 12. of the thirteenth). Wherefore FK or GI' the side of the Octohedron, is sesquialter to the righ●line HL. And by the same reason may we prove that the sides of the Octohedron are sesquialter to the rest of the right lines which make the pyramid HNCI, namely, to the right line●●, N, NC, CI, LN, and CH: wherefore those right lines are equal, and therefore the triangle● which are described of them, namely, the triangles HCN, HNL, NCL, and CHL. which make the pyramid HNCL, are equal and equilater. And forasmuch as the angles of the same pyramid, namely, the angles H, C, N, L, do end in the centres of the bases of the Octohedron, therefore it is inscribed ●o the same Octohedron, by the first definition of this book. Wherefore in an Octohedron ●euen, is inscribed a trilater equila●●●●●●amis● which was required to ●e don●.  

A Corollary.  
The bases of a Pyramid inscribed in an Octohedron, are parallels to the bases of the Octohedron. For forasmuch as the sides of the bases of the Pyramid touching the one the other, are parallels to the sides of the Octohedron which also touch the one the other, as for example, HL was proved to be a parallel to GI', and LC to DIEGO, therefore, by the 15. of the eleventh, the plain superficies which is drawn by the lines HL and LC, is a parallel to the plain superficies drawn by the lines GI' and DI. And so likewise of the rest.   

Second Corollary.  
A right line joining together the centres of the opposite bases of the Octohedron, is sesquialter to the perpendicular line drawn from the angle of the inscribed pyramid to the base thereof. For forasmuch as the pyramid and the cube which containeth it do in the self same points end their angles (by the 1. of this book): therefore they shall both be enclosed in one and the self same Octohedron (by the 4. of this book). But the diameter of the cube joineth together the centres of the opposite bases of the Octohedron, and therefore is the diameter of the Sphere which containeth the cube and the pyramid inscribed in the cube (by the 13. and 14. of the thirteenth): which diameter is sesquialter to the perpendicular which is drawn from the angle of the pyramid to the base thereof: for the line which is drawn from the centre of the sphere to the base of the pyramid, is the sixth part of the diameter (by the 3. Corollary of the 13. of the thirteenth). Wherefore of what parts the diameter containeth six, of the same parts the perpendicular containeth four.    

¶ The 7. Proposition. The 7. Problem. In a dodecahedron given, to inscribe an Icosahedron.  
SVppose that the dodecahedron given, be ABCDE. And let the centres of the circles which contain six bases of the same dodecahedron be the polnes L, M, N, P, Q, O. 
Construction.  And draw these right lines OL, OM, ON, OP, OQ, and moreover these right lines LM, MN, NP, PQ, QL. And now forasmuch as equal and equilater pentagons are contained in equal circles, therefore perpendicular lines drawn from their centres to the sides shall be equal (by the 14. of the third), and shall divide the sides of the dodecahedron into two equal parts (by the 3. of the same). Wherefore the foresaid perpendicular lines shall co●outre in the point of the section, 
Demonstration.  wherein the sides are divided into two equal parts, as LF and MF do. And they also contain equal angles, namely, the inclination of the bases of the dodecahedron, (by the 2. corollary of the 18. of the thirteenth). Wherefore the right lines LM, MN, NP, PQ, QL, and the rest of the right lines which join together two centres of the bases, and which subtende the equal angles contained under the said equal perpendicular lines, are equal the one to the other (by the 4. of the first). Wherefore the triangles OLM, OMN, ONP, OPQ, OQL, and the rest of the triangles which are set at the centres of the pentagons, are equilater and equal. Now forasmuch as the 12. pentagons of a dodecahedron contain 60. plain superficial angles, of which 60. every ●hre make one solid angle of the dodecahedron, it followeth that a dodecahedron hath 20. solid angles: but each of those solid angles is subtended of each of the triangles of the Icosahedron, namely, of each of those triangles which join together the centres of the pentagons which make the solid angle, as we have before proved. Wherefore the 20, equal and equilater triangles which subtende the 20. solid angles of the dodecahedron, and have their sides which are drawn from the centres of the pentagons common, do make an Icosahedron (by the 25. definition of the eleventh): and it is inscribed in the dodecahedron given (by the first definition of this book) for that the angles thereof do all at one time touch the bases of the dodecahedron. Wherefore in a dodecahedron geuen● i● inscribed an Icosahedron: which was required to be done.   

¶ The 8. Proposition. The 8. Problem. In a dodecahedron given, to include a cube.  
DEscribe (by the 17. of the thirteenth) a dodecahedron. 
Construction.  And by the same, take the 12. sides of the cube, each of which subtend one angle of each of the 12. bases of the dodecahedron: for the side of the cube subtendeth the angle of the pentagon of the dodecahedron, by the 2. corollary of the 17. of the thirteenth. If therefore in the dodecahedron described (by the self same 17. proposition) we draw the 12. right lines subtended under the foresaid 12. angles, and ending in 8. angles of the dodecahedron, and concurring together in such sort that they be in like sort situate, as it was plainly proved in that proposition, then shall it be manifest, that the right lines drawn in this dodecahedron from the foresaid 8. angles thereof do make the foresaid cube, which therefore is included in the dodecahedron, for that the sides of the cube are drawn in the sides of the dodecahedron, and the angles of the same cube are set in the angles of the said dodecahedron. As for example take 4. pentagons of a dodecahedron, 
Demonstration.  namely AGIBO, BHCNO, CKEDN and DFAON. And draw these right lines AB, BC, CD, DA. Which four right lines make a square: for that each of those right lines do subtend equal angles of equal pentagons, & the angles which those 4. right lines contain are right angles, as we proved in the construction of the dodecahedron, in the 17. proposition before alleged. Wherefore the six bases being squares, do make a cube (by the 21. definition of the eleventh) and for that the 8. angles of the said cube are set in 8. angles of the dodecaheeron, therefore is the said cube inscribed in the dodecahedron (by the first definition of this book). Wherefore in a dodecahedron is inscribed a cube: which was required to be done.   

¶ The 9 Proposition. The 9 Problem. In a Dodecahedron given to include an Octohedron.  
SVppose that the dodecahedron given be ABGD. 
Construction.  Now (by the 3. corollary of the 17. of the thirteenth take the 6. sides which are opposite the one to the other, those 6. sides, I say whose sections wherein they are divided into two equal parts, are coupled by three right lines which in the centre of the sphere, wherein the Dodecahedron is contained, do cut the one the other perpendicularly. And let the points wherein the foresaid sides are cut into two equal parts be A, B, G, D, C, I And let the foresaid three right lines joining together the said sections be AB, GD and CI. And let the centre of the sphere be E. 
Demonstration.  Now forasmuch as (by the foresaid corollary) those three right lines are equal, it followeth (by the 4. of the first) that the right lines subtending the right angles which they make at the centre of the sphere, which right angles are contained under the halves of the said three right lines, are equal the one to the other: that is, the right lines AG, GB, BD, DA, CA, CG, CB, CD, and IA, IG, IB, ID are equal the one to the other. Wherefore also the 8. triangles CAG, CGB, CBD, CDA, JAG, IGB, IBD & IDA are equal and equilater. And therefore AGBDCI is an Octohedron by the 23. definition of the eleventh.) And the said Octahedron is included in the dodecahedron (by the first definition of this book:) for that all the angles thereof do at one time touch the sides of the dodecahedron. Wherefore in the dodecahedron given, is included an Octohedron: which was required to be done.   

¶ The 10. Proposition. The 10. Problem: In a Dodecahedron given, to inscribe an equilater trilater Pyramid.  
SVppose that the Dodecahedron given, be ABCD, of which Dodecahedron take three bases meeting at the point S, namely these three bases ALSIK, DNSLE and SIBRN: 
Construction.  and of those three bases take the three angles at the points A, B, D: and draw these right lines AB, BD and DA: and let the diameter of the sphere containing the dodecahedron, be SO, and then draw these right lines AO, BO and DO: Now forasmuch as (by the 17. of the thirteenth) the angles of the dodecahedron are set in the superficies of the sphere described about the Dodecahedron● 
Demonstration.  therefore if upon the diameter SO, and by the point A, be described a semicircle, it shall make the angle SAO a right angle (by the 31. of the third.) And likewise if the same semicircle be drawn by the points D and B, it shall also make the angles SBO, and SDO right angles. Wherefore the diameter SO containeth in power both the lines SA, AO, or the lines SB, BO, or else SD, DO, but the lines SA, SD, SB are equal the one to the other, for they each subtend one of the angles of equal pentagons. Wherefore the other lines remaining, namely, AO, BO, DO are equal the one to the other. And by the same reason may be proved that the diameter HD which subtendeth the two right lines HA, AD, containeth in power both the said two right lines, and also containeth in power both the right lines HB and BD, which two right lines it also suhtendeth. And moreover by the same reason the diameter AC, which subtendeth the right lines CB and BA, containeth in power both the said right lines C● and BA. But the right lines HA, HB and CB are equal the one to the other, for that each of them also subtendeth one of the angles of equal pentagons● wherefore the right lines remaining, namely, AD, BD, and BA are equal the one to the other. And by the same reason may be proved that each of those right lines AD, BD and BA is equal to each of the right lines AO, BO and DO. Wherefore the six right lines AB, BD, DA, AO, BO, & DO are equal the one to the other. And therefore the triangles which are made of them, namely, the triangles ABD, AOB, AOD and BOD are equal and equilater: which triangles therefore do make a pyramid ABDO, whose base is ABD and top the point O. Each of the angles of which pyramid, namely, the angles at the points A, B, D, O, do in the self same points touch the angles of the Dodecahedron. Wherefore the said pyramid is inscribed in the Dodecahedron, (by the first definition of this book.) Wherefore in a Dodecahedron given, is inscribed a trilater equilater pyramid: which was required to be done.   

¶ The 11. Proposition. The 11. Problem. In an Icosahedron given, to inscribe a cube.  
IT was manifest by the 7. of this book, that the angles of a Dodecahedron are set in the centres of the bases of the Icosahedron. And by the 8. of this book, it was proved, that the angles of a cube are set in the angles of a Dodecahedron. Wherefore the self same angles of the cube, shall of necessity be set in the centres of the bases of Icosahedron. Wherefore the cube shallbe inscribed in the Icosahedron (by the first definition of this book). Wherefore in an Icosahedron given, is included a cube: which was required to be done.   

¶ The 12. Proposition. The 12. Problem. In an Icosahedron given, to inscribe a trilater equilater pyramid.  
BY the former proposition it was manifest, that the angles of a cube are set in the centres of the bases of the Icos●hedron. And (by the first of this book) it was plain that the four angles of a pyramid are set in four angles of a cube. Wherefore it is evident, by the first definition of this book, that a pyramid described of right lines joining together these four centres of the bases of the Icosahedron, shallbe inscribed in the same Icosahedron. Wherefore in an Icosadron given, is inscribed an equilater trilater pyramid: which was required to be done.   

¶ The 13. Problem The 13. Proposition. In a Cube given, to inscribe a Dodecahedron.  
TAke a Cube ADFL. 
First part of the construction.  And divide every one of the sides thereof into two equal parts in the points T, H, K, P: G, L, M, F: and pkQs. And draw these right lines TK, GF, pQ, Hk, Ps, and LM: which lines again divide into two equal parts in the points N, V, Y, I, Z, X. And draw these right lines NY, VX, and IZ: Now the three lines NY, VX, and IZ, together with the diameter of the cube, shall cut the one the other into two equal parts in the centre of the cube, by the 3●. of the eleventh: let that centre be the point O. And not to stand long about the demonstration, understand all these right lines to be equal and parallels to the sides of the cube and to cut the one the other right angled wise, by the 29. of the first. Let their halves, namely, FV, GV, high, and kI, and the rest such like, be divided by an extreme and mean proportion, by the 30. of the sixth: whose greater segments let be the lines FS, GB, HC, and kE, etc. and draw these right lines GI', GE, BC, and BE. Now forasmuch as the line GI' is equal to the whole line GV, which is the half of the side of the cube: 
First part of the demonstration.  and the line IE is equal to the line BV, that is, to the less segment: therefore, the squares of the lines GI' and IE, are triple to the square of the line GB, by the 4. of the thirteenth: But unto the squares of the lines GI' and IE, the square of the line GE is equal, by the 47. of the first● for the angle GIE is a right angle: Wherefore the square of the line GE is triple to the square of the line GB. And forasmuch as the line FG is erected perpendicularly to the plain AGkL, by the 4. of the eleventh: for it is erected perpendicularly to the two lines AG and GI': therefore the angle BGE is a right angle: for the line GE is drawn in the plain AGkL. Wherefore the line BE, containing in power the two lines BG and GE, by the 47. of the first, is in power quadruple to the line GB (for the line GE was proved to be in power triple to the same line GB): Wherefore the line BE is in length double to the line BG, by the ●0. of the sixth. But (by construction) the line CE is double the line IE: Wherefore the halves GB and IE, are in proportion the one to the other, as their doubles BE and CE: by the 15. of the fifth. Wherefore the line CE is the greater segment of the line BE divided by an extreme and mean proportion. And forasmuch as the self same thing may be proved touching the line BC: therefore the lines BE and BC, are equal, making an Isosceles triangle. Now let us prove that three angles of the Pentagon of the Dodecahedron are set at the points B, C, E: and the other two angles are set between the lines BC and BE.  
Forasmuch as the circle which containeth the triangle BCE circumscribeth the Pentagon whose side is the line CE, by the 11. of the fourth: 
Second part of the construction.  Extend the plain of the triangle BCE, by the parallel lines dB and HE, cutting the line AD, namely, the diameter of AD the base of the cube in the point I: and let it cut the line Ah the diameter of the cube in the point m. And by the point I draw in the base AD, a parallel line unto the line Ad: which let be Il. And forasmuch as from the triangle AHN is, by the parallel line lI, taken away the triangle All, like unto the whole triangle AHN, 
Second part of the Demonstration.  by the Corollary of the 2. of the sixth: the lines All, and lI, shall be equal. But as the line HA is to the line Ad, so (by the 2. of the sixth) is the line Hl to the line lI, or to the line lA, which is equal to the line lI. And the greater segment of the line HA. (which is half the side of the cube) is, as before hath been proved, the line Ad, that is, the line GB, which is equal to the line Ad (by the 33. of the first). Wherefore the 〈◊〉 segment of the line Hl is the line ●A. And as the whole line Hl is to the greater segment, so shall the same greater segment Hl be to the less segment lA, by the 5. of the thirteenth. Wherefore the line HA is divided by an extreme and mean proportion in the point l. But in the triangle AHN, the line NA, which is drawn from the centre of the base AD, is in the point I cut like unto the line All, by thou parallel line lI (by the same second of the sixth): for the lines HN and Il, are parallels, by construction. Wherefore the line NA is in the point I divided by an extreme and mean proportion by the superficies dBEH. And forasmuch as the line YON which coupleth the centres of the opposite bases, is a parallel to the line HE: A plain superficies extended by the line YON, parallel wise to the plain dBEH: the two plains shall cut the lines AO and AN (the semidiameter of the cube, and the semidiameter of the base AD) into the self same proportions in the points m and I, by the 17. of the eleventh. But the line AN is in the point I divided by an extreme & mean proportion: Wherefore the semidiameter of the cube is in the point m divided by an extreme and mean proportion by the plain of the triangle BCE. And forasmuch as the rest of the triangles described in the cube after the like manner, may by the same reasons be proved to be in a plain which cutteth the semidiameter of the cube by an extreme and means proportion: it is manifest that three plains of the Dodecahedron shall under every angle of the cube concur in one & the self same point o● the semidiameter being cut by an extreme and mean proportion. Now resteth to prove that the right lines which couple that point of the semidiameter with the angles of the triangle BEC, are equal: whereby may be proved that the Pentagons' are equilater, and equiangle.  
Take the two bases of the cube. Whereon are set the triangle BCE, 
Third part of the construction.  namely, the bases AF and Ak, take also the same diameter of the cube that was before, namely, Ah: and let the side set at the point n, of the section of the diameter by an extreme & mean proportion, be the line Cn or Bn: and let the centre of the cube be as before the point O. And extend the line Cn to the line Bd, and let it concur with it in the point a. And forasmuch as the plain which passeth by the line HCE and the centre O (cutting the cube into two equal parts) is parallel to AF the base of the cube by construction: imagine that by the point n, be extended a plain superficies parallel to the former parallel plains, 
Third part of the demonstration.  which shall cut the semidiameter OA & the line Ca, proportionally in the point n, by the 17. of the eleventh: For those lines do touch the extreme parallel plains extended by the lines HE and EO, and by the lines Add and dB. But it is proved that the line OA is divided by an extreme and mean proportion in the point n: wherefore the line Ca, is also 〈◊〉 by an extreme and mean proportion in the point n. Again forasmuch as BCE is an Isoscels triangle, and it is proved that the line BY cutteth the base CE into two equal parts in the point I, the angles BIC and BIE● shall be right angles. Imagine by the line BY and the centre O a plain to pass (cutting the cube into two equal parts) parallel to the base AD. And unto those plains let there be imagined an other parallel plain passing by the point n: which let be ne: which shall cut the semidiameter AO and the half side of the cube, namely, the line lH, like, in the points n and c by the 17. of the eleventh. Wherefore the line IH is in the point e divided by an extreme 〈◊〉 mean proportion. Wherefore the line He is equal to the line CI or IE: namely, each are less segments. And forasmuch as the line je is to the line IC (which is equal ●o the ●ine EH) as the whole is to the greater segment, take away from the whole line je th● greater segm●● IC: there shall remain the less segment Ce by the 5. of the thirteenth: Wherefore the line I● is divided by an extreme & mean proportion in the point C. Again unto the same plains imagine an other plain to pass by the point a, parallel wise, and let the same be ag. Now then (by the same 17. of the eleventh) the lines Ca and Cg are in like sort cut in the points n and e. But the line Ca was in the point n cut by an extreme and mean proportion, wherefore the line Cg shall be cut in the point e, by an extreme & mean proportion. But the line IC is to the line Ce, as the greater segment is to the less: wherefore the line Ce, is to the line egg, as the greater segment to the less: and therefore their proportion is as the whole line IC is to the greater segment Ce, and as the greater segment Ce is to the less segment egg: wherefore the whole line Ceg which, maketh the greater segment and the less, is equal to the whole line IC or IE. And forasmuch as two parallel plain superficieces (namely, that which is extended by JOB and that which is extended by the line ag) are cut by the plain of the triangle BCE, which passeth by the lines ag and IB, their common sections ag and IB shall be parallels (by the 16. of the eleventh). But the angle BUY or BIC is a right angle, wherefore the angle agC is also a right angle (by the 29. of the first) and those right angles are contained under equal sides, namely, the line gC is equal to the line CI, and the line ag to the line BY, by the 33. of the first: wherefore the bases Ca and CB are equal, by the 4. of the first. But of the line CB the line CE was proved to be the greater segment: wherefore the same line CE is also the greater segment of the line Ca: but cn was also the greater segment of the same line Ca Wherefore unto the line CE, the line cn which is the side of the dodecahedron, and is set at the diameter, is equal. And by the same reason the rest of the sides, which are set at the diameter may be proved eguall to lines equal to the line CE. Wherefore the pentagon inscribed in the circle where in is contained the triangle BCE is, by the 11. of the fourth equiangle, and equilater. And forasmch as two pentagons, set upon every one of the bases of the cube do make a dodecahedron, and six bases of the cube do receive twelve angles of the dodecahedron: and the 8. semidiameters do in the points where they are cut by an extreme and mean proportion receive the rest: therefore the 12. pentagon bases containing 20. solid angles do inscribe the dodecahedron in the cube: by the 1. definition of this book. Wherefore in a cube given is inscribed a dodecahedron: which was required to be done.  

First Corollary.  
The diameter of the sphere, which containeth the dodecahedron, containeth in power these two sides, namely, the side of the Dodecahedron, and the side of the cube wherein the Dodecahedron is inscribed. For in the first figure a line drawn from the centre O, to the point B the angle of the Dodecahedron, namely the line OB, containeth in power these two lines OV the half side of the cube, and VB the half side of the dodecahedron, by the 47. of the first. Wherefore by the 15. of the fifth, the double of the line OB, which is the diameter of the sphere containing the Dodecahedron, containeth in power the double of the other lines OV and VB, which are the sides of the cube, and of the dodecahedron.   

¶ Second Corollary.  
The side of a cube divided by an extreme and mean proportion, maketh the less segment the side of the dodecahedron inscribed in it: and the greater segment the side of the cube inscribed in the same Dodecahedron: For it was before proved, that the side of the dodecahedron is the greater segment of BE the side of the triangle BEC● but the side BE (which is equal to the line● GB and SF) is the greater segment of GF the side of the cube: which line ●E (subtending th● angle of the pentagon) was (by the ●. of this book) the side of the cube inscribed in the dodecahedron.   

Third Corollary.  
The side of a cube, is equal to the sides of a Dodecahedron inscribed in it, and circumscribed about it. For it was manifest by this proposition, that the side of a cube maketh the less segment, the side of a Dodecahedron inscribed in it, namely, as in the first figure the line BS the side of the Dodecahedron inscribed, is the less segment of the line GF the side of the cube. And it was proved in the 17. of the thirteenth, that the same side of the cube subtendeth the angle of the pentagon of the Dodecahedron circumscribed: and therefore it maketh the greater segment the side of the Dodecahedron or of the pentagon, by the first corollary of the same. Wherefore it is equal to both those segments.    

The 14. Problem. The 14. Proposition. In a cube given, to inscribe an Icosahedron.  
SVppose that the cube given be ABC, 
Construction.  the Centres of whose bases let be the points D, E, G, H, I, K: by which points draw in the bases unto the other sides parallels not touching the one the other. And divide the lines drawn from the centres, as the line DT. etc. by an extreme and mean proportion in the points A, F: L, M: N, B: P, Q: R, S: C, O: by the 30. of the sixth: and let the greater segments be about the centres. And draw these right lines, AL, AG, AM, and TG. And forasmuch as the lines cut are parallels to the sides of the cube: 
Demonstration.  they shall make right angles the one with the other by the 29. of the first: and forasmuch as they are equal: their sections shall be equal, for that the sections are like by the 2. of the fourteenth. Wherefore the line TG is equal to the line DT, for they are each, half sides of the cube. Wherefore the square of the whole line TG, and of the less segment TA, is triple to the square of the line AD the greater segment (by the 4. of the thirteenth). But the line AG containeth in power the lines AT & TG, for the angle ATG is a right angle. Wherefore the square of the line AG is triple to the square of the line AD. And forasmuch as the line MGL is erected perpendicularly to the plain passing by the lines AT, & which is parallel to the bases of the cube (by the corollary of the 14. of the eleventh) therefore the angle AGL is a right angle. But the line LG is equal to the line AD, for they are the greater segments of equal lines: Wherefore the line AG (which is in power triple to the line AD) is in power triple to the line LG. Wherefore adding unto the same square of the line AG, the square of the line LG, the square of the line AL, which (by the 47. of the first) containeth in power the two lines AG and GL, shallbe quadruple to the line AD or LG. Wherefore the line AL is double to the line AD (by the 20. of the sixth:) and therefore is equal to the line AF, or to the line LM. And by the same reason may we prove that every one of the other lines which couple the next sections of the lines cut, as the lines AM, PF, PM, MQ, and the rest are equal. Wherefore the triangles ALM, APF, AMP, PMQ and the rest such like, are equal, equiangle, and equilater, by the 4. and eight of the first. And forasmuch as upon every one of the lines cut of the cube are set two triangles, as the triangles ALM, and BLM● there shallbe made 12. triangles. And forasmuch as under every one of the ●. angles of the cube, are subtended the other 8. triangles, as the triangle AMP. etc. of 1●. and 8. triangles, shall be produced 20. triangles equal and equilater containing the solid of an Icosahedron, by the 25. definition of the eleventh, which shallbe inscribed in the cube given ABC by the first definition of this book. The invention of the demonstration of this dependeth of the ground of the former. Wherefore in a cube given, we have described an Icosahedron: which was required to be done.  

First Corollary.  
The diameter of a sphere which containeth an Icosahedron, containeth two sides, namely the side of the Icosahedron, and the side of the cube which containeth the Icosahedron. For if we draw the line AB, it shall make the angles at the point A right angles: for that it is a parallel to the sides of the cube: wherefore the lin● which coupleth the opposite angle● of the Icosahedron, at the points F and B, containeth in power the line AB (the sid● of th● cube) and the line AF (the side of the Icosahedron) by the 47. of the first. Which line FB is equal to the ●iameter of the sphere, which containeth the Icosahedron, by the demonstration of the ●●. of the thirteenth.   

Second Corollary.  
The six opposite sides of the Icosahedron divided into two equal parts: their sections are coupled by three equal right lines, cutting the one the other into two equal parts, and perpendicularly in the centre of the sphere which containeth the Icosahedron. For those three lines are the three lines which couple the centres of the bases of the cube, which do in such sort in the centre of the cube, cut the one the other, by the corollary of the third of this book, and therefore are equal to the sides of the cube. But right lines drawn from the centre of the cube to the angles of the Icosahedron, every one of them shall subtend the half side of the cube, and the half side of the Icosahedron (which half sides contain a right angle) wherefore those lines are equal. Whereby it is manifest that the foresaid centre is the centre of the sphere which containeth the Icosahedron.   

Third Corollary.  
The side of a cube divided by an extreme and mean proportion, maketh the greater segment the side of an Icosahedron described in it. For the half side of the cube maketh the half of the side of the Icosahedron the greater segment: wherefore also the whole side of the cube, maketh the whole side of the Icosahedron the greater segment by the 15. of the fifth, for the sections are like by the ●. of the fourteenth.   

¶ Fourth Corollary.  
The sides and bases of the Icosahedron, which are opposite the one to the other, are parallels. Forasmuch as every one of the opposite sides of the Icosahedron, may be in the parallel lines of the cube, namely, in those parallels which are opposite in the cube: and the triangles which are made of parallel lines, are parallels, by the 15. of the eleventh: therefore the opposite tri●ngle● of the Icosahedron, as also the sides, are parallels the one to the other.    

¶ The 15. Problem. The 15. Proposition. In an Icosahedron given, to inscribe an Octohedron.  
SVppose that the Icosahedron given be ACDF: and by the former second Corollary, 
Construction.  let there be taken the three right lines which cut the one the other into two equal parts perpendicularly, and which couple the sections into two equal parts of the sides of the Icosahedron: which let be BE, GH, and KL, cutting the one the other in the point I And draw● these ri●ht lines ●G, GE, EH, and HB. And forasmuch as the angles at the point I are (by construction) right angles, 
Demonstration.  ●nd are con●●ined under equal lines● the 〈◊〉 G● and ●● shall 〈…〉 squ●re, by the ●. of the ●irst. Likewise ●nto tho●e 〈◊〉 shall be squall the lines dr●w●n from 〈◊〉 points K and ● to every one of the poin●●s ●, G. ●, H: And therefore the triangles which 〈◊〉 the ●●ramis DGENK, shall be equal 〈…〉. And by 〈…〉.   

¶ The 16. Problem. The 16. Proposition. In an Octohedron given, to inscribe an Icosahedron.  
LEt there be taken an Octohedron, whose 6. angles, let be A, B, C, F, P, L. And draw the lines AC, BF, PL, 
Construction.  cutting the one the other perpendicularly in the point R (by the 2. Corollary of the 14. of the thirteenth). And let every one of the 12. sides of the Octohedron be divided by an extreme and mean proportion, in the points H, X, M, K, D, S, N, G, V, E, Q, T. And let the greater segments be the lines BH, BX, FM, FK, AD, AQ, CS, CT, PN, PG, LV, LE: And draw these lines HK, XM, GE, NV, DS, QT. Now forasmuch as in the triangle ABF, the sides are cut proportionally, namely, as the line BH is to the line HA, so is the line FK to the line KA (by the 2. of the fourteenth): 
Demonstration.  therefore the line HK shall be a parallel to the line BF (by the 2. of the sixth). And forasmuch as the line AC cutteth the line HK in the point Z, and the line ZK is a parallel unto the line RF, the line RA shall be cut by an extreme and mean proportion in the point Z: by the 2. of the sixth: namely, shall be cut like unto the line FA: and the greater segment thereof shall be the line ZR. Unto the line ZK put the line RO equal, by the 3. of the first: and draw the line KO: now then, the line KO shall be equal to the line ZR, by the 33. of the ●irst. Draw the lines KG, KE, and KI. And forasmuch as the triangles ARF, and AZK, are equiangle (by the 6. of the sixth) the sides AZ and ZK, shall be equal the one to the other, by the 4. of the sixth, for the sides AR and RF, are equal. Wherefore the line ZK shall be the less segment of the line RA. But if the greater segment RZ be divided by an extreme & mean proportion, the greater segment thereof shall be the line ZK, which was the less segment of the whole line RA, by the 5. of the thir●enth. And forasmuch as the two lines FE and FG, are equal to the two lines AH and AK, namely, each are less segments of equal sides of the Octohedron, and the angles HAK and EFG are equal, namely, are right angles, by the 14. of the thirteenth: the bases HK and GF shall be equal, by the 4. of the first: And by the same reason unto them may be proved equal the lines XM, NV, DS, and QT. And forasmuch as the lines AC, BF, and PL, do cut the one the other into two equal parts, and perpendicularly, by construction: the lines HK and GE (which subtend angles of triangles like unto the triangles whose angles the lines AC, BF, and PL subtend) are cut into two equal parts in the points Z and I, by the 4. of the sixth, so also are the other lines NV, XM, DS, QT (which are equal unto the lines HK & GE) cut in like sort, and they shall cut the lines AC, BF, and PL like. Wherefore the line KO (which is equal to RZ) shall make the greater segment the line RO, which is equal to the line ZK (for the greater segment of the RZ was the line ZK): and therefore the line HEY shall be the less segment, when as the whole line RI is equal to the whole line RZ. Wherefore the squares of the whole line KO, and of the less segment HEY, are triple to the square of the greater segment RO, by the 4. of the thirteenth. Wherefore the line KI, which containeth in power the two lines KO and HEY, is in power triple to the line RO (by the 47. of the first): for the angle KOI is a right angle. And forasmuch as the lines FE and FG (which are the less segments of the sides of the Octohedron) are equal: and the line FK is common to them both: and the angles KFG and KFE (of the triangles of the Octohedron) are equal: the bases KG and KE shall (by the 4 of the first) be equal: and therefore the angles KIE and KIG which they subtend, are equal (by the 8. of the first): Wherefore they are right angles, by the 13. of the first. Wherefore the right line KE (which containeth in power the two lines, KI and ●E by the 47. of the first) is in power quadruple to the line RO (or IE): for the line RI is proved to be in power triple to the same line RO: But the line GE is double to the line IE: Wherefore the line GE is also in power 〈…〉 PF. And by the same reason may be proved, that the ●est of the eleven solid angles of the 〈◊〉, are 〈…〉 the sections of every one of the sides of the Octohedron, namely, in the points E, N, V, H, ●, M, ●, D, S, Q, T. Wherefore there are 12. angles of the Icosahedron. Moreover, forasmuch as every one of the bases of the Octohedron, do each contain triangles of the Icosahedron, 〈…〉 pyramids ABC●FP (which is the half of the Octohedron) the triangle FCP receiveth in th● section of his sides the ● triangle GMS: and the triangle CPB containeth the triangle NXS: and th● triangle ●AP containeth the triangle HND: and moreover the triangle APF containeth the triangle EDGE, and the same may be proved in the opposite pyramid ABCFL: Wherefore there shall be eight triangle●. And forasmuch as besides these triangles, to every one of the solid angles of the Octohedron 〈◊〉 subtended two triangles, as the triangles KEG and MEG, to the angle F: and the triangles HNV and XNV, to the angle B: also the triangles NDS and ●DS, to the angle P: likewise the triangle● DHK and QHK, to the angle A: Moreover the triangles EQT and VQT, to the angle L: and finally the triangles SXM and TXM, to the angle C: these 12. triangles being added to th●● for 〈◊〉 triangles, shall produce ●0. triangles equal and equilater coupled together: which shall male an Icosahedron, by the 25. definition of the eleventh: and it shall be inscribed in the Octohedron given ABC●●L, by the first definition of this book: for the 1●. angles thereof are set in 1●. like sections of the sides of the Octohedron. Wherefore in an Octohedron given, is inscribed an Icosahedron.  

¶ First Corollary.  
The side of an equilater triangle being divided by an extreme and mean proportion: a right line subtending within the triangle, the angle which is contained under the greater segment and the less: is in power duple to the less segment of the same side. For the line KE, which subtendeth the angle KFE of the triangle AFL, which angle KFE is contained under the two segments KF & FE, was proved equal 〈◊〉 the line HK, which containeth in power the two less segments HA and AK, by the 47. of the ●●rst fo● 〈◊〉 angle HAK is 〈…〉.   

Second Corollary.  
The bases of the Icosahedron are concentrical (that is, have one and the self same centre) with the bases of the Octohedron which containeth it.  
For suppose that 〈…〉 Octohedron 〈◊〉 ECD the base of an Icosahedron: and let the centre of the base ABG be the point F. And draw these right lines FA, FB, FC, and FE. Now than the 〈…〉 to the two lines FB and BC: for they are lines drawn from the centre, and are also less segments: and they contain the 〈…〉.    

¶ The 17. Problem. The 17. Proposition. In an Octohedron given, to inscribe a Dodecahedron.  

Construction.  SVppose that the Octohedron given be ABGDEC: whose 12. ●ides let be cut by an extreme and mean proportion, as in the former Proposition. It was manifest that of the right lines which couple th●se sections, are made 20. triangles, of which 8. are concentrical with the bases of the Octohedron, by the second Corollary of the former Proposition. If therefore in every one of the centres of the 20. triangles be inscribed (by the 1. of this book) every one of the ●●. ●●gles of the Dodecahedron, 
Demonstration.  we shall find, that ●. angles of the Dodecahedron are set in the 8. centres of the bases of the Octohedron: namely, these angles I, u, ct, O, M, a, P, and X: and of the other 12. solid angles there are two in the centres of the two triangles which have one side common under every one of the solid angles of the Octohedron: namely, under the solid angle A, the two solid angles, K, Z: under the solid angle B, the two solid angles H, T: under the solid angle G, the two solid angles Y, V: under the solid angle D, the two solid angles F, L: under the solid angle E, the two solid angles S, N: under the solid angle C, the two solid angles Q, R: and forasmuch as in the Octohedron are six solid angles, under them shall be subtended 12. solid angles of the Dod●cahedron: and so are m●de 20. solid angles composed of 12. equal and equilater superficial pentagons (as it was 〈◊〉, by the 5. of this book) which therefore contain a Dodecahedron (by the 24. definition of the eleventh). And it is inscribed in the Octohedron (by the 1. definition of this book): for that every one of the bases of the Octohedron do receive angles thereof. Wherefore in an Octohedron given, is inscribed a Dodecahedron.   

¶ The 18. Problem. The 18. Proposition. In a trilater and equilater Pyramid, to inscribe a Cube.  

Construction.  SVppose that there be a trilater equilater Pyramid, whose base let be ABC, and ●oppe the point D. And let it be comprehended in a Sphere● by the 13. of the 〈◊〉, And l●● the centre of that Sphere be the point E. And from the solid angles A, B, C, D, draw right lines passing by the centre E, unto the opposite bases of the pyramis● and they shall ●all perpendicularly upon the bases, and shall also fall upon the centres of the circles which contain the bases, by the Corollary of the 13. of the thirteenth. Let the cen●re of the triangle ABC, be the point G, and let the centre of the triangle ADC be the point H● and of the triangle ADB let the point N be the centre, and finally, let the point F be the centre of the other triangle DBC. And let the right lines falling upon those centres be DEG, BEH, CEN, & AEF. And by those centres G, H, N, F, let there be drawn from the angles to the opposite sides these right lines, AGL, DHK, BNM, and DFL, which shall fall perpendicularly upon the sides BC, CA, AD, and CB, by the Corollary of the 1●. of the thirteenth, and therefore they shall cu● them into two equal parts in the points K, L, M, by the 3. of the third. Again let the lines which were drawn from the solid angles to the opposite bases be divided into two equal parts, namely, the line DG in the point T, the line CN in the point O, the line AF in the point P, and the line BH in ●he point R: and draw the lines HT, FT, HO, and FO. Now forasmuch as the lines GK, 
Demonstration.  and GL, which are drawn from the centre of one and the self same triangle ABC to the sides, are equal, and the lines DK and DL are equal, for they are the perpendiculars of equal & like triangles: 
Produce in the figure the line TF to the point B.  and the line DG is common to them. Wherefore, by the 8. of the first, the angles KDG & LDG, are equal. And forasmuch as the lines HD & DF are drawn from the centre of equal circles which contain the equal triangles ADC & DBC, therefore they are equal, & the line DT is common to them both, and they contain equal angles, as before hath been proved. Wherefore the bases HT and FT are equal by the 4. of the first. And by the same reason if we draw the lines CF and CH, may we prove that the other lines HO and FO, are equal to the same lines HT and FT, and also the one to the other. Wherefore also after the same manner may be proved that the rest of the lines, which couple the centres of the triangles and the sections of the perpendiculars into two equal parts, as the lines NP, GR, GP, RN: NT, PH, GO, and RF, are equal. And forasmuch as from every one of the centres of the bases are drawn three right lines to the sections into two equal parts of the perpendiculers, and there are four centres, it followeth, that these equal right lines so drawn, are twelve in number, of which every three and three make a solid angle in the four centres of the bases, and in the four sections into two equal parts of the perpendiculars: wherefore that solid hath 8. angles, contained under 12. equal sides, which make six quadrangled figures, namely, HOFT, PGRN, PHOG, GOFR, FRNT, and TNPH. Now let us prove that those quadrangled figures are rectangle.  
Forasmuch as upon DC the common base of the triangles ADC and BDC falleth the perpendiculars AS and BS, which are drawn by the centres H and F: either of these lines HS and SF shallbe the third part of either of these lines AS and SB: for the line AH is duple to the line HS, and divideth the base DC into two equal parts by the corollary of the 12. of the thirteenth. Wherefore in the triangle ABS, the sides AS and BS are cut proportionally in the points H and F: and therefore the line HF is a parallel to the side AB, by the 2. of the sixth. Wherefore the triangles ASB and HSF are equiangle, by the 6. of the sixth. Wherefore the base HF shallbe the third part of the base AB, by the 4. of the sixth. We may also prove that the line TO is the third part of the line DC, for the lines EC and ED, which are drawn from the centre of the sphere which containeth the pyramid are equal: and the line EN, (which is drawn from the centre to the base) is the third part of the line EC, so also is the line GE the third part of the line ED (by the corollary of the 13. of the thirteenth) for it is the sixth part of the diameter of the sphere which containeth the pyramid: And the line ON, is the half of the whole line NC wherefore the residue EO is the third part of the line EC● and so also is the line ET the third part of the line ED. Wherefore the line TO in the triangle DEC is a parallel to the line DC, and is a third part of the same, by the former 2. and 4. of the sixth, as the line HF was proved the third part of the line AB. But AB and DC being sides of the pyramid are equal. Wherefore the lines HF and TO, being the third parts of equal lines, are equal, by the 15. of the fifth. Wherefore by the 8. of the first the angles HTF, and TFO are equal: and by the same reason, the angles opposite unto them, namely, the angles FOH and OHT are equal the one to the other, and also are equal to the said angles HT● and TFO: but these four angles are equal to 4. right angles by the corollary of the 32. of the first: wherefore the angles of the quadrangle HOFT are right angles. And by the same reason may the angles of the other five quadrangled figures be proved right angles● Now resteth to prove that the foresaid quadrangles are each in one and the self same plain.  
Take the quadrangle HOFT: and forasmuch as in the triangle ASB, the line HF is proved a parallel to the line AB, therefore it cutteth the lines SV, and SB proportionally in the points I and F. by the 2. of the sixth: Now then forasmuch as SF was proved the third part of the line SB, the line SI, shall also be the third part of the line SU. Moreover forasmuch as the line VS, which coupleth the sections into equal parts of the opposite ●ides of the pyramid, namely, of the sides AB and DC, is by the centre E divided into two equal parts, by the corollary of the second of this book (for it is the diameter of the octohedron inscribed in the pyramid): therefore the line SI is two third parts of the half line SE. And by the same reason, forasmuch as in the triangle DEC the line TO is proved to be a parallel to the side DC, it shall in the self same triangle cut the lines CE and SE, proportionally in the points O and I by the same 2. of the sixth: but the line EO is proved to be a third part of the line EC. Wherefore the line EI is also a third part of the line ES. Wherefore the residue IS shallbe two third parts of the whole line ES. Wherefore the point I cutteth of the lines TO and HF. Wherefore the two lines HIF and TIO cutting the one the other, are in one and the self same 'plaine, by the 2. of the eleventh. And therefore the points H, T, F, O are in one & the self same plain. Wherfore● the rectangle figure HOFT being quadrilater and equilater, and in one and the self same plain, is a square, by the definition of a square. And by the same reason may the rest of the bases of the solid be proved to be squares equal and plain or superficial: Now than the solid is comprehended of 6. equal squares (which are contained of 12. equal sides) which squares make 8. solid angles, of which four are in the centres of the bases o● the pyramid, and the other 4. are in the middle sections of the four perdendiculars. Wherefore the solid HOFTPGRN, is a cube by the 21. definition of the eleventh, and is inscribed in the pyramid, by the first definition of this book. Wherefore in a trilater equilater pyramid given, is inscribed a cube.  

¶ A corollary.  
The line which cutteth into two equal parts the opposite sides of the Pyramid, is triple to the side of the cube inscribed in the pyramid, and passeth by the centre of the cube. For the line SEV, whose third part the line SI is, cutteth the opposite sides CD and AB into two equll parts: but the line EI (which is drawn from the centre of the cube to the base is proved to be a third part of the line ES: wherefore the side of the cube which is double to the line EI shall be a third part of the whole line VS, which is (as hath been proved) double to the line ES.    

The 19 Problem The 19 Proposition. In a trilater equilater Pyramid given, to inscribe an Icosahedron.  
SVppose that the pyramid is given, 〈◊〉 AB●D● every one of whose s●des 〈◊〉 be divided into two equal parts in the poy●●●●●, M, K, L, P, N. 
Construction.  And i● every one of the bases of that pyramid, descried the triangl●● L●●, PMN, NKL, and 〈…〉 which triangles shall be equilater by the 4. of the fir●t, ●or the sides subtend equal angles of the pyramid, contained under the halves of the sides of the same pyramis● wherefore the sides of the said triangles are equal. Let those sides be divided by an extreme and mean proportion (by the 30. of the sixth) in the points C, E, Q, R, S, T, H, I, O, V, Y, X. Now then those sides are cut into the self same proportions, by the 2. of the fourteenth: and therefore they make the li●e sections equal, by the ●. part of the ninth of the fifth. Now I say, that the foresaid points do● receive the angles of the Icosahedron inscribed in the pyramid AB●D. In the foresaid triangles let there again be made other triangles by coupling the sections, and let those triangles be TRS, JOH, CEQ, and VXY, which shall be equilater: for every one of their sides do sub●●●d equal angles of equilater triangles, and those said equal angles are contained under equal side● (namely, under the greater segment and the less) ● and therefore the sides which subtend those angles are equal by the 4. of the first. Now let us prove that at each of the foresaid points, as for example at T, is set the solid angle of an Icosah●dron● 
Demonstration.  Forasmuch as the triangles TRS and TQO are equilater and equal, the 4. right lines TR, TS, TQ and TO shall be equal. And forasmuch as ●PNK is a square cutting the pyramid AB●D into two equal pa●●●●, by the corollay of the second of this booke● the line TH shall be in power duple to the line TN or NH by the 47. of the first. For the lines TN or NH are equal, for that by construction they are each less segments: and the line RT or TS is in power duple to the same line TN or NH (by the corollary of the 16. of this book) for it subtendeth the angle of the triangle contained under the two segments. Wherefore the lines TH, TS, TR, TQ, and TO are equal: and so also are the lines HS, SR, RQ, QO, and OH, which subtend the angles at the point T, equal. For the line QR containeth in power the two lines PQ and PR the less segments, which two lines the line TH also contained in power. And the rest of the lines do subtend angles (of equilater triangles) contained under the greater segment and the less. Wherefore the five triangles TRS, TSH, THO, TOQ, TQR are equilater and equal making the solid angle of an Icosahedron at the point T, by the 16. of the thirteenth, in the side PN of the triangle P NM. And by the same reason in the other sides of the 4. triangles PNM, NKL, FMK, & LFP (which are inscribed in the bases of the pyramid) which sides are 12● in number shall be set 12. angles of the Icosahedron contained under 20. equal & equilater triangles of which fowere are set in the 4. bases of the pyramid, namely, these four triangles, TRS, HOI, CEQ, VXY: 4. triangles are under 4. angles of the pyramid: that is, the four triangles CIX, YSH, ERV, TQO: and under every one of the six sides of the pyramid are set two triangles, namely, under the side, of the triangles THS and THO● under the side DB the triangles RQE and RQT: under the side DA the triangles COQ, and COI: under the side AB, the triangles EXC and EXV● under the side BG the triangles SVR and SVY: and under the side AG the triangles JYH and JYX. Wherefore the solid being contained under 20. equilater and equal triangles shall be an Icosahedron by the 23. definition of the eleventh: and shall be inscribed in the pyramid AB●D by the first definition of this book, for all his angles do at one time touch the bases of the pyramid. Wherefore in a trilater equilater pyramid given, we have inscribed an Icosahedron.   

¶ The 20. Proposition. The 20. Problem. In a trilater equilater Pyramid given, to inscribe a dodecahedron.  
SVppose that the pyramid given be ABGD, ●che of whose sides let be cut into two equal parts: and draw the lines which couple the sections, which being divided by an extreme and mean proportion, and right lines being drawn by the sections, shall receive 20. triangles making an Icosahedron, as in the former proposition it was manifest. Now than if we take the centres of those triangles, we shall there find the 20. angles of the dodecahedron inscribed in it by the 5. of this book. And forasmuch as 4. bases of the foresaid Icosahedron are concentricall with the bases of the pyramid, as it was proved in the 2. corollary of the 6. of this book: there shall be placed 4● angles of the dodecahedron, namely, the 4. angles E, F, H, D, in the 4. centres of the bases: and of the other 16. angles, under every one of the 6. sides of the pyramid are subtended two; namely, under the side AD, the angles CK: under the side BD the angles LI: under the side GD the angles M, N: under the side AB the angles T, S: under the side BG the angles P, O: and under the side AG the angles R, Q: so there rest 4. angles, whose true place we will now appoint. Forasmuch as a cube contained in one and the self same sphere with a dodecahedron, is inscribed in the same dodecahedron, as it was manifest by the 17. of the thirteenth, and 8. of this book: it followeth that a cube and a dodecahedron circumscribed about it, are contained in one and the self same bodies, for that their angles concur in one and the self same points. And it was proved in the 18. of this book, that 4. angles of the cube inscribed in the pyramid are set in the middle sections of the perpendicular● which are drawn from the solid angles of the pyramid to the opposite bases: wherefore the other 4. angles of the dodecahedron are also, as the angles of the cube, set in those middle sections of the perpendiculars. Namely, the angle V is set in the midst of the perpendicular AH● the angle Y in the midst of the perpendicular BF: the angle X in the midst of the perpendicular GE: and lastly the angle D in the midst of the perpendicular D which is drawn from the top of the pyramid to the opposite base. Wherefore those 4. angles of the dodecahedron may be said to be directly under the solid angles of the pyramid, or they may be said to be set at the perpendiculars. Wherefore the dodecahedron after this manner set, is inscribed in the pyramid given (by the first definition of this book) for that upon every one of the bases of the pyramid are set an angle of the dodecahedron inscribed. Wherefore in a trilater equilater pyramid is inscribed a dodecahedron.   

The 21. Problem. The 21. Proposition. In every one of the regular solids to inscribe a Sphere.  
IN the 13. of th● thirteenth and th● other 4. propositions following, 
This proposition Campane hath, & is the last also in order of the 15. book with him.  i● was declared that ●he ●● regular solides●●re so contained in a sphere, that ●ight lin●● drawn from the cen●●● o● the 〈…〉 of 〈◊〉 solid inscribed, are equal. Which right lines therefore make pyramids, whose ●oppes are the centre of the sphere, or of the solid, and the bas●●●●e cu●●● one of the bases of those solids. And 〈…〉 solid squall and like the one to the other, and described in equal circles: those circle's shall cut the sphere: for the angles which touch the circumference of the circle, touch also the superficies of the sphere. Wherefore perpendiculars drawn from the centre of the sphere to the bases, or to the plain superficieces of the equal circles, are equal, by the corollary of the assumpt of the 1●. of the twelfth. Wherefore making the centre the 〈◊〉 of the sphere which 〈◊〉 the solid, and th● space some one of the equal perpendicular●, d●scrib● a sphere, and it shall touch every one of the bases of 〈◊〉 solid: 〈…〉 perficies of the sphere pass beyond those bases: when as those p●●pe●diculars 〈…〉 are drawn from the centre to the bases, by the 3. corollary of the sa●●●●●umpt. Wher●fore ●e have i● every one of the regular bodies inscribed a sphere: which regular bo●●● are in number one i● 〈◊〉 by the corollary of the 1●. of the 〈◊〉.  

A Corollary.  
The regular figures inscribed in spheres, and also the spheres circumscribed about them, or containing them, have one and the self same centre. Namely, their pyramids, the ●ngles of whose bases touch the super●●●●●● of th●●●here, do from those angles 'cause equal right lines to be draw●● to one and ●he self 〈◊〉 poyn● making the top●●● of the pyramids in the same point: and therefore they 〈…〉 th● c●●tres of the spheres in the self same tops when 〈◊〉 the right lines drawn from those angles to the cro●●ed superficies, wherein are 〈◊〉 the angles of the bases of the pyramids, are equally   

An advertisement of Flussas ●  
Of these solids, only the Octohedron receiveth the other solids inscribed one with 〈…〉 other. For the Octohedron containeth the Icosahedron inscribed in it: and the same Icosahedron containeth the Dodecahedron inscribed in the same Icosahedron: and the same dodecahedron containeth the cube inscribed in the same Octohedron, and 〈…〉 ●●r●●mscribeth the Pyramid inscribed in the said Octohedron. But this happeneth not in the other solids.    
The end of the fifteenth Book of Euclides Elemen●●● after Ca●pa●● and 〈◊〉.   

¶ The sixteenth book of the Elements of Geometry added by Flussas.  
IN the former fifteenth book hath been taught how to inscribe the five regular solids one with in an other. Now seemeth to rest, to compare those solid so inscribed, one to an other, and to set forth their passion's and proprieties: which thing, Flussas considering, in this sixteenth book added by him, 
The argument of the 16. book.  hath excellently well and most cunningly performed. For which undoubtedly he hath of all them which have a love to the Mathematicals, deserved much praise and commendation: both for the great tra●ailes and payn●s (which it is most likely) he hath ta●●n in inventing such strange and wonderful propositions with their demonstrations, in this book contained, as also for participating and communicating abroad the same to others. Which book also, that the reader should want nothing conducing to the perfection of Euclides Elements: I have with some travail translated, & for the worthiness ●hereof have added it, a● a sixteenth book to the 15. books of Euclid. Vouchsafe therefore gentle reader diligently to read and poise it, for in it shall you find no● only matter strange and delectable, but also occasion of invention of greater things pertaining to the natures of the five regular solid●s●  

¶ The 1. Proposition. A Dodecahedron, and a cube inscribed in it, and a Pyramid inscribed in the same cube, are contained in one and the self same sphere.  
FOr the angles of the pyrami● are se● in the angel's of the cube wherein it is inscribed (by the first of the fiuetenth● and all the angles of the cube are set in the angles of the dodecahed●●● circumscribed 〈…〉 (〈◊〉 the 8. of the fifteenth): And all the angles of the Dodecahedron, are set in the superficies of the sphere, by the 17. of the thirteenth. Wherefore those three solids inscribed one within an other, are contained in one and the self same sphere, by the first definition of the fifteenth. A dodecahedron therefore and a cube inscribed in it, and a pyramid inscribed in the same cube, are contained 〈…〉 ●●lfe same sphere.  

〈…〉  
These three solids li 〈…〉 elf same Icosahedron, or Octohedron, or Pyramid. 〈…〉 me Icosahedron, by the, 5.11. & 12. of the fifteenth: and they are 〈…〉 ctohedron, by the 4. 6. and 16. of the same: lastly they are inscribed in 〈…〉 the first, 18. and 19 of the same. For the angles of all these solid 〈…〉 the circumscribed Icosahedron, or octohedron, or pyramid.    

¶ The 〈…〉 The proportion of a Dodecahedron circumscribed about a cube, to a Dodecahedron inscribed in the same cube, is triple to an extreme & mean propartion.  
FOrasmuch as in the ●● corollary of the 13. of the fiu●●enth, it was proved, that the side of a Dodecahedron inscribed in a cube, is the less segment of the side of that cube divided by an extreme and mean proportion: and the side of the dodecahedron circumscribed about the same cube, is the greater segment of the side of the same cube (which thing also was taught in the 13. of the fifteenth) the side of the Dodecahedron circumscribed, shallbe to the side of the Dodecahedron inscribed, as the greater segment of a right line divided by an extreme and mean proportion, is to the less segment of the same, which proportion is called an extreme and mean proportion by the definition, and by the 30. of sixth. But the proportion of like solid prolihedrons, is triple to the proportion of the side● of like proportion, by the corollary of the 17. of the twelfth. Wherefore the proportion of the Dodecahedron circumscribed about the cube, is to the dodecahedron inscribed in the same cube, in triple proportion of the sides joined together by an extreme and mean proportion: The proportion therefore of ● Dodecahedron circumscribed about a cube to a dodecahedron inscribed in the same cube, is triple to an extreme and mean proportion.   

The 3. Proposition. In every equiangle, and equilater Pentagon, a perpendicular drawn from one of the angles to the base, is divided by an extreme and mean proportion by a right line subtending the same angle.  
SVppose that ABCDF be a● equiangle and equilater pentagon: 
Construction.  and from one of the angles namely, from A, let there be drawn to the base CD a perpendicular AG: and let the line BF subtend the angle BAP, 
Demonstration.  which line BF let the line AD cut in the point I Then I say that the line BF, cutteth the line AG by an extreme and mean proportion. For forasmuch as the angles GAF and GAB are equal by the 27. of the third, and the angles A●F and AFB, are equal by the 5. of the first● therefore the ●●gles remaining at the point E, of the triangles AEB and AEF are equal: for that they are the residues of two right angles by the corollary of the 32. of the first. But the angle EGC, is by construction a right angle● wherefore the lines BF & CD are parallels by the 28. of the first. Wherefore as the line DIEGO is to the line IA, so i● the line GE to the line EA, by the 2. of the sixth. But the line DA, is in the poin● I divided by an extreme and mean proportion, by the 8. of the thirteenth. Wherefore the line GA is in the point E, divided by an extreme and mean proportion (by the ●. of the fourteenth). Wherefore in every equiangle and equilater pentagon, a perpendicular drawn from one of the angles to the base, is divided by an ex●reme and mean proportion by a right line subtending the same angle●  

¶ A Corollary.  
The line which subtendeth the angle of a pentagon, is a parallel to the side opposite unto the angle. As it was manifest in the lines ●F and CD.    

The 4. Proposition. If from the angles of the base of a 
* By a Pyramid understand a Tetrahedron throughout all this book.  Pyramid, be drawn to the opposite sides, right lines cutting the said sides by an extreme and mean proportion: they shall contain the bise of the Icosahedron inscribed in the Pyramid, which base shallbe inscribed in an equilater triangle, whose angles cut the sides of the base of the Pyramid by an extreme and mean proportion.  

Construction.  SVppose that ABG be the base of a pyramid, in which let be inscribed an equilater triangle FKH, which is done by dividing the sides into two equal parts. And in ●his triangle let there be inscribed the base of the Icosahedron inscribed in the pyramid: which is described by dividing the sides FK, KH, HF, by an extreme & mean proportion in the points C, D, E, by the 19 of the fiuetenth. Again let the sides of the pyramid, namely, AB, BG, and GA be divided by an extreme and me●ne proportion in the points I, M, L, by the 30. of the sixth. And draw these right lines AM, BL, GI'. 
Demonstration.  Then I say that those lines describe the triangle CDE of the Icosahedron. For forasmuch as the lines BG and FH are parallels, by the 2. of the sixth: by the point D let the line ODN be drawn parallel to either of the lines BG & FH. Wherefore the triangle HDN shallbe like to the triangle HKG, by the corollary of the 2. of the sixth. Wherefore either of these lines DN and NH shall be equal to the line DH, the greater segment of the line KH or FH. And forasmuch as the line FO is a parallel to the line HK, and the line OD to the line FH● the line OD shall be equal to the whole line FH in the parallelogram FODH, by the 34. of the ●irst. Wherefore as the whole line FH is to the grea●er segment FE, so shall the lines equal to them be, namely, the line OD to the line DN, by the 7. of the fifth. Wherefore the line ON is divided by an extreme and mean proportion in the point D, by the 2. of the fourteenth. But the triangles AOD, AFE, and ABM, are like the one to the other, and so also are the triangles ADN, AEH, and AMG, by the corollary of the second of the sixths' Wherefore as FE is to EH, so is OD to DN, and BM to MG. Wherefore the line AM cutting the lines FH and ON, like unto the line BG in the points E, D, M, describeth ED the side of the triangle of the Icosahedron ECD, which is described in the sections E, C, D, by supposition. And by the same reason the lines BL and GI' shall describe the other sides EC and CD of the same triangle. By the point E, let there be drawn to GI' a parallel line PEQ. Now forasmuch as the lines BM and FE are parallels, the line AM is in the point E, cut like to the line AB in the point F, by the 2. of the sixth. Wherefore the line AE is equal to the line EM: and unto the line EM also are equal either of the lines GD and DIEGO: which ●re cut l●ke unto the foresaid lines. Again forasmuch as in the triangle ADI the lines DIEGO and EP are parallels, as the line DIEGO is to the line EP, so is the line AD to the line AE: but as the line AD is to the line AE, so is the line DG to the line EQ by the 2. of the sixth: wherefore as the line DIEGO is to the line EP, so is the line DG to the line EQ: and alternately as the line DIEGO is to the line DG, so is the line EP to the line EQ: but the lines DIEGO and IG are equal: wherefore also the lines EP and EQ are equal. And forasmuch as the line AH is equal to the line FH, whose greater segment is the line HN● therefore the whole line AN, is divided by an extreme and mean proportion in the point H, by the ●. of the thirteenth. But as the line AN is to the line AH, so is the line AD to the line AE, by the 2. of sixth (for the line● FH and ON are parallels:) and again as the line AD is to the line AE, so (by the same) is the line AG to the line AQ, and the line AI to the line AP: for the lines PQ, and GI' are parallels: Wherefore the lines AG and AI are divided by an extreme and mean proportion in the points Q & P: & the line AQ shallbe the greater segment of the line AG or AB. And forasmuch as the whol● line AG is to the greater segment AQ, as the greater segment AI is to the residue AP: the line A● shallbe the less segment of the whole line A● or AG. Wherefore the li●● PEQ (which by the point E passeth parallelwise to the line GI') cutteth the lines AG and BA by an extreme and mean proportion in the points Q and P. And by the same reason the line ●R (which by the point C, passeth parallelwise to the line AM) shall fall upon the sections P and R: so also shall the line RQ (which by the point D passeth parallelwise to the line BL) fall vpo● the sections RQ. Wherefore either of the lines PE and EQ shallbe equal to the line CD, in the parallelograms PD, and QC, by the 34. of the first. And forasmuch as the lines PE and EQ are equal, the lines PC, CR, RD and DQ shallbe likewise equal. Wherefore the triangle PRQ i●●quilater, and cutteth the sides of the base of the pyrami● in the points P, Q, R, by an extreme and mean proportion. And in it is inscribed the base ECD of the Icosahedron contained in the for●sayd pyramid. If therefore from the angles of the base of a pyramid, be drawn to the opposite sides, right lines cutting the said sides by an extreme and mean proportion: they shall contain the base of the Icosahedron inscribed in the pyramid, which base shall be inscribed in an equilater triangle, whose angles cut the sides of the base of the pyramid by an extreme & mean proportion.  

¶ A Corollary.  
The side of an Icosahedron inscribed in an Octohedron, is the greater segment of the line, which being drawn from the angle of the base of the Octohedron cutteth the opposite side by an extreme and mean proportion. For, by the 16. of the fifteenth, FKH is the base of the Octohedron, which containeth the base of the Icosahedron CDE: unto which triangle FKH, the triangle HKG is equal, as hath been proved. By the point H draw unto the line ME a parallel line HT, cutting the line DN in the point S. Wherefore ES, DT, and ET, are parallelograms: and therefore the lines EH and MT are equal: and the lines EM and HT are like cut in the points D and S, by the 34. of the first. Wherefore the greater segment of the line HT is the line HS, which is equal to ED the side of the Icosahedron. But (by the 2. of the sixth) the line TK is cut like to the line HK by the parallel DM. And therefore (by the 2. of the fourteenth) it is divided by an extreme and mean proportion. But the line TM is equal to the line EH. Wherefore also the line TK is equal to the line EF or DH. Wherefore the residues EH and TG are equal. For the whole lines FH and KG are equal. Wherefore KG the side of the triangle HKG is in the point T divided by an extreme and mean proportion in the point T, by the right line HT, and the greater segment thereof is the line ED the side of the Icosahedron inscribed in the Octohedron, whose base is the triangle HKG (or the triangle FKH which is equal to the triangle HKG) by the 16. of the fifteenth.    

¶ The 5. Proposition. The side of a Pyramid divided by an extreme and mean proportion, maketh the less segment in power double to the side of the Icosahedron inscribed in it.  
SVppose that ABG be the base of a pyramid: 
Construction.  and let the base of the Icosahedron inscribed in it, be CDE, described of three right lines, which being drawn from the angles of the base ABG cut the opposite sides by an extreme and mean proportion, by the former Proposition: namely, of these three lines AM, BY, and GI'. Then I say, that AI the less segment of the side A●, is in power duple to CE the side of the Icosahedron. For, forasmuch as by the former Proposition, it was proved that the triangle CDE is inscribed in an equilater triangle, 
Demonstration.  whose angles cut the sides of ABG the base of the pyramid by an extreme and mean proportion, let that triangle be FHK, cutting the line AB in the point F. Wherefore the less segment FA is equal to the segment AI, by the 2. of the fourteenth: (for the lines AB and AG are cut like). Moreover the side FH of the triangle FHK is in the point D cut into two equal parts, as in the former Proposition it was proved, and FCED also by the same is a parallelogram: Wherefore the lines CE and FD are equal, by the 33, of the first. And forasmuch as the line FH subtendeth the angle BAG of an equilater triangle, which angle is contained under the greater segment AH and the less segment AF● therefore the line FH is in power double to the line AF or to the line AI the less segment, by the Corollary of the 16. of the fifteenth. But the same line FH is in power quadruple to the line CE, by the 4. of the second: (for the line FH is double to the line CE). Wherefore the line AI being the half of the square of the line FH is in power duple to the line CE, to which the line FH was in power quadruple. Wherefore the side AG of the pyramid being divided by an extreme and mean proportion, maketh th● less segment AI in power duple to the side CE of the Icosahedron inscribed in it.  

¶ A Corollary.  
The side of an Icosahedron inscribed in a pyramid, is a residual line. For the diameter of the Sphere which containeth the five regular bodies, being rational, is in power sesquialtera to the side of the pyramid, by the 13. of the thirteenth: and therefore the side of the pyramid is rational, by the definition: which side being divided by an extreme and mean proportion, maketh the less segment a residual line, by the 6. of the thirteenth. Wherefore the side of the Icosahedron being commensurable to the same less segment (for the square of the side of the Icosahedron is the half of the square of the said less segment) is a residual line, by that which was added after the 103. of the tenth book.    

¶ The 6. Proposition. The side of a Cube containeth in power half the side of an equilater triangular Pyramid inscribed in the said Cube.  
FOr forasmuch as the side of the pyramid inscribed in the cube subtendeth two sides of the cube which contain a right angle, by the 1. of the fifteenth: it is manifest, by the 47. of the first, that the side of the pyramid subtending the said sides, is in power duple to the side of the cube: Wherefore also the square of the side of the cube is the half of the square of the side of the pyramid. The side therefore of a cube containeth in power half the side of an equilater triangular pyramid inscribed in the said cube.   

¶ The 7. Proposition. The side of a Pyramid is duple to the side of an Octohedron inscribed in it.  
FOrasmuch as by the 2. of the fifteenth it was proved, that the side of the Octohedron inscribed in a pyramid coupleth the middle sections of the sides of the pyramid. Wherefore the sides of the pyramid and of the Octohedron are parallels, by the Corollary of the 39 of the first: and therefore, by the Corollary of the 2. of the sixth, they subtend like triangles. Wherefore (by the 4. of the sixth) the side of the pyramid is double to the side of the Octohedron, namely, in the proportion of the sides. The side therefore of a pyramid is duple to the side of an Octohedron inscribed in it.   

¶ The 8. Proposition. The side of a Cube is in power duple to the side of an Octohedron inscribed in it.  
IT was proved in the 3. of the fifteenth, that the diameter of the Octohedron inscribed in the cube, coupleth the centres of the opposite bases of the cube. Wherefore the said diameter is equal to the side of the cube. But the same is also the diameter of the square made of the sides of the Octohedron, namely, is the diameter of the Sphere which containeth it, by the 14. of the thirteenth. Wherefore that diameter being equal to the side of the cube, is in power double to the side of that square, or to the side of the Octohedron inscribed in it, by the 47. of the first. The side therefore of a Cube, is in power duple to the side of an Octohedron inscribed in it: which was required to be proved.   

¶ The 9 Proposition. The side of a Dodecahedron, is the greater segment of the line which containeth in power half the side of the Pyramid inscribed in the said Dodecahedron.  
SVppose that of the Dodecahedron ABGD the side be AB: and let the base of the cube inscribed in the Dodecahedron be ECFH, by the ●● of the fifteenth. And let the side of the pyramid inscribed in the cube be CH, by the 1. of the fifteenth. 
Construction.  Wherefore the same pyramid is inscribed in the Dodecahedron, by the 10. of the fifteenth. Then I say, that AB the side of the Dodecahedron is the greater segment of the line which containeth in power half the line CH, which is the side of the pyramid inscribed in the Dodecahedron. 
Demonstration.  For forasmuch as EC the side of the cube being divided by an extreme and mean proportion maketh the greater segment the line AB, the side of the Dodecahedron, by the ●●rst Corollary of the 17. of the thirteenth: (For they are contained in one and the self same Sphere (by the first of this book): and the line EC the side of the cube containeth in power the half of the side CH, by the 6. of this book. Wherefore AB the side of the Dodecahedron, is the greater segment of the line EC, which containeth in power the half of the line CH, which is the side of the Dodecahedron inscribed in the pyramid. The side therefore of a Dodecahedron, is the greater segment of the line which containeth in power half the side of the Pyramid inscribed in the said Dodecahedron.   

¶ The 10. Proposition. The side of an Icosahedron, is the mean proportional between the side of the Cube circumscribed about the Icosahedron, and the side of the Dodecahedron inscribed in the same Cube.  
SVppose that there be a cube ABFD, in which let there be inscribed an icosahedron CLIGOR, by the 14. of the fifteenth. 
Construction.  Let also the Dodecahedron inscribed in the same be EDMNPS, by the 13. of the same. Now forasmuch as CL the side of the Icosahedron is the greater segment of AB the side of the cube circumscribed about it, by the 3. Corollary of the 14. of the fifteenth: 
Demonstration.  and the side ED of the Dodecahedron inscribed in the same cube is the less segment of the same side AB of the cube, by the 2. Corollary of the 13. of the fifteenth: it followeth that AB the side of the cube being divided by an extreme and mean proportion, maketh the greater segment CL the side of the Icosahedron inscribed in it, and the less segment ED the side of the Dodecahedron likewise inscribed in it. Wherefore as the whole line AB the side of the cube, is to the greater segment CL the side of the Icosahedron, so is the greater segment CL the side of the Icosahedron, to the less segment ED● the side of the Dodecahedron, by the third definition of the sixth. Wherefore the side of an Icosahedron, is the mean proportional between the side of the cube circumscribed about the Icosahedron, and the side of the Dodecahedron inscribed in the same cube.   

¶ The 11. Proposition. The side of a Pyramid, is in power 
† That is, a● 18. to 1.  Octodecuple to the side of the cube inscribed in it.  
FOr, by that which was demonstrated in the 18. of the fifteenth, the side of the pyramid is triple to the diameter of the base of the cube inscribed in it: 
Demonstration.  and therefore it is in power nonecuple to the same diameter (by the 20. of the sixth). But the diamer is in power double to the side of the cube, by the 47. of the first. And the double of nonecuple maketh Octodecuple. Wherefore the side of the pyramid is in power Octodecuple to the side of the cube inscribed in it.   

¶ The 12. Proposition. The side of a Pyramid, is in power Octodecuple to that right line, whose greater segment is the side of the Dodecahedron inscribed in the Pyramid.  
FOrasmuch as the Dodecahedron and the cube inscribed in it, are set in one and the s●lf● same pyramid, by the Corollary of the first of this book: and the side of the pyramid circumscribed about the cube is in power octodecuple to the side of the cube inscribed, by the former Proposition: but the greater segment of the self same side of the cube, is the side of the Dodecahedron which containeth the cube, by the Corollary of the 17. of the thirteenth. Wherefore the side of the pyramid is in power octodecuple to that right line, namely, to the side of the cube, whose greater segment is the side of the Dodecahedron inscribed in the pyramid.   

¶ The 13. Proposition. The side of an Icosahedron inscribed in an Octohedron, is in power duple to the less segment of the side of the same Octohedron.  
FOrasmuch as in the 17. of the fifteenth, it was proved, that the side of an Icosahedron inscribed in a pyramid, coupleth together the two sections (which are produced by an extreme and mean proportion) of the side of the Octohedron which make a right angle: and that right angle is contained under the less segments of the sides of the Octohedron, and is subtended of the side of the Icosahedron inscribed: it followeth therefore, that the side of the Icosahedron which subtendeth the right angle, being in power equal to the two lines which contain the said angle, by the 47. of the first, is in power duple to every one of the less segments of the side of the Octohedron which contain a right angle. Wherefore the side of an Icosahedron inscribed in an Octohedron, is in power duple to the less segment of the ●ide of the same Octohedron.   

¶ The 14. Proposition. The sides of the Octohedron, and of the Cube inscribed in it, are in power the one to the other 
† That i●, as 9 to 2.  in quadrupla sesquialter proportion.  
SVppose that ABGDE be an Octohedron, and let the cube inscribed in it be FCHI. Then I say, that AB the side of the Octohedron, is in power quadruple sesquialter to FI the ●ide of the cube. Let there be drawn to BE the base of the triangle ABE a perpendicular AN: and again let there be drawn to the same base in the triangle G●E the perpendicular GN: which AN & GN shall pass by the centres F and I: and the line AF is duple to the line FN, by the Corollary of the 12. of the thirteenth. Wherefore the line AO is duple to the line OE, by the 2. of the sixth. For the lines FO and NE are parallels. And therefore the diameter AG is triple to the line FI. Wherefore the power of AG is 
* That is, as 18. to 2. or 9 to 1.  noncuple to the power of FI. But the line AG is in power duple to the side AB, by the 14. of the thirteenth. Wherefore the square of the line AB, being ing the half of the square of the line AG, which is noncuple to the square of the line FI, i● quadruple sesquialter to the square of the line FI. The sides therefore of the Octohed●●●●nd of the cube inscribed in it● are in power the one to the other, in quadruple sesquialter proportion.   

¶ The 1●. Proposition. The side of the Octohedron, is in power quadruple sesquialter to that right line, whose greater segment is the side of the Dodecahedron inscribed in the same Octohedron.  
FOrasmuch as in the 14. of this book, it was proved, that the side of the Octohedron is in power quadruple sesquialter to the side of the cube inscribed in it: but the side of the cube being cut by an extreme and mean proportion, maketh the greater segment the side of the Dodecahedron circumscribed about it, by the 3. Corollary of the 13. of the fifteenth: therefore the side of the Octohedron is in power quadruple sesquialter to that right line (namely, to the side of the cube) whose greater segment is the side of the Dodecahedron inscribed in the cube. But the Dodecahedron and the cube inscribed one within an other, ar● inscribed in one and the self same Octohedron, by the Corollary of the first of this book. The side therefore of the Octohedron, is in power quadruple sesquialter to that right line, whose greater segment is the side of the Dodecahedron inscribed in the same Octohedron.   

¶ The 16. Proposition. The side of an Icosahedron, is the greater segment of that right line, which is in power duple to the side of the Octohedron inscribed in the same Icosahedron.  
SVppose that there be an Icosahedron ABGDFHEC: whose side let be BG or ●C● and let the Octohedron inscribed in it be AKD●: and let the side thereof be AL. Then I say, that the side ●C is the greater segment of that right line which is in power duple to the side AL. For forasmuch as figures inscribed and circumscribed have o●e & the self-same centre, by the Corollary of the ●1. of the fifteenth, let the same be the point I Now right line● drawn by th●● 〈◊〉 to the middle sections of the opposite sides, namely, the lines AID and KIL, do in the point I ●ut 〈…〉 the other in●● two squall 〈◊〉, and perpendicularly, by the Corollary of the 14. of the fifteenth: and forasmuch as they couple the middle sections of the opposite lines BG and HF, therefore they cut them perpendiularly: 
Draw in the figure a line from B to H.  wherefore also the lines BG 〈…〉, are parallels, by the 4. Corollary of the 14. of the 〈…〉. Now then draw a line from B to H: and the said ●●ne BH shall be equal and parallel to the line KL, by the 33. of the first. But the line BH subtendeth ●w● sides of the pentagon which is composed of the sides of the Icosahedron, namely, the sides BA and AH: Wherefore the line BH being cut by an extreme and mean proportion maketh the greater segment the side of the pentagon, by the 8. of the thirteenth: which side is also the side of the Icosahedron, namely, EC. And unto the line BH the line KL● is equal: and the line KL is in power duple to AL the side of the Octohedron, by the 47. of the first: for in the square AKDL the angle KAL is a right angle. Wherefore EC the side of the Icosahedron, is the greater segment of the line BH or KL, which is in power duple to AL ●he side of the Octohedron inscribed in the Icosahedron. Wherefore the side of an Icosahedron, is the greater segment of that right line, which is in power duple to the side of the Octohedron inscribed in the same Icosahedron.   

¶ The 17. Proposition. The side of a Cube is to the side of a Dodecahedron inscribed in it, in duple proportion of an extreme and mean proportion.  
FOr it was manifest by the ●. corollary of the 13. of the fifteenth, that the side of a cube divided by an extreme and mean proportion, maketh the less segment, the side of the dodecahedron inscribed in it: but the whole is to the less segment in duple proportion of that in which it is to the greater, by the 10. definition of the fifth. For the whole, the greater segment, and the less, are lines in continual proportion, by the 3. definition of the sixth. Wherefore the whole namely the side of the cube, is to the side of the dodecahedron inscribed in it, namely, to his less segment, in duple proportion of an extreme and mean proportion', namely, 
* What the duple of an extreme and mean proportion is.  of that which the whole hath ●o the greater segment, by the 2. of the fourteenth.   

¶ The 18. Proposition. The side of a Dodecahedron is, to the side of a Cube inscribed in it, in converse proportion of an extreme and mean proportion.  
IT was proved in the 3. corollary of the 13. of the fifteenth, that the side of a Dodecahedron circumscribed about a Cube, is the greater segment of the side of the same Cube. Wherefore the whole side of the Cube inscribed is to the greater segment, namely, to the side of the dodecahedron circumscribed, in an extreme and mean proportion: wherefore by conversion, the greater segment, that is, the side of the dodecahedron, is to the whole, namely, to the side of the Cube inscribed, in the converse proportion of an extreme and mean proportion, by the 13. definition of the fifth.   

¶ The 19 Proposition. The side of an Octohedron, is sesquialter to the side of a Pyramid inscribed in it.  
FOr (by the corollary of the 14. of the thirteenth) the Octohedron is custe into two quadrilater pyramids, one of which let be ABGDF: 
Construction.  and let the centres of the circles which contain the 4. bases of the Octohedron be K, E, I, C. And dr●w these right lines KE, ●I, IC, CK, and EC. Wherefor● K●IC is a square, and one of the bases of the cube inscribed in the Octohedron, by the 4. of the fifteenth. And forasmuch as the angles of a cube and of the pyramid inscribed in it, are for in the centres of the bases of the Octohedron circumscribed about the cube, by the 6● of the fifteenth: and the side of the pyramid coupleth the opposite angle● of the base of th● cube, by the 1. of the fifteenth, it is manifest that the line EC is the side of the pyramid inscribed in the Octohedron ABGDF. Then I say that GD the side of the Octohedron, is sesquialter to EC the side of the pyramid inscribed in it. From the point A draw to the bases BG and FD perpendiculars AN and AM● which (by the corollary of the 12. of the thirteenth) shall pass by the centres E and C. 
Demonstration.  And draw the line NM. Now forasmuch, as BGDF is a square, by the 14. of the thirteenth, the lines NG and MD shall be parallels and equal. For the lines BG and FD are by the perpendiculars cut into two equal parts in the points N and M (by the 3. of the third). Wherefore the lines NM and GD shall be parallels and equal, by the 33. of the first. And forasmuch as the lines AN and AM which are the perpendiculars of equal and like● triangles are c●t a like in the points ● and C, the lines EC and NM● shall be parallels, by the 2. of the sixth: and therefore by the corollary of the same, the triangles AEC, and ANM shall be like. Wherefore as the line AN is to the line AE, so is the line NM to the line EC by the 4. of the sixth. But the line AN is sesquialter to the line AE, for the line AE is duple to the line EN, by the corollary of the 12● of the thirtenth● wherefore the line NM, or the line GD which is equal unto it, is sesquialter to the line EC. Wherefore GD the side of the Octohedron, is sesquialter to EC the sid● of the pyramid inscribed in it.   

¶ The ●0. Proposition. If from the power of the diameter of an Icosahedron, be taken away the power tripled of the side of the cube inscribed in the Icosahedron: the power remaining shall be sesquitertia to the power of the side of the Icosahedron.  
LEt there be taken an Icosahedron ABGD: and l●● two bases of the cube inscribed in it, joined together be EHKL and LKFC: and let the diameter of the cube be FH and the side be EH, and let the diameter of the Icosahedron be ●G, and the side be AB. Then I say, that if from the power of the diameter GB, be taken away the power tripled of EH the side of the cube: 
Demonstration.  the power remaining, shall be sesquetertia to the power of AB the side of the Icosahedron. For forasmuch as the centres of inscribed and circumscribed figures, are in one & the self same point, by the corollary of the 21. of the 〈◊〉 the diameters BG and FH shall in one and the self same point cut the one the other into two equal parts: for we have before by the same corollary taught, that the tops of equal and like pyramids do in that point concur, let that point be the centre I Now the angles of the cube, which are at the points F and H are set at the centres of the bases of the Icosahedron by the 11. of the fiuetenth● Wherefore the line FH shall be perpendicular to both the bases of the Icosahedron, by the corollary of the assumpt of the 16. of the twelfth. Wherefore the line IB containeth in power the two lines IH and HB, by the 47. of the first. But the line HB, is drawn from the centre of the circle which containeth the base of the Icosahedron namely, the angle B is placed in the circumference, and the point H is the centre. Wherefore the whole line BG containeth in power the whole lines FH and the diameter of the circle (namely, the double of the line BH) by the 15● of the fifth. But the diameter which is double to the line HB is in power sesquiterti● to the side of the equilater triangle inscribed in the same circle● by the corollary of the ●●. of the thirteenth. For it is in proportion to the side● as the side is to the perpendicular, by the corollary of the 8. of th●●ixth. And FH the diameter of the cube, is in power triple to EH the side of the same cube, by the 15● of the thirteenth. If therefore from the power of the diameter BG, be taken away the power tripled of EH the side of the cube inscribed● that is● the power of the line FH: the residue (namely, the power of the diameter of the circle which is duple to the line HB shall be sesquiterti● to the side of the triangle inscribed in that circle: which self side is AB the side of the Icosahedron. If therefore from the power of the diameter of an Icosahedron, be taken away the power tripled of the side of the cube inscribed in the Icosahedron, the power remaining shall be s●squitertia ●o the power of the side of the Icosahedron.  

A Corollary.  
The diameter of the Icosahedron, containeth in power two lines, namely, the diameter of the cube inscribed, which coupleth the centres of the opposite bases, and the diameter of the circle which containeth the base of the Icosahedron. For it was manifest, that BG the diameter containeth to power the line FH which doupleth the centres, and the double of the line BH, that is, the diameter of the circle containing the bas● wherein i● the centre H●    

¶ The 21. Proposition. The side of a Dodeca●edron is the less segment of that right line, which is in power duple to the side of the Octohedron inscribed in the same Dodecahedron.  
LEt there be taken a Dodecahedron AB●DCT, one of whose sides let be AB. And let the Octohedron inscribed in the Dodecahedron be EFLKI: one of whose sides let be EF. Then I say that AB the side of the Dodecahedron, is the less segment of a certain right ●ine ● cut by an extreme and mean proportion) which is in power duple to EF the side of the Octohedron inscribed in the Dodecahedron. Draw the diameters EL and FK of the Octohedron. Now they couple the middle sections of the opposite sides of the dodecahedron AB and GD, (by the 9 of the fifteenth, & 3. corollary of the 17. of the thirteenth) & every one of those diameters being divided by an extreme and mean proportion, do make the less segment, the side of the dodecahedron, by the 4. corollary of the same. Wherefore the side AB is the less segment of the line FK. But the line FK containeth in power the two equal lines EF & EK, by the 47. of the first: for the angle FEK is a right angle of the square FEKL of the Octohedron. Wherefore the line FK is in power duple to the line EF. Wherefore the line AB (the side of the dodecahedron) is the less segment of the line FK, which is in power duple to EF the sid● of the Octohedron. The side therefore of a Dodecahedron i● the less segment of that right line, which is in power duple to the side of the Octohedron inscribed in the same Dod●cahedron.   

¶ The 22. Proposition. The diameter of an Icosahedron is in power sesquitertia to the side of the same Icosahedron, and also is in power sesquialter to the side of the Pyramid inscribed in the Icosahedron.  
FOr forasmuch as it hath been proved (by the 10. of this book) that if from the power of the diameter of the Icosahedron be taken away the triple of the power of the side of the cube inscribed in it, there shallbe left a square sesquitertia to the square of the side of the Icosahedron: But the power of the side of the cube tripled, is the diameter of the same cube, by the 15. of the thirteenth: And the cube, & the pyramid inscribed in it are contained in one & the self same sphere, by the first of this book, and in one & the self same Icosahedron by the corollary of the same. Wherefore one and the self same diameter of the cube, or of the sphere which containeth the cube and the pyramid, is in power sesquialter to the side of the pyramid by the 13. of the thirteenth. Wherefore it followeth, that if from the diameter of the Icosahedron, be taken away the triple power of the side of the cube, or the sesquialter power of the side of the pyramid, which are the powers of one and the self same diameter, there shall be left the sesquitertia power of the side of the Icosahedron. The diameter therefore of an Icosahedron is in power sesquitertia to the side of the same Icosahedron, and also is in power sesquialter to the side of the Pyramid inscribed in the Icosahedron.   

The 23. Proposition. The side of a Dodecahedron is to the side of an Icosahedron inscribed in it, as the less segment of the perpendicular of the Pentagon, is to that line which is drawn from the centre to the side of the same pentagon.  

Constrution.  LEt there be taken a Dodecahedron ABGDFSO. Whose side let be AS or SO: and let the Icosahedron inscribed in it be KLNMNE, whose side let be KL. From the two angles of the pentagon● BAS and FAS of the Dodecahedron, namely, from the angle●● and F, let there be drawn to the common base AS perpendicular lines BC & FC: which shall pass by the centres K & L of the said pentagons, by the corollary of the 10. of the thirteenth. Draw the lines BF and RO. Now forasmuch as the line RO subtendeth the angle OFR of th● pentagon of the dodecahedron, it shall cut the line FC by an extreme and mean proportion, by the 3. of this book, let it cut it in the point I And forasmuch as the line KL is the side of the Icosahedron inscribed in the Dodecahedron, it coupleth the centres of the bases of the dodecahedron: for the angles of the Icosahedron are set in the centres of the bases of the dodecahedron, by the 7. of the fifteenth. Now I say that SO, the side of the dodecahedron is to KL the side of the Icosahedron, as the less segment IF of the perpendicular line CF, is to the line LC which is drawn from the centre L to AS the side of the pentagon. For forasmuch as in the triangle BCF the two sides CB and CF are in the centres L and K cut like proportionally, 
Demonstration.  the lines BF and KL shallbe parellels, by the 2. of the sixth. Wherefore the triangles BCF, and KCL shall be equiangle, by the corollary of the same. Wherefore as the line CL is to the line KL● so is the line CF to the line BF, by the 4. of the sixth. But CF maketh the less segment the line IF, by the 3. of this book, and the lin● BF maketh the less segment the line SO, namely, the side of the Dodecahedron, by the 2. corollary of the 13. of the fifteenth. For the line BF which coupleth the angles B and F of the bases of the dodecahedron, is equal to the side of the cube, which containeth the dodecahedron, (by the .13. of the fifteenth). Wherefore as the whole line C●, is to the whole line BF, so is the less segment IF to the less segment SO (by the 2. of the 14). But as the line CF is to the line BF, so is the line CL proved to be to the line KL. Wherefore as the line IF is to the line SO, so is the line CL to the line KL. Wherefore alternately by the 16. of the fifth, as the line IF the less segment of the perpendicular of the pentagon FAS, is to the line LC which is drawn from the centre of the pentagon, to the base, so is the line SO the side of the Dodecahedron to th● line KL the side of the Icosahedron inscribed in it. The side therefore of a Dodecahedron is to the side of an Icosahedron inscribed in it, as the less segment of the perpendicular of the pentagon, is to that line which is drawn from the cen●re to the side of the same pentagon.   

¶ The 24. Proposition. If half of the side of an Icosahedron be divided by an extreme & mean proportion: and if the less segment thereof be taken away from the whole side, and again from the residue be taken away the third part: that which remaineth shall be equal to the side of the Dodecahedron inscribed in the same Icosahedron.  
SVppose that ABGDF be a pentagon, 
Construction.  containing five sides of the Icosahedron by the 16. of the thirteenth, and let it be inscribed in a circle, whose centre let be the point E. And upon the sides of the pentagon, let there be reared up triangles, making a solid angle of the Icosahedron at the point I, by the 16. of the thirteenth. And in the circle ABD, inscribe an equilater triangle AHK. From the centre E draw to HK the side of the triangle, and GD the side of the pentagon, a perpendicular line, which let be ECNM. And draw these right lines EG, ED, IG and ID. And divide the line BG into two equal parts in the point T. And draw these lines IN, IT, TN, ET. And forasmuch as in the perpendiculars IT & IN are the centres of the circles which contain the equilater triangles IBG, & IGD, by the corollary of the first of the third. Let those centres be the points S and O. And draw the line SO. Divide the line TB the half of BG the side of the Icosahedron by an extreme and mean proportion in the point R, by the 30. of the sixth, and let the less segment thereof be RB. And forasmuch as the line SO coupleth the centres of the triangles IBG, & IGD, it is by the 5. of the fifteenth, the side of the Dodecahedron inscribed in the Icosahedron, whose side is the line BG. From the side BG take away ●R the less segment of the half side. And from the residue GR take away the third part GV (by the 9 of the sixth.) Then I say that the residue RV is equal to SO the side of the Dodecahedron inscribed. 
Demonstration.  For forasmuch as the perpendicular EN is in the point C divided by an extreme and mean proportion, by the corollary of the first of the fourteenth, and the greater segment thereof is the line EC, and unto the line EC the line CM is equal, by the corollary of the 12. of the thirteenth: wherefore the line EC is to the line CN, as the line CM is to the same line CN, by the 7. of the fifth. But as the line EC is to the line CN, so is the whole line ●N, to the greater segment EC, by the 3. definition of the sixth. Wherefore (by the 11. of the fifth), as the whole line EN is to the greater segment EC, so is the line CM to the line CN. Wherefore the line CM, is divided by an extreme and mean proportion in the point N, namely, is divided like unto the line EN, by the 2. of the fourteenth. Wherefore the line EM exceedeth the line EN by the less segment of his half, namely, by MN. And forasmuch as EGD is the triangle of an equilater and equiangle pentagon ABGDF, and ETN is likewise the triangle of the like pentagon inscribed in the pentagon ABGDF: Therefore by the 20. of the sixth, the triangle ETN is like to the triangle EGD● Wherefore as the line EG is to the line EN, so by the 4. of the sixth, is the line GD to the line NT. Wherefore the line GD (or BG which is equal unto it) exceedeth the line NT by the less segment of the half of BG. For the line EG did in like sort exceed the line EN. But that less segment is the line BR. Wherefore the residue RG is equal to the line TN. And forasmuch as IBG is an equilater triangle: the perpendicular ST shallbe the half of the line SI which is drawn from the centre, by the corollary of the 12. of the thirteenth: wherefore the line IT exceedeth the line IS by his third part. And forasmuch as the line SO which coupleth the sections, is a parallel to the line TN, by the 2. of the sixth. For the equal perpendiculars IT, and IN are cut like in the points S & O: therefore the triangles ITN & I●O, are like by the corollary of the second of the sixth. Wherefore as the line IT is to the line IS, so by the 4. of the sixth is the line TN to the line SO. But the line IT exceedeth the line IS by a third part: wherefore the line TN, exceedeth the line SO by a third part: but the line TN is proved equal to the line RG. Wherefore the line RG exceedeth the line SO by a third part of himself, which is GU. Wherefore the residue RV, is equal to the line SO, which is the side of the dodecahedron inscribed in the Icosahedron, whose side is the line BG. If therefore half of the side of an Icosahedron, be divided by an extreme & mean proportion: and if the less segment thereof be taken away from the whole side, and again from the residue be taken away the third part: that which remaineth shall be equal to the side of the dodecahedron inscribed in the same Icosahedron.   

The 25. Proposition. To prove that a cube given, is to a trilater equilater pyramid inscribed in it, triple.  
SVppose that the cube given, be ABCH: and let the pyramid inscribed in it be AGDF. Then I say that the cube ABCH is triple to the pyramid AGDF. For forasmuch as the base AFD is common to the pyramid AFDB and AFDG, the pyramid AFDB shallbe set without the pyramid AFDG. Likewise the rest of the bases of the inscribed pyramid are common to the rest of the pyramids sort without: which are these: the pyramid AGDC upon the base AGD: the pyramid AGF● upon the base AGF● and the pyramid GDFH upon the base GDF. Which pyramids taken without, are four in number, equal and like the one to the other, by the ●. definition of the eleu●th. For every one of them is contained under thr● half squares of the cube, and one of the bases of the pyramid inscribed. Wherefore every one of them is contained under the half base of the cube, & the altitude of the cube. As the pyramid ALGF, hath to his base half of the square EH, namely, the triangle EGF, & hath to his altitude, the altitude of the cube, namely, the line AE. Wherefore the said pyramid is the sixth part of the cube. For if the cube be divided into two prisms, by the plain CBFG, the prism ACBGEF, shallbe triple to the pyramid AEGF, having one & the self same base with it EGF, and one and the self same altitude EA, by the first corollary of the 7. of the twelfth. Wherefore the said outward pyramid AEGF is the sixth part of the whole cube. Wherefore also the same pyramid together with the other three outward pyramids AFDB, AGDC, and GDFH, ●hal contain two third parts of the cube. Wherefore the residue, namely, the pyramid inscribed AGDF, shall contain one third part of the cube. And therefore conversedly the cube shall be triple to it: wherefore we have proved ●hat a cube given triple to a trilater & equilater pyramid inscribed in it.   

¶ The 26. Proposition. To prove that a trilater equilater Pyramid is duple to an Octohedron inscribed in it.  
LEt there be taken a trilater Pyramid ABCD: whose six sides let be cut into two equal parts, in the points E, K, F, L, G, and H● inscribing thereby an Octohedron in the pyramid, by the 2. of the fifteenth. Wherefore the pyramids AEGH, BEFK, CFGL, & DKHL, fall without the Octohedron inscribed, by the same second of the fifteenth. But the outward Pyramids (namely, AEGH, and the three other) are like unto the whole pyramid, by the 7. definition of the eleventh. For the bases of the whole pyramid are by parallel lines drawn in them cut into like triangles, by the Corollary of the 2. of the sixth, of which the foresaid pyramids are made. Wherefore the whole pyramid is to every one of them in triple proportion of that in which the sides of like proportion are, by the 8. of the twelfth. But by construction, the proportion of the side A● to the side A● is duple. Wherefore the whole pyramid ABCD is octuple to the pyramid AEGH, and so is it to every one of the pyramids which are equal to AEGH. For duple proportion multiplied into itself twice maketh octuple. Wherefore it followeth that the 4. pyramids AEGH, ●EFK, CFGL, and DKHL, taken together, make the half of the whole pyramid ABCD. Wherefore the residue, namely, the Octohedron EGLKHF, is the other half of the pyramid. Wherefore the pyramid is duple to the Octohedron. Wherefore we have proved that a trilater equilater pyramid is duple to an Octohedron inscribed in it.   

¶ The 27. Proposition. To prove that a Cube is sextuple to an Octohedron inscribed in it.  
LEt there be taken a cube ABCD, EFGH: whose 4. standing lines AE, BF, CH, & DG, let be cut into two equal parts in the points I, K, M, L: and● by those points let there be extended a plain KLMI: which shall be a square, and parallel to the squares BC & FH, by the 15. of the eleventh. Wherefore in it shall be the base which is common to the two pyramids of the Octohedron inscribed in the cube, by that which was demonstrated in the third of the fifteenth. Let that base be NPRQ, coupling the centres of the bases of the cube: and upon that base let be set the two pyramids of the Octohedron, which let be NPQRS, and NPQRT. And forasmuch as those two pyramids taken together, have their altitude equal with the altitude of the whole cube, 
Demonstration.  each of them a part hath to his altitude half the altitude of the cube, namely, half of the side of the cube, as the line KB. And forasmuch as the square KLMI is double to the square NRQP, by the 47. of the first: the other squares of the cube shall also be double to the square NRQP. And forasmuch as the cube, as it was manifest by the last of the fifteenth, is resolved into six pyramids, whose bases are the bases of the cube, & the altitudes the lines drawn from the centre to the bases, which are equal to half the side of the cube● it followeth that every one of the six pyramids of the cube, having his base double to the base of each of the pyramids of the Octohedron, and the self same altitude that the said pyramids of the Octohedron have, is double to either of the pyramids of the octohedron, by the 6. of the twelfth. And forasmuch as every one of the pyramids of the cube is equal to the two pyramids of the Octohedron, the six pyramids of the cube shall be sextuple to the whole Octohedron. Wherefore it is manifest, that a cube is sextuple to an Octohedron inscribed in it.   

¶ The 28. Proposition. To prove that an Octohedron is quadruple sesquialter to a Cube inscribed in it.  
SVppose that the Octohedron given be ABCDEF: and let the cube inscribed in it be GHIK, VQRS. Then I say, that the Octohedron is quadruple sesquialter to the cube inscribed in it. Forasmuch as the lines drawn from the centre of the Octohedron, or of the Sphere which containeth it, unto the centres of the bases of the Octohedron, are proved equal, by the 21. of the fifteenth: and the angles of the cube are set in the centres of those bases, by the 4. of the fifteenth: it followeth, that the self same right lines are drawn from one and the self same centre of the cube and of the Octohedron: for they have each one and the self same centre, by the Corollary of the 21. of the fifteenth. Let that centre be the point T. Wherefore the base BDFC, which cutteth the Octohedron into two equal and quadrilater pyramids, by the Corollary of the 14. of the thirteenth, shall also cut the cube into two equal parts, by the Corollary of the 39 of the eleventh. For it passeth by the centre T, by that which was demonstrated in the 14. of the thirteenth. And forasmuch as the base of the cube is in the 4. centres G, H, I, K, of the bases of the pyramid ABDFC, a plain LNOM, extended by those points, shall be parallel to the plain BDFC, by that which was demonstrated in the 4. of the fifteenth, and shall cut the pyramid in the points L, N, O, M: and the lines LN, BD, and NO, DF, shall be parallels, so also shall the lines OM, FC, and LM, BC: and the square GHIK of the cube shall be inscribed in the square LNOM, by the same. Wherefore the square LNOM is duple to the square GHIK, by the 47. of the first. From the solid angle A, let there be drawn to the plain superficies BDFC, a perpendicular, which let fall upon it in the point T, and let the same perpendicular be AT, cutting the plain LNOM in the point P. And it shall also be a perpendicular to the plain LNOM, by the Corollary of the 14. of the eleventh. Again from the angle BAD of the triangle ADB, let there be drawn by the centre H of the triangle, to the base a line AHX. Wherefore the line AX is sesquialter to the line AH, by the Corollary of the 12. of the thirteenth. Wherefore the line AH is duple to the line HX. But the other lines AB, AD, AF, AC, and the perpendicular APT, are cut like unto the line AHX, by the 17. of the eleventh: Wherefore the line AP is double to the line PT. Wherefore the line AP is the altitude of the cube, for the line PT is the half thereof. And forasmuch as upon the base GHIK of the cube, and under the altitude AP of the same cube, is set the pyramid AGHIK: the said pyramid is the third part of the cube, by the Corollary of the 7. of the twelfth. But unto the pyramid AGHIK the pyramid ALNOM is duple, by the 6. of the twelfth, for the base of the one is double to the base of the other. Wherefore the pyramid ALNOM is two third parts of the cube. And forasmuch as the pyramids ALNOM, and ABDFC, are like, by the 7. definition of the eleventh: therefore they are in triple proportion of that in which the sides of like proportion AH to AX, or AL to AB, are, by the Corollary of the 8. of the twelfth. But the side AB is proved to be sesquialter to the side AL. Wherefore the pyramid ABCDF is to pyramid ALNOM, as 27. is to 8. (that is, in sesquialter proportion tripled: for the quantity or denomination of sesquialter proportion, namely, 1 ½ multiplied into itself once maketh 2¼, which again multiplied by 1½ maketh 3 1/●, that is, 27. to 8.). But of what parts the pyramid ALNOM containeth 8, of the same the cube containeth 12: namely, is sesquialter to the pyramid. Wherefore of what parts the cube containeth 12, of the same the whole Octohedron (which is double to the pyramid ABDFC) containeth 54. Which 54. hath to 12. quadruple sesquialter proportion. Wherefore the whole Octohedron is to the cube inscribed in it, in quadruple sesquialter proportion. Wherefore we have proved that an Octohedron given is quadruple sesquialter to a cube inscribed in it.  

¶ A Corollary.  
An Octohedron is to a cube inscribed in it, in that proportion that the squares of their sides are. For by the 14. of this book, the side of the Octohedron is in power quadruple sesquialter to the side of the cube inscribed in it.    

¶ The 29. Proposition. To prove that an octohedron given, is 
* That is at 13. 1/● is to ●.  tredecuple sesquialter to a trilater equilater pyramid inscribed in it.  
LEt the octohedron ge●en, be AB: in which let there be inscribed a cube FCED, by the 4. of the fifteenth, and in the cube let there be inscribed a pyramid FEGD, by the ●. of the fifteenth. And forasmuch as the angles of the pyramid are (by the same first of the fifteenth) set in the angles of the cube: and the angles of the cube are set in the centres of the bases of the Octohedron, namely, in the points F, E, C, D, G by the 4. of the fifteenth. Wherefore the angles of the pyramid, are set in the centres F, C, E, D of the octohedron. Wherefore the pyramid FEDG is inscribed in the octohedron (by the 6. of the fifteenth.) And forasmuch as the octohedron AB is to the cube FCED, inscribed in it quadruple sesquialter (by the former proposition): and the cube CDEF is to the pyramid FEDG inscribed in it triple, by the 25. of book: wherefore three magnitudes being given, namely, the octohedron, the cube and the pyramid, the proportion of the extremes (namely, of the octohedron to the pyramis) is made of the proportions of the means, (namely, of the octohedron to the cube, and of the cube to the pyramid,) as it is easy to see by the declaration upon the 10. definition of the fifth. Now then multiplying the quantities or denominations of the proportions (namely, of the octohedron to the cube which is 4 1/●, and of the cube to the pyramid, which is 3) as was taught in the definition of the sixth, there shallbe produced 13 1/●, namely, the proportion of the octohedron to the pyramid inscribed in it. For 4 ½, multiplied by 3. produce 13 ½. Wherefore the Octohedron is to the pyramid inscribed in it in tredecuple sesquialter proportion. Wherefore we have proved that an Octohedron is to a trilater equilater pyramid inscribed in it, in tredecuple sesquialter proportion.   

¶ The 30. Proposition. To prove that a trilater equilater Pyramid, is noncuple to a cube inscribed in it.  
SVppose that the pyramid given, be ABCD, whose two bases let be ABC, and DBC, and let their centres be the points G and I And from the angle A, draw unto the base BC a perpendicular AE: likewise from the angle D draw unto the same base BC, a perpendicular DE: and they shall concur in the section E by the 3. of the third and in them shallbe the centres G and I, by the corollary of the first of the third. And forasmuch as the line AD is the side of the pyramid, the same AD shall be the diameter of the base of the cube which containeth the pyramid, by the 1 of the fiuetenth. 
Demonstration.  Draw the line GI'. And forasmuch as the line GI' coupleth the centre● of the bases of the pyramid: the said line GI' shallbe the diameter of the base of the cube inscribed in the pyramid by the 18. of the fifteenth. And forasmuch as the line AG is double to the line GE, by the corollary of the twelfth of the thirteenth: the whole line AE shall be triple to the line GE: and so is also the line DE to the line IE. Wherefore the lines AD and GI' are parallels, by the 2. of the sixth. And therefore the triangles AED, and GEIS are like● by the corollary of the same. And forasmuch as the triangles AED, and GEIS are like, the line AD● shallbe triple to the line GI', by the 4. of the sixth. But the line AD is the diameter of the base of the cube circumscribed about the pyramid ABCD, and the line GI' is the diameter of the base of the cube inscribed in the pyramid ABCD: but the diameters of the bases are equemultiplices to the sides (namely, are in power duple). Wherefore the side of the cube circumscribed about the pyramid ABCD, is triple to the side of the cube, inscribed in the same pyramis, by the 15. of the fifth: but like cubes are in triple proportion the one to the other of that in which their sides are, by the 33. of the eleventh: and the sides are in triple proportion the one to the other: Wherefore triple taken three times bringeth forth twenty sevencuple, which is 27. to 1: for the 4. terms 27.9.3.1, being set in triple proportion: the proportion of the first to the fourth, namely, of 27. to 1. shallbe triple to the proportion of the first to the second, namely, of 27. to 9, by the 10. definition of the fifth: which proportion of 27. to 1. is the proportion of the sides tripled, which proportion also is found in like solides. Wherefore of what parts the cube circumscribed containeth 27. of the same, the cube inscribed containeth one: but of what parts the cube circumscribed, containeth 27. of the same, the pyramid inscribed in it, containeth 9 by the 25. of this book: wherefore of what parts the pyramid AB CD containeth 9 of the same, the cube inscribed in the pyramid, containeth one. Wherefore we have proved that a trilater and equilater pyramid, is non●cuple to a cube inscribed in it.   

¶ The 31. Proposition. An Octohedron hath to an Icosohedron inscribed in it, that proportion, which two bases of the Octohedron have to five bases of the Icosahedron.  
SVppose that the octohedron given be ABCD, and let the Icosahedron inscribed in it, be FGHMKLIO. Then I say that the octohedron is to the Icosahedron, as two bases of the octohedron, are to five bases of the Icosahedron. For forasmuch as the solid of the octohedron consisteth of eight pyramids, set upon the bases of the octohedron, 
Demonstration.  and having to their altitude a perpendicular line drawn from the centre to the base: let that perpendicular be ER, or ES, being drawn from the centre E (which centre is common to either of the solids, by the corollary of the 21. of the fifteenth) to the centres of the bases, namely, to the points R and S. Wherefore for that three pyramids are equal and like, they shallbe equal to a prism set upon the self same base, and under the self same altitude, by the corollary of the seventh of the twelfth. But unto this prism is double that prism which is set upon the self same base, and hath his altitude duple, namely, the whole line RS by the corollary of the 25. of the eleventh: for it is equal to the two equal and like prisms whereof it is composed. Wherefore the prism set upon the base of the octohedron, and having to his altitude the line RS is equal to six pyramids, set upon six bases of the Octohedron, and having to their altitude the line ER. So there remain two pyramids (for in the octohedron are 8. bases) which shall be equal to the prism which is set upon the third part of the base of the octohedron, and under the altitude RS. For prisms under one and the self same altitude, are in proportion the one to the other, as are their bases, by the corollary of the 7 of the twelfth. Wherefore the two prisms which are set upon the base of the octohedron, and upon a third part thereof, and under the altitude RS, are equal to the 8. pyramids of the Octohedron, or to the whole solid of the octohedron. And forasmuch as the Icosahedron inscribed in the octohedron, hath his bases set in the bases of the Octohedron, by the 17. of the fifteenth: it followeth that the pyramids set upon the bases of the Icosahedron, & having to their tops one and the self same centre E, are contained under the self same altitude, that the pyramids of the octohedron are contained under. namely, under the line ER, or ES. And therefore a prism, set upon the base of the Icosahedron, and having his altitude double to the altitude of the pyramid, namely, the whole line RS, is equal to six pyramids set upon the base of the Icosahedron, and under the altitude ER or ES, as we have proved in the octohedron. Wherefore the 20. pyramids, set upon the 20. bases of the Icosahedron, are equal to three prisms set upon the base of the Icosahedron, and under the altitude RS, and moreover to an other prism set upon a third part of the base of the Icosahedron and under the same altitude RS, which prism is a third part of the former prism, by the corollary of the 7. of the twelfth: for their proportion is as the proportion of the bases. Wherefore two prisms set upon the base of the octohedron, and a third part thereof, and under the altitude RS, is to 4. prisms set upon three bases of the Icosahedron, and a third part thereof, and under the same altitude RS, in the same proportion that the bases are, that is, as 4. third parts of the base of the Octodron (which are equal to one base, and 1/●;) to ten third parts of the base of the Icosahedron (which are equal to three bases & 1/●;) or as two third parts of the base of the Octohedron, are to five third parts of the base of the Icosahedron. But two third parts of the base of the Octohedron, are to five third parts of the base of the Icosahedron, as two bases are to five bases (by the 15. of the fifth, for they are parts of equemultiplices:) And two prisms of the Octohedron are to 4. prisms of the Icosahedron, as the solid of the Octohedron is to the solid of the Icosahedron, when as each are equal to each of the solids: Wherefore (by the ●●. of the fifth) the solid of the Octohedron, is to the solid of the Icos●hedron inscribed in it, as two bases of the Octohedron, are to five bases of the Icosahedron. An Octohedron therefore is to an Ico●ahedron inscribed in it, in that proportion, that two bases of the Octohedron, are to five bases of the Icosahedron.   

¶ The 32. Proposition. The proportion of the solid of an Icosahedron to the solid of a Dodecahedron inscribed in it, consisteth of the proportion of the side of the Icosahedron to the side of the Cube contained in the same sphere, and of the proportion tripled of the diameter to the line which conpleth the centres of the opposite bases of the Icosahedron.  

Construction.  SVppose that there be ● Dodecahedron, whose diameter let be high, and let the Icosahedron contained in the same sphere be ABGC, whose dimetient let be AC. And let the right line which coupleth the centres of the opposite base● be BG. And let the dodecahedron inscribed in the Icos●hedron be that which is set upon the diameter BG, by the 5. of the fifteenth. And let the side of the cube be DE, and let the side of the Icosahedron be D●, both the said solids being described in one and the self same sphere. Then I say that the proportion of the solid of the Icosahedron ABCG to the solid of the dodecahedron set upon the diameter BG, inscribed in it, consisteth of the proportion of the line DF to the line DE, and of the proportion tripled of the line AC to the line BG. 
Demonstration.  For forasmuch as the solid of the Icosahedron ABGC is to the solid of the dodecahedron high, being contained in one and the self same sphere, as DF is to D●, by the 8. of the fourteenth But the dodecahedron whose diameter is high, is to the dodecahedron whose diamer is BG, in triple proportion of that in which the diameter high is to the diameter BG, by the corollary of the 17. of the twelfth: & the lines high and AC are equal by supposition (namely, the diameters of one and the self same sphere). Wherefore as high is to BG, so is AC to BG. Wherefore the proportion of the extremes, namely, of the Icosahedron ABGC to the Dodecahedron set upon the diameter BG which coupleth the c●ntres, consisteth (by the 5. definition of the sixth) of the proportions of the me●nes, namely, of the proportion of the Icos●hedron ABCG to the dodecahedron high (which is one and the same with the proportion of DF to DE) and of the proportion of the same high to the other dodecahedron set upon the diameter BG, inscribed in the same Icosahedron ABGC, by the same 5. of the fifteenth: which proportion is triple to the proportion of the line high (or the line AC) to GB which coupleth the centres of the opposite bases of the Icosahedron. The proportion therefore of the solid of the Icosahedron to the solid of a D●●ecahedron inscribed in it, consisteth of the proportion of the side of the Icosahedron to the side of the Cube contained in the same sphere, and of the proportion tripled of the diameter to the lin● which coupleth the centres of the opposite bases of the Icosahedron.   

¶ The 33. Proposition. The solid of a Dodecahedron exceedeth the solid of a Cube inscribed in it, by a parallelipipedon, whose base wanteth of the base of the Cube by a third part of the less segment, and whose altitude wanteth of the altitude of the Cube, by the less segment of the less segment, of half the side of the Cube.  
FOrasmuch as by the 17. of the thirteenth, and 8. of the fifteenth, it w●s manifest, that the base of a cube inscribed in a dodecahedron, doth with his sides subtend t●● angles of 4. pentagons concurring at one and the self same side of the dodecahedron: 
Construction.  let that base of the cube be ABCD: and let the side whereat 4. bases of the dodecahedron circumscribed concur, be EG: which shall contain a solid AEBDGC set upon the base ABCD. Divide the sides AB and DC into two equal parts in the points L and N. And draw the line LN, which is a parallel to the side EG, as it was manifest by the 17. of the thirteenth. The perpendiculars also ER and GO which couple those parallels, are each equal, to half of the side EG, and each is the greater segment of half the side of the cube, and therefore the whole line EG is the greater segment of the whole line LN the side of the cube (by the foresaid 17. of the thirteenth). By the points R and O, draw unto the sides AB and CD parallel lines FH and IK. And draw these right lines EF, EH, GI' and GK. Now forasmuch as the two lines FH & ER touching the one the other are parallels to the two lines IK and GO touching also the one the other, 
Demonstration.  & not being in the self same plain with the two first lines: therefore the plain super●icieces EFH and GIK passing by those lines are parallels, by the 15. of the eleventh: which plains do cut the solid AEBDGC. Whererefore there are made four quadrangled pyramids set upon the rectangle parallelograms LH, LF, NK, and NI, and having their tops the points E and G. And forasmuch as the triangles GOK and ERH are equal and like, by the 4. of the first, namely, 
Extend in the figure a line from the point E to the point B.  they contain equal angles comprehended under equal sides, and they are parallels by construction, being set in the plains GIK and EFH: the figures GKHE, OKHR, and GORE shall be parallelograms, by the definition of a parallelogram, and therefore the solid GOKERH is a prism, by the 11. definition of the eleventh. And by the same reason may the solid GOIERF be proved to be a prism. And forasmuch as upon equal bases NOKC, and RLBH, and under equal altitudes OG and RE are set pyramids: those pyramids shall be equal to ●hat pyramid which is set upon the ●ase CKID (which is double to either of the bases NOKC, and RLBH) and under the same altitude OG, by the 6. of the twelfth. And forasmuch as the side GE is the greater segment of the line CB, the line KH, which by the 33. of the first, is equal to the line GE, shall be the greater segment of the same line CB, by the 2. of the fourteenth. Wherefore the residues CK and HB shall make the less segment of the whole line CB. But as the greater segment KH is to the two lines CK and HB the less segment, so is the rectangle parallelogram OH to the two rectangle parallelograms OC and HL, by the 1. of the sixth. Wherefore the less segment of the parallelogram NB, shall be the two parallelograms OC and HL. Put the line KM double to the line KC and draw the line MS parallel to the line CN. Wherefore the parallelogram OKMS is equal to the parallelograms OC and HL, by the 1. of the sixth. Wherefore the pyramid set upon the base OKMS containeth two third parts of the prism set upon the self same base, by the 4. corollary of the 7. of the twelfth. Wherefore the prism which is set upon two third parts of the base OKMS is equal to the two pyramids NOKCG and RLBHE. For the sections of a prism are one to the other, as the sections of the base are, by the first corollary of the 25. of the eleventh. But the sections of the base are as the sections of the line CB or KM, by the 1. of the sixth. Wherefore the two pyramids NOKCG and RLBHE, add unto the prism GOKERH two third parts of the prism set upon the base OKMS. And forasmuch as the line KM is the less segment of the whole line BC (for it is equal to to the two lines CK and H●), and the prism set upon the base OKHR is cut like unto the line KM, namely, in each are taken two thirds, as hath before been proved: the prism equal to the two pyramids, shall add unto the prism GOKERH, which is set upon the greater segment KH, two th●●ds of the less segment. Wherefore in the line BC there shall remain one third part of the less segment: and therefore in the rectangle parallelogram NB which is half the base of the cube, there shall remain the same third part of the less segment. And by the same reason may we prove that in the other pyramids ONDIG, and RLAFE, and in the prism GOIERF is left the self same excess of the base LAND, namely, the third part of the less segment. Wherefore the whole prism contained between the triangles IGK and FEH, and under the length of the greater 
Extend in the figure a line from the point E to the point B.  segment and two third parts of the less segment of BC the side of the cube, is equal to the solid composed of 4. bases of the dodecahedron and set upon the base of the cube. Wherefore the base of that prism wanteth of the whole base of the cube only a third part of the less segment: and the altitude of that prism was the line GO, which is the greater segment of half the side of the cube. And forasmuch as unto the triangle IGK, is double the rctangle parallelogram set upon the same base IK, (the side of the cube) and under the altitude GO, by the 41. of the first: it followeth that three rectangle parallelograms set upon the same base IK, the side of the cube, and under the altitude OG the greater segment of half the side of the cube, are sextuple to the triangle IGK. Wherefore those three rectangle parallelograms do make one rectangle parallelogram set upon the base IK and under the altitude of the line GO tripled. But by the 7. definition of the eleventh, there are six prisms equal and like unto the foresaid prism, being set upon every one of the six bases of the cube: which prisms are in proportion the one to the other as their bases are by the 3. corollary of the 7. of twelfth. Wherefore the solid composed of these six prisms, shall want of the base ABCD the third part of the less segment, and taking his altitude of the foresaid rectangle parallelogram, the said altitude shall be equal to three greater segments (one of which is GO) of half the side of the cube.  
Now resteth to prove that these three segments want of the side of the cube by the less segment of the less segment of half the sid● of the cube. Suppose that AB the side of the cube be divided into the greater segment AC, and into the less segment CB (by the 30. of the sixth). And divide into two equal parts the line AC in the point G, and the line CB in the point E. And unto the line CG put the line CL equal. Now forasmuch as the lines AG and GC are the greater seg●●●tes of half the line AB, for ●che of them is the half of the greater segment of the whole line AB: the lines EB and EC shall be the less segments of half the line AB. Wherefore the whole line C● is the greater segment, and the line CE is the less segment. But as the line CL is to the line CE, so is the line CE to the residue EL. Wherefore the line EL is the greater segment of the line CE, or of the line EB which is equal unto it. Wherefore the residue LB is the less segment of the same EB (which is the les●e segment of halfa the side of the cube). But the lines AG, GC, and CL are three greater segments of the half of the whole line AB: which three greater segments make the altitude of the foresaid solid: wherefore the altitude of the said solid wanteth of AB the side of the cube by the line LB, which is the less segment of the line BE. Which line BE again is the less segment of half the side AB of the cube. Wherefore the foresaid solid consisting of the six solids, whereby the dodecahedron exceedeth the cube inscribed in it, is set upon a base which wanteth of the base of the cube by a third part of the less segment, and is under an altitude wanting of the side of the cube by the less segment of the less segment of half the side of the cube. The solid therefore of a dodecahedron exceedeth the solid of a cube inscribed in it, by a parallelipipedon, whose base wanteth of the base of the cube by a third part of the less segment, and whose altitude wanteth of the altitude of the cube, by the less segment of the less segment of half the side of the cube.  

¶ A Corollary.  
A Dodecahedron is double to a Cube inscribed in it, taking away the third part of the less segment of the cube, and moreover the less segment of the less segment of half of that excess. For if there be given a cube, from which is cut of a solid set upon a third part of the less segment of the base and under one and the same altitude with the cube: that solid taken away hath to the whole solid the proportion of the section of the base to the base, by the 32. of the eleventh. Wherefore from the cube is taken away a third ●art of the less segment. Farther, forasmuch as the residue wanteth of the altitude of the cube, by the less segment of the less segment of half the altitude or side, and that residue is a parallelipipedon, if it be cut by a plain superficies parallel to the opposite plain superficieces, cutting the altitude of the cube by a point, it shall take away from that parallelipipedon a solid, having to the whole the proportion of the section to the altitude, by the 3. Corollary of the 25. of the eleventh. Wherefore the excess wanteth of the same cube by the thi●d part of the less segment, and moreover by the less segment of the less segment of half of that excess.    

¶ The 34. Proposition. The proportion of the solid of a Dodecahedron to the solid of an Icosahedron inscribed in it, consisteth of the proportion tripled of the diameter to that line which coupleth the opposite bases of the Dodecahedron, and of the proportion of the side of the Cube to the side of the Icosahedron inscribed in one and the self same Sphere.  
SVppose that AHBCK be a Dodecahedron● whose diameter let be AB: and let the line which coupleth the centres of the opposite bases be KH● and let the Icosahedron inscribed in the Dodecahedron ABC, be Dee: whose diameter let be DE. Now forasmuch a● o●e and the self same circle containeth the pentagon of a Dodecahedron, & the triangle of an Icosahedro● described in one and the self same Sphere, by the 14. of the fourteenth: Let that circle be IGO. Wherefore IO is the side of the cube, and IG the side of the Icosahedron, by the same. Then I say, that the proportion of the Dodecahedron AHBCK to the Icosahedron DEF inscribed in it, consisteth of the proportion tripled of the line AB to the line KH, and of the proportion of the line IO to the line IG. For forasmuch as the Icosahedron DEF is inscribed in the Dodecahedron ABC, 
Demonstration.  by supposition, the diameter DE shallbe equal to the line KH, by the 7. of the fifteenth. Wherefore the Dodecahedron set upon the diameter KH shall be inscribed in the same Sphere, wherein the Icosahedron DEF is inscribed: but the Dodecahedron AHBCK is to the Dodecahedron upon the diameter KH in triple proportion of that in which the diameter AB is to the diameter KH, by the Corollary of the 17. of the twelfth: and the same Dodecahedron which is set upon the diameter KH, hath to the Icosahedron DEF (which is set upon the same diameter, or upon a diameter equal unto it, namely, DE) that proportion which IO the side of the cube hath to● IG the side of the Icosahedron, inscribed in one & the self same Sphere, by the 8 of the fourteenth. Wherefore the proportion of the Dodecahedron AHBCK to the Icosahedron DEF inscribed in it, consisteth of the proportion tripled of the diameter AB to the line KH, which coupleth the centres of the opposite bases of the Dodecahedron (which proportion is that which the Dodecahedron AHBCK hath to the Dodecahedron set upon the diameter KH) and of the proportion of IO the side of the cube to IG the side of the Icosahedron (which is the proportion of the Dodecahedron set upon the diameter KH to the Icosahedron DEF described in one and the self same Sphere) by the 5. definition of the sixth. The proportion therefore of the solid of a Dodecahedron to the solid of an Icosahedron inscribed in it, con●isteth of the proportion tripled of the diameter to that line which coupleth the opposite bases of the Dodecahedron, and of the proportion of the side of the cube to the side of the Icosahedron inscribed in one and the self same Sphere.   

The 35. Proposition. The solid of a Dodecahedron containeth of a Pyramid circumscribed about it two ninth parts, taking away a third part of one ninth part of the less segment (of a line divided by an extreme and mean proportion) and moreover the less segment of the less segment of half the residue.  
IT hath been proved that the Dodecahedron, together with the cube inscribed in it is contained in one and the self same pyramid, by the Corollary of the first of this book. And by the Corollary of the 33. of this book, it is manifest, that the Dodecahedron is double to the same cube, taking away the third part of the less segment, and moreover the less segment of the less segment of half the residue, or of this excess. But a pyramid is to the same cube inscribed in it nonecuple, by the 30. of this book. Wherefore the Dodecahedron inscribed in the pyramid, and containing the same cube twice, taking away the self same third of the less segment, and moreover the less segment of the less segment of half the residue, shall contain two ninth parts of the solid of the pyramid (of which ninth parts each is equal unto the cube) taking away this self same excess. The solid therefore of a Dodecahedron containeth of a Pyramid circumscribed about it two ninth parts, taking away a third part of one ninth part of the less segment (of a line divided by an extmere and mean proportion) and moreover the less segment of the less segment of half the residue.   

¶ The 36. Proposition. An Octohedron exceedeth an Icosahedron inscribed in it, by a parallelipipedon set upon the square of the side of the Icosahedron, and having to his altitude the line which is the greater segment of half the semidiameter of the Octohedron.  
SVppose that there be an Octohedron ABCFPL: 
Construction.  in which let there be inscribed an Icosahedron HKEGMXNVDSQT● by the ●6. of the fifteenth. And draw the diameters AZRCBROIF, and the perpendicular KO parallel to the line AZR. Then I say, that the Octohedron ABCFPL is greater th●n the Icosahedron inscribed in it, by a parallelipipedon set upon the square of the side HK or GE, and having to his altitude the line KO or RZ: which is the greater segment of the semidiameter AR. Forasmuch as in the same 16. it hath been proved, that the triangles KDG and KEQ are described in the bases APF and ALF of the Octohedron: 
Demonstration.  therefore about the solid angle there remain upon the base FEG three triangles KEG, KFE, and KFG, which contain a pyramid KEFG. Unto which pyramid shall be equal and like the opposite pyramid MEFG set upon the same base FEG, by the 8. definition of the eleventh. And by the ●ame reason shall there at every solid angle of the Octohedron remain two pyramids equal and like: namely, two upon the base AHK, two upon the base BNV, two upon the base DPS, and moreover two upon the base QLT. Now then there shall be made twelve pyramids, set upon a base contained of the side of the Icosahedron, and under two le●●e segments of the side of the Octohedron containing a right angle, as for example the base GEF, And forasmuch as the side GE subtending a right angle, is, by the 47. of the ●irst, in power duple to either of the lines EF and FG, and so the ●●de● KH is in power duple to either of the sides AH and AK: and either of the lines AH, AK, or EF, FG, is in power duple to either of the lines AZ or ZK which contain a right angle, made in the triangle or base AHK by the perpendicular AZ. Wherefore it followeth that the side GE or HK, is in power quadruple to the triangle EFG or AHK. But the pyramid KEFG, having his base EFG in the plain FLBP of the Octohedron, shall have to his altitude the perpendicular KO (by the 4. definition of the sixth) which is the greater segment of the semidiameter of the Octohedron, by the 16. of the fifteenth. Wherefore three pyramids set under the same altitude and upon equal bases, shall be equal to one prism set upon the same base, and under the same altitude, by the 1. Corollary of the 7. of the twelfth. Wherefore 4. prisms set upon the base GEF quadrupled (which is equal to the square of the side GE) and under the altitude KO (or RZ the greater segment which is equal to KO) shall contain a solid equal to the twelve pyramids, which twelve pyramids make the excess of the Octohedron above the Icosahedron inscribed in it. An Octohedron therefore exceedeth an Icosahedron inscribed in it, by a parallelipipedon set upon the square of the side of the Icosahedron, and having to his altitude the line which is the greater segment of half the semidiameter of the Octohedron.  

¶ A Corollary.  
A Pyramid exceedeth the double of an Icosahedron inscribed in it, by a solid, set upon the square of the side of the Icosahedron inscribed in it, and having to his altitude that whole line of which the side of the Icosahedron is the greater segment. For it is manifest by the 19 of the fiuetenth, that an octohedron & an Icosahedron inscribed in it are inscribed in one & the self same pyramid. It hath moreover been proved in the 26. of this book, that a pyramid is double to an octohedron inscribed in it. Wherefore the two excesses of the two octohedrons (unto which the pyramid is equal) above the two Icosahedrons' (inscribed in the said two octohedrons) being brought into an solid, the said solid shallbe set upon the self same square of the side of the Icosahedron, and shall have to his altitude the perpendicular KO doubled: whose double coupling the opposite sides HK and XM maketh the greater segment the same side of the Icosahedron, by the first and second corollary of the 14. of the fiu●●en●h.    

The 37. Proposition. If in a triangle having to his base a rational line set, the sides be commensurable in power to the base, and from the top be drawn to the base a perpendicular line cutting the base: The sections of the base shall be commensurable in length to the whole base, and the perpendicular shall be commensurable in power to the said whole base.  
And now that the perpendicular AP is commensurable in power to the base BG, 
Second part of the Demonstration.  i● thus proved. Forasmuch as the square of AB is by supposition, commensurable to the square of BG: and unto the rational square of AB is commensurable the rational square of BP (by the 12. of the eleventh) Wherefore the residue, namely, the square of PA is commensurable to the same square of BP, by the 2. part of the 15. of the eleventh. Wherefore by the 12. of the tenth, the square of PA is commensurable to the whole square of BG. Wherefore the perpendicular AP is commensurable in power to the base BG, by the 3. definition of the tenth: which was required to be proved.  
In demonstrating of this, we made no mention at all of the length of the sides AB and AG, but only of the length of the base BG: for that the line BG is the rational line first set: and the other lines AB and AG are supposed to be commensurable in power only to the line BG. Wherefore if that be plainly demonstrated, when the sides are commensurable in power only to the base, much more easily will it follow, if the same sides be supposed to be commensurable both in length and in power to the base: that is, if their lengths be expressed by the roots of square numbers.  

¶ A Corollary. 1.  
By the former things demonstrated, it is manifest that if from the powers of the base, and of one of the sides, be taken away the power of the other side, and if the half of the power remaining, be applied upon the whole base, it shall make the breadth that section of the base which is coupled to the first side. For from the powers of the base BG, and of one of the sides AG, that is, from the squares BD and GL, the power of the other side AB, namely the square BK (or the parallelogram EM) is taken away. And of the residue, (namely, of the square BD, and of the parallelogram OD, or DR, which by supposition is equal unto OD) the half (namely of the whole FR, which is PD, for the lines GR and PB are equal to the lines GO and PO) is applied to the whole line BG or GD: and maketh the bredthe the line PG the section of the base BG, which section is coupled to the first side AG. And by the same reason in the other side, if from the squares BD and BK be taken away the square GL, there shall remain the rectangle parallelogram FO: For the parallalelogramme EM is equal to the square BK, and the parallelogram GM to the parallelogram OD. Wherefore FP the half of the residue FO, maketh the breadth BP, which is coupled to the first side taken AB.   

A Corollary. 2.  
If a perpendicular drawn from an angle of a triangle do cut the base: the sections are to the other sides in power proportional by an Arithmetical proportion. For it was proved that the excess of the powers of the lines AG and AB is one and the same with the excess of the powers of the lines PG and PB. If therefore the powers do equally exceed the one the other, they shall by an Arithmetical proportion, be proportional.    
The end of the sixteenth Book of the Elements of Geometry added 〈…〉.   

A brief treatise, added by Flussas, of mixed and composed regular solids.  
REgular solids are said to be composed and mixed, when each of them is transformed into other solids, keeping still the form, number, and inclination of the bases, which they before had one to the other: some of which yet are transformed into mixed solids, and other some into simple. Into mixed, as a Dodecahedron and an Icosahedron: which are transformed or altered, if ye divide their sides into two equal parts, and take away the solid angles subtended of plain superficial figures made by the lines coupling those middle sections: for the solid remaining after the taking away of those solid angles, is called an Icosidodecahedron. 
Icosidodecahedron.  If ye divide the sides of a cube and of an Octohedron into two equal parts, and couple the sections, the solid angles subtended of the plain superficieces made by the coupling lines, being taken away, there shall be left a solid, which is called an Exoctohedron. 
Exoctohedron.  So that both of a Dodecahedron and also of an Icosahedron, the solid which is made, shall be called an Icosidodecahedron: and likewise the solid made of a Cube & also of an Octohedron, shall be called an Exoctohedron. But the other solid, namely, a Pyramid (or Tetrahedron) is transformed into a simple solid: for if ye divide into two equal parts every one of the sides of the pyramid, triangles described of the lines which couple the sections, and subtending, and taking away solid angles of the pyramid, are equal and like unto the equilater triangles left in every one of the bases: of all which triangles is produced an Octohedron, namely, a simple and not a composed solid. For the Octohedron hath four bases, like in number, fonne, and mutual inclination with the bases of the pyramid: and hath the other four bases with like situation opposite and parallel to the former. Wherefore the application of the pyramid taken twice, maketh a simple Octohedron, as the other solids make a mixed compound solid.  

¶ First Definition.  
An Exoctohedron is a solid figure contained of six equal squares, and eight equilater and equal triangles.   

¶ Second Definition.  
An Icosidodecahedron is a solid figure, contained under twelve equilater equal and equiangle Pentagons', and twenty equal and equilater triangles.  
For the better understanding of the two former definitions, and also of the two Propositions following, I have here set two figures, whose forms, if ye first describe upon pasted paper or such like matter, and then cut them and fold them accordingly, they will represent unto you the perfect forms of an Exoctohedron and of an Icosidodecahedron.   

¶ The first Problem. To describe an equilater and equiangle exoctohedron, and to contain it in a sphere given: and to prove that the diameter of the sphere is double to the side of the said exoctohedron.  
And now forasmuch as the opposite sides and diameters of the bases of the cube are parallels, 
That the exoctohedron is contained in a sphere.  the plain extended by the right lines QT, VR, shall be a parallelogram. And for that also in that plain lieth QR the diameter of the cube, and in the same plain also is the line MH, which divideth the said plain into two equal parts, and also coupleth the opposite angles of the exoctohedron this line MH therefore divideth the diameter into two equal parts, by the corollary of the 34. of the first, and also divideth itself in the same point, which let be S, into two equal parts, by the 4 of the first. And by the same reason may we prove that the rest of the lines, which couple the opposite angles of the exoctohedron, do in S the centre of the cube divide th● one the other into two equal parts. For every one of the angles of the exoctohedron are set in every one of the bases of the cube. Whe●●●ore making the centre the point S, and the space SH or SM, describe a sphere, and it shall touch every one of the angles eq●edistant from the point S.  
And forasmuch as AB the diameter of the sphere given, is put equal to the diameter of the base of the cube, 
That the exoctohedron is contained in the sphere given.  namely, to the line RT, and the same line RT is equal to the line MH, by the 33. of the first: which line MH coupling the opposite angles of the exoctohedron, is drawn by the centre: wherefore it is the diameter of the sphere given which containeth the exoctohedron.  
Finally forasmuch as in the triangle RFT, the line PO doth cut the sides into two equal parts, 
That the dia●●●ter of the sphere is do●ble to the side ●f the exoctohedron.  it shall cut them proportionally with the bases namely, as FR is to FP, so shall RT be to PO, by the 2. of the sixth. But FR is double to FP, by supposition: wherefore RT, or the diameter HM, is also double to the line PO the side of the exoctohedron. Wherefore we have described an equilater & equiangle exoctohedron, and comprehended it in a sphere given, and have proved that the diameter of the sphere is double to the side of the exoctohedron.   

¶ The 2. Problem. To describe an equilater & equiangle Icosidodecahedron, & to comprehend it in a sphere given: and to prove that the diameter being divided by an extreme and mean proportion, maketh the greater segment double to the side of the Icosidodecahedron.  
Now let us prove that it is contained in the Sphere given, whose diameter is NL. 
That the Icosidodecahedron is contained in the sphere given.  Forasmuch as perpendiculars drawn from the centres of the Dodecahedron, to the middle sections of his sides, are the halves of the lines, which couple the opposite middle sections of the sides of the Dodecahedron, by the 3. Corollary of the 17. of the thirteenth: which lines also, by the same Corollary, do in the centre divide the one the other into two equal parts: therefore right lines drawn from that point to the angles of the Icosidodecahedron (which are set in those middle sections) are equal: which lines are 30. in number according to the number of the sides of the Dodecahedron: for every one of the angles of the Icosidodecahedron are set in the middle sections of every one of the sides of the Dodecahedron. Wherefore making the centre the centre of the Dodecahedron, and the space any one of the lines drawn from the centre to the middle sections, describe a Sphere, and it shall pass by all the angles of the Icosidodecahedron, and shall contain it.  
And forasmuch as the diameter of this solid, is that right line, whose greater segment is the side of the cube inscribed in the Dodecahedron, by the 4. Corollary of the 17. of the thirteenth, which side is NI, by supposition. Wherefore that solid is contained in the Sphere given whose diameter is put to be the line NL.  

¶ An advertisement of Flussas.  
To the understanding of the nature of this Icosidodecahedron, ye must well conceive the passions and proprieties of both those solids, of whose bases i● consisteth, namely, of the Icosahedron and of the Dodecahedron. And although in it the bases are placed oppositely, yet h●u● they one to the other one & the s●me inclination. By reason whereof there he hidden in it the actions and p●●●●ons of the other regular solids. And I would have thought i● not impertinent to the purpose to have set forth the inscriptions and circumscriptions of this solid, if w●nt of t●h● had not hindered. But to the end the reader may the better a●●aine to the understanding thereof, I have here following briefly set forth, how it may in o● about every one of the five regular solids be inscribed or circumscribed: by the help whereof ●e may, with small travail or rather none at all, so that he have well p●ysed and considered the d●monstrations pertaining to the foresaid fi●e regular solids, demonstrate both the inscription of the said solids in it, and the inscription of it in the said solids.  

¶ Of the inscriptions and circumscriptions of an Icosidodec●hedron.  
An Icosidodecahedron may contain the other five regular bodies. For it will receive the angles of a Dodecahedron, in the centres of the triangles which subtend the solid angles of the Dodecahedron: which solid angles are 20. in number, and are placed in the same order in which the solid angles of the Dodecahedron taken away or subtended by them, are. And by that reason it shall receive a Cube and a Pyramid contained in the Dodecahedron: when as the angles of the one are set in the angles of the other.  
An Icosidodecahedron receiveth an Octohedron, in the angles cutting the six opposite sections of the Dodecahedron, even as if it were a simple Dodecahedron.  
And it containeth an Icosahedron, placing the 12. angles of the Icosahedron in the self same centres of the 12. Pentagons'.  
It may also by the same reason be inscribed in every one of the five regular bodies: namely, in a Pyramid, if ye place 4. triangular bases concentrical with 4. bases of the Pyramid, after the same manner, that ye inscribed an Icosahedron in a Pyrami●. So likewise may it be inscribed in an Octohedron, if ye make 8. bases thereof concentrical with the 8. bases of the Octohedron. It shall also be inscribed in a Cube, if ye place the angles which receive the Octohedron inscribed in it, in the centres of the bases of the Cube. Moreover, ye shall inscribe it in an Icosahedron, when the triangles compassed in of the Pentagon bases, are concentrical with the triangles, which make a solid angle of the Icosahedron. Finally, it shall be inscribed in a Dodecahedron, if ye place every one of the angles thereof in the middle sections of the sides of the Dodecahedron, according to the order of the construction thereof.  
The opposite plain superficieces also of this solid are parallels. For the opposite solid angles are subtended of parallel plain superficieces, as well in the angles of the Dodecahedron subtended by ●ri●ngle●, a● in the angles of the Icosahedron subtended of Pentagons', which thing may easily be demonstrated. Moreover in thi● solid are infinit● properti●● & passion's, springing of the solide● whereof ●t is composed.  
Wherefor● it is manifest that a Dodecahedron & an Icosahedron, mixed, are transformed into one & the self same solid of an Icosidodecahedron. A cube also and an octohedron are mixed and altered into an other solid, namely, into one and the same Exoctohedron. But a pyramid is transformed into a simple and perfect solid, namely into an Octohedron.  
If we will frame these two solids joined together into one solid, this only must we observe.  
In the pentagon of a dodecahedron inscribe a like pentagon, so that let the angles of the pentagon inscribed be set in the middle sections of the sides of the pentagon circumscribed, and then upon the said pentagon inscribed, let there be set a solid angle of an Icosahedron, and so observe the self same order in every one of the bases of the Dodecahedron: and the solid angles of the Icosahedron set upon these pentagons shall produce a solid consisting of the whole Dodecahedron, and of the whole Icosahedron. In like sort, if in every base of the Icosahedron, the sides being divided into two equal parts be inscribed an equilater triangle, and upon every one of those equilater triangles be set a solid angle of a Dodecahedron: there shall be produced the self same solid consisting of the whole Icosahedron, & of the whole Dodecahedron.  
And after the same order, if in the bases of a cube, be inscribed squares subtending the solid angles of an Octohedron, or in the bases of an Octohedron, be inscribed equilater triangles subtending the solid angles of a cube, there shall be produced a solid consisting of either of the whole solids, namely, of the whole cube and of the whole Octohedron.  
But equilater triangles inscribed in the bases of a pyramid, having their angles set in the middle sections of the sides of the pyramid, and the solid angles of a pyramid set upon the said equilater triangles, there shall be produced a solid, consisting of two equal and like pyramids.  
And now if in these solids thus composed, ye take away the solid angles, there shallbe restored again the first composed solids: namely, the solid angles taken away from a Dodecahedron and an Icosahedron composed into one, there shallbe left an Icosidodecahed●on: the solid angles taken away from a cube, and an octohedron composed into one solid, there shallbe left an exocthedron. Moreover the solid angles taken away from two pyramids composed into one solid, there shall be left an Octohedron.  
Flussas after this setteth forth certain passions and properties of the five simple regular bodies: which although he demonstrateth not, yet are they not hard to be demonstrated, if we well pease and conceive that, which in the former books hath been taught touching those solids.   

Of the nature of a trilater and equilater Pyramid.  
A trilater equilater Pyramid, is divided into two equal parts, by three equal squares, which in the centre of the pyramid cut the one the other into two equal parts, and perpendicularly, and whose angles are set in the middle sections of the sides of the pyramid. From a pyramid are taken away 4. pyramids like unto the whole, which utterly take away the sides of the pyramid, and that which is left is an octohedron inscribed in the pyramies in which all the solids inscribed in the pyramid are contained. A perpendicular drawn from the angle of the pyramid to the base, is double to the diameter of the cube inscribed in it. And a right line coupling the middle sections of the opposite sides of the pyramid, is triple to the side of the self same cube. The side also of the pyramid is triple to the diameter of the base of the cube. Wherefore the same side of the pyramid is in power duple to the right line which coupleth the middle sections of the opposite sides. And it is in power sesquialter to the perpendicular which is drawn from the angle to the base. Wherefore the perpendicular is in power sesquitertia to the line which coupleth the middle sections of the opposite sides. A pyramid, and an Octohedron inscribed in it, also an Icosahedron inscribed in the same Octohedron, do contain one and the self same sphere.   

Of the nature of an Octohedron.  
Four perpendiculars of an Octohedron, drawn in 4. bases thereof from two opposite angles of the said Octohedron, and coupled together by those 4. bases, describe a Rhombus, or diamond figure: one of whose diameters is in power duple to the other diameter. For it hath the same proportion that the diameter of the Octohedron, hath to the side of the Octohedron. An Octohedron & an Icosahedron inscribed in it, do contain one and the self same sphere. The diameter of the solid of the Octohedron, is in power sesquialter to the diameter of the circle which containeth the base: and is in power triple to the right line which coupleth the centres of the opposite bases: and is in power 
* That is as 8. 103 ●  duple superbipartiens tercias to the perpendicular or side of the foresaid Rhombus: and moreover is in length triple to the line which coupleth the centres of the next bases. The angle of the inclination of the bases of the Octohedron, doth with the angle of the inclination of the bases of the pyramid, make angles equal to two right angles.   

Of the nature of a Cube.  
The diameter of a cube, is in power sesquialter to the diameter of his base: and is in power triple to his side: and unto the line which coupleth the centres of the next bases, it is in power sextuple. Moreover the side of the cube is to the side of the Icosahedron inscribed in it, as the whole is to the greater segment: unto the side of the Dodecahedron, it is as the whole is to the less segment: unto the side of the Octohedron, it is in power duple: and unto the side of the pyramid, it is in power subduple. Moreover the cube is triple to the pyramid: but to the cube the Dodecahedron is in a manner duple. Wherefore the same Dodecahedron is in a manner sextuple to the said pyramid.   

Of the nature of an Icosahedron.  
Five triangles of an Icosahedron, do make a solid angle, the bases of which triangles make a pentagon. If therefore from the opposite bases of the Icosahedron be taken the other pentagon by them described, these pentagons shall in such sort cut the diameter of the Icosahedron which coupleth the foresaid opposite angles, that that part which is contained between the plains of those two pentagons, shallbe the greater segment: and the residue which is drawn from the plain to the angle shall be the less segment. If the opposite angles of two bases joined together, be coupled by a right line, the greater segment of that right line is the side of the Icosahedron. A line drawn from the centre of the Icosahedron to the angles, is in power quintuple, to half that line which is taken between the pentagons, or of the half of that line which is drawn from the centre of the circle which containeth the foresaid pentagon: which two lines are therefore equal. The side of the Icosahedron containeth in power either of them, and also the less segment, namely, the line which falleth from the solid angle to the pentagon. The diameter of the Icosahedron containeth in power the whole line, which coupleth the opposite angles of the bases joined together, and the greater segment thereof, namely, the side of the Icosahedron. The diameter also is in power quintuple to the line which was taken between the pentagons, or to the line which is drawn from the centre to the circumference of the circle which containeth the pentagon composed of the sides of the Icosahedron. The dimetient containeth in power the right line which coupleth the centres of the opposite bases of the Icosahedron, and the diameter of the circle which containeth the base. Moreover the said dimetient containeth in power the diameter of the circle, which containeth the pentagon, and also the line which is drawn from the centre of the same circle to the circumference: That is, it is quintuple to the line drawn from the centre to the circumference. The line which coupleth the centres of the opposite bases, containeth in power the line which coupleth the centres of the next bases, and also the rest of that line of which the side of the cube inscribed in the Icosahedron is the greater segment. The line which coupleth the middle sections of the opposite sides, is triple to the side of the dodecahedron inscribed in it. Wherefore if the side of the Icosahedron, and the greater segment thereof be made one line, the third part of the whole, is the side of the dodecahedron inscribed in the Icosahedron.   

Of the nature of a Dodecahedron.  
The diameter of a dodecahedron containeth in power the side of the dodecahedron, and also that right line, unto which the side of the dodecahedron is the less segment, and the side of the cube inscribed in it is the greater segment: which line is that which subtendeth the angle of the inclination of the bases, contained under two perpendiculars of the bases of the dodecahedron. If there be taken two bases of the dodecahedron distant the one from the other by the length of one of the sides, a right line coupling their centres, being divided by an extreme and mean proportion, maketh the greater segment the right line which coupleth the centres of the next bases. If by the centres of five bases set upon one base, be drawn a plain superficies, and by the centres of the bases which are set upon the opposite base be drawn also a plain superficies, and then be drawn a right line coupling the centres of the opposite bases, that right line is so cut, that each of his parts set without the plain superficies, is the greater segment of that part which is contained between the plains. The side of the dodecahedron is the greater segment of the line which subtendeth the angle of the pentagon. A perpendicular line drawn from the centre of the dodecahedron to one of the bases, is in power quintuple to half the line which is between the plains: And therefore the whole line which coupleth the centres of the opposite bases, is in power quintuple to the whole line which is between the said plains. The line which subt●deth the angle of the base of the dodecahedron, together with the side of the base, are in power quintuple to the line which is drawn from the centre of the circle, which containeth the base, to the circumference. A section of a sphere containing three bases of the dodecahedron taketh a third part of the diameter of the said sphere. The side of the dodecahedron, and the line which subtendeth the angle of the pentagon, are equal to the right line which coupleth the middle sections of the opposite sides of the dodecahedron.     
¶ The end of the Elements of Geometry, of the most ancient Philosopher 〈◊〉 of Megara.   

 

  

  

  

  

  

  

  

  

  

  

  

  

  

  

  

  

  

  

   

Faults escaped.  
 
●cl. ●ag. Line. Faultes● Co 〈◊〉 〈◊〉.    Errata Lib. 1.  1 2 41 point B. at Campane point C, a● Campane 3 1 22 a●l lines drawn all righ● 〈…〉 3 1 28 lines drawn to the superficies right lines drawn to the circumference 9 1 42 li●es AB and AC, lines AB and BC 15 1 35 are equal are proved equal 20 2 28 (by the first (by the fourth 21 1 39 t●e centre C. the centre E 2● 2 ● I●●ower right If two right 25 2 3 f●●● petition fifth petition 49 2 7 14. ●●. 32.64 etc. 4.8.16.32.64. & 53 1 39 the triangle NG the triangle K● 54 2 25 (by the 44 (by the 42 57 2 23 and C●G in the and CGB is th●    In stead of ●lussates through out 〈◊〉 whole book read ●lus●as.    Errata Lib. 2.  60 2 29 Gnomon FGEH Gnomon AHKD   30 Gnomon EHFG Gnomon ●CKD 69 1 18 the whole line the whole ●igure 76 2 9 the diameter CD the diameter AHF    Errata Lib. 3.  82 2 36 angle equal to the angle 92 1 last the line AC is the line AF 〈◊〉    Errata Lib. 4.  110 2 10 CD toucheth the ED toucheth the   12 side of the other angle of the other 115 1 21 and HB and HE 117 2 44 the angle ACD the angle ACB 118 1 2 into ten equal into two equal 121 1 3● CD, and EA, CD, DE, & EA, ●●● 1 29 the first the third    Errata Lib. 5.  126 1 43 it maketh 12. more than 17. by 5. it maketh 24. more than 17. by 7. 129 1  In stead of the figure of the 6. definition draw in the mag●● a figure like unto th●s. 134 2 4 As AB is to A, so is CD to C As AB is to B, so is CD to D 141 2 last But if K exceed M, But if H exceed M,   

LIEF IS DEATH AND DEATH IS LIEF: AETATIS SVAE: XXXX  
AT LONDON Printed by john day, dwelling over Aldersgate beneath Saint Martins.  
¶ These Books are to be sold at his shop under the gate. 1570.