MATHESIS ENUCLEATA: OR, THE ELEMENTS OF THE MATHEMATICS. By J. CHRIST. STURMIUS, Professor of Philosophy and Mathematics in the University of Altorf. Made English by J. R. A. M. and R. S. S. LONDON, Printed for Robert Knaplock at the Angel, and Dan. Midwinter and Tho. Leigh at the Rose and Crown, in St. Paul's Churchyard. 1700. The AUTHOR's PREFACE TO THE READER, Containing a SYNOPSIS of his Method. I THAT the Reader may the better apprehended our design and aim, we have thought fit to premise some things concerning the Methods, both general and particular, we make use of in the following Treatise. For as heretofore a sort of a blind deference to, and superstitious Veneration of Antiquity, and especially of Aristotle, has hindered the growth and progress of Natural Philosophy, which of late has made such considerable advances, since it has ventured to stand upon its own Bottom, to make new Additions to former Inventions, to essay new and unknown Objects, to substitute Things instead of Names, Certainties instead of Doubts, and Experience in the room of dull Credulities; not derogating in the mean while from the deserved Praises of the Ingenious among the Ancients: So without doubt Mathematics also, unless our Predecessors had imagined that it had long ago been brought to its utmost Perfection by Euclid, Archimedes, Apollonius, and other Ingenious Ancients, would have arrived long since to a higher pitch, and by this time have surpassed those Limits, which now we admire its arrival to. II It is confessed by all, that no Human Knowledge whatsoever can lay a more just claim to an unshaken Evidence and Certainty, or boast an higher necessity of its Demonstrations, or a greater multitude of undeniable Truths, than the Mathematics; and that those Propositions we have, found out by Archimedes, demonstrated by Euclid, Apollonius, and others, are at the same time unquestionable and altogether wondered. But we may with Truth affirm, that most of their Propositions may either be disposed in a better order, or propounded easier, or demonstrated more evidently and directly, or taught after a more short and compendious way, now at lest after they are already found out, and with a great deal of Pains demonstrated by their first Inventors; and of this Opinion are several of the best and most celebrated Mathematicians of the present Age. III It is certain Euclid has demonstrated several Propositions, (as Prop. 2, 3, 20, 30, lib. 1. 2, 5, 6, 10, 15, 28, 29, lib. 3. etc.) whose Truth to any attentive Person appears from the very terms, more clearly and certainly than the truth of Axiom 13. lib. 1. which his Interpreters dare not admit without a Demonstration. And though those superfluous Demonstrations derogate nothing from the certitude of the thing, yet by an unnecessary increase of the number of Propositions, and (which frequently follows thence) an inverting the order of things, they breed Tediousness and Confusion. IV. There are none, unless those who are bigoted to Antiquity, but must own that the Elements of Euclid are destitute of a just and orderly Disposition of things. For to omit, that in the first Book there are handled several sorts of Subjects, and a great variety of Properties demonstrated of them promiscuously without any respect to similitude or conveniency; there is this never to be excused, that, after he has in the first Book deduced and demonstrated some particular Affections of Magnitude, he proceeds, in the second, to those things which are universal and common to any quantity; than in the third and fourth, he contemplates the Circle and the Properties of Figures inscribed in it, or circumscribed about it; in the fifth again he treats of the universal Doctrine of Reasons and Proportions; and yet not so universally, but in the seventh again he is obliged to demonstrate the same of Numbers particularly, which might have been done for all Quantities whatsoever, by one general Demonstration. V Than as for the method of Demonstrating used by the Ancients; it is true that it nicely regarded the certainty of its Conclusions, nor would it admit any thing into its Demonstrations, which was not either a first Principle and so self evident, (called by them an Axiom) or might not be supposed, beyond all Contradiction, possible to be effected (and on that account named a Postulate) or, thirdly, an arbitrary Denomination of the thing proposed which needed no Demonstration, (and was called a Definition or Explication of a Term;) or, lastly, which had not been evidently demonstrated before: Yet I believe none will deny, but that this Method would have been more deservedly esteemed, if with the certainty of its Conclusions it had joined a greater Easiness, Brevity, and Evidence, which is wanting in most of the Demonstrations of the Ancients; who thought it enough, firmly and infallibly to establish the truth of their Theorems, and extort the Assent of their Readers; little regarding by what Ambages, by how many circumambulatory Propositions, and almost whole Volumes, it was done; that thereby they might be forced to acknowledge the thing to be so, while how it came to be so, or from what intrinsic Cause o● Condition of the Subject requiring it, such and such an Attribute agreed to it, remained in the mean while obscure, or altogether unknown. VI Hence they made such frequent use of Apagogical Demonstrations, or deductive ad absurdum & impossibile, which aught not to be done, but where no ostensive Demonstration can be had, or for illustrating negative Propositions rather than demostrating them; for the method of Deduction ad impossibile, does not so much demonstrate the Truth itself directly, as the consequent Absurdity of the opposite Supposition; whence it follows very indirectly, (though most certainly) that the Proposition is true, while in the mean time the original Reason of its Truth remains altogether hid and in the dark. VII. But that we may not seem unjustly to reject the particular Method of the Ancients, made use of by Euclid, as in lib. 12. Prop. 2, 10. etc. and by others, but especially by Archimedes, who peculiarly addicted himself to it, whence it has been by some called the Archimedean Method, and by Renaldinus the Method per Explosum excessum atque defectum; besides its Deduction doubly add absurdum, whereon it always relies, as e. g. it infers the equality of two Magnitudes A and B by a far fetch round about way, by showing, that if B be supposed greater or lesle than A, from either Position there would follow an absurdity; and thence as it were begging the Equality by a new Inference, which though it may pass free from Suspicion, yet it neither aught, nor can be admitted, without this Limitation: In comparing those things whose Natures are capable of Equality, if it can be demonstrated that the one is neither greater nor lesle than the other, we may thence justly infer their Equality. VIII. The Learned are now generally agreed, that, besides that Synthetick Method, whereby the Ancients either ostensively deduced their Problems and Theorems from evident and common Principles, or apagogically demonstrated them by Deductions ad absurdum, they also made use of a certain sort of Analysis, whereby they found out those Theorems and Problems; and which, to raise the greater Admiration in their Readers, they afterwards studiously concealed and kept to themselves: Which Method is undoubtedly preferable to the other, as not only demonstrating the certainty of the Propositions so found, but at the same time showing the invention of them too; and this is that Method that Vieta, Harriot, and Des Cartes, and their Followers have not only brought to light in this last Age, but to a great degree of Perfection too, and whereof Carolus Renaldinus, in that vast Work he has entitled Ars Analytica Mathematum, has given us a large Treatise. IX. There has appeared moreover of late another particular Method invented by Bonaventura Cavallerius, which is called the Method of indivisibles, whereby the most difficult and abstruse Problems of Geometry are found out and Demonstrated with an incredible ease, which is the abovementioned Renaldinus' Opinion of it, lib. 1. Resol. & Comp. p. 239. which, to demonstrate the Equality or Proportions of Figures and Bodies that may be compared with one another, goes to work after a way which seems to be more natural than any other, by supposing plane Figures to consist of innumerable lines, and solids of innumerable Plans (called their indivisible Parts or Elements, because the Lines are conceived without latitude, and the Plans without any thickness,) and relying on this self-evident Axiom, That if all the Indivisibles of one Magnitude collectively taken, be equal or proportional to all the correspondent Indivisibles of another, or taken separately each to each, than also those Magnitudes will be equal or proportional among themselves. Which Inference can be guilty of no Fallacy, nor liable to any Error, as long as those Elements are taken and conceived in that sense their Author's design them; which is sufficiently demonstrated by Renaldinus, lib. 1. Compos. & Resol. p. 245. towards the bottom, and at the beginning of p. 306. and also by Honoratus Faber, in his Synopsis, p. 24. and Dr. Barrow in Lect. Geom. p. 24. and the following, against Tacquet and the other Adversaries of this method. X. There is another Method akin to this, which may be properly named Generative, very much followed by Faber in his Synopsis, and Barrow in his Lect. Geom. whose Author Renaldinus tells us was Guldinus, lib. Cit p. 253. showing at large its Rules and Foundations in the following pages, viz. The rise of Lines from certain motions of points; of Plane and Curvilinear Surfaces, from the determinate progress or rotation of a given Line, and of Solids by the various motions of various Surfaces; the Productions whereof are so represented to the Imagination, that the intrinsic nature of the magnitudes thence arising may become known, and their Properties and Affections may, from their Natures thus known, be easily and briefly deduced. XI. Near akin to this Method of Cavellerius is that other of Infinite Progressions, wherein having found a certain Progression of like Parts circumscribed about, or inscribed in any given magnitude, which may be continued by Bisection ad infinitum; and than at length (by virtue of the Doctrine of Exhaustions, founded on Prop. 1. lib. 10. Eucl.) will terminate in the magnitude itself, I say, wherein the sum of those infinite Terms, collected by Rules on purpose for the addition of those Progressions, and consequently the quantity or proportion of the proposed Magnitude, to any other given one, may be expressed or defined. But this termination of Figures infinitely circumscribed or inscribed in a Circle, not pleasing Renaldinus (altho' his Dissension seems only to consist in Words,) he exhibits another Method like it, (which he peculiarly calls his own) built on twelve fundamental Theorems, and illustrated by several Examples, lib. 1. de Resol. & Comp. p 277 & seqq. XII. Of late also, the most ingenious Mr. Isaac Newton, to demonstrate his Philosophiae Naturalis Principia Mathematica, lib. 1. sect. 1. premises some Lemmas of his method of Rationes primae & ultimae, or evanescent quantities, thereby to avoid the tediousness of deducing long and perplexed apagogical Demonstrations after the way of the Ancients. For finding that his Demonstrations might be very much contracted by the method of Indivisibles, and knowing at the same time, that that method was scrupled by some, and thought not very Geometrical, he rather chose to found his Method on the sums and proportions of quantities which he calls Evanescent, which performs the same as the Method of Indivisibles, and may be more safely used, which he inculcates in these very words, and others such like, in Scholar of Lemma. 11. And also answers several objections which might seem to make against it. XIII. But it would be in vain for us to attempt, in this place, to explain all and each of those various methods at length; having only proposed to ourselves, to demonstrate the chief and principal Theorems and Inventions of the Mathematics, and to use sometimes one of them, and sometimes another, (having first Demonstrated their Foundations) according as we shall judge this or that of them, fittest to Demonstrate the thing in hand, and so show the reasons, and use of each of them in the process of this discourse. And altho' H. Faber in his Synopsis, p. 8. Insinuates, that Analytick terms aught not to be made use of in Geometrical Demonstrations, because that Algebraick method seems to be too difficult for young beginners; yet we are of the quite contrary opinion (nay we can scarce doubt but that that Ingenious man would also agreed with us herein, if he saw the way we make use of those foundations of Algebra, which is only of the most simple and general principles of it) especially in this case, where the said method is by little and little instilled with the Demonstrations themselves; and the literal Computations taught from their first Principles, than which nothing is more easy: And this is that which we design to do, and so use the learner by degrees to this sort of Demonstration, thereby to prepare him the better for the Analytick Geometry of the Moderns, which is the highest apex of the Mathematics. But we had rather our Reader should himself found, than we trouble ourselves any further to tell him here, how compendiously we Demonstrate the Propositions of Geometry, by the help of these Analytick notes, without the tedious Concatenation of a long Chain of Consequences, which would be otherwise unavoidable. FOURTEEN. After this way we design to go through the following Scheme. 1. We shall deduce several propositions of Euclid, Archimedes, and Apollonius from our definitions, and the generations of Magnitudes therein proposed; as Corollaries necessarily flowing from them, and confirmed only by an immediate and simple consequence. 2. We shall demonstrate their chief Theorems (for the sake of which they were forced to Demonstrate several others before hand, the knowledge whereof for their own sake was not so necessary or valuable) without any long series of antecedent Propositions, or Foreign principles, from a few direct and intrinsic Principles of their own. Whence 3. It will follow, that after this Method we shall propose things, and treat of them; in a more natural Order, and first of all deliver those which are most universal and common to all quantities, and than descend to those which in a more special manner regard Magnitude, and distribute and dispose all according to certain general distinct Classes of the things to be treated of, and their affections. Hence also 4. We deduce from those universal Theorems, by way of Corollary, the Precepts of vulgar Arithmetic, and specious Computation, which afterwards we make use of in particular Demonstrations after a very short and compendious way; and, for this very reason, some Learned men of the present Age are of opinion, that the Ancients often fell into that tedious and intricate prolixity in their Demonstrations, because they would not acknowledge the great affinity there was between Arithmetic and Geometry, taking particular care not to introduce the Terms and Operations of Arithmetic into Geometry; though at the same time they never scrupled to transfer the names of Plan, Square, Cube, and such like to numbers. 5. Lastly, Having first Demonstrated the first and Fundamental Theorems of Elementary Geometry, we may safely build on them the Praxes of all kinds of Mathematical Arts, that are most useful and requisite to several Exigencies of human Life, as, first, Trigonometry both Plain and Spherical, the Construction and use of the Tables of Sines and Tangents. 2. The Construction of Logarithms, and a compendious application of them to Trigonometry: And in the 3. and last place, the fundamental Precepts of Algebra, or the Analytic Art; by the help whereof the learner may at length arrive to the higher and more recluse parts of Geometry, and become master thereof: Not to mention several Geometrical and Arithmetical Problems, which we have all along derived from several of our Theorems, by way of Corollary, which it may be some other time, may make an Appendix of this work. XU. And thus when we shall have Demonstrated not only the chief Theorems of the Ancient Mathematicians, omitting the unnecessary crowd of those that are only Subsidiary, but also have demonstratively deduced the fundamental Precepts of the most necessary and useful Arts that flow from them, and that are abstracted from matter, as of Arithmetic, Trigonometry, and Algebra; I hope none will doubt but that in this little Volume we have exhibited, as it were the Nucleus or Kernel of the pure and genuine Mathematics (for those other Sciences and Arts which go by the name of mixed Mathematic, are most of them parts of natural Philosophy, from the application of Mathematics to the Phaenomena of nature) and so may justly bear the name of Mathesis Enucleata. XVI. Nor are we ignorant, nor shall we conceal what several Learned men have both proposed and already done, for removing those difficulties and blemishes of the Ancient Mathematical Methods we have just now mentioned. The late Admonishments of the anonymous Author of L'Art de Penser, not lesle ingeniously than modestly delivered, Part. 4. Chap 9 10. of the said Treatise, are sufficiently known; as also the laudable endeavours of A. Tacquet and Honoratus Faber and several others above mentioned for contracting, new ordering, and more easily and directly Demonstrating the chief Geometrical Inventions of the Ancients. There are moreover extant of a certain anonymous Author, Elementa Geometrica novo ordine ac methodo feré Demonstrata, Printed at London about 26. Years ago. There are also F. Ignatius Gaston Pardies Elemens' de La Geometry, etc. Translated into Latin after the third Edition, by the Famous Schmidtius Professor at Geneva: As also of F. Mich. Mourgues', of the Society of J. Nouveaux Elemens' de Geometry, abreges par des Methodes particulieres en moins de Cinquante propositions, etc. There are also several other Essays of reducing the Mathematics into a better Order and Method, the titles whereof we have only as yet seen; and even while these papers were in the Press, there happened into our hands a Treatise of F. Lamy's Entitled Les Elemens' de Geometry, ou de la mesure des Corpse, etc. Printed at Paris in 1685; so that we may only seem to some to do what has been done already, in endeavouring to show our Reader a new and shorter way to the Mathematics. XVII. But as none can blame Jacobus le Maire, because, after the happy discovery of the Magellanick Passage from the Atlantic into the Pacifick Sea, he would needs yet endeavour to found another shorter, which he accordingly did; nor can they be blamed, who now a days consult about finding one from these parts of the World, by the North to the East India's. Thus also, in an affair of that moment, that one or a few are not sufficient to bring it to Perfection, if any one who comes after, not only invited, but also assisted by the ingenious Endeavours of those who have go before him, shall undertake to add to their Inventions, to help on the business by his advices, and show what things are capable of a further Polish, and the method how to perform it, doubtless such an one aught not to be blamed, nor accused of arrogance, unless at the same time he endeavours to depretiate the essays of others, and cry up his own as the only valuable; Which how far it is from our design, the work itself will abundantly show. Moreover as the senses of men are differently affected by different Objects, and their Palates have different Relishes of the same thing, according to different Preparations of them: So the same truth takes and insinuates itself more easily with one proposed and demonstrated after this way, more with another after that way; and we are so much the more likely to suit the different genii of different People, by how many more and different methods and ways we show them, leading to the same end, of which every one may take that which he likes best. XVIII. We therefore Publish, by the Divine assistance, these our Endeavours also, after so many other ingenious and elaborate ones in the same kind; nor can we doubt the approbation of some of our Readers. This at lest we can experimentally affirm: That not a few of those to whom these our thoughts were partly publicly read in Lectures, and partly privately taught (for they were only designed for Learners) were not a little taken with the Concise brevity and facility of the Demonstrations; so that we may reasonably hope to be acceptable to those, to whom either time, or sufficient force of genius is wanting, to run over the vast Volumes of the Ancient Mathematicians, and comprehend their prolix Demonstrations, and long series' of far fetched Consequences; and as for those who have both leisure and genius to do so, this may serve for an Encouragement towards it; that after they have go through the chief truths and propositions they contain, Demonstrated in a more easy and shorter way, they may so much the more confidently adventure upon those celebrated and ingenious Treatises, from the reading whereof they were before deterred by the length and almost insuperable difficulty of their tedious and perplexed Demonstrations. XIX. Being now about to enter upon the matter itself, we will only further hint these few things. 1. Since our whole design is for the advantage of young Students (which aught to be a Professors chief Care and Study) we must not omit the Explication of the most simple terms; especially since we design to deduce some Corollaries immediately from them, which heretofore have unnecessarily increased the number of Propositions and Demonstrations. 2. To encumber our work as little as we can with words we have made use, especially in our Analytick Calculus, of some Symbols, as = for Equality, as also of □ and ▭ for Square and Rectangle, and of √ the common Radical sign for the square Root, with the Line √ ¯ on the top for connecting of quantities together, the Root whereof is jointly taken, √ 3 for the Cube Root, √ √ for the Biquadratick Root. XX. That the Reader may at one view see the Contents of the following Treatise, we have thought fit to present him here with it, by way of Synopsis. It is divided in two Books. I The first whereof contains the chief and most select Propositions of Euc●id's Elements, of Archimedes' Treatises of the Sphere and Cylinder, as also of the dimension of the Circle, etc. Wherein that which these author's have demonstrated by a long and tedious Series of Consequences, and for the most part indirectly, we have here endeavoured to Demonstrate directly, and so that the Demonstration of each Proposition depends, either on no other, and so is evident by its own light, or on a very few of the antecedent ones. II After the same way in the second Book we treat of the Conic Sections, and Demonstrate the chief properties of the Conoid, Spheroid, Cycloid, Conchoid and Spiral Lines, which are extant either in Apollonius or Archimedes and others, and what they have Demonstrated by long and tedious process' we have here exhibited in a short and easy Compendium. And that, III In such a method, as does not so much require intent and severe thinking, as a bore and easy inspection, and application of the Principles of specious Algebra, and method of Indivisibles. which yet, IU. We don't barely suppose, and remit our Readers to other Books to learn (which would be too troublesome) but in the Process of the Work itself, they are gradually, and as occasion presents, derived frrm their Original Fountain, and first Principles. V By the same way also the most useful and necessary Mathematical Praxes are laid down under the names of Corollaries and Scholia, the Construction of the Tables of Sines and Tangents taught, the Original and use of the Logarithms Demonstrated, and the Precepts both of Plain and Spherical Trigonometry deduced from their first Principles, etc. VI The Praxis also of pure Arithmetic, and of common Decimal, and (which is seldom used) of Tetractical, as also the Doctrine of Surds, are derived from their first Original Whereunto, VII. As a Compliment of the whole Work, we have added an Introduction to the Specious Analysis, or new Geometry of the Moderns, particularly according to the Method of Des Cartes, but much facilitated by later Inventions, and Comprising the Precepts of the Art in six or seven Pages, but illustrated with above forty Examples in the different degrees of Equations. What the Reader's opinion will be of these our Endeavours, designed only for the use of young Students, time must teach us. The Author himself at lest, amongst his other performances, allows these the first place. ADVERTISEMENT. At the Hand and Pen in Barbican are Taught, Viz. Writing, Arithmetic, Book-keeping, Algebra, Geometry, Measuring, Surveying, Gauging, Astronomy, Geography, Navigation and Dialling. By Robt. Arnold. People Taught abroad. Elementa Arithmeticae Numerosae & Speciosae. In usum Juventutis Oxoniae è Theatro Sheldoniano. Prostant Venales Londini apud Dan Midwinter & Tho. Leigh ad Insigne Rosae Coronatae in Caemeterio Divi Pauli. Mathesis Enucleata: OR, The Elements of the Mathematics. Book I Explaining the First Principles of the Mathematics; among which are (in the first place) Definitions, and some Consectaries that flow from them. CHAP. I Containing the Definitions or Explications of the Terms which relate to the Object of Mathematics. DEFINITION I Mathematics is the Science or Knowledge of Quantity, and of Being's, as far as they are subject to it, or measurable; and may justly claim the Name of Universal, while it is employed in Demonstrating those Properties which are common to all or most Quantities: But when it descends to the different Species of Quantity, and is busied in contemplating the Affections belonging particularly to this or that Quantity, it is distinguished by various Names, and distributed into various Parts, according to the various diversities of the Objects. DEFINITION II QUantity may be defined in General, whatever is capable of any sort of Estimate or Mensuration, as immediately the Habitudes and Qualities of Things, as e. g. the multitude of Stars in the Heaven, or of Soldiers in an Army, the length of a Rope, or Way, the weight of a Stone, the swiftness or slowness of Motion, the Price of Commodities, etc. but mediately, the very things themselves wherein those Estimable Qualities are inherent. Whence with the ingenious Weigelius we may not incongruously reduce them all to these four Kind's or Genders, viz. 1. to Quanta Naturalia, Natural Quantities, or such as Nature has furnished us with, as Matter with its Extension and Parts, the Powers and Forces of Natural Bodies, as Gravity, Motion, Place, Light, Opacity, Perspicuity, Heat, Cold, etc. 2. to Moral Quanta or Quantities, depending for the most part on the Manners of Men, and arbitrarious Determinations of the Will; as for Example, the Values and Price of Things, the Dignity and Power of People, the Good or Evil of Actions, Merits and Demerits, Rewards and Punishments, etc. 3. to Quanta Notionalia, arising from the Notions and Operations of the Understanding, as e. g. the amplitude or narrowness of our Conceptions, universality or particularity, etc. in Logic; the length or brevity of Syllables, Accent, Tone, etc. in Grammar: And lastly, to Quanta Transcendentia, Transcendent Quantities, such as are obvious in Moral, Notional, and Natural Being's; as Duration, i e. the Continuation of the Existence of any Being; which in Physics especially is named Time, and may be conceived as a Line, etc. To these you may moreover add Unity, Multitude or Number, Necessity and Contingency. DEFINITION III NVmber (whereon we shall make some special Remarks) if it be taken in the Concrete, is nothing else than an Aggregate or Multitude of any sort of Being's; taken abstractedly, it is, as Euclid calls it, 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉, a multitude, or (as they call it) Quotity of Unities, on the one hand Number, i e. many are opposed to one; and in that sense Unity is not a Number: On the other hand Unity may be esteemed a Number, since it is no lesle (if I may be allowed that term) some Quotity than two or three. But as we denote or signify particular Things, when we speak of them Universally, by the Letters of the Alphabet, A, B, C, (a, b, c,) etc. as universal Signs or Symbols of them; so for distinctly and compendiously Expressing the innumerable Variety of Numbers, Men have found out various Notes, the most natural whereof, are Points disposed in particular extended Orders, as ... to denote Three, 9 dots arranged in 3 rows and 3 columns to denote Nine, etc. But that way which is most commodious for Practise, is by the common Notation, or Ciphers, 1, 2, 3, 4, 5, 6, 7, 8, 9 the invention whereof, as we have it by Vulgar Tradition, is owing to the Arabians. By a very few of these we express any number though never so great, by a wondered, though now adays familiar, Artifice; the first Inventor of them having Established this as an arbitrary Law, that the first of them shall signify unity or one; the 2d two, etc. as often as they stand alone; but placed in a row with others, or on the left hand of one or more 0, or naughts, (which of themselves stand for nothing, but fill up empty places) if in the second, before a naught, they denote Ten; if in the 3d. Hundreds; in the 4th Thousand; in the 5th Myriads or Ten of Thousand; in the 7th so many Thousand of Thousand, or Million; in the 8th Ten of Million, etc. and so onwards, increasing always in decuple Proportion, by Ten, Hundred, Thousand, etc. COROLLARY I HEnce you have a way of expressing or writing any Sum by these Notes, which you may hear expressed in Words; as if we were to express in Notes the year of our Lord, One Thousand Six Hundred Ninety and Nine, it is manifest, that according to the method above described, by placing 9 on the right hand in the first place, and nine again in the second towards the left, six in the third, and one or unity in the fourth, the business will be done. Thus it will be easy to any one with a little attention, to express any Number whatsoever by these Notes; (as suppose that which Swenterus proposes, in Delic-Physico-Math. Part. 1. Probl. 75.) Eleven Thousand, Eleven Hundred, and Eleven. COROLLARY II HEnce you have also the Foundation and Reason of the Rule of Numeration in Arithmetic, or expressing any Numbers or Sum in Words, which you see written in Ciphers; which for greater ease may be done thus, viz. beginning from the first Figure towards the right hand, over every fourth Figure note a Point, (including always that which was last pointed) and at every second Punctation or Point, draw short strokes thus, one over the 2d, two over the fourth, etc. the first denoting Million, the second Million of Million, or Bimillions, the third Trimillions, etc. and the Intercepted Points the Thousand, in their kinds, etc. SCHOLION. HEre I cannot omit, on this occasion, what the forementioned Weigelius has hinted about another way of Numeration, and which Dr. Wallis mentions, Oper Mathemat. Part 1. p. 25. & 66. showing there a way (and illustrating it by Examples) of Numeration, and of Expressing the Figures; which proceeds thus; whereas now adays in numbering we ascend from uni y or 1 to ten (the reason whereof; after which Aristotle makes a prolix Inquiry, Probl. 3 Sect. 15. was taken without doubt from the denary Number of our Fingers) if from unity we proceed only to four, (which Aristotle in the same place tells us some of the Thracians used to do of old,) and thence returning back again to Unity, we should proceed again after the same way; we might after that way obtain a vastly more simple and easy Arithmetic, than we have now adays. Which, even hence we may conclude; because for Multiplication and Division there would need no other Table (or Pythagorick Abacus) than this easy and short one: 1.1.1 once one is one; 2.2.10 i e. twice two are four; 2.3.12 twice three are four and two. 3.3.21 thrice three are twice four and one Pag. 5. Fig. 1. 2 3 4 5 6 7 8 9 10 11 12 13 And although it is pity that we can't hope now a-days to substistute this vastly easier way of Computing, in room of the other now in use, because the other is universally received, and most sorts of Measures and other Quantities are fitted and accommodated to the decuple Proportion; yet it aught not to be altogether neglected in Mathematics, which might receive very great advantage hereby, especially in Trigonometry, if the ingenious Invention of the Logarithms had not already supplied its use therein. The whole Foundation of this Tetractys or Quaternary Arithmetic, is placed only in these three Notes, 1, 2, 3; ●o that any one of them alone, or in the first place, should denote Units, in the second place, Tetrads, or so many Fours (or Quaternions,) in the third place so many Sixteens, in the fourth place so many times Four Sixteens, or 64. s. etc. always proceeding in a Quadruple Proportion. For which way of Numeration there might be found out terms as commodious as those we now use, and which are thereby grown Familiar to us, as one, ten, twenty, a hundred, a thousand, etc. which will be evident by what follows: One, Vnum 1 One. Ten, Decem 10 Quatuor, Tetras, a Quaternion or Four. Twenty, Viginti 20 a Biquaternion Thirty, Triginta 30 a Triquaternion Hundred, Centum 100 a Tetraquaternion Thousand, Mille 1000 a Quartan Tenterhook Thousand, 10000 etc. a Tetraquartan, etc. DEFINITION IU. A Magnitude is whatever is conceived to be Extended or Continuous, or has parts one without another, and contained within some common Term or Terms: wherein that is called a Point which is conceived (as indivisible, or) to have no Parts, and so no Magnitude, but is notwithstanding the beginning or first Principle of all Magnitude. DEFINITION V IF we conceive a Point (A) (Fig. 1.) to be moved towards B, by this motion it will leave a trace, or describe the Magnitude AB of one only Dimension, that is Length without Latitude, or which at lest we are to conceive so, and is called a Line: If that Line AB be conceived again so to be moved, as that its extreme Points AB shall describe other Lines BC and AD, it will describe by that Motion the Magnitude AB CD, or (to denote it more compendiously by the Diagonal Letters) AC or BD, having both length and breadth, but without any depth or thickness, or at lest so to be conceived, and this is called Superficies or Surface: Lastly, if this Surface AC be conceived so to move, e. g. upwards or downwards, that its opposite Points A and C again describe other Lines OF and CH, and consequently each of its Lines other Surfaces, etc. by this Motion there will be form a Magnitude of three Dimensions, which we call a Solid or Body, which we will also denote by the two diametrically opposite Letters AH and DG. But as this Motion of the Point, Line, or Surface, may be various, so there will be produced by them various sorts of Lines, Surfaces, and Solids: But these Productions stop here, and proceed not further; for the Motion of a Body can only produce another Body greater than the first, but no more new Dimensions. CONSECTARYS. I POints therefore being moved through equal Intervals in the same or a like way or trace (e. g. in a straight or the shortest trace) describe equal Lines; and II The same or equal Lines moved through the same Right-lined or Curvilinear Paths, describe equal Surfaces; and III Equal Surfaces moved according to the same Methods and Conditions describe equal Solids: which, if rightly understood, are the first certain and infallible Foundations of the Method of Indivisibles. But here you must take care to distinguish between the way which the Line itself describes, and that which its Ends or extreme Points describe: For although e. g. the Point a (Fig. 24.) moves along in a more obliqne way than A, and so describes a longer Line a c; yet the Line a b describes by a parallel Motion, an equal space with the Line A B, (viz.) the same which the whole Line A b, whereof they are parts, would describe. See Faber's Synopsis, p. m. 13. DEFINITION VI. BUT that we may a little further prosecute this Genesis of Magnitudes (as very much conducing to understand their Nature and Properties) if the Point A moves to B the shortest way, it describes the Right Line AB; but if in any (one) other it will describe the Curve or Compounded Line ACB: From whence, with F. Morgues, we may infer these CONSECTARIES. I THat two Right Lines (a) Eucl. Ax. 14. beginning from the same Point A, and ending in the same Point B, will necessarily coincide, nor can they comprehend or enclose Space; for if they did, one must deviate, and so would cease to be a Right Line. II In a Space comprehended by three Right Lines AB, BC, CA, (a) Eucl. Ax. 14. any two taken together, must needs be greater than any one alone. Moreover we may add this before hand; III In a Circle a Right Line drawn from A to B (Fig. 3) will fall within the Circle, because the Curve Line ADB described, as we shall hereafter show, being longer than a Right Line, must necessarily fall beyond it, or on the outside of it. And lastly, IV. A Tangent, or Line (b) which does not cut or enter into the Circle, touches it only in one Point. Moreover if a Right Line AB (Fig. 4.) move on another Right Line BC, remaining in the same Position to it, it will generate a Plan Surface, to which a Right Line being any way applied, will touch it with all its point, as Faber rightly describes it; if a Right Line be moved on a Curve, or a Curve on a Right Line, etc. they will generate a Curve Surface, called Gibbous, or Convex without, and Concave within. DEFINITION VII. IF a Right Line be fixed at one of its ends A, and the other be moved round (Fig. 5.) it will describe in this Motion a Circular Plane, or a Circle; and by the motion of its end or extreme Point B, (a) Eucl. l. 1. Prop. 20. the Periphery or Circumference of that Circle BEF. (b) Eucl. Prop. 2. lib. 3. The fixed Point A is called the Centre of that Circle; the Lines AB, AC, etc. its Radii or Semidiameters; all of which are equal one to another. Any Right Line BC drawn from one part of the Circumference through the Centre to another, is called the Diameter, and divides the Circle into two Semicircles BECB and BFCB. The Circumference of a Circle, whether great or small, is divided into 360 equal parts called Degrees, and each Degree into 60 Minutes, etc. From this Geniture of the Circle presupposed, there evidently follow these CONSECTARIES. I THat 2 Circles which cut one another cannot have the same common Centre; for if they had, the Radii ED and EA drawn from the common Centre E (Fig. 6.) would be equal to the common Radius EBB that is the part to the whole. II Nor can two Circles touching one another within side, have one and the same Centre, for the same reason. III Of Lines falling from any given Point without the (Fig. 7.) Circle, (b) & 6 of the same Book. and (c) Eucl. 8. l. 3. passing through the Periphery to the opposite Concave part of it, that which passes through the Centre of it, is the longest, viz. AB; and of the other that which is nearest to it is longer than that which is more remote: But on the contrary of those which fall on the Convex Periphery, that which tends towards the Centre, as Ab, is the lest, and the rest gradually greater, and there can be but two, as A and OF, or A and Of, equal: All which will appear very evident by drawing other Circles from the Centre A through B, D, E, and b, d, e. Or thus; having drawn two other Circles, from the Radii AB and Ab, if we conceive the Radii Ab and Cb to move towards the right hand, their ends will always recede further from one another; the same is also evident of the Radii AB and CB, moved also to the right together. IV. Moreover (Fig. 8.) of all the Lines drawn within the Circle (a) Eucl. 5 l. 3 the Diameter is the greatest, and the rest gradually lesle, by how much the more remote they are from the Centre, etc. Which will be very evident to any one who contemplates a Circle inscribed in a Square, as also the Genesis of Curvity itself; as also many other ways which I shall now omit; or to mention one more thus; because the two Radii CA and CB being moved, in order to meet together, necessarily approach nearer to one another in their extreme Points. DEFINITION VIII. THE Aperture or opening of two Lines (Fig. 9) AB, AD, etc. that are both fixed at one end at A, and the other ends opened or removed farther and farther from one another, is called an Angle, (a) Prop. 15. lib. 3. and usually denoted by 3 Letters, D, A, B, (whereof that which denotes the Angular Point, always stands in the middle,) and measured by the Arch of a Circle BD, or a certain number of Degrees which it intercepts. The greatest Aperture of all BAC is when the 2 Legs of the Angle AB and AC make one Right Line, and is measured by a Semicircle, or 180 Degrees. The mean or middle Aperture BAE or CAESAR, when one Leg EA is erected on the other AB or AC at Right Angles, so that it inclines neither one way nor the other, (thence called a Perpendicular) is named a Right Angle, whose measure is consequently a Quadrant (or quarter part) of a Circle or 90 Gr. Wherhfore a Semicircle is the measure of two Right Angles: An Aperture or Angle BAD lesle than a Right Angle (and so measured by lesle than 90 degrees) is called an Acute Angle; and that which is greater than a Right Angle, as DAC (and so consisting of more than 90 degrees) is called an Obtuse Angle. Whence me may now draw these COROLLARIES. I TWO or more Contiguous Angles (a) Eucl. 13. lib. I with the Coral. constituted on the same Right Line BC, and at the same Point A (as DAB, and DAC or DAB, DAE and EAC) make two Right Angles, as filling the Semicircle; and consequently, II All the Angles that can be constituted about the Point A (as filling the whole Circle) are equal to 4 Right ones: As also on the other side (a) Euclid. l. 1. Prop. 14. if two Right Lines AB and AD meet on the same point A of another Right Line AC, and make the Contiguous Angles equal to 2 Right ones, that is, if they fill a Semicircle, BC will necessarily be the Diameter of a Circle, and consequently a Right line. III If one of the Contiguous Angles BAE be a Right one, the other CAESAR will be so also. IV. If two Right Lines AB, CD, cut one another in E, the 4 Angles they make will be equal to 4 Right ones. V And as it is evident at first sight (Fig. 10.) that any Circle having one half (or Semicircle) folded on the other, at the Diameter ECD, the two Semicircles EHD, and EID, must needs agreed, or every where coincide one with the other; so if the Angle ACD be supposed equal to the Angle BCD, that is, the Arch AD to the Arch BD, having one Leg CK or CL common; the others AC and BC being supposed before equal, 1. The Bases BL and ALL, KB and KA, will be also equal; for these will coincide too, and therefore the Angles also. 2. The Line AB being bisected in KING, the two Angles (b) Eucl. l. 1. 4 & 3 (c) lib. III 3. at KING will also coincide and be equal, and consequently Right Angles: and contrariwise, 3. The Angles at the Base of equal (d) lib. 1. 5. Legs, CAB, CBA, and also those below the Legs, the Legs being produced to FLETCHER and G, are equal. 4. Consequently the Spaces ACL and BCL, ACK and BCK are equal to one another. 5. The Contiguous Angles AED and BED insisting on equal Arches AD and BD are equal, and è contra; as also those that are not Contiguous, if their Vertex's are equidistant from E, etc. 6. It is hence also manifest, that a Perpendicular erected on the middle of any Line AB, inscribed in any Circle, passes through its Centre, by what we have just now said; and if you likewise erect Perpendiculars on the middle of any 2 Lines, ab and bm (Fig. 11.) connecting any 2 Arches, or any 3 Points, a, b, m, that are not placed all in the same Right Line, those 2 Perpendiculars ke, not, will determine (by their Intersection) the Centre of a Circle that shall pass through these 3 Points. DEFINITION IX. IF one Right Line DE cut or pass through another AB (Fig. 12) the opposite Angles at the top or intersection ACD and ECB are called Vertical; as also the other two ACE and DCB: Whence follow these COROLLARIES. I THat the Vertical Angles are always (a) Eucl. lib. 1. 15. Equal; for both ACD and ECB with the third, ACE, which is common to both, fill or are equal to a Semicircle; as likewise both ACE and DCB with the third ECB, which is common. II Contrariwise, if at (a) The same Prop. Scholar 1. the Point C of the Right Line DE, the 2 opposite Lines AC and CB make the Vertical Angles x and z equal, than will AC and CB make one Right Line; for, since x and oh make a Semicircle, and z and x are equal, by Hypoth. oh and z will also make or fill a Semicircle, whose Diameter will be ACB. III By the same Argument it will appear, that of 4 Lines (b) Scholar 2. proceeding from the same Point so as to make the opposite Vertical Angles equal, the 2 opposite ones AC and CB, as also DC and CE, will make each but one Right Line; for since all the 4 Angles together make a whole Circle, or 4 Right Angles, and the sum of x and oh is equal (by Hypoth.) to the sum of oh and z, it follows, that both the one and the other will make Semicircles, whose Diameter will be AB and DE, and so Right Lines. DEFINITION X. IN any Circle, a Right Line, as D. G, that subtends any Arch of it DGB, is called the Chord of that Arch (Fig. 13.) BF (a part cut of from the Semidiameter BC passing through the middle of the Chord) is called the Sagitta or Intercepted Axe, but most commonly the Versed Sine; and DF let fall from the other extremity of the given Arch BD, on the Semidiameter at Right Angles, is called the Right Sine of that Arch BD, or of the Angle BCD; also DIEGO is called the Right Sine of the Compliment (or for brevity sake, Sine Compl.) of that Arch DH, or Angle DCH, etc. but the greatest of all the Right Sins HC let fall from the other extremity (or end) of the Quadrant (which is indeed the same as the Semidiameter of the Circle) is called the whole Sine or Radius; lastly, BE is called the Tangent of the Arch BD or Angle BCD, and CE its Secant: Whence Mathematicians, for the sake of Trigonometrical Calculations, have divided the whole Sine or Radius of the Circle into 1000, 10000, 100000, 1000 000, 10000 0000, etc. parts, thence to make a proportionable Estimate of the number of Parts in the Sine, Tangent, or Secant of any Arch, etc. as may be seen in the Tables of Sines, Tangents, and Secants. From these Suppositions and Explications of the Terms, we shall now infer from this Definition the following COROLLARIES. I IN equal Circles (and so much more (a) Among the rest Vid. Eucl. 28 & 29 lib. 3, and also partly the 26 and 27. in one and the same) as the Radii or Semidiameters BC and bc are equal, so also it is evident, that the Right Sines DF & Df, of equal Arches BD and bd, or equal Angles BCD and bcd, also the Tangents BE and be, and Secants CE and ce, and Subtenses or Chords DG and dg, also the Sagittae or intercepted Axes BF and bf, of double the Arches DBG and dbg, etc. will be equal, and so consist of an equal number of Parts of the whole Sine or Radius, etc. which both is evident from what we have said before, and may be further evinced, if one Circle be conceived to be put on the other, and the Radius BC on the Radius bc, that so they may coincide, by reason of the equality of the Arches BD and bd; and so of all the rest. And è contra, II In unequal Circles, the Sins, Tangents, etc. of equal Angles BCD or bcd (Fig. 14.) or similar Arches, or Arches of an equal number of Degrees, BD and bd, will be also similar or like, etc. i e. the Sine df consists of as many parts of its Radius bC, as the Sine DF does of its Radius BC, etc. e. g. if the Radius BC be double of the Radius bc, each thousandth part of the one, will be double of each thousandth Pag. 12. 14 15 16 17 18 19 20 21 22 23 24 part of the other, but they are alike 1000 in each; because the degrees in the Circumference of the little one, particularly in the Arch bd are but half as big as those in the Arch BD, and yet equal in number in both. Thus also if the Sine DF contains 700 of the 1000 parts of its Radius BC, df will also contain 700 of the 1000 parts of its smaller Radius bc, and in like manner the Chords DG and dg, and the Tangents BE and be, etc. contain a like number of parts, each of its own Radius. SCHOLION. IT may not be amiss here to note by the by (although it may seem more proper to be taught after the Doctrine of Proportions) that if, u g. the degrees of a greater Circle be each of them respectively double, or triple, or quadruple, etc. of the degrees of a lesle Circle, according as the Radius of the one is double or triple to the Radius of the other, than, at lest as far as Mechanical Practice can require, you may found the Arch of a greater Circle equal to the whole Periphery of a lesle, viz. if you take reciprocally that part of the greater Periphery, which shall be as the Radius of the lesle to the Radius of the greater, or as one degree of the lesle Periphery to one degree of the greater. e. g. if the lesle Radius bc be half the greater BC, and so also the Periphery, and each of the degrees of the one, be one half of the Periphery, and of each of the degrees of the other, one half of the greater Periphery will reciprocally be equal to the whole lesle Periphery, or 180 degrees of the one to 360 of the other, etc. 2. The same (at lest in this case where the Radius cb is double of the Radius CB) may be done also Geometrically by the same reason. Having described Circles on each Radius, suppose the Radius CB (Fig. 15.) so to move with an equable motion about its Centre c, as to take or move the Radius of the greater Circle cb along with it, and coming, u g. to I stops that also at 1, and going forward, to II stops that again at 2, etc. Hence it will be manifest to any attentive Reader, that when the lesle Radius CB shall have described the Semicircumference B. II III the greater Radius cb having moved to 3, will have described precisely a quarter of its circumference; and if still the lesle Radius C. IU. moves on to the right hand, and continueth to carry the greater c 4 along with it onwards the same way, it will necessarily follow, that in the same moment as the Radius C. IU. (together with c 4.) shall come to its first situation in B, having described a whole Circle, the opposite Radius c 4 will be come to 5, and have described half its Circle, having moved all along with an equal Motion. Hence it is evident, that the whole lest Circle answers exactly to half the greater, and half of the first to a quarter of the last; as also the Quadrant B. II to the Octant (or 8th. part) b 2, etc. whence any Arch being given, as B. I in the lest Circle, if you draw through I the Radius of the greater Circle c 1 you'll cut of an Arch b 1. equal to the given Arch in magnitude, but only half in the number of degrees. 3. Hence follows naturally that celebrated Proposition of Euclid, that the Angle at the Centre BC. I or BC. II is double of the corresponding Angle at the Periphery bc. 1. or bc. 2, etc. which in this case is manifest, and in the other 2 (Fig. 16.) of the wholes or remainders DCD and DPD it is also (a) Eucl. p. 20. l. 3. certain; which is true also of the parts BCD and BPD to be added or subtracted by the first Case. (b) Eucl. p. 9 l. 1. 4. Hence we have a new way of bisecting any given Angle CDE, or Arch CE (Fig. 17.) viz. if you make C B equal to the Leg DC, and from this, as Radius, describe an Arch BF equal to the Arch CE, and draw DF. And with the same facility we might obtain the Trisection, if the greater Radius being triple to the lesle, was thus carried along by an equable Motion, as we have shown how to do already in a double Radius; and this at first sight may seem very probable. But whether the triple Radius be immediately carried round by the simple Radius C B, or by means of the double Radius cb, neither the one nor the other will 'cause an equable Motion. For in the latter Case, while the Radius cb describes the quadrant Bb, the Radius de will not describe so much as a Quadrant; but while cb with the same velocity describes the other Quadrant bf, the Radius de will come to g, describing an Arch as much greater than a Quadrant as the former was lesle. In the former case on the contrary, the Radius C B moved on to D beyond a Quadrant, while the was carried from B to e, but if the Radius CD moving on, should again carry the along with it, the one would describe the same Arch ebb, while the other would describe one lesle than before. 5. Hence the Angle at the Centre ACE (FLETCHER g. 19) upon the Arch A, is equal to the Angle ADB at the Periphery, upon double that Arch AB. 6. Hence the Angle ADB in the Semicircle (Num. 1.) (a) Eucl. 31. lib. 2. is a Right Angle, in a Segment lesle than a Semicircle (Num. 2.) is an obtuse Angle, and in a greater (Num. 3) an Acute one, because the Angle at the Centre ACE upon the half Arch, is equal to the Angle ADB pr. praeced. 5. and is a Right Angle in the first Case, Obtuse in the second, and Acute in the third. 7. Hence Angles in the same Segment, or (b) Eucl. 26 and 27. lib. 3. on equal Segments of equal Circles, or on the same or equal Arches, are all equal and è contra. DEFINITION XI. WHen 2 or more Lines AB and CD are so continued as to keep always the same distance from one another (whose Genesis may be conceived to proceed from the uniform Motion of 2 Points A and C, always keeping the same distance from each other) they are called Parallels: But as it evidently follows from this Definition, that (a) Eucl. lib. 1. 30. those Lines which are Parallel to one third, are parallel to one another (since adding or subtracting equal Intervals to or from other equal ones, the sums or remainders must needs be equal;) so if the Parallels are Right Lines and cut transversly (or slopingly across) by another Right Line OF, you'll have these COROLLARIES. I THE Angles (b) Eucl. lib. 1. 29. which we call Alternate ones, GHK and HGI (Fig. 21.) are equal by Corollary I Definition X. since the distances GK and HI, which are the Right Sins of the said Angles, are supposed equal. II The External Angle EGA is also equal to the Internal opposite Angle GHK, by Consect. 1. definite. 9 because that External Angle EGA is equal to the alternate Vertical one HGI. III The same Internal Angle GHK, with the other internal opposite one on the same side AGH (as well as the External one EGA, by Coral. 1. definite. 8.) are equal to two Right ones. IV. On the contrary, If any Right Line OF (a) lib. 1. Prop. 27 & 28 cutting 2 others AB and CD transversly, makes the alternate Angel's GHK and HGI equal, their Right Sins, by Consect. 1. definite. 10. will be equal, and consequently the Lines AB and CD parallel: and the same will follow, if the External Angle be supposed equal to the Internal, or the 2 Internal ones on the same side equal to 2 Right ones; since from either Hypothesis the former will immediately follow. V From whence it appears more than one way (b) That the 3 Internal Angles of any Triangle (e. g. H, G, KING, which will serve for all) taken together, (a) lib. 1. Prop. 32, are equal to two Right ones, and the External one GHD is equal to the two Internal opposite ones. For we might either conclude with Euclid, that 1, 2, 3, together make 2 Right ones, by Consect. 1. definite. 8. but 2 = TWO and 3 = III pr. 1 and 2 of this, therefore I, TWO, III = 2 Right ones; or with others, 1, TWO, 4 are = 2 R. but 1 = I and 4 = III pr. 1st of this. Therefore, etc. or more briefly with F. Pardies, 1 = I pr. 1st of this, but 1, TWO, III, together = to 2 Right ones, by the 3d of this; therefore I, TWO, III = 2 R. Q. E D. DEFINITION XII. IF a Right Line AB (Fig. 22.) be conceived to move from the top of a plain Angle GOD with a motion always parallel to its self, so that at one end A it shall always touch the Leg AC, and all along cut the Leg AD, while at length being come to FLETCHER, it shall only touch that Leg with its other end B, and so fall at length wholly within the Angle GOD: It will describe by this motion within the Legs GOD the Triangular Figure EAF, and without them the Triangular Figure BAF; its parts within them a f continually increasing, and the other without fb continually decreasing; but with all its Parts, or the whole Line, it will describe the Quadrangular Figure AEFB: Consequently if the other Leg AD of the given Angle GOD (Fig. 23.) or any part of it AB, be moved along the other Leg remaining parallel to itself, it will also describe a Quadrilateral Figure, which will be also equilateral, if the Line describing it AB, be equal to the Line A according to which it is directed; but if either of the Lines, as AD be greater than the other, the opposite Sides will be only equal; for the describent or describing Line is always necessarily equal to its self, and the Points A, B, D, moved with an equal Motion, describe also in the same time equal Lines A, BF, DG. From these Geneses of Quadrangles and Triangles we have the following CONSECTARIES. I THese Quadrilateral Figures are also Parallellograms, i e. they have their opposite Sides Parallel; (a) Scholar Prop. 34. lib. I because the Line that describes them is supposed to remain always parallel to its self, and the Points A and D, or A and B, to be always equidistant. II Because the 2 Internal opposite Angles (b) The first part of the same Proposit. A and E, and also E and F, etc. are equal to 2 Right ones, by Consect. 3. definite. 11. if one Angle u g. that at A be a Right one, all the others must necessarily be so too [in which case the quadrilateral and equilateral Figure OF is called a Square, and the other AG an Oblong, or Rectangle:] if there be no Right Angle, the opposite Angles transversly or cross-ways, a and f, or a and g are equal, because both the one and the other, with the third (e) make 2 Right ones [in which case the quadrilateral Equilateral of is called a Rhombus, but the other ag a Rhomboid. III The Transversal (or Diagonal) Line (c) the latter part of the same Prop. in any Parallelogram, divides it into two equal Triangles AEF and FAB.; for all the Lines and Angles on each side are equal, and as the describent (Line) AB moved through the Angle EAF upon the Line A described the Triangle AEF; so the Line OF, equal to the former, moved after the same way, through the Angle AFB also equal to the former Angle, upon the equal Line FB, must necessarily describe an equal Triangle; or, in short, all the Indivisibles of, or their whole increasing Series, are necessarily equal to the like number of Indivisibles fb, increasing reciprocally after the same way. IV. All Parallellograms that are between the same Parallels AB and CF (Fig. 24.) i e. having the same Altitude (a) lib. 1. Prop. 35 & 36 and the same or equal Bases, as CD or CD and cd, are equal among themselves; for they may be conceived to be described by the equal Lines AB and ab equally moved through the same or equal Intervals of the Parallel Lines; so that all or each of the Indivisibles or Elements AB will necessarily be equal to all and each of the Indivisibles ab; for they all along answer one to the other both in number and magnitude. SCHOLIUM. HEre you have a Specimen of the Method of Indivisibles, introduced first by Bonaventura Cavallerius, and since much facilitated; and although these Indivisibles placed one by another, or as it were laid upon an heap, cannot compose any Magnitude, yet by an imaginary Motion they may measure it, and as it were, after a negative way, demonstrate the Equality of two Magnitudes compared together, viz. if we conceive a certain number of such Elements in any given Magnitude, and thence conclude that in another consisting of the like Elements, ordered or ranked after the same way, there can be neither more nor lesle in number than in the first; thence follows their Equality, etc. V Hence therefore it is also Evident, that Triangles upon the same and equal Bases as CD and cd, and placed between the same Parallels, are necessarily equal, because they are the half of equal Parallellograms AD and ad, by the 3d Consectary of this Definition. Pag. 18. 25 26 27 28 29 30 31 32 VII. Because it is manifest in Rectangular Parallellograms, if the Altitude AB, and Base BC (Fig. 26.) measured and divided by the same common Measure, be conceived to be (multiplied or) drawn one cross the other, that the * This is manifest from its Genesis; for the five parts of the Line AB by its motion along the part of the Base BE, describes 5 little Squares, and moving along the following part, describes five more, etc. Area AC, thereby described, will be divided into as many little square Measures or Area's, as the number of their Sides multiplied together would produce Units; therefore the Area of any other Parallelogram will be after the like manner produced, if the Base be multiplied by the Perpendicular height, equally as if it were a Rectangle of the same Base and Altitude. VIII. Consequently also you may have the Area of any Triangle, by Consectary 3 and 5. if the Base be multiplied by half the Perpendicular height; or, the whole Base being multiplied by the height, if you take the half of the product. DEFINITION XIII. BUT as there are various Species of Triangles, while first with relation to their Sides, one is called Equilateral, as ABC (Fig. 27.) because all its Sides are equal; another Equicrural or Isosceles, as DEF, because it has two equal Sides DE and OF, while its Base DF may be either longer or shorter; and a third is called Scalenum, as GHI, because it has all its Sides unequal; than again in respect to their Angles, one is called Rectangled, as a, b, c, because it has one Right Angle at a; Another Obtusangled, as d, e, f, because it has one obtuse Angle at d; a third is called Acuteangled, as g, h, i, because all its three Angles are Acute: So each of these kinds has its peculiar properties, which we shall partly hereafter demonstrate in their proper places, and partly deduce here as CONSECTARYS. I ALL Equilateral Triangles are also Equiangular, and consequently Acutangled; for, having found a Centre for three Points, and the Periphery A, B, C, (see Fig. 28.) by Consect. 6. definite. 8. the three Arches AB, BC, and AC, answering to equal Chords; and consequently the three Angles at the Centre OH are equal, by Consect. 1. of definite. 10. and therefore the three Angles at the Periphery also, as being half of the other, by the 3d Consectary of the same Definition. Each Angle therefore is one third part of two Right ones, by Consect. 6. definite. 12. two thirds of one Right Angle, i e. 60 degrees, and consequently Acute. II It follows also by the same Reason in an Isosceles Triangle, that the Angles at the Base opposed to equal Sides are equal, and (a) lib. I Prop. 5. the same otherwise demonstrated. consequently Acute; for having circumscribed a Circle about it, equal Arches will correspond to the equal Chords DE and OF, and equal Angles at the Centre DO and FOE will correspond to them, and equal ones at the Periphery DFE, and FDE to these again. And it is evident that each of these are lesle than a Right Angle h. e. an Acute one, because all three are equal to two Right ones. Wherhfore if the third is a Right Angle, the other two at the Base will necessarily be half Right ones. SCHOLIUM. WE will here (a) show by way of Anticipation, the truth of the Pythagorick Theorem, esteemed worth an Hecatomb: Which hereafter we will demonstrate after other different ways; viz. In a Right Angled Triangle BAC (Fig. 29.) the Square of the greatest Side opposite to the Right Angle, is equal to the Squares of the other two Sides taken together. For having described the Squares of the other two Sides, AC dE, DE ab (taking ED = AB) and the Square of the greatest BC cb, it will be evident, that the parts X and Z are common to each, and that the two other Triangles in the greatest Square BAC and BDb, are equal to the two Triangles bac and Cdc which remain in the two Squares of the lesser Sides; and so the whole truth of the Proposition will be evident, while these two things are undoubtedly true: 1. That the Side of the greatest Square Bb will necessarily concur with the Extremity of the lesle Db, and the other Side of the greatest Square Cc with its Extremity c, will precisely touch the Continuation of the Sides of the two lest Squares day; as you'll see them both expressed in the Figure. 2. The said two Triangles are every way equal; for the Angles at C with the intermediate one at Z, make two Right ones, therefore they are equal; but the Side CA is equal to the Side cd, and CB to Cc, and the Angles at A and d Right ones. Wherhfore if we conceive the Triangle ABC to to be turned about C, as a Centre to the right hand, it will exactly agreed with the Triangle Cdc, and the Point B will necessarily fall on the continued Line d E, as agreeing with the Line AB. Hence it is now evident, that Ca = BD, and because ba is also = bD, and the Angles at a and D Right ones. Where, if we conceive the Triangle bac to be moved about b as a Centre, until ba coincides with bD, and ac with DB, bc will also necessarily coincide with bB Q. E. D. To this Demonstration of Van Schooten's, which we have thus illustrated and abbreviated, we will add another of our own, more like Euclids, but somewhat easier, which is this: Having drawn the Lines (as the other Figure 29 directs) the ▵ ACD being on the same Base AC with the Square AI, and between the same Parallels, is necessarily one half of it, but it is also half of the Parallelogram CF being on the same Base with it, viz. DC; therefore this Parallelogram = ▭ AI. In like manner ▵ ABE is half the ▭ ALL, and also half the Parallelogram BF, therefore BF = ▭ ALL: therefore CF + BF that is the ▭ of BD = to the two ▭ ▭ AI + AL. Q. E. D. For because the Side BE occurs to, or meets the Side LK, and the Side CD the Side IH continued, it yet more apparently follows; because the Angles a and b, and also c and d, are manifestly equal, as making both ways, with the Intermediate x or z, Right Angles. Therefore the ▵ BAC being turned on the Centre B and laid on BLE will exactly agreed with it, and turned on the Centre C and laid on CID, will agreed with that also, etc. DEFINITION FOURTEEN. ALL Rectilinear or Right Lined Figures that have more than three or four Sides (to the latter sort of which, there remains to be added another Species besides Parallellograms, called Trapeziums, whose Angles and Sides are unequal, as KING, L, M, N, Fig. 30.) are called by one common Name Poligons, or Many-sided and Many-Angled Figures, and particularly according to the Number of their Sides and Angles, Pentagons, Hexagons, Heptagons, etc. All whereof, as also Trapezia, being resolvible into Triangles by Diagonal Lines, (as may be seen in the 31 and foregoing Fig.) you have these CONSECTARIES. I YOU have the Area of any Polygon by resolving it into Triangles, and than adding the Area's of each Triangle found by Consect. of Definition 12 into one Sum. II The Area of the Trapezium KLMN (in the first of the Fig. 30) whose two opposite Sides, at lest KL and MN, are Parallel, may be had more compendiously, if the Sum of the Sides be multiplied by half the common height KO. SCHOLIUM. HEnce we have the foundation of Epipedometry or Masuring of Figures that stand on the same Base, and Ichnography; in the Practice whereof this deserves to be taken special Notice of, that to work so much the more Compendiously, you aught to divide your Figure into Triangles, so that (Fig. 31.) 2 of their Perpendiculars may (as conveniently can be) fall on one and the same Base. For thus you'll have but one Base to measure, and 2 Perpendiculars to found the Area of both: But for Ichnography, the distance of the Perpendiculars from the nearest end of the Base must be taken; which we shall supersede in this Place and Discourse more largely on hereafter. 2. This resolution of a Polygon into Triangles may be performed by assuming a point any where about the middle, and making the sides of the Polygon the Bases of so many Triangles; (see the 2d Figure marked 31) wherein it is evident; 1 That all the Angles of any Polygon are equal to twice so many right ones, excepting 4, as the Polygon has sides; for it will be resolved into as many Triangles as it has sides, and each of these has its Angel's equal to 2 right ones. Subtracting therefore all the Angles about the Point M (which always make 4 right ones by Cons. 2. Def. 8.) there remain the rest which make the Angles of the Polygon. 2 All the external Angles of any right lined Figure (e, e, e, etc.) are always equal to 4 right ones; for any one of them with its Contiguous internal Angle is equal to 2 right ones pr. Consect. 1 of the said Def. and so altogether equal to twice so many right ones as there are Sides or internal Angles of the Figure. But all the internal One's make also twice so many right One's, excepting 4 therefore the external One's make those 4. DEFINITIOF XU. AMong all these plain Figures those are called Regular whose Angles and Sides are all equal, as among trilateral Figures the Equilateral Triangle, among Quadrilateral ones, the Square, and in other kinds, several Species which are not particularised by Names; but all others in whose Angles or Sides there is any inequality, are called Irregular: Thou some of these also, and all the other may be inscribed in a Circle. Whence you have these CONSECTARIES. I THE Areas of the Regular Figures may be obtained yet easier, if having found their Centre (by Consect. 6. definite. 8.) you draw from thence the Right Lines CB, CA, etc. (Fig. 32.) till there be formed as many Triangles ACB, ACF, etc. as the Figure has Sides; for since all these Triangles have their Bases AB, BF, as so many Chords, and their Altitudes CD, CG, as so many parts of intercepted Axes DE and GH, and also equal pr. Consect. 1. definite. 10. and so by Consect. 5. definite. 10. are equal among themselves; one of their Area's being found and multiplied by the number of Sides, or half the Altitude by the Sum of all the Sides, you'll have the Area of the whole Polygon: For it is manifest from what we have already said, and very elegantly Demonstrated by F. Pardies, That any Regular Polygon inscribed in or circumscribed to a Circle, is equal to the Triangle Aza, one Leg whereof is equal to the Perpendicular height let fall from the Centre upon any Side, and the other to the whole Periphery of the Polygon. Now if the Triangles into which the Polygon is resolved, do all stand on the same Right Line Aa, (Fig. 32) and are all equal and of the same height, to which the Perpendicular AZ is equal, it will necessarily follow, that each pair of Triangles ABZ and ABC, BZF and BCF, etc. are equal among themselves, pr. Consect. 5. definite. 12. and consequently the Sum of all the former will be equal to the Sum of all the latter, that is, the Triangle Aza to the Polygon given. II Since Regular Figures inscribed in a Circle, by bisecting their Arches AB, BF, etc. may be easily conceived to be changed into others of double the number of Sides, (as a Pentagon into a Decagon, etc.) and that ad Infinitum; a Circle may be justly esteemed a Polygon of infinite Sides, or consisting of an infinite Number of equal Triangles, whose common Altitude is the Semidiameter of the Circle: So that the Area of any Circle is equal to a Right Angled Triangle (as AZa) one of whose Sides AZ is equal to its (a) Archimedes of the Dimension of the Circle, Prop. 1. Semidiameter, and the other Aa to its whole Circumference. SCHOLIUM. IT may not be amiss to note these few things here, concerning the Inscription of Regular Figures in a Circle. I Having described a Circle on any Semidiameter AC, (a) Euclid. Prop. 15. lib. 4. (Fig. 33. N. 1.) that Semidiameter being placed in the Circumference, will precisely cut of one sixth part of it, and so become the Side of a Regular Hexagon: and so the Triangle ABC will be an Equilateral one, and consequently the Angle ACB and the Arch AB 60 Degrees, by Cons. 1. definite. 13. II Hence a Right Line AD, omitting one point of the division B, and drawn (b) Eucl. 2d Coral. of the same. to the next D, giveth you the Side of a Regular Triangle inscribed in the Circle, and subtends twice 60, i e. 120 Degrees. Pag. 24. 33 34 35 36 37 38 39 40 III If the Diameters of the Circle AD and DE (N. 2.) cut one another at Right Angles in the Centre C, the Right Lines AB, BD, etc. will be the Sides of an inscribed Square ABDE: For (c) Eucl. 6. lib. 4. the Sides AB, BD, etc. are the equal Chords of Quadrants, or Quadrantal Arches, and the Angles ABDELLA, BDE, etc. will be all Right ones, as being Angles in a Semicircle (per Scholar 6. definite. 10.) composed each of two half Right ones, by Consect. 2. definite. 13. IV. Euclid very ingeniously shows us how to Inscribe a Regular Pentagon also, Lib. 4. Prop. 10 & 11. and also a Quindecagon (or Polygon of 15 Sides) Prop. 16. But though the first is too far fetched to be shown here, yet (supposing that) the second will easily and briefly follow: In a given Circle from the same point A (N. 3.) inscribe a Regular Pentagon AEFGHA, and also a Regular Triangle ABC; than will BF be the Side of the Quindecagon, or 15 Sided Figure. For the two Arches A and OF make together 144 Degrees, and AB 120: (a) Therefore the difference BF will be 24, which is the 15th. part of the Circumference. V The Invention of Renaldinus would be very happy, if it could be rightly Demonstrated; (as he supposes it to be in his Book of the Circle) which gives an Universal Rule of dividing the Periphery of the Circle into any number of equal Parts required, in his 2d Book De Resol. & Comp. Mathem. p. 367. which in short is this: Upon the Diameter of a given Circle AB Fig. 34.) make an Equilateral Triangle ABDELLA, and having divided the Diameter AB into as many equal Parts, as you design there shall be Sides of the Polygon to be Inscribed, and omitting two, e. g. from B to A, draw through the beginning of the third from D, a Right Line, to the opposite Concave Circumference, and thence another Right Line to the end of the Diameter B, which the two parts you omitted shall touch thus, e. g. for the Triangle, having divided AB into three equal parts, if omitting the two B2, through this beginning of the 3d you draw the Right Line DIII, and thence the Right Line III B, which will be the Side of the Triangle; and so IU. B will be the Side of the Square, VB the Side of the Pentagon, etc. N. B. The Demonstration of these (Renaldinus adds, p. 368.) we have several ways prosecuted in our Treatise of the Circle: Some of the most noted Ancient Geometricians, have spent a great deal of pains in the Investigation and Effection of this Problem, and several of the Moderns have lost both time and pains therein: Whence, we hope, without the imputation of Vain Glory, we may have somewhat obliged Posterity in this point. DEFINITION VI. IF the Plane of any Parallelogram AC (Fig. 25.) be conceived to move along a Right Line A, or another Plane OF downwards, remaining always Parallel to its self; there will be generated after this way a Solid having six opposite Planes Parallel, two whereof, at lest, will be equal to one another, whence it is called a Parallelepiped; and particularly a Cube or Hexaedrum, if the Parallelogram ABCD that describes it be a Square, and the Line along which it is moved, A, equal to the Side of that Square, and Perpendicular to the describing Plane, and consequently all the six Parallel Planes comprehending this Solid, equal to one another. But if the describing or Plane Describent (Fig. 36.) be a Triangle or Polygon, the Solid is called a Prism, if a Circle, it is called a Cylinder. Now from the Genesis of these Solids you have the following CONSECTARIES. I IF the Planes or Parallelograms Describent (a) Eucl. l. 11. p. 29, 30, 31 ABCD and abcd (Fig. 37.) are equal, and their Lines of Motion A and ae also equal; the Solids thereby described, viz. Parallelepipeds, Cylinders, and Prisms, (which will therefore have their Bases and heigths equal) will be equal among themselves; because the describent Indivisibles of the one, will exactly answer, both in number and position, to those of the other, as we have already shown in Parallellograms; Consequently therefore, II Any Parallelepiped (b) Eucl. l. 11. p. 28. may be divided by a Diagonal Plane BDHF (or a Plane passing through its Diagonals) into two equal Prisms; for by Consect. 3. definite. 12. the Triangles ABDELLA and BCD, are equal, and are supposed to be moved by an equal Motion through equal spaces. III And since it is evident, even by this Genesis of them, that in Right Angled Cubes and Parallelepipeds, if the Base ABCD (Fig. 38.) being divided into little square Area's, be multiplied by the height A, divided by a like measure for length, after this way you may conceive as many equal little Cubes to be generated in the whole Solid, as is the number of the little Area's of the Base multiplied by the number of Divisions of the side A; you may moreover obtain the Solidity of any other Parallelepipeds, that are not Right Angled ones, by multiplying their Bases and Perpendicular Heigths together. IV. Moreover since every Triangular Prism is the half of a Parallelepiped, and any Multangular Prism may be resolved into as many Triangular ones, as its Base contains Triangles; you may obtain the Solidity (or Solid Contents) either of the one or the other, if you multiply the Triangular, or Multangular Base of them into their Perpendicular Height. V After the same manner you may likewise have the Solidity of a Cylinder, which may be considered as an Infinite Angled Prism, just as the Circle is as an Infinit-Angled Polygon. DEFINITION XVII. IF any Triangle ABC (Fig. 39 N. 1.) be conceived to move with one of its Plane Angles C, from the Vertex or top of a Solid Angle (determined by two Planes aAb and cAa joined together in the common Line Aa) with a motion always parallel to itself; so that its extreme Angular Point A shall always remain in the Line Aa, but with its Sides AB and AC shall all along raze on the two Angular Planes, till at length it falls wholly within the Solid Angle: by this its motion it will describe within the Solid Angle, the Figure we call Pyramidal, whose Base will be the Triangle abc, and its Vertex A will also describe without it another Quadrangular Pyramid, whose common Vertex will be the same A, but the Base the Quadrangle Cb, described by the Side of the movable Triangle BC: The first Pyramid it will describe with its Triangular Parts, 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 continually increasing from the point A, and ending in the Triangle abc; but the latter Pyramid will be described by the remaining parts, continually decreasing downwards from the whole ABC, and the Quadrangular Trapeiza 〈◊〉 〈◊〉 〈◊〉 〈◊〉 〈◊〉 at length ending in the Right Line bc: So that in the mean while the said Triangle with its whole space describes, according to what we have said before, the Triangular Prism composed of those two Pyramids. From this Geeesis of Pyramids you'll have the following CONSECTARIES. I OF what sort soever the Describing Triangles are ABC, ABC (N. 2.) so they are equal; and whatever the Solid Angles are, comprehended under the Planes abA, and acA, abA and acA, so they are accommodated to the Plane Angles A and A, and such that, etc. DEFINITION XVIII. THere may be exhibited another easier Genesis of Cones and Pyramids, but it respects only the Dimension of the Surface, and not of the Solidity of them, viz. If you have a fixed point A that is not in the Angular Plane BCDEF (Fig. 41.) and a Rght Line OF let fall from that point to any Angle of the Plane, be conceived to move round the sides BC, CD, etc. This Plane by its motion will describe as many Triangles ABC, GOD, etc. as the Angular Plane has Sides. And these Triangles all meeting at the point A, make that Solid which we call a Pyramid. Now if instead of the Angular Plane there be supposed a Circular one, (or an Angular one of Infinite Sides) the Solid thence produced is called a Cone, whose Surface is equal to Infinite Triangles, constituted on the Base BCDE, and whose Solidity would consequently equal an Infinite Angled Pyramid of the same height. And after the same manner by the motion of the Line OF, remaining always Parallel to itself about Parallellograms or Triangular Planes, will be generated Parallelipipeds, Prisms, and Cylinders. But as one Pyramid will be produced more upright than another, according as the point A stands more over the middle of the Plane BCD, etc. (Fig. 42.) or Pag. 28. 41 42 43 44 45 46 47 Schemes for rai-sing the five Regular Bodies' respects it more obliquely: So in particular a Cone is called a right Cone, when the line AO being let fall to the Centre of the circular Plane (otherwise called the Axis of the Cone) constitutes on all sides right Angles with it; but it is called an Obliqne or Scalene Cone, when the Axe stands obliquely on the Base. Which distinction may also be easily understood when applied to Cylinders, though a right Cone and Cylinder may also be conceived to be generated after another way, as if for the one a Triangle and for the other a right angled Parallelogram AOB and LAOB be be conceived to be moved round a line AO considered as immovable (whence it is called an Axe;) and also a truncated Cone may be form if a right angled Trapezium, (a) Archimedes lib. I de Con. and Cylin. prop. 7. and 8. 2 of whose Sides are parallel be moved, etc. And as we have deduced the Solidity of these Bodies from the foregoing Genesis, so their external Surfaces, as also of Prisms and Parallelepipeds may easily be found from the present Genesis, by any one who attentively considers the following COROLLARIES. I SInce the whole external Surface, except the Base, of any Pyramid is nothing but a System of as many Triangles ABC, GOD etc. as the Pyramid has Sides; if the Area's of those Triangles separately found by Consect. 8. Def. 12. be added into one Sum, you'll have the Superficial Area of the whole Pyramid. II If a Pyramid be cut with a Plane b, c, d, e, parallel to its Base BCDE (Fig. 41.) The Surface of that truncated Pyramid comprehended between the prallel Planes may be obtained if having found the Surface of the Pyramid A bcde cut of from the rest by Consect. 1. you subtract it from the Surface of the whole Pyramid. II The external Surface of a right Pyramid that stands on a regular Polygon Base is equal to a Triangle, whose Altitude is equal to the Altitude of one of the Triangles which compose it, and its Base to the whole Circumference of the Base of the Pyramid. IV. Therefore the Surface of a right Cone, by what we have already said, is equal to a Triangle whose height is the side of the Cone, and the Base equal to the Circumference of the Base of the Cone. V The Surface of a truncated right Cone, or Pyramid is equal to a Trapezium which has 2 parallel Sides the lowest of which is equal to the Perifery of the Base, and the other to the Perifery of the Top or upper Part, and the height, to the intercepted Part. VI The Surface of a right Cylinder or Prism is equal to a Parallelogram which has the same height with them, and for its Base a right line equal to the Perifery of that Cylinder or Prism. DEFINITION XIX. IF a semicircular Plane ADB (Fig. 43. N. 1.) be conceived to move round its Diameter AB which is fixed, as an Axis, by this Motion it will describe a Sphere, and with its Semicircumference the Surface of that Sphere; every part whereof is equally distant from the middle Point of that Axis C (which is therefore called the Centre of that Sphere.) Now if (N. 2.) this semicircular Ambitus (a) Archimedes lib. I de Cono & Cyl. Prop. 22. Coral. & Prop. 23. be conceived to be divided before that revolution, first into 2 Quadrants AD, and BD, and than each of those again into as many parts equal in Number and Magnitude as you please, and having drawn the Chords OF, FE, ED, etc. let the Polygon AFEDGHB Inscribed in the Semicircle be conceived together with it to be turned about the Axis AB; than will A 1 FLETCHER, and B 4 FLETCHER describe 2 Cones about the Diameters F f, and H h; and the Trapezia about the Axes 1, 2, 2 C, C 3, and 34, will describe so many truncated Cones, and the lines OF, FE, etc. so many Conical Superficies, by the Antecedent Def. and so the whole Polygonal Plane AFEDGHB a Conical Body inscribed in the Sphere, and contained under only Conical Surfaces. And as any attentive Person may easily perceive such a Body to be lesle than the ambient Sphere, and its whole Surface lesle than the Surface o● the ambient Sphere; so he may as easily trace these following CONSECTAYS. I IF the Arches OF, FE, etc. be further bisected, and a Polygonal Figure of double the number of Sides inscribed in the Semicircle, and conceived to be moved round after the way we have already shown, the Pseudoconical Body hence arising, will approach nearer and nearer to the Solidity of the Sphere, and the Surface of the one to the Surface of the other, and hence (if we continued this Bisection, or conceive it to be continued ad Infinitum) you may infer. II That a Sphere may be looked upon as much a Pseudoconical Body, consisting of infinite Sides, and its Surface will be equal to the infinite Conical Surfaces of that Body; which we will take further notice of below. DEFINITION XX. IF the Diameter AB of the Semicircular Plane ADB (Fig. 43. N. 3.) be conceived to be divided into equal Parts (as here the Semidiameter AC into 3.) and if the circumscribing Parallelograms CE, 2 E, 1 G on the transverse Parallels CD, 2 e, 1 f be conceived together with the Semicircle itself to revolve about the fixed Axe AB; it is evident that there will be form from the Semicircle a Sphere as before, and from the Circumscribed Parallellograms, so many circumscribed Cylinders of equal height: but if all the Altitudes or heights of these are bisected or divided into two, and so make the number of circumscribing Parallellograms double, there will be form (by moving them round as before) double the Number of Cylinders of half the height, but which yet being taken together, approach much nearer the solidity and roundness of the Sphere, than the former, which were fewer in Number (viz. the six latter Parallellograms approach nearer to the Plane of the Circle than the three former) and thus if that bisection of the Altitudes be conceived to be continued ad infinitum, the innumerable Number of those infinitely little Cylinders will coincide with the Sphere itself. Moreover if you conceive any Polyedrous or Multilateral Figure to be circumscribed about the Sphere (which we here endeavour to delineate by the Polygon ABCD N. 4. circumscribed about the Circle) and the solid Angles thereof to be cut by other Planes ab, which shall touch the Sphere; it is manifest there will thence arise another Polyedrous Figure, the Solidity whereof will approach nearer to the Solidity of the Sphere, and its Surface to the Spherical Surface than the former, and if the Angles of this be again in like manner cut of, there will still arise another new Solid, and new Surface approaching yet nearer to the Solidity and Surface of the Sphere than the former, etc. and so after an infinite Process they will coincide with the Sphere and its Surface themselves. Whence flow these COROLLARIES. I THE Sphere may be considered as a Polyedrous Figure, or as consisting of innumerable Bases, i e. composed of an innumerable Number of Pyramids, all whose Vertex's meet in the Centre, and so whose common height is the Semidiameter of the Sphere, and the sum of all the Bases equal to the Superficies of the Sphere. II If you can found a Proportion between a Cylinder of the same height with any Sphere, and whose Base is equal to the greatest Circle of that Sphere, and innumerable Cylinders circumscribed about it, as we have just now shown; than you may also obtain the Proportion between the said circumscribed Cylinder and the inscribed Sphere: Which to have here hinted may be of service hereafter in its proper place. DEFINITION XXI. THere remain those Bodies to be considered which are called Regular, which correspond to the Regular Plane Figures; and as those consist of equal Lines and Angles, so these likewise are comprehended under Regular and Equal Planes meeting in equal solid Angles; and as those may be Inscribed and Circumscribed about a Circle, so may the latter likewise in and about a Sphere. But whereas there are infinite Species of Regular Plane Figures, there are only five of Regular Solids; the first whereof is contained under four Equal and Equilateral Triangles, whence it is named a Tetraedrum; the second is terminated by six equal Squares, and thence is called Hexaedrum, and otherwise a Cube; the third being comprehended under eight Equal and Equilateral Triangles, is called an Octaëdrum; the fourth is contained under twelve Regular and Equal Pentagons, and so is named a Dodecaëdrum; the fifth, lastly, is contained under twenty Regular and Equal Triangles, and is thence nominated an Icosaëdrum. Besides these five sorts of Regular Bodies there can be no other; for from the concourse of three Equilateral Triangles arises the Solid Angle of a Tetraëdrum, from four the Solid Angle of an Octaëdrum, from five the Solid Angle of an Icosaëdrum; from the concourse of four Squares you have the Solid Angle of an Hexaëdrum; from that of three Pentagons you have the Solid Angle of a Dodecaëdrum; and in all this Collection of Plane Angles, the Sum does not arise so high as to four Right ones. But four Squares, or three Hexagons meeting in one Point, make precisely four Right Angles, and so by Consect. 2. definite 8. would constitute a Plane Surface, and not a Solid Angle. Much lesle therefore could three Heptagons or Octagons, or four Pentagons meet in a Solid Angle, to form a new Regular Body; for those added together would be greater than four Right Angles. But now, for the Measures of these five Regular Bodies, take the three following CONSECTARIES. I SInce a Tetraëdrum is nothing else but a Triangular Pyramid, and an Octaëdrum a double Quadrangular one, their Dimension is the same as of the Pyramids in Scholar of definite. 17. II The Solidity of an Hexaëdrum may be had from Consect. 3. definite. 13. III A Dodecaëdrum consists of twelve Quinquangular Pyramids, and an Icosaëdrum of twenty Triangular ones, all the Vertex's or tops whereof meet in the Centre of a Sphere that is conceived to circumscribe the respective Solids, and consequently they have their Altitudes and Bases equal: Wherhfore having found the Solidity of one of those Pyramids, and multiplied it by the number of Bases (in the one Solid 12, in the other 20) you have the Solidity of the whole respective Solids. DEFINITION XXII. BEsides these Definitions of the Regular Bodies, we may also form like Ideas of them from their Genesis, which particularly Honoratus Fabri has given us a short and ingenious System of, in his Synopsis Geometrica, p. 149. and the following. I Suppose an Equilateral Triangle ABDELLA to be inscribed in a Circle (Fig. 44. N. 1.) whose Centre is C, whence having conceived the Radii CA, CB, CD, to be drawn, imagine them to be lifted up together with the common Centre C, so that the point C ascending Perpendicularly, at length you'll have the Line● EA, EBB, ED, equal to the Lines AB, BD, DA, After this way there will be generated, or made a Space consisting of fou● Equal and Equilateral Triangles, which is called a Tetraëdrum Hence we shall by and by easily demonstrate, the quantity o● the Elevation CE, and the Proportion of the Diameter of th● Sphere OF to be Circumscribed to the remaining part CF and so the reason of the Euclidean Genesis proposed lib. 13 Prop. 13. II Much like this, but somewhat easier to be conceived, is th● Genesis of the Octaëdrum, where by a mental raising of the Centre C (Fig. 44. N. 2.) of the Square ABDE inscribed in th● Circle, together with the Semidiameters CA, CB, CD, CE until being more and more extended they at length become th● Lines OF, BF, DF, OF, all equal among themselves, and 〈◊〉 the side of the Square AB or BD; and its manifest, that by th● like extension conceived to be made downwards to G, the● will be form eight equal and regular Triangles, which w● all concur in the two opposite Points F and G. We might also deduce another Genesis of the Octaëdrum from a certain Section of a Sphere, and also give the like of a Hexaëdrum o● Cube: but we have already given the easiest, of the one, vi● that which is also common to Parallelepipeds; and that of th● other just now given is sufficient to our purpose. CHAP. II Containing the Explication of those terms, which relate to the affections of the Objects of the Mathematics. DEFINITION XXIII. EVery Magnitude is said to be either Finite if it has any bounds or terms of its Quantity; or Infinite if it has none, or at lest Indefinite if those bounds are not determined, or at lest not considered as so; as Euclid often supposes an Infinite Line, or rather perhaps, an Indefinite one, i e. considered without any relation to its bounds or Ends: By a like distinction, and in reality the same with the former, all quantity is either Measurable, or such that some Measure or other repeated some number of Times, either exactly measures and so equals it, (which Euclid and other Geometricians emphatically or particularly call Measuring) or else is greater; or on the other side Immense, whose Amplitude or Extension no Finite Measure whatsoever, or how many times soever repeated, can ever equal: In the first Case, on the one Hand, the Measure (viz. which exactly measures any quantity) is called by Euclid an aliquot Part (a) lib. 5. Def. 1. or simply a Part of the thing measured: as e. g. the Length of one Foot is an aliquot Part of a Length or Line of 10 Foot. In the latter Case the Mea●ure (which does not exactly measure any Quantity) is called an Aliquant Part, as a line of 3 or 4 Foot is an Aliquant part of a Line of 10 Foot. Now therefore, omitting ●hat perplexed Question, whether or not there may be an infinite Magnitude, we shall here, respecting what is to our purpose, deduce the following CONSECTARY. EVery Measure, or part strictly so taken, is to the thing Measured, or its whole, as Unity to a whole number, for that (which is one) repeated a certain number of times, is supposed exactly to measure the other. DEFINITION XXIV. IF the same Measure measures 2 different quantities (whether the one can exactly Measure the other or not) those Quantities are said to be absolutely Commensurable; but if they can have no common Measure; they are called Incommensurable▪ Notwithstanding which they both retain one to the other a certain relation of Quantity, which is called Reason or Proportion, a● we shall further show hereafter. In the mean while we hav● hence, as an infallible Rule to try whether Quantities can admit o● a Common Measure or not, this CONSECTARY. THose Quantities are Commensurable, whereof (a) Euclid lib. 10. Prop. 5, 6, 7, 8. one 〈◊〉 to the other, either as Unity to an whole Number, or a● one whole Number to another, for either one o● them is the Measure of the other, as also of i● self, and than it is to that other as Unity to 〈◊〉 Number by the Consect. of the preced. or els● they admit of some third Quantity for a common Measure which will be to either of them separately as Unity to some Number: therefore they are one to another as Number to Number. DEFINITION XXV. IF 2 Quantities of the same Kind, considered as Measures on● of the other, being applied one to the other, exactly agreed or are exactly equal every way, (as e. g. 2 Squares on the sam● common Side, or two Triangles whose Lines Angles and Space exactly agreed and conicide) or at lest may be equally measured by a common Measure applied to both) as e. g. a Square and an Oblong, or a Rhombus, or Triangle, each of whose Area's were 20 square Inches, altho' they do not agreed in Line● and Angles; the first may be called Simply Equal, and the other totally equal, or equal as to their wholes: But if one be greater and the other lesle, they are Unequal, and that which exceeds i● called the greater, and that which is deficient the lesle, and tha● part by which the lesle is exceeded by the greater, in respect to the greater is called Excess, in respect to the lesle Defect, and by a common Name they are called the Difference. All which as they are plain and easy, so they afford us a great many self-evident Truths, which are used to be called Axioms, as these and the like CONSECTARYS. I THe whole is greater than its Part, whether it be an Aliquot or aliquant Part. II Those Quantities which are equal to a third are equal betwixt themselves. III That which is greater or lesle than one of the equal Quantities is also greater or lesle than the other. IV. Those Quantities which, being applied one to the other, or placed one upon the other, either really or mentally, agreed; may be esteemed as totally equal: And on the Contrary, V Those Quantities which are totally equal will agreed, etc. To which might be added several others which we have already made use of and supposed as such in the preceding Definitions. DEFINITION XXVI. THere are moreover Addition, Subtraction, Multiplication and Division, which are common affections of all Quantities as well as of Numbers. Addition is the Collection of several Quantities (for the most part of one kind) into one total or Sum; which is either done so, that the whole (which is commonly called the Sum or Aggregate) obtains a new Name, or else by a bore connexion of the Quantities to be added by the Copulative and, or the usual Sign + (i e. plus or more) as for Example 2 Numbers ... and .... (suppose 3 and 4) added together make the Sum ....... (i e. 7, or which is the same thing 3+4;) and this Line— added to this other— gives the Sum— which is nothing but the 2 Lines joined, or taken together. But now if we would treat of these Lines, or any other 2 Quantities to be added, more generally; by calling the first a (a) and the latter (b) we may fitly writ their Sum a + b. SCHOLIUM. HAving thus explained the Term of Addition, these and the like Axioms emerge of themselves: If to equal Quantities you add Equal the Sum will be Equal; but if to Equal you add unequal the Aggregate will be unequal, etc. Moreover it may not be amiss to admonish the Tyro of these 2 things. 1. In Addition may be see● the vast usefulness of that very Ingenious though familiar Invention mentioned in definite. 3. for hereby we may collect into one Su● not only Ten, and Hundred, but Thousand, Million, My riads, as though they were only Units; which we will Illustrate by an Example. DEFINITION XXVII. SUbtraction is the taking one Quantity from another (of th● same kind;) which is so performed that either the remained obtains a new Name, or by a bore separation of the Subtrahend by the privative Particle lesle, or the usual Sign − which stand for it, as e. g. ... or three being subtracted from ......▪ or 7, the remainder or difference is .... or 4 and this Lin●— Subtracted from that— leaves— Now if we would signify this more generally either of the● Lines, or the Number above, or any 2 Quantities whatsoever that are to be Subtracted one from the other, by naming th● first (a) and the latter (b) we shall have the remainder a— ● Herein are evident these and the like Axioms: If from equ●● Quantities you Subtract Equal ones, the Remainders or Differences 〈◊〉 be equal. Here it will be worth while to take notice of, from this and the preced. definite. the following CONSECTARIES. I IF a negative Quantity be added to itself considered a positive (as − 3 to + 3 or − a to + a) the Sum wi●● be 〈◊〉 for to add a Privation or Negative is the same thing a● to Subtract a Positive, wherefore to join a Negative and Positive together, is to make the one to destroy the other. II If a negative be subtracted from its positive (− a from + a) the remainder will be double of that positive (+2 a) for to subtract or take away a privation or negative, is to add that very thing, the privation of which you take away; for really that which in words is called the addition of a Privation, is in reality a Subtraction, and a subtraction of it, is really an addition; and what is here called a Remainder, is indeed a Sum or Aggregate; and what is there called a Sum, is truly a Remainder. Thus, III If the positive Quantity (+ a) be taken from the privative one (− a) the remainder is double the privative one (− 2 a) since, taking away a positive one, there necessarily arises a new Privation which will double that you had before. Hence, IU. You have the Original of the Vulgar Rules in Literal Addition and Subtraction: If the Signs of the unequal Quantities are different, in the room of Addition you must subtract, and in room of Subtraction add, and to the sum or remainder, prefix the Sign in the first place of the greatest, in the next of that from which you Subtract: but if the Signs are both the same, and the greatest quantity to be subtracted from the lesle, you must, on the contrary, subtract according to the natural Way, the lest from the greater, and prefix the contrary Sign to the remainder: Which Rules you may see Illustrated in the following Examples: Addition Subtraction. 4 b − 2 a from 2 a + b from 3 a+ + b 3 b+ + a Subst. a − b Subst. 2 a+ + b 7 b+ + a R. a+ + b R. a − b NOTE. ☞ Instead of the Authors 4th Consect. as far as it relates to Subtraction, which may seem a little perplexed, take this general Rule for Subtraction in Species, viz. Change all the Signs of the lower Line, or Subtrahend, and than add the Quantities, and you have the true Remainder. SCHOLIUM. IN this Literal Subtraction, we have not that conveniency which the invention of Vulgar Notes supplies us with, that from the next foregoing Note we may borrow Unity, which in the following Series goes for 10, etc. This is done in Tetractycal Subtraction only with this difference, that an Unite here borrowed goes only for 4. That the easiness of this Operation may appear, we will add one Example, wherein from this number,— you are to subtract this, 1232002310232 321012321223 310323323003 Wherever therefore the inferior Note is greater than the superior one, the facility is much greater here than in common Subtraction, because never a greater number than 3 is to be subtracted out of a greater, than 4 and 2: but if the inferior number be greater than the superior, you borrow unity from the left hand, which is equivalent to 4; the rest is performed as in common Subtraction. DEFINITION XXVIII. MVltiplication, generally Speaking, is nothing else but a Complex or manifold Addition of the same quantity, wherein that which is produced is peculiarly called the Product, and those quantities by which it is produced, are called the Multiplicand and the Multiplier: The first denotes the Quantity which is to be multiplied, or added so many times to its self; and the other the Number by which it is to be multiplied, or determins how many times it is to be added to itself. The same terms are applied moreover to Lines and other Quantities. But here are two things to be chief noted; 1. That the Multiplication of one number by another, or of a Line by a Line, may be considered as having a double Event; for the Product may be either of the same or a different kind, as, e. g. when .... 4 is multiplied by 3 ... the product may be considered either as a Line, thus, ............ or as a Plane Surface in this Form, 12 dots arranged in three rows and four columns Whence it is also named a Plane Number, and the product is conceived to be form by the motion of an erect Line AB, consisting of 3 equal parts, along another BC, consisting of 4 equal parts, and conceived as lying along. So also the Multiplication of Lines (e. g. of the Line A— B by the Line B— C) may be conceived to be so performed, that the Product also shall be a Line, e. g. C— D (concerning the usefulness of which Multiplication in Geometry, we shall have occasion to speak more hereafter;) or so, that the Product shall be a Plane or Surface, arising from the motion of the erect Line AB, along AC, conceived as lying along; as we have already shown. But as for the most part these Planes so produced are called Rectangles, if the Lines that form them are unequal; but if they are equal they are called Squares, (otherwise the Powers of the given Quantities;) and in this case the Lines that form them are called Square Roots; so also if those Planes are multiplied again into a third Quantity (as either a Line or a Number) there will arise Solids, and particularly if that third Quantity be the Root of the Square, the Product is called a Cube, etc. The other thing to be noted is, That both these ways of Multiplying either Numbers or Lines, are expressed by a very compendious, though arbitrary way, of Notation, viz. by a bore Juxtaposition of the Letters which denote such and such Species of Quantities, as, e. g. if for the forementioned Number or Line AB we put a, and for BC b, the Product will be ab; or if the Efficients are equal, as a and a the Square thence produced, will be aa or a; and if this Square be further multiplied by its Root a, than the Cube thence produced will be aaa or a, etc. Which being premised, you have these following CONSECTARIES. I IF a Positive Quantity be multiplied by a Positive one, the Product will be also Positive; since to multiply is to repeat the Quantity according as the Multiplier directs: Wherhfore to multiply by a Positive Quantity, is to repeat the Quantities positively; as on the other side, to multiply by a Privative, is so many times to repeat the Privation of that Thing: Which we shall show further hereafter. II Equal Quantities (a and a) multiplied by the same (b), or contrariwise, will give equal Products (ab and ab or ba). III The same Quantity (z) multiplied by the whole Quantity (a + b + c) or by (a) Eucl. lib. 2. prop. 1. all its parts separately, will give equal Products. Also IV. The whole (a + b) whether it be multiplied by (b) lib. 2. prop. 2. itself, or by its parts separately, will give equal Products. SCHOLIUM I THe Vulgar Praxis of Numeral Multiplication, is founded on these two last Consectarys, as e. g. to multiply 126 by 3; you first multiply 6 by 3, then 2, i e. 20 by 3, then 1, i e. 100 by the same, and than add each of those partial Products into one Sum: In like manner being to multiply 348 by 23, you first multiply each Note of the Multiplicand by the first of the Multiplier (3) and than by the second (2) (i e. 20) etc. which is to be done likewise after the same manner in Tetractical Multiplication; only in this latter, which is more easy, you have nothing to reserve in your mind, but all is immediately writ down, (which might also be done in Vulgar Multiplication) as may be seen by this Example underneath, as also the great easiness of this sort of Multiplication, beyond the common way, because there is no need of any longer Table than that we have shown page 7. SCHOLIUM II It is manifest from what we have said, I IF the Base of a Parallelogram be called (b) and its Altitude a, it's Area may be expressed by the Product ab, by Cons. 7. definite. 12. II If the Base of a ▵ be b or ebb, and its Altitude a its Area will be half ab or half eab, by Consectary 8. of the same Definition. III If the Base of a Prism or Parallelepiped or Pyramid be half ab or ab, and its Altitude c, the solid Contents of that Prism will be half abc, and of the Parallelepiped abc, by Consect. 3 & 4. Def. 16. and of the Pyramid ⅙ abc, by Cons. 3. Def. 17. DEFINITION XXIX. DIvision, in general, is a manifold or complicated Subtraction of one quantity (which is called the Divisor) out of another (which is called the Dividend) whose multiplicity, or how many times the one is contained in the other, is shown by another quantity arising from that Division, which is therefore called the Quote or Quotient. Here also the Divisor is of the same kind with the Dividend, or of a different kind, e. g. of the same kind if the product ............ (12) be divided by (3) whence you'll have the Quotient .... (4) or dividing the aforementioned Line CD by the Line AB you'll again have the Line BC; but of a different kind, if the plane number above found 12 dots arranged in 3 rows and 4 columns or the Rectangle ABCD be divided by a Retroduction, or a moving backwards again the erect Side AB, by whose motion the Rectangle was first form, that so the Line BC may remain alone again. But both these kinds of Division as they have their peculiar Difficulties in Arithmetic and Geometry, which we shall further elucidate in their proper places; so they may be universally and very easily performed in Species (or by Letters) which will be sufficient to our present purpose; or by a bore separation of the Divisor from the Dividend, if it be actually therein included; or by placing the Divisor underneath the Dividend with a Line between. Thus if ab be to be divided by (b) the Quotient will be a; if by a, the Quotient will be (b); but if a or ab be to be divided by c which Letter since it is not found in the Dividend, cannot be taken out of it) the Quotients are a/ c and ab/ c i e. a or ab divided by c, after the same manner as if 2 were to be divided by 3; which Divisor, since it is not contained in the Dividend, is usually placed under it, by a separating Line thus, ⅔, 2 divided by 3. SCHOLION. HOW difficult Common Division is, especially of a large Dividend by a large Divisor, is sufficiently known: but how easily it is performed by Tetractical Arithmetic, we will barely bring one Example to show. If the Product found in Scholar 1. of the preceding Definition, 1200 203 22 be again to be divided by its Multiplier 133, it may be performed after the usual way, but with much more ease, as the following Operation will show; or according to a particular way of Weigelius, by writing down the Divisor, and its double and triple, in a piece of paper by itself, after this way: 123 312 1101 Divisor, Double, Triple. and than moving that piece of Paper to the Dividend, note, which of those three Numbers comes nearest to the first Figures of the Dividend; for that barely subtracted gives the Remainder, and will denote the Quotient to be writ down in its proper place; as the operation itself will show better than any words can. Thus after Weigelius' way: DEFINITION XXX. EXtraction of Roots is a Species of Division, wherein the Quotient is the Root of the given Square or Cube, etc. But the Divisor is not given, neither is it all along the same (as it is in Division) but must be perpetually found, and they are several. And as the Squares of Simple Numbers 1, 2, 3, etc. viz. 1, 4, 9, 16, etc. and their Cubes 1, 8, 27, 64, etc. may be had immediately out of a Multiplication Table, as also their Roots, without any further trouble; and likewise in Species, as the Roots of the Square aa or a, or of the Cube aaa or a, are without doubt (a); so if the Square Root be to be extracted out of de, or the Cube Root out of fgm (because the letters are different, and no one can be taken for the Root) the Square Root is commonly noted by this Sign √ de, the Cube Root by this √ C, or 3 √ fgm, etc. as also in Numbers that are not perfect Squares (as e. g. 2, 3, 5, 6, 7, 8, 10, 11, 15, 17, 19, etc.) we can not otherwise express the Square Roots, than after this manner √ 2, 7, √ 19, etc. and in those that are not perfectly Cubical (as all between 1, 8, 27, 64, etc.) we can only express their Cube Roots after some such manner, √ c. 7, or 3 √ 7. √ c. 61, or 3 √ 61 etc. Which forms of Roots in specious Computation, we call Surd Quantities, in Vulgar Arithmetic Surd Numbers, i e. such as cannot be perfectly expressed by any Numbers; although we have Rules at hand to determine their Values nearer and nearer ad Infinitum. These Rules accommodated to Square and Cube Numbers, etc. which otherwise are more difficult to be comprehended, appear plain and easy to him, who multiplies a Root expressed by 2 Letters (called therefore commonly a Binomial) first Quadratically, than Cubically, etc. For he will have as CONSECTARYS. I THE Square of any assumed Root, as also, Prop. 4. lib. 2. Eucl. and at the same time a general Rule for Extracting the Square Root, all expressed in these few Notes: aa+ + ab + bb. II The Cube of the same Root, a New Theorem, and at the same time a Rule for Extracting any Cube Root, contained in this Theorem: a +3 aab+ + abb + bbb. SCHOLIUM I WHich that we may more plainly show, especially as far as it relates to the Rules of Extraction, consider, 1. That the Root of the Square aa+ + ab + bb is already known (for we assumed for the Root the Quantity a + b) so that now we are to see which way this Root is to be obtained out of that Square by Division. It will presently appear, that the first Note of the Root a, will come out of the first part of the Square aa, and the other part b must be obtained out of the remainder 2 ab + bb; and so as there are 2 Notes of the Root, the Square must be distinguished as it were into 2 Classes, each of which gives a particular note of the Root. Than it is manifest, that the first Note of the Root (a) may be obtained out of the Square aa by a simple Extraction. Now it is evident, if I would have by Division the other Note of the Root, the next following part of the remaining Classis must be divided by 2 a, the double of the Quotient just now found, and that nothing should remain after this Division (for now we have the whole Root a + b) you must not only subtract the Product of the Divisor and this new Quotient, but also the Square of this new Quotient: Which is the Vulgar Method and Rule for the Extraction of Square Roots taught in common Arithmetic. Likewise if you would extract the Root of the abovementioned Cube, which we already know, having form it from a + b, it is manifest, that the first Note of the Root a will come out of the first part of the Cube a, and the other b, must be obtained out of the remainder 3 abb + bbb, and so, as there are two Notes of the Root, the Cube must be distinguished, as it were into two Classes, each of which will give a particular Note of the Root. Now it is manifest, the first Note of the Root a is obtained by simple Extraction of the Root out of the Cube aaa. It is moreover evident, if I would have by Division the other Note of the Root b, the next remaining part must be divided by 3 aa (the triple Square of the precedent Quotient, or thrice the precedent Quotient multiplied by itself) and, that nothing should remain after this division (for now we have the whole Cube Root a + b) you must not only subtract from the remaining Dividend the Product of the Divisor, and the new Quotient (3 aab) but also the Product of the Square of the new Quotient, and thrice the precedent Quotient (3 abb) and moreover the Cube of that new Quotient b: Which is the Method of extracting Cube Roots in Vulgar Arithmetic. SCHOLIUM II FRom what we have said you have also the Reason of another rule in Arithmetic which teaches how to approach continually nearer and nearer to the Square and Cube Roots of numbers that are not exact Squares and Cubes; viz. by adding to the given Number perpetually new Classes and Ciphers or o's, two at a time, to the Square, and three to the Cube, and so continued on the operation as before; which will add Decimal Parts to the Integrals before found; and the next operation (if you add a second Class of Ciphers) will exhibit Centesimal Parts, and so on ad Infinitum. For Example, If I would have the Square Root of 2 pretty near, I can assign not nearer whole Number than 1. But by adding a new Class of 2 Ciphers, i e. multiplying the given Number by 100 (whereby the Root is multiplied by 10) you'll have 14, nearly the Root of 200, that is, 14 14/10 10 or 1 4 4/10 10 much nearer the Root of 2 than the former; and thus you may always come nearer and nearer ad Infinitum, but never to an exact Root. For if you could have the exact Root of 2, or 3, or 5, etc. in any Fraction whatsoever, that Fraction must be of such sort, that its Numerator and Denominator being squared, the Fraction thence arising must exactly equal 2 or 3, or 5, etc. that is, its Numerator must be exactly double, or trible, or Quadruple, etc. of of the Denominator; which can never be, because both are Squares, and in a Series of Squares no such thing can hap. Hence you have these CONSECTARIES. III THat it is a certain mark of Incommensurability, if on● quantity is 1, and other the √ 2, or √ 3, or √ 5, &c▪ IU. That these sorts of Quantities are notwithstanding Commensurable in their Powers, i e. their Squares are as 1 and 2 or as a number to a number. V Those Quantities which are to one another, as 1 an● √ √ 2, or as √ 2 and the √ √ 3 are incommensurable in Powe● also. Which being rightly understood, you may easily comprehend several (a) Eucl. l. 10 Prop. 9, 10, 11 12, 13 etc. Propositions of lib. 10 Eucl. especially aft●● some few things premised concerning Reason and Proportion. SCHOLIUM. III FRom what we have shown may easily be concluded, th● to any proposed Quantity whatsoever, which Euclid cal● (b) lib. 10. Def. 5, 6, ● 8, 9, 10, 11. Rational, and for which we may always put I, there may be several others both commensurable and incommensurable, and that either simply or in power so; those which are commensurable to a Rational given Quantity, either Simply or only in Power (which, e. g. are to it, as 2, 3, 4, etc. ½, ⅓, ¼, etc. or as √ 2, √ 3, √ 4, √ ½, √ ⅓, √ ¼, etc.) are called also Rational: but those which are Incommensurable both ways (i e. both simply and in power) as (√ √ ● √ √ ⅓, etc.) are called Irrational. In like manner the Square of a given Rational Quantity (as I) is called Rational, an● Quantities commensurable to it (as 2, 3, 4, 5, etc. ½, ⅓, ● etc. □) are called also Rational; but incommensurable on● (√ √ 2, √ √ 5, etc.) Irrational, and the Sides and Roots of the● more Irrational. DEFINITION XXXI. ANy Quantities whatsoever of the same kind, whether commensurable or incommensurable, equal or unequal, admit of a twofold respect or relation of their magnitude, one whereof, when only the difference or excess of one above another is respected (as 10 which is 3 more than 7) is called an Arithmetical Reason or Respect; the other, wherein respect is rather had to the Amplitude, whereby one is contained once, or a certain number of times in the other (as 3 is contained thrice in 10 and ⅓ part more) is called Geometrical Reason, or by way of Emphasis, only Reason, and by other's Proportion; and this Reason or Proportion, if the lesle is exactly contained a certain number of times in the greater (as 3 in 6, or 4 in 12) is generally called, on the part of the greatest term, Multiple, and on part of the lesle, Submultiple, and particularly in the first Example double, when 6 is taken in respect to 3, and subduple when 3 is taken in respect to 6; in the other triple and subtriple, etc. If the lesle be contained in the greater once or more times, unity only remaining over and above (as 3 in 4 and 4 in 9) the Reason or Proportion is called Superparticular and Subsuperparticular, and is noted by the terms Sesqui & Subsesqui, joining the ordinal Name of the lesser Term; as the the Reason of 4 to 3 is called Sesquitertian, and contrariwise Subsesquitertian; the Reason of 9 to 4 is called double Sesquiquartan, and contrariwise Subduple Subsequiquartan, etc. If lastly, the lesle be contained in the greater once, or a certain number of times, several units remaining over and above, it is commonly called Superpartient Reason and is expressed by the word Super or Subsuper, joined with the adverbial Name of the remaining Parts, and the ordinal Name of the lesser Term; thus, e. g. the Reason of 7 to 4, is called Supertriquartan, 12 to 5 double Superbiquintan, etc. but when the Quotient arises by the division of the greater term by the lesle, and is commonly expressed in the same words, it is also commonly called by the name of Reason (e. g. 2 is the name of the Reason of 6 to 3, 2¼ of 9 to 4, or contrariwise, etc.) as also the quotient arising by division of the Consequent by the Antecedent (as ½ in the first case, 4 4/9 9 in the latter) by which name the antecedent Term of the Reason being multiplied, produces its Consequent; which is evident by naming any Reason e or i, or oh, etc. Thus if the antecedent Term be called a or b etc. the Consequent may be rightly called ea or ebb, oa or ob, etc. and because in an Arithmetical Relation we only respect the excess of the first above the following, or of the following above the foregoing (which may be called x or z) if the antecedent (which may be called a or b) be lesle, the consequent may properly be called a + x or b + z; but if it be greater, the other will be a − x or b − z. CONSECTARYS. I WE may hence readily infer, that if the Diameter of any Circle be called a the Circumference may be called ea, (for whatever the proportion is between them, i● may be expressed by the Letter e) and the Area, according to Consectary 2. Definition 15, will be ¼ eaa. II If for the Base of any Cylinder or Cone you put ¼ eaa and the Altitude (b) the Solidity of that Cylinder may be rightly expressed by ¼ eaab, by Consect. 5. definite. 16, and of the Cone by ½ eaab, by Consect. 4. definite. 17. DEFINITION XXXII. AS the Identity (or sameness) of several Geometrical Reasons used to be called Geometrical Proportionality, or emphatically Proportion; so the similitude (or likeness) of severa● Arithmetical Reasons, is deservedly called Arithmetical Proportionality, or by a particular Name Progression; and consequently those Progressionals, or Arithmetical Proportionals, which exceed one another by the same difference, either uninterruptedly or continually as 2, 5, 8, 11, 14, etc. ascending, or 30, 28, 26, 24, 22, 20, etc. descending; or interruptedly, as 2 and 5, 7 and 10, 11 and 14, etc. ascending; or 30 and 26, 24 and 20, 1● and 13, etc. descending: For which, and all other in what cas●● soever, we may universally put this (or such like) continual Progression, u g. a, a + x, a+ + x, a+3 +, etc. ascending; o● a, a − x, a − 2x, a − 3x, etc. descending, but in an interrupted Progression, u g. b and b + z and c and c − z, d and d − z, etc. descending. Whence you have this CONSECTARY. ANY Difference being given, the following Terms of me Progression, continually proceeding from the first assumed or given one, may be found; as also several Antecedents that interruptedly follow the given or assumed ones, viz. by adding or subtracting the given Difference to or from the former Terms to found the latter. DEFINITION XXXIII. IN like manner, since Reasons are said to be the same, which have the same Denomination of Reason, those quantities will be proportional which continually ascend by the same denomination of Reason, as 2, 4, 8, 16, 32, 64, etc. or descend, ●s 81, 27, 9, 3, 1. there by the Denomination of the Reason 2, here 3; or that ascend interruptedly, as 2, 4; 3, 〈◊〉: 5, 10, etc. or descend, as 40, 10; 28, 7; 20, 5; 8, 〈◊〉, etc. Whence you have these CONSECTAYS. HAving two Terms given, or only one with the Denomination of the Reason (e. g. the Term 2 with the Denomination of the Reason 3, or universally the first Term a with the Denomination of the Reason e) it will be easy to found ●s many more Terms of the Geometrical Progression or Proportion as you please, viz. by always multiplying the Antecedent by the Denomination of the Reason, that you may have 2, 〈◊〉, 18, 54, etc. or a, ea, e a, e a, etc. in continued, or 2 and 〈◊〉, 4 and 12, 5 and 15, etc. and aea, beb, dead, etc. in discontinued or interrupted Proportion. Thus having rightly understood what we have said in this ●3 and 31 Definition, there will follow these Corollaries as so ma●y Axioms. II That equal Quantities have the same proportion to the same Quantity (α) Eucl. lib. Prop. 7. and the same has the like to equal Quantities. III But a greater quantity has a greater Reason to the same (β) prop. 8. than a lesle, and the same has a greater proportion to a lesle Quantity than to a greater. IV. On the contrary, those that have the (γ) prop. 9 same proportion to the same quantity, and that likewise the same to them are equal. V But that which bears a (δ) prop. 10 greater proportion to the same is greater; but that to which the same bears a greate● proportion is lesle. VI Proportions equal to one third (ε) 16. are also equal amoner themselves, etc. DEFINITION XXXIV. HEre remain two things to be taken notice of; First th● If any whole (quantiy) be so divided into two equ●● parts (α) Eucl. definite. 3. lib. 6 that the whole, the greater part an● the lesle are in a continual proportion; th● (whole) is said to be cut in extreme and me● Reason. 2. In a continual Series of that kind 〈◊〉 Proportionals (e. g. 2. 4. 8. 16. 32, etc. or a, 〈◊〉 e a, e a, e a, etc.) the Reason of the first Ter● to the third (β) Eucl. definite. 10. l. 5. (2 to 8, or a to e a) is particularly called Duplicate, and to the 4th (〈◊〉 or e a) Triplicate, etc. of that Reason which the same first Te● has to its second, or any other antecedent of that Series to 〈◊〉 Consequent: But generally these Duplicate and Triplicate Reason's, etc. as others also of the first Term to the third or four●● of Proportions continually cohering together, (whether the are the same as in the foregoing Examples, or different as 〈◊〉 these, 2, 4, 6, 18, or a, ea, eia, eioa, etc. viz. if the nam● of the first Reason be e, of the second i, 〈◊〉 the third oh, etc.) I say, the Reasons of the fir●● Term (2 or a) to the third (6 or eia) 〈◊〉 to the 4th (18 or eioa) are said to be compounded of the continual intermediate Reasons. Now from our general Example, what Eucl● says, is manifest, CONSECTARY I THat the denomination of a compounded Reason arises from the Multiplication of the denominations of the given Simple (α) Eucl. l. 6. Def. 5. Reasons; as the denomination of the reason compounded of both (viz. a to eia) is produced by multiplying the denomination of the first Reason e by the denomination of the second Reason i, and the denomination of the Reason compounded of the three (viz. a to eioa) is produced by the denomination of the first Reason e, multiplied by the denomination of the second Reason i; and the Product of these by the denomination of the third Reason oh, etc. CONSECTARY II SO that it is very easy after this way, having never so many Reasons given, whether continued (as 2 to 3, 3 to 6, or ●a, ea, eia,) or interrupted or discrete (as 2 to 3, and 5 to 10, or a to ea, and b to i b) to express their compounded Reason: ●n the first case it easily obtained by the bore omission of the intermediate Term or Terms (2 to 6, or a to eia;) and in the other by multiplying first of all the Names of the compounding Reasons among themselves (CITIZEN ½ and 2, e. and i.) and by the Product (3 or ei) as the name of the Reason compounding the first Term (2 or a) that you may have the oh her 6 or eia) or (if any one had rather do so in this latter case) by turning the discrete or interrupted Reasons into continued ones, by making as 5 to 10 in the second Reason, so is the Consequent of the first 3 to 6, or as b to ib, so ea to eia,) and than by referring the first 2 to the third 6, or the first a to the third eia, etc. In a word therefore, any Duplicate Reason may be appositely expressed by a to e a, and Triplicate by a to e a, the one immediately discernible by a double, the other by a triple Multiplication into itself; as you may also commodiously, and denote others compounded, e. g. of 2 by a to eia, of 3 by a to eioa, etc. SCHOLIUM. WE will here advertise the Reader, that though the Names 〈◊〉 duplicate & triplicate Reasons, etc. are chief appropriate to Geometrical Proportionality, yet the Moderns have also accommodated them to Arithmetical also; as e. g. That Arithmetic● Progression is called Duplicate, whose Terms are the Squares 〈◊〉 Numbers Arithmetically Proportional (e. g. 1, 4, 9, 16, 25 etc.) and Triplicate, whose Terms are Cubes, (etc. as 1, ● 27, 64, etc. DEFINITION XXXV. AND now at length we may understand what Magnitude Geometers particularly call like, or similar. Whereas ● General one number may be said to be like another, one rig●● Line to another, one obtuse Angle to another, a Triangle ●● a Triangle, and the like; but an Acute Angle is not like ● Obtuse one, nor a Triangle like a Parallelogram, or a rig●● Line like a Curve one; or a Square like an Oblong, etc. Yet ●mong those Figures which may after that rate in general be sa●● to be like, there is notwithstanding a great deal of dissimilitude; therefore in a strict Sense we call only those Right Li●●ed Figures similar or like (α) which have each of their Angl● respectively equal to each of the other (as A and A) B and B C and C, etc. Fig. 48.) and the Sides about those equal Angel's Proportional, viz. as BASILIUS to AC, so BA to AC, etc. (〈◊〉 and among Solid Figures those are said to be Similar, each o● whose Planes are respectively Similar one to the other, and equ●● in number on both sides; as, e. g. the Plane AC is similar to th● Plane AC, and CG to CG, etc. and six in number on bo●● sides. Pag. 54. 47 48 49 50 51 52 53 Book I Section II Containing several Propositions demonstrated from the foregoing Foundations. CHAP. I Of the Composition and Division of Quantities. Proposition I THE Sum and Difference of two unequal Quantities added together, make double of the greatest. Demonstration. Suppose a be the greatest, b the least, than will their Sum be a + b Andrea their Difference a − b Their Sum 2 a, by Consectary 1. Definition 27, Q. E. D. CONSECTARY. HEnce by a bore Subsumption (α) Eucl. lib. 6. definite. 1. you have the truth of Consect. 1. definite. 8. that 2 unequal contiguous Angles on the same Right Line, ACD and ACE (Fig. 49. (β) lib. II Def. 9 ) i e. if we call the Right Angle BCD or BE (a) and the difference between the one and the other (b) a + b and a − b, make 2 a, i e. are equal to 2 right ones. (α) Eucl. 13. l. 1. Proposition II IF the Difference of two unequal Quantities be subtracted from their Sum, the Remainder will be double of the lest. Demonstration. If from the Aggregate or Sum a + b You subtract the Difference a − b The Remainder will be 0+2 b by Consectary 2. Definition 27. Q. E. D. Proposition III BUt if the Sum or Aggregate be subtracteed from the Difference, the remainder is so much lesle than nothing, as is the double of the last Quantity. Demonstration. For if from the Difference a − b You subtract the Sum or Aggregate a + b The Remainder will be— 0 − 2 b by Consectary 3 of the aforesaid Definition. Q. E. D. Proposition IU. IF a Positive Quantity be multiplied by a Negative one, or contrariwise, the Product will be a Negative Quantity. Exposition. If a − b be to be multiplied by c; it is certain, that a multiplied by c, makes ac a Positive Quantity, by Consect. 1. definite. 28. Moreover b by the same c (a Negative by a Positive) will make − bc; and so the whole Product of a − b by + c, will be ac − bc. Demonstration. Suppose a − b = e; therefore ec will be = to the Product of a − b by c: and since a − b is = e by the Hypoth. adding on both sides h, you'll have a = e + b. by Scholar definite 26. and multiplying both sides by c, ac = ec + bc, by Consect. 2. definite. 28. and by further subtracting from each side bc, you'll have ac − bc = ec, that is, to the Product of a − b by c. Q. E D. CONSECTARY. SInce ac − bc is the Product of a − b by c, it is manifest also, that if ac − bc be divided by c, you'll have a − b for the Quotient; and so always a Positive Quantity (as ac) divided by a Positive one, c, will give a Positive Quotient; but a Negative Quantity − bc divided by a Positive one, will give a Negative Quotient. Proposition V IF a Negative Quantity be multiplied by a Negative one, the Product will be Positive. Exposition. Suppose a − b be to be multiplied by − c; it is certain, that 〈◊〉 multiplied by − c will give the Negative Quantity − ac, by Prop. 4. but − b multiplied by the same − c will produce + bc, and so the whole Product will be − ac + bc. A Demonstration like the former. Suppose a − b = e, than will − ec = the Product of a − b by − c: and since a − b is = e, adding b on both sides you'll ●ave a = e + b, by Scholar definite 26. and multiplying both sides by − c, you'll have ac = − ec − bc, by Prop. 4. and Consect. ●●. definite. 28. and by adding bc on both sides, you'll have − ac + bc = − ec, i e. to the Product of a − b by − c Q. E. D. CONSECTARYS. I SInce therefore a − c + bc arises from a − b by − c, it is manifest, that if − ac + bc be divided again by − c, you will again have a − b, and consequently a Negative Quantity divided by Negative, will give a Positive Quotient but a Positive Quantity + bc divided by a Negative one, wil● give a negative Quotient − b. II We have therefore the Foundation and Demonstration o● the Rules of Specious Computation, in the multiplication an● division of Compounded Quantities, viz. that the same Sig● multiplied together (as + by + or − by −) give + but different (as + by − or − by +) give the Sign − Which Rules the following Examples will Illustrate, as als● several other we shall meet with in the following Chapter. Multiplication. Division. CAAP. II Of the Powers of QUANTITIES. Containing (after a compendious Way) most part of the 2d Book of Euclid; and the Appendix of Clavius to Lib. 9 Prop. 14. Proposition VI. IF any whole Quantity be divided into two parts (α) Eucl. lib. 2. prop. 3. A. C. also the third. the Rectangle contained under the whole, and one of its parts, is equal to the Square of the same part, and the Rectangle contained under both the parts. Demonstration. Let a + b represent the whole a + b b one part of it, or a the other. ab + bb the Rectangle, aa + ab the Rectangle. (See Fig. 50.) Q. E. D. Proposition VII. IF a whole Quantity be divided into two parts (β) Eucl. Prop. 4. lib. 2. the Square of the whole is equal to the Squares of both those parts and 2 Rectangles contained under them. Demonstration. This is evident from the preceding, and may moreover thus appear further. Let the Parts be a and b, than will the whole be Which if you multiply by itself You have the Square (See Fig. 51. N. 1.) Q.E.D. CONSECTARYS. I HEnce you have the Original Rule for Extracting of Square Roots, as we have shown after Definition 30. and here have further Illustrated in Scheme Nᵒ 2. II Hence it naturally follows, that the Square of double any Side is Quadruple of the Square of that Side taken singly. III Hence also you have the addition of furred Numbers, or in general of furred Quantities, by help of the following Rule (supposing in the mean while their Multiplication:) Suppose these 2 Surds √ 8 and √ 18, or more generally √ 75 aa and √ 27 aa, are to be added together; first add their Squares 8 and 18, etc. than double their Rectangle (√ 144) that is, multiply it by the √ 4, and than the double of this √ 576, i e. having extracted the Square Root, (24) and added it to the Sum of the first Squares (26) the Root of the whole Sum (50) viz. √ 50, is the Sum of the two furred Quantities first proposed. SCHOLIUM. BUT if it happens that the Root of the double Product cannot be expressed by a Rational Number (as, when the proposed Quantities are Surds, as √ 3 and √ 7, to whose Squares 3+7, i e. 10, you must add the double Product of √ 7 by √ 3, i e. √ 84, which cannot be expressed by a Rational Number) than that double Product must be joined under a Surd Form, or Radical Sign, to the Sum of the Squares (thus, viz. 10+ √ 84) and to this whole Aggregate prefix another Radical Sign, thus, ; or also you may only simply join the Surd Quantities proposed by the Sign + thus, √ 3+ √ 7. Here also you may note, that the two Surd Quantities proposed in the first case of Consectary 2. are called Communicants; in the other case of this Scholium, Non-Communicants: For in this case each quantity under the Radical Sign may be divided by some Square, and have the same Quotient (e. g. 8 and 18, may be divided the first by 4, the other by 9, and the Quotient of both will be 2; likewise 75 aa and 27 aa may be divided, the one by 25 aa, the other by 9 aa, the Quotient of both being 3; and than if the Quotient on both sides be left under the Radical Sign, and the Root of the dividing Square set before it, the same quantities will be rightly expressed under this form: 2 √ 〈◊〉 and 3 √ 2, also 5 a √ 3 and 3 a √ 3; and than the addition is easy, viz. only collecting or adding together the Quantities prefixed to the Radical Sign; so that the Sums will be of the one 5 √ 〈◊〉 and of the other 8 a √ 3, which are indeed the same we have shown in Consect. 2. For if contrariwise we square the Quantities that stand without, or are prefixed to the Radical Sign, and than set those Squares (25 and 64 aa) under the Radical Sign, multiplying by the Number prefixed to it, you'll have for the one √ 50, for the other √ 192 aa (Consect. after Schol-Prop. 22.) Proposition VIII. IF any whole Quantity (viz. Line or Number) be divided (α) Eucl. & Clau. 5. into two equal parts, and two unequal ones, the Rectangle of the unequal ones, together with the Square of (the intermediate part or) the difference of the equal part from the unequal one, is equal the Square of the half. An Universal Demonstration. Suppose the parts to be a and a, and the whole 2 a; let one of the unequal Parts be b, the other will be 2 a − b, and the difference between the equal and unequal part a − b. The equal ones Rectangle Difference The Sum will be aa (the other parts destroying one another) Q.E.D. (Vid. Fig. 52.) Proposition IX. IF to any whole Quantity divided into two equal parts (α) Eucl. & Clau. 6. you add another Quantity of the same kind, the Rectangle or Product made of the whole and the part added, multiplied by that part added, together with th● square of the half, will be equal to the Square o● the Quantity compounded of that half, and th● part added. Demonstration. Let the whole be called 2 a, the part added b, than th● quantity compounded of the whole and the part added will b● 2 a + b; and that compounded of the half and the part added a + b. The Quantity compounded of the whole, and the part added is, the half a Comp. Multip. by the part added 2 ab + bb □ of the half aa = □ aa+ + ab + bb (Vid. Fig. 53.) Q. E. D. Proposition X. IF a Quantity be divided any how into (b) Eucl. & Clau. prop. 7. l. 2 two parts, the Square of the whole, together with that of one of its parts, is equal to two Rectangles contained under the whole and the first part, together with the Square of the other part. Pag. 63. 54 55 56 57 58 59 60 The Universal Demonstration. Let a be one part and b the other, the whole a + b a + b the whole. The whole a + b a the first part * a + b aa + ab aa + ab 2 ab + bb 2 aa+ + ab the double rectangle □ of the whole aa+ + ab + bb add bb the □ of the other part * add aa Sum 2 aa+ + ab + bb = to the Sum .... 2 aa+ + ab + bb (Vid. Fig. 54. Nᵒ 1.) Q.E.D. CONSECTARY. HEnce you have the Subtraction of Surd Numbers, or more generally of Surd Quantities, by help of the following Rule. Add the Squares of the given Roots according to Consect. 3. Prop. 7. and from their Sum subtract the double Rectangle of their Roots; the Root of the Remainder will be the difference sought of the given Quantities. As, if the √ 8 (BC) is to be subtracted from √ 50AC (Fig. 54, Nᵒ 1.) you must add 50, i e. the whole Square AD) and 8, (i e. the other Square superadded DE,) and the Sum will be 58, equal to the two Rectangles OF and FE+ □ GH, by ●his Prop. I found therefore those two Rectangles by multiplying √ 50 by √ 8, and than the Product √ 400 by 2 or √ 4, thereby to obtain the double Rectangle √ 1600, i e. (having actually Extracted the Root) 40. This double Rectangle therefore ●or 40, being subtracted out of the superior Sum, the remainder 18 will be the □ GH, and so its Root (viz. √ 18) gives ●he required Difference between the given Surd Quantities. SCHOLION. BUT this Subtraction may be performed yet a shorter way, if each quantity under its Radical can be divided by some square, so that the same Quotient may come out on both sides that is, if the Surd Quantities are Communicants, as e. g. √ 5● (the number 50 being divided by 25) is equal to 5 √ 2 a● √ 8 to 2 √ 〈◊〉; for than the numbers prefixed to the Radical Sig● being subtracted from one another (viz. 2 √ 〈◊〉 from 5 √ 2) yo● have immediately the remainder or difference 3 √ 2, i e. √ 1● But if the proposed Quantities are not Communicants (as if th● √ 3 is to be subtracted from √ 7) the remainder may be briefs expressed by means of the Sign − thus, , or according to the foregoing Consectary, thus, . Proposition XI. IF any Quantity be divided into two parts, (α) Eucl. & Clau. prop. 8. the Quadru● Rectangle contained under the whole and one of its parts, together with the Square of its other parts, will be equal to the Square of 〈◊〉 Quantity compounded of the whole and the other part. Demonstration. Suppose a + b the whole. b one part. ab + bb the Rectangle of these two. mult. by 4 4 ab+ + bb the Quadruple Rectangle. Add aa the Square of the other part. Sum aa+ + ab + bb The Quantity compounded of the whole and the first part Square of the Compound Quantity Q. E ● (Vid. Fig. 55.) Proposition XII. IF any Quantity be divided into two equal parts (β) Eucl & Clau. prop. 9 and into two other unequal ones, the Squares of the unequal parts taken together will be double the Square of half the quantity, and the Square of the difference, viz. of the equal and unequal part, taken together. Demonstration. Suppose the equal parts to be a and a, the difference (b) the greater of the unequal Parts to be a + b, the lesle a − b. The greater part The lesle Half Difference Sum of these 2 aa+ + abb Sum aa + bb Q. E. D. (Vid. Fig. 56.) Proposition XIII. IF to any whole Quantity (α) Eucl. & Clau. x. divided into two equal parts there be added another Quantity of the same kind, the Square of the Quantity compounded of the whole, and the quantity added, together with the square of the quantity added, will be double the square of the half the quantity, and the square of the Sum of the half and the part added taken together. Demonstration. Suppose the whole to be 2 a, the half parts a and a, the quantity added b; than the quantity compounded of the whole and the quantity added, will be 2 a + b, and that of the half and the quantity added a + b. Comp. of the whole and quantity added Sum Half Qu. compouded of hal● and qu. added, Sum 2 aa+ + ab+● + Manifestly the half of the former Sum. Q.E.D. CHAP. III Of Progression, or Arithmetical Proportionals. Proposition FOURTEEN. IF there are 3 Quantities in continued Progression, or ● Arithmetical continued Proportion, the Sum of the Extrem● is double of the middle Term. Demonstration. Such are e. g. a, a + x, a+ + x ascending, or a, a − x, a − 2 x descending. By Definition 32. the Sum of the Extremes in the first 2 a+ + x, in the latter 2 a − 2 x; in both manifestly double the middle Term Q.E.D. Proposition XU. IF there are 4 of these Continued Proportionals, the Sum ● the Extreme Terms is equal to the Sum of the me●● Terms. Demonstration. Such are e. g. a, a + x, a+ + x, a+ + x, etc. ascending, or a, a − x, a − 2 x, a − 3 x, &c descending; in the one the Sum of the Extremes is 2 a+ + x, in the other 2 a − 3 x; and also of the means 2 a+ + x and 2 a − 3 x Q. E D. Proposition XVI. IF there are never so many of these continued Proportionals, the Sum of the Extremes is always equal to the Sum of any 2 other of them, equally remote from the Extremes, or also double of the middle Term, if the number of the Terms is odd. Demonstration. Suppose a, a + x, a+ + x, a+ + x, a+ + x, a+ + x, a+ + x, etc. or a, a − x, a − 2 x, a − 3 x, a − 4 x, a − 5 x, a − 6 x, and the Sum of the Extremes, as also of any 2 equally remote from the Extremes, and the double of the middle Term is in the first Series 2 a+ + x, in the latter 2 a − 6 x, etc. Q.E.D. SCHOLIUM I NOR can we doubt but that this will always be so, how far soever the Progression be continued; if you consider that the last Term contains in itself the first, and moreover the difference so many times taken, as is the number of Terms excepting one, but that the first has no difference added to it; and therefore though the last since one contains one difference lesle than the last; the second on the contrary has one more than the first, and consequently the Sum of the one will necessarily be equal to the Sum of the other; and in like manner the last except two, contains two Differences lesle than the last; but, on the contrary, the third exceeds the first by a double Difference, the double Difference being added to it, etc. as is obvious to the Eye in our first universal Example. Hence you have these CONSECTARIES. I YOU may obtain the Sum of any Terms in Arithmetical Proportion, if the Sum of the Extremes be multiplied by half the number of Terms, or (which is the same thing) half the Sum by the number of Terms, II To obtain therefore the Sum of 600, or never so many such Terms, you need only have the Extremes and the number of Terms: So that you have a very compendious Way of proceeding in Questions that are solvible by these Progressions, if, having the first Term and Difference of the Progression given, you can obtain the last, neglecting the intermediate one's. III But you may obtain the last Term, by Multiplying the given given Difference by the given Number of Terms lessened by Unity, and than adding the first Term to the Product; as i● evident from the preceding Scholium. IV. Hence we may easily deduce this Theorem, that the Sum of any Arithmetical Progression beginning from oh, is subduple of the Sum of so many Terms, equal to the greatest, as is the number of Terms of that Progression. For if the first Term is oh and the last x, and the given number of Terms a, the Sum of the Progression will be ½ axe, by Consect. 1. but the Sum of so many Terms equal to the greatest, axe. Q.E.D. SCHOLIUM II NOw if any one would be satisfied of the truth of this last Consect. without the literal or specious Notes, let him consider, that if the first Term be supposed to be oh, the last (whatever it is) will be the sum of the Extremes. The last therefore multiplied by half the number of Terms, gives the Sum of the Progression, by Consect. 1. and the same last Term multiplied by the whole number of Terms, gives the Sum of so many Terms equal to the greatest. But that this must needs be double of the precedent 'tis evident, because any Multiplicand being multiplied by a double multiplier, must needs give a double Product. Now as this Consectary will be of singular Use to us hereafter for Demonstrating several Propositions, so the three former are the very same Practical Rules of Arithmetic, which are commonly made use of in Arithmetical Progressions; for the Illustration whereof Swenterus gives us several Ingenious Examples in his Delic. part 1. Quest. 70. etc. CHAP. IU. Of Geometrical Proportion in General. Proposition XVII. IF there are three Quantities continually (α) Proportional, the Rectangle of the Extremes, is equal to the Square of the mean Term. Demonstration. Such are e. g. a, ea, e a, The mean Term, The Extremes Rectangle Square Q.E.D. SCHOLIUM. Whence by the way may appear that Proposition of Archimedes (α) Eucl. 17 l. 6 & 20 l. 7. (β) lib. I de Sphaer. & Cyl. Prop. 14. That the Surface of a Right Cone is equal to the Circle, whose Radius is a mean Proportional between the Side of that Cone and the Semidiameter of the Base. For suppose OF to be a mean Proportional between the side of the Cone BC (Fig. 57) and the Semidiameter of the Base CD, since an equal number of Peripheries answer to an equal number of Radii in the same Proportion; half the Product of the first Line BC into the last Periphery, ½ e ab (that is, by Consect. 4. definite. 18. the Surface of the given Cone) will be equal to half Product of the mean Line into the mean Periphery, ½ e ab (i e. by Consect. 2. definite. 15.) to the Area of the Circle of the mean Proportional EF. Q. E. D. The same Proposition of Archimedes may also be Demonstrated after this Way: If the side of the Cone BC be called b, and the Semidiameter of the Base a) so that the Periphery may, by Consect. 1. definite. 31. be 2 ea, and so the Surface of the Cone, by Consect. 4. definite. 18. eab) the √ ab will be a mean proportional between b and a, by this 17th Proposition; which being taken for Radius, the whole Diameter will be √ 2 ab, and the Periphery 2 e √ ab; therefore by Consect. 2. definite. 15. half the Radius ½ √ ab multiplied by the Periphery (since √ ab multiplied, by √ ab necessarily produces ab) will give you the Area of the Circle by that mean (a) lib. I de Sphaer. & Cyl. prop. 15. Proportional, equal to the Surface of the given Cone, which before was expressed in the same Terms. Q. E. D. Hence also naturally flows this other Proposition, That the Surface of the Cone (½ e ab) is to its Base (½ ab) as the Side of the Cone (e b) is to the Radius of the Base b, as may appear from the Terms. Proposition XVIII. IF (b) E●cl. 16 l. 6 & 19 l. 7. 4. Quantities are Proportional, either continuedly or dircretely, the Product of the Extremes is equal to the Product of the Means. Demonstration. Suppose one Continual Proportional, a, ea, e a, e a. Extremes e a Means e a a ea Prod. e aa = Prod. e aa. Q.E.D. SCHOLIUM. ON this Theorem is founded the Rule of Three in Arithmetic; so called because having 3 Numbers, (2. 5. 8.) it finds an unknown fourth Proportional. For although this fourth be, as we have said, unknown, yets its Product by 2 is known, because the same with the Product of the Means, 5 and 8. Wherhfore the Rule directs to multiply the third by the second, that you may thereby obtain the Product of the Extremes: which divided by one of the Extremes, viz. the first, necessarily gives the other, i e. the fourth sought. Proposition XIX. IF 2 Products (on the other side) arising from the Multiplication of 2 Quantities, are equal, those 4 Quantities will be at lest directly Proportional. Demonstration. Suppose eba be the equal Product of the Extremes, and eab of the Means; the Extremes will either be ebb and a, or e and ba, or b and ea, as also the Means. But what way soever either is taken, there can be no other Disposition or placing of them, than one of the following. 1 ebb ebb a a — ee ab — ea b; or inversly. — a ebb — ab e — b ea 〈◊〉 e e e ba ba — ebb a — ea b; or inversly. — ba e — a ebb — b ea 3 b b ea ea — a ebb — ba e; or inversly. — ea b — ebb a — e ba; or inverting the Order of them all. In all these Dispositions there may be immediately seen a Geometrical Proportion, by what we have in Definition 3● and 33. CONSECTARSY. I AS we have shown one Sign of Proportionality in the Definition of it, viz. That the same Quotient will arise by dividing the Consequents by the Antecedents; so now we have another Sign of it, viz. The Equality of the Products of the Extremes and Means. II By a bore Subsumption may hence appear the Truth of Prop. 14. lib. 6. Euclid. at lest partly: Which we shall yet more commodiously show hereafter. Proposition XX. IF there are never so many Continual Proportionals, the Product of the Extremes is equal to the Product of any 2 of the Means that are equally distant from the Extremes, as also to the Square of the mean or middle Term, if the Terms are odd. Demonstration. Such are e. g. a, ea, e a, e a, e a, e a, e a, etc. and the Product of the Extremes, and of any two Terms equally remote them, and the Square of the mean or middle Term, every where e aa. Q. E. D. SCHOLIUM I NOR can there be any doubt but this will always be so, how far soever the Progression is continued; if 〈◊〉 consider that the last Term always contains the first, 〈◊〉 way of Reason, so many times multiplied as is the place of that Term in the rank of Terms, excepting one. Although therefore the last Term but one is in one degree of its Reason lesle than 〈◊〉 last, ●he second on the contrary, is in one more than the 〈…〉 re●ore the Product of the one will necessarily 〈…〉 ●e Product of the other. Thus also the las● 〈…〉 Degrees of Proportion lower than the 〈…〉 being to be multiplied into that, exc●●ds 〈◊〉 first 〈…〉 of the Proportion, etc. as may be seen thou Universal Ex●●●●● Hence you have the following CONSECTARIES. ●. HAving some of the Terms given in a Continual Proportion (e. g. suppose 10) you may easily found any other that shall be required (e. g. the 17th) as the last; If ●he 2 Terms given, being equally remote from the first and ●hat required (as are e. g. the vl and tenth) be multiplied by one another, and this Product, like that also of the Extremes, be divided by the first. II But this may be performed easier, if you moreover take ●n this Observation, That if, e. g. never so many places of proportionals, passing over the the first, be noted or marked by Ordinals or Numbers according to their places (as in this universal Example) a, ea, e a, e a, e a, e a, e a, 1 II III IV. V VI The place of the 7th Term is (e. g.) VI. (and so the place of any other of them being lesle by Unity than its number is among the Terms) and also composed of the places of any other equally distant from the Extremes, e. g. V and I IU. and II or twice III etc. III Here you have the Foundation of the Logarithms, i. ● of a Compendious Way of Arithmetic, never enough to b● praised. For if, e. g. a rank of Numbers from Unity, continually Proportional, be signed or noted with their Ordinals, as w● have said, as Logarithms, 1. 2. 4. 8. 16. 32. 64. 128. 256, etc. 1 II III IV. V VI VII. VIII. and any two of them (as 8 and 32) are to be multiplied together; add their Logarithms III and FIVE, and their Sum VII● gives you the Logarithm of their Product 256, as the Te●● equally remote from the 2 given one's and the first, and 〈◊〉 whose Product with the first (which is Unity) i e. itself w●●● be equal to the Product of the Numbers to be multiplied: A●● contrariwise, if, e. g. 128 is to be divided by 4, subtracting t●● Logarithm of the first TWO from the Logarithm of the second V● the remaining Logarithm FIVE points out the number sought 3● so that after this way the Multiplication of Proportionals 〈◊〉 by a wondered Compendium, turned into Addition, and the Division into Subtraction, and Extraction of the Square Ro●● into Bisecting or Halving, (for the Logarithm of the Squa●● Number 16 being Bisected, the half TWO gives the Root sought 4) of the Cube Root into Trisection (for the Logarithm of th● Cube 64 being Trisected, the third part gives the Cubi● Root sought 4). SCHOLIUM II THat we may exhibit the whole Reason of this admirabl● Artifice (which about 35 years ago was found out b● the Honourable Lord John Naper Baron of Merchiston in Scotland and published something difficult, but afterwards rendered much easier and brought to perfection by Henry Briggs, the first S●vilian Professor of Geometry at Oxford.) I say that we may exhibit the whole Reason of it in a Synopsis, after an easy way when its use appeared so very Considerable in the great Numbers in the Tables of Sines and Tangents, nor yet could they be useful without mixing vulgar Numbers with them, especially in the Practical Parts of Geometry, the business was to accommodate this Logarithmical Artifice to them both. First ●herefore that Artists might assign Logarithms to all the common Numbers proceeding from 1 to 1000 and 10000, & c. ●hey first of all pick out those which proceed in continued Geometrical Proportion, and particularly, though arbitrariously, those which increase in a Decuple Proportion, e. g. 1. 10. 100 1000 10000, etc. But now to fit them according to the Foundation of Consect. 8. a Series of Ordinals in Arithmetical Progression, we don't only substitute the simple Number 1, 2, 3, etc. but augmented with several Ciphers after them, that so we may also assign ●heir Logarithms in whole Numbers to the intermediate Numbers between 1 and 10, 10 and 100, etc. Wherhfore, by ●his first Supposition, Logarithms in Arithmetical Proportion, answer to those Numbers in Geometrical Proportion, after the ●ay we here see, 1 10 100 Log. 0000000 10000000 20000000 1000 10000 30000000 40000000, etc. As that they also exhibit certain Characteristical initial Notes, whereby you may see, that all the Logarithms between 1 and ●0 begin from 0, the rest between 10 and 100 from 1, the ●ext from 100 to 1000 from 2, etc. The Logarithms of the Primary Proportional Numbers being ●hus found, there remained the Logarithms of the intermediate Numbers between these to be found: For the making of which, ●fter different ways, several Rules might be given drawn from ●he Nature of Logarithms, and already shown in Consect. 3. See Briggs' Arithmetica Logarithmica, and Gellibrand's Trigonometria Britannica; the first whereof, chap. 5. and the following, shows ●t length both ways delivered by Neper in his Appendix. But ●he business is done more simply by A. Vlacq. in his Tables of Sines etc. whose mind we will yet further explain thus: If you are ●o found, e. g. the Logarithm of the Number 9, between 1 and ●0, augmented by as many Ciphers as you added to the Logarithm of 10, or the rest of the Proportionals (h. e. between 10000000 and 100000000) you must found a Geometric Mean Proportional, viz. by multiplying these Numbers together and extracting the Square Root out of the Product, by Pr● 17. Now if this Mean Proportional be lesle than 9 augments by as many Ciphers, between it and the former Denary Number you must found a second mean Proportional, than between this and that same a third; and so a fourth, etc. but if it 〈◊〉 greater, than you must found a mean Proportional between and the next lesle, etc. till at length after several Operation you obtain the number 9999998, approaching near 90000000. Now if between the Logarithm of Unity a● Ten (i e. between 0 and 10000000) you take an Arithmetical Mean Proportional (05000000) by Bisecting their S● by Prop. 14. and than between this and the same Logarithm Tenterhook, you take another mean, and so a third and a fourth, 〈◊〉 at length you will obtain that which answers to the last abo● mentioned, viz. 9 See the following Specimen. A TABLE of the Geometrical Proportionals between 1 and 10, augmented by 7 Ciphers, and of t●● Arithmetical Proportionals between 0 and 10000000 being the Logarithms corresponding to them. Geometrical Mean Proportionals. Arithmetical Logar. mean Proportionals 31622777 First, 05000000 56234132 Second, 07500000 74989426 Third, 08750000 86596435 Fourth, 09375000 93057205 Fifth, 09687500 89768698 Sixth, 09531250 91398327 Seventh, 39609375 90579847 Eighth, 09570312 90173360 Ninth, 09550781 89970801 Tenth. 09541015 Which is thus made: In the first Table a Geometrical Mean proportional between 10 000 000 and 100 000 000 the first Number of it; than another Mean between that and ●e same last 100000000, gives the second; and so to the ●●th, 93057205. Which, since it is already greater than the novenary, another Mean between it and the precedent fourth, ●●comes in order a sixth, but sensibly lesle than the Novenary. ●herefore between it and the fifth you will have a seventh ●ean yet greater than the Novenary; and between the sixth ●●d seventh, an eighth, somewhat nearer to the Novenary, but ●t yet sensibly equal, but somewhat bigger; moreover between ●●e sixth and eighth you will have a ninth, between the ninth ●●d sixth a tenth gradually approaching nearer the Novenary, but ●●t somewhat sensibly differing from it. Now if you conque this inquiry of a mean Proportional between this tenth, 〈◊〉 somewhat too little, and the precedent ninth as somewhat ●o big, and so onwards, you will at length obtain the Num●●r 8999 9998, only differing two in the last place from the novenary Number augmented by seven Ciphers, and conse●ently insensibly from the Novenary itself. But for the Logarithm of this in the second Column, by the same process you ●●e to found Arithmetical Mean Proportionals between every 2 ●ogarithms answering to every two of the superior ones, till you ●nd, e. g. the Logarithm of the tenth Number 09541015, ●d so at length the Logarithm of the last, not sensibly differing from the Novenary, 09542425. Thus having found, with a great deal of labour, but also ●ith a great deal of advantage to those that make use of them, ●●e Logarithms of some of the numbers between 1 and 10, and ●0 and 100, etc. you may found innumerable ones of the other intermediate Numbers with much lesle labour, viz. by the help ●f some Rules, which may be thus obtained from Consect. 3 of ●e precedent Proposition. The Sum of the Logarithms of the ●umber Multiplying and the Multiplicand, gives the Logarithm of the product. 2. The Logarithm of the Divisor subtracted from the Logarithm of the Dividend, leaves the Logarithm of the Quotient: ●he Logarithm of any number doubled, is the Logarithm of the Square, tripled of the Cube, etc. 4 The half Logarithm of any number is ●he Logarithm of the Square Root of that number, the third part of 〈◊〉 the Cube Root, etc. Thus, e. g. if you have found the Logarithm of the number 9, after the way we have shown, by th● same reason you may found the Logarithm of the number 5 (vi● by finding mean Proportionals between the second and the fi● number of our Table, and between their Logarithms, etc. and by means of these 2 Logarithms you may obtain several others: First, since 10 divided by 5 giveth 2; if the Logarithm of 5 be subtracted from the Logarithm of 10, you'll have th● Logarithm of 2, by Rule the second. Secondly, since 10 multiplied by 2 makes 20, and by 9 makes 90, by adding th● Logarithms of 10 and 2, and 10 and 9, you'll have the L●garithms of the numbers 90 and 20, by Rule 1 Thirdly Since 9 is a Square, and its Root 3, half the Logarithm of 〈◊〉 gives the Logarithm of 3, by Rule 4. since 90 divided by 〈◊〉 gives 30, the Logarithm of this number may be had by subtracting the Logarithm of 3 from the Logarithm of 90, b● Rule the second. Fifthly, 5 and 9 squared make 25 and 8▪ the Logarithms of 5 and 9 doubled, give the Logarithms 〈◊〉 these numbers, by Rule 3 In like manner, sixthly, the Su● of the Logarithms of 2 and 3, or the Difference of the L●garithms of 5 and 30, give the Logarithm of 6, and the Su● of the Logarithms of 3 and 6, or 2 and 9, giveth the Logarithm of 18; the Logarithm of 6 doubled, gives the Logarithm of 36, etc. And after this way you may found and reduce it to Tables, the Logarithms of Vulgar Numbers from 1 to 100LS (as in the Tables of Strauch. p. 182, and the following) or 〈◊〉 100000 (as in the Chiliads of Briggs) But as to the manner ●deducing the Tables of Sines and Tangents from these Logarithms of Vulgar Numbers, we will show it in Scholar of Pr●● 55, only hinting this one thing beforehand; that this Artifi● of making Logarithms is elegantly set forth by Pardies in hi● Elements of Geometry, pt 112. by a certain Curve Line then●● called the Logarithmical Line; by the help whereof he suppose Logarithms may be easily made; and having found those o● the numbers between 1000 and 10000, he shows, that all others may be easily had between 1 and 1000 Wherhfore w● shall Discourse more largely in Scholar definite 15. lib. 2. Proposition XXI. IF the first Term of never so many Continual Proportionals, be subtracted from the last, and the Remainder divided by the name of the Reason or Proportion lessened by Unity, the Quotient will be equal to the Sum of all except the last. Demonstration. ea e a e a e a e a The last Term lesle the first e a − a Divided by the name of the Reason lessened by unity. e − 1 * Quote. e a + e a + e a + e a + ea + a; And it is evident from the Operation, that the same will always hap though the number of Terms be continued never so far. e − 1 e − 1 e − 1 e − 1 e − 1 e − 1 CONSECTARYS. I WHerefore in adding never so great a Series of Geometrical Proportionals, since it is enough that the first and last Term, and the Name of the Reason be known, by this Prop. and having found at lest some of the Terms of the Proportion, any other may be afterwards found, whose place will be compounded of the places of the two Antecedent ones, according to Consect. 2. Prop. 20. viz. by Multiplying the Terms answering to the two abovementioned places, and dividing the Product by the first Term; thence it will be very easy to add a great Series of Proportionals into one Sum, though the particular separate Terms remain almost all of them unknown. SCHOLIUM. THese are the same Practical Arithmetical Rules concerning Geometrical Progressions; for the illustration of whic● Swenterus in Delic. has given us so many pleasant Examples, li● 1 Prop. 59 and folio First of all, that famous Example is of th● kind which relates to the Chequer-worked Table or Board t● fling Dices on, with its 64 little Squares, which Dr. Wa●● has translated out of the Arabic of Ebn Chalecan, into Latin in Oper. Mathem. part. 1. Chap. 31. for the illustration of whic● we have heretofore composed an Exercitation, and shall he● only note these few things: If there are supposed 64 Terms 〈◊〉 double Proportion from Unity, and the first of them, note● with their local Numbers, are these that follow; 1 2 4 8 16 32 64 128 I TWO III IV FIVE VI VII You may have the Term of the 13th place, 8192, by multiplying together the VIth and VIIth place; and the Ter● of the XXVIth place, by squaring or multiplying this new Product again by itself, and moreover the Term of the L●● place, by multiplying that Product again by itself; and further more the Term of the LIXth place, by multiplication of the number last found by the number of the VIIth place, and lastly the Term of the LXIIId place (i e. the last in the proposed Series) by multiplying this last of all by the number of the IV●● place. II Moreover you may, by this Art, collect infinite Seri● of Proportional Terms into one Sum, although it is impossible 〈◊〉 run over all the Terms separately, because infinite. e. g. in 〈◊〉 continued Series of Fractions, decreasing in a double Proportion ½ ¼, ⅛ , , etc. ad infinitum, if you take them backward, you may justly reckon a cipher or 0, for the fi●● Term (for between ½ and 0 there may be an infinite Numbe● of such Terms) and the infinite Sum of these Terms will b● precisely equal to Unity; for subtracting the first 0, from the last ½, and the remainder ½ being divided by the name of th● Reason lessened by 1, i e. by I which divides nothing, th● Quotient ½ is the Sum of all the Terms excepting the last, by Prop. 21. and so the last ½ being added, the Sum of all in that Series will be I Now if the last is not ½ but I, the Sum of all will necessarily be 2; if 2 be the last, the Sum of all will be 4; in a word, it will be always double the last Term. III And since in this case the Sum of all the precedent Terms is equal to the last Term, the one being subtracted from the other, there will remain nothing, i e. ½ − ¼ − ⅛ − − , etc. in Infinitum, is = 0, and also 1 − ½ − ¼, etc. or 2 − 1 − ½ − ¼. etc. = 0. IV. In like manner the Sum of infinite Fractions decreasing in triple Reason in an infinite Series (⅓+ + + , etc.) will be equal to ½: for if from the last ⅓ (again in an inverted Order) you subtract the first 0, and the Remainder ⅓ be divided by the name of the Reason lessened by Unit, i e. by 2, the Quotient ⅙ will be the Sum of all the antecedent Terms, and adding to this last ⅓ or the Sum of all will be or ½. V Thus an infinite Series of Fractions decreasing from ¼ in a Quadruple Proportion (¼+ + etc.) is equal to ⅓; for subtracting the first 0 from the last ¼, and the remainder ¼ being divided by the name of the Proportion, i e. by 3, you will have the sum of all except the last, and adding also the last ¼ or , you'll have the whole Sum or ⅓. VI Thus also an infinite Series decreasing from ⅕ in a Quintuple Proportion (⅕+ + , etc.) is equal to ¼: ⅙+ + , etc. is equal to ⅕ etc. and so any Series of this kind is equal to a Fraction, whose Denominator is lesle by an Unit than the Denominator of the last Fraction in that Series. VII. Generally also, any infinite Series of Fractions decreasing according to the Proportion of the Denominator of the last Term, and having a common Denominator lesle by an unit than the Denominator of the last Term (e. g. ⅔+ + , etc. or ¾+ + , etc. or ⅘+ + , etc.) is equal to Unity, after the same way as the Series Consect. 2. which may be comprehended under this kind, and which may be demonstrated in all its particular cases by the same method we have hitherto made use of, or also barely subsumed from Consect. 4, 5, and 6. For since ⅓+ + , etc. is equal to ½; ⅔+ + will be equal to , or 1, and so in the rest. VIII. Particularly the sum of , &c, decreasing in a Quadruple Proportion, is equal to ; and the sum of , etc. is equal to ; and the sum of , etc. decreasing in Octuple Proportion, is equal to ⅛: For subtracting the first Term 0, and dividing the remainder by the name of the Reason lessened by 1, i e. by 3, the Quotient gives the sum of all except the last. This therefore (viz. ) being added, the sum of all will be or : In like manner being divided by the name of the Reason lessened by Unity, the Quotient will give , and adding the last, the sum of all will be i. e. ⅛. So that hence it is evident, that , etc. or − , etc. in Infinitum, will be equal to nothing; also ⅛ − − − etc. = 0. IX. The Sum of a simple Arithmetical Progression (i e. ascending by the Cardinal Numbers) continued from 1, ad Infinitum, is Subduple of the Sum of the same number of Terms, each of which is equ●● to the greatest; or on the contrary, this latter Sum is double of the fo●mer. We might have subsumed this in Consect. 4. Prop. 16. for, prefixing a cipher before Unity, it will be a case of that Consectary, the Sum of the Progression remaining still the same. B●● that this is true, in an infinite Series beginning from Unity (f●● in a finite or determinate one, the proportion of the Sum is always lesle than double, though it always approaches to it, and come so much the nearer by how much greater the Series is) 〈◊〉 shall now thus Demonstrate: To the Sum of three Terms, 〈◊〉 2, 3, i e. 6, the sum of as many equal in number to the greatest, i e. 9, has the same Proportion as 3 to 2; but t● the sum of six Terms, 1, 2, 3, 4, 5, 6, i e. 21, the su● of as many equal to the greatest, i e. 36, has the same proportion as 3 to 1+¾, that is, as 3 to 2 − ¼, the decrease being ¼: but to the sum of 12 Terms, which may be found b● Consect. 1. Prop. 16. = 78, the sum of so many equal to the greatest, viz. 144. has the same proportion (dividing both sid● by 48) as 3 to 1 , i e. 3 to 1+½+⅛ (for 24 make ½, a●● the remainder is the same as ⅛) that is, as 3 to 2 − ¼ − ●● the decrement being now ⅛. Since therefore, by doubling th● number of Terms onward, you'll found the decrement to be , an● so onwards in double Proportion; the sum of an infinite Number of such Terms, in Arithmetical Progression, equal to th● greatest will be to the sum of the Progression from 1, ad Infinitum, as 3 to 2 − ¼ − ⅛ − , etc. that is, by Consect. 2 and 3, as 3 to 2 − ½, that is, as 3 to 1 ½, or as 2 to 1. Q.E.D. X. The Sum of any Duplicate Arithmetical Progression (i e. a Progression of Squares of whole numbers ascending) continued from 1 ad Infinitum, is subtriple of the Sum of as many Terms equal to the greatest as is the number of Terms: For any such finite Progression is greater than the subtriple Proportion, but approaches nearer and nearer to it continually, by how much the farther the Series of the Progression is carried on. Thus the Sum of 3 Terms 1, 4, 9 = 14 is to thrice 9 = 27 as 1 , or 1 , or 1+½+ to 3 (dividing both sides by 9,) the Sum of six Terms, 1, 4, 9, 16, 25, 36, viz. 91. to six times 36, i e. to 216 (dividing both sides by 72) is as 1+¼+ to 3; and the Sum of 12 Terms 650, to 12 times 144, i e. to 1728 (dividing both sides by 576) is as 1+⅛+ to 3, etc. the Fractions adhering to them thus constantly decreasing, some by their half parts, others by three quarters (for is ; therefore the first decrement is and , is ; therefore the second decrement is , etc.) Wherhfore the Sum of the Infinite Progression will be to the Sum of the like number of Terms equal to the greatest, as , etc. to 3, that is, by Consect. 3 and 8, as 1 to 3. Q.E.D. XI. The Sum of a triplicate Arithmetical Progression (i e. ascending by the Cubes of the Cardinal Numbers) proceeding from 1 through 27, 64, etc. ad Infinitum, is Subquadruple of ●he Sum of the like number of Terms equal to the greatest. For the Sum of 4 Terms, 1, 8, 27, 64, i 100, to 4 times 64, i e. 256 (dividing both sides by 64) will be found to be as 1+½+ to 4; but the Sum of 8 Terms, 1, 8, 27, 64, 125, 216, 343, 582, i e. 1296 to 8 times 512, that is, 4096 (dividing both Sides by 1024.) will be found to be as 1+¼+ to 4, etc. The adhering Fractions thus constantly decreasing, the one by their ½ part, the others by ●● (for is , and is , etc. Wherhfore the Sum of the Infinite Progression will be to the Sum of a like (Infinite) number of Terms, equal to the greatest, as , etc. , etc. to 4; that is, by Consect. 3 and 8, as 1 to 4. Q. E. D. XII. The Sum of an Infinite Progression, whose greatest Term is a Square Number, the others decreasing according to the odd numbers 1, 3, 5, 7, etc. is in Subsesquialteran Proportion of the Sum of the like number of equal Terms, i e. as 2 to 3. For the Sum of three such Terms, e. g. 9, 8, 5, i e. 22 to thrice 9, i e. 27. is (dividing both sides by 9) 〈◊〉 2 , viz. to 3, or 2+½ − to 3. But the Sum of s●● such Terms, 36, 35, 32, 27, 20, 11, i e. 161, to six time 36, i e. 216 (dividing both sides by 72) is as 2+¼ − , etc. the adhering Fractions thus always decreasing, some by ½, others by ¾, as above in Consect. 10. Wherhfore the Sum of the Infinite Progression will be to the Sum of the like number of Terms equal to the greatest, as , etc. to 3, i e. by Consect. 3 and 8, as 2 to 3. Q E. D. SCHOLIUM II THus we have, after our method, demonstrated the chie● Foundations of the Science or Method, or Arithmetic ●● Infinites, first found out by Dr. John Wallis, Savilian Professor of Geometry at Oxford, and afterwards carried further by Det●lerus Cluverus, and Ishmael Bullialdus. And from these Foundations we will in the following Treatise demonstrate, and that directly and à priori, in a few Lines, the chief Propositions o● Geometry, which the Ancients have spent so much labour, and composed such large Volumes to demonstrate, and that but indirectly neither. Proposition XXII. THe Powers of Proportionals whether continuedly or discreetly, such as the Squares, Cubes, etc. are also Proportional. Demonstration. Continual Proportionals. Discrete Proportionals. a ea e a e a a ea b ebb Squares aa e a e a e a a e a b e b Cubes a e a e a e●a³ a e a b e b Q E. D. SCHOLIUM. YOu founded in this Truth, 1. the Reason of the Multiplication and Division of Surd Quantities: For since from the Nature and Definition of Multiplication, it is certain, that 1 is to the Multiplier as the Multiplicand to the Product (for the multiplicand being added as many times to itself as there are Units in the Multiplier, makes the Product) if the √ 5 is to be multiplied by √ 3, then as 1 to the √ 3, so the √ 5 to the Product; and, by the present Proposition, as 1 to 3, so 5 to the □ Product, i e. to 15. Wherhfore the Product is √ 15; and so the Rule for Multiplying Surd Quantities is this: Multiply the Quantity under the Radical Signs, and prefix a Radical Sign to the Product. (α) Eucl. lib. 6. prop. 22. Likewise since it is certain from the Nature of Division, that the Divisor is to the Dividend as 1 to the Quotient (for the Quotient expresses by its Units how many times the Divisor is contained in the Dividend) if the √ 15 is to be divided by √ 5, you'll have √ 5 to the √ 15 as 1 to the Quotient, and, by the present Scholium, 5 to 15, as 1 to the □ of the Quotient, i e, to 3. Therefore the Quotient is the Root of 3, and so the Rule of dividing Surd Quantities this; viz. Divide the Quantities themselves under the Radical Signs, and prefix the Radical Sign to the Quotient. II Hence also flows the usual Reduction in the Arithmetic of Surds, of Surd Quantities to others partly Rational, and on the contrary, of those to the form of Surds, e. g. If you would reduce this mixed Quantity 2 a √ b, i e. 2 a multiplied by the √ b, to the form of a Surd Quantity; which shall all be contained under a Radical Sign; The Square of a Rational Quantity without a Sign 4 aa, if it be put under a Radical Sign, in this form √ 4 aa, it equivalent to the Rational Quantity 2 a; but the √ 4 aa being multiplied by √ b maketh √ 4 aab. by N● 1. of this Scholium. Therefore √ 4 aab is also equivalent to the Quantity first proposed 2 a √ b. Reciprocally therefore, if th● form of a mere Surd Quantity √ 4 aab, is to be reduced to on● more Simple, which may contain without the Radical Sig● whatever is therein Rational, by dividing the Quantity comprehended under the sign √ by some Square or Cube, etc. as here by 4 aa, (i e. √ 4 aab by √ 4 aa, i e. 2 a) the Quotient wil● be √ b, which multiplied by the Divisor 2 a, will rightly express the proposed Quantity under this more simple Form 2 a √ ● Which may also serve further to illustrate the Scholia of Prop. 7. and 10. Proposition XXIII. IF there are four Quantities Proportional, (a, ea, b, ebb) they will be also Proportional, 1. Inversly. ea to a as ebb to b. 2. Alternatively, (α) Eucl. 15, 16. u 9 10, 13, vi●. a to b as ea to ebb. 3. Compoundedly, (β) 18, v. a + ea to ea, so b + ebb to ebb. 4. Conversly, a + ea to a as b + ebb to b. 5. Dividedly, (γ) 17, v. a − ea to ea as b − ebb to or a ebb or 〈◊〉 6. (α) Eucl. 15, 16. u 9 10, 13, vi●. By a Syllepsis, a to ea as a + b to ea + ebb. 7. By a Dialepsis, a to ea as a − b to ea − ebb. Which are all manifest, by comparing the Rectangles of the Means and Extremes according to to Prop. 19 and its Consect. 1. or by dividing any of the Consequents by their Antecedents, according to Def. 31. Proposition XXIV. IF in a (β) Eucl. 3, 20, 22. lib. u 14. seven. double Rank of Quantities you have as a to ea, so b to ebb, and also as ea to oa, so ebb to ob, than you'll have also by proportion of Equality orderly placed, as the first a, to the last oa, in the first Series; so the first b, to the last ob, in the second Series. Which is manifest from the Terms themselves. Proposition XXV. BUt (γ) Eucl. 21, 23. lib. v. if they are disorderly placed as oa to ea † so eob to ob * as ea to a † so ob to ebb, * you'll have here again by proportion of Equality, as the first oa to the last a, in the first Series; so the first eob to the last ebb, in the second Series. As is evident from the Rectangles of the Extremes and Means, as also from the very Terms. Proposition XXVI. IF (α) Eucl. 1. 12. u 5. 6. 12. seven. as the whole ea to the whole a, so the part ebb to the part b; than also will the Remainder Remainder Whole Whole ea − ebb to the a − b, as the ea to the a. This is evident from the Rectangle of the Extremes and Means, both which are eaa − eab. Q.E.D. Proposition XXVII. REctangles or Products having one common Efficient or Side, are one to another as the other Efficients or Sides. Demonstration. Suppose the Products to be ab and ac, having the common Efficient a; I say they are as b to c, so ab to ac. Which is evident at first sight, by comparing the Products of the Extremes and Means, and also fully shows, that other way of proving Proportionality, whereby by dividing the Consequents by their Antecedents, the identity or sameness of the Quotients are want to be demonstrated. SCHOLIUM I I THe Reduction of Fractions either to more compounded or more simple ones is founded on this Theorem; on the one hand by multiplying, on the other by dividing, by the same quantity, both the Numerator and the Denominator, as, e. g, and and , ⅓, , , etc. are in reality the same Fractions. And II The Reduction of Fractions to the same Denomination, as if and are to be changed into two others that shall have same Denominator; (α) Eucl. 5 & 19 lib. u 7 & 11. l. 7. this is to be done by multiplying the Denominators together for a new Denominator, (β) Besides several other Prop. see also the 17 & 18 lib. seven. and each Numerator by the Denominator of the other for a new Numerator, and you'll have for the two Fractions above— and SCHOLIUM II WE will here for a conclusion of Proportionals, show the way of cutting or dividing any Quantity in Mean and Extreme Reason, viz. if for the greater Part you put x, the lesle will be a − x; and so by Hypoth. these three, a, x and a − x, will be proportional, by Def. 34. Therefore by Prop. 17. the Product of the Extremes aa − axe = to the Square of the Mean xx, and (adding on both sides axe) aa = xx + axe; and moreover adding on both sides ¼ aa, you'll have aa = xx + ax+ + aa. Now this last Quantity, since it is an exact Square, whose Root is x+ + a, you'll have √ aa = x+ + a, and (subtracting from both sides a) √ aa − ½ a = x. Now therefore we have a Rule to determine the greater part of a given Quantity to be divided in Mean and Extreme Reason, viz. if the given Quantity be a Line, e. g. AB = a (Fig. 58.) join to it (α) Eucl. 11. lib. II & 30. lib. vi. at Right Angles AC = ½ a: Wherhfore by the Theorem of Pythagoras from Scholar definite 13. the Hypothenuse CB, or, which is equal to it, CD = √ ¾ aa; and consequently AC = ½ a being taken out of CD, the Remainder AD, or A, which is equal to it, will be = x, the greatest part sought; according to Euclid, whose Invention this first Specimen of Analysis, by way of Anticipation, reduces to its original Fountain. As for Numbers (though none accurately admits of this Section) the sense of the Rule, or which is all one as to the thing itself, is this: Add the Squares of a whole Number and its half, and subtract the said half from the Root of the Sum (which can't be had exactly, since it is √ . CAAP. V Of the Proportion or Reasons of Magnitudes of the same kind in particular. Proposition XXVIII. TRiangles and Parallellograms, also Pyramids and Prisms and Parallelepipeds, lastly Cones and Cylinders, each kind compared among themselves, if they have the same Altitude, are in the same Proportion to one another as their Bases. Demonstration. This and the following Proposition might have been by ● bore Subsumption added, as Consectaries to the precedent; fo● the Altitudes in the one, and Bases in the other, may be looke● on as common Efficients, and the Magnitudes mentioned as their Products: But for the greater distinction sake, we will thus Demonstrate them more particularly. I If the equal Altitudes of two Triangles, o● (α) Eucl. Prop. 1. lib. vi. two Parallellograms, are called b and 〈◊〉 Base of the one a, and of the other ea; these Products will be ba and bea, the other ½ ba and ½ bea, by Def. 28. Scholar 2. II Likewise the equal Altitudes of two Priso● (β) Prop. 5.6. lib. xii. 25, 32. xi. & Cons. 30 & 31 of the same or Pyramids, may be called b, and the Proportion of their Bases expressed by a and ea; a●● the Prisms will be among themselves as ba to be● and the Pyramids as ⅓ ba to ⅓ bea, (γ) Prop. 11. lib. xii. by the sai● Scholar Num. 3. (δ) Prop. 35, 36, 37, 38, 39, 40, lib. lib. 1 & 29, 30, 31. lib. xi. III There is also the same Proportion of Cylinders and Cones as of Pyramids and Prisms, by Consect. 4. definite. 17. But, as a to ea so is ba to bea. − ½ ba to ½ bea. − ⅓ ba to ⅓ bea. Q.E.D. CONSECTARY. THerefore Magnitudes of the same kind upon the same or equal Bases (dgr;) and of the same height, are equal among themselves, and the contrary. Proposition XXIX. TRiangles and Parallellograms, Pyramids and Prisms and Parallelepipeds, Cones and Cylinders, being on equal Bases, are in the same Proportion as their heights. (*) Schol prop. 1. l. 6, 12, 13, 14 Demonstration. Let all their Bases be called a, and the Proportions of their heights be as b to ebb: Therefore, 1. the Parallellograms, Parallelepipeds and Cylinders, are one to the other of the same kind, as ba to eba; the Triangles as ½ ba to ½ eba; the Pyramids and Cones as ⅓ ba to 3 eba, by Def. 28. Scholar 2. But, as b to ebb, so is ba to eba. and ½ ba to ½ eba. and ⅓ ba to ⅓ eba. Q.E.D. Proposition XXX. EQual ‖ lib. 6. prop. 14. 15. l. 11.34. l. 12.11, and its Coral. also Prop. 15. Triangles, Parallellograms, Prisms, Parallelepipeds, also equal Pyramids, Cones, and Cylinders, have their Bases and heights reciprocally Proportional. Demonstration. For if for the equal Triangles you put ½ ab, for the Cones and Pyramids ⅓ ab, and for the rest ab; whether the Bases of the equal Quantities are supposed to be a, and so the Altitudes on both sides b; or if the Base of the one be a and b the Altitude, but the Base of the other b and the Altitude a, you'll certainly have eitherways, as a to a, so Reciprocally b to b; the Base of the former to the Base of the latter, as the Altitude of the latter to the Altitude of the former, or, as a to b, so Reciprocally a to b. Q.E.D. CONSECTARY. AND those Magnitudes of the same kind, whose Bases and Altitudes are thus Reciprocal, are equal by Prop. 18. for the Product or Rectangle of the Extremes is ab, and that of the Means ba. Proposition XXXI. TRiangles, Parallellograms, Prisms, Parallelepipeds, Pyramids, Co●●● and Cylinders, each kind compared among themselves, are in the Proportion compounded of the Proportion of their Altitudes and Bases. (α) Prop. 23. lib. 6. Demonstration. Suppose the Base of the one to be a, and the other ea, and the Altitude of the one b, of the other ib; therefore the one will be to the other, as a b to eiab, or ½ ab to ½ eiab, or ⅓ ab to ⅓ eiab; i e. every where as a to ei● i e. in Proportion compounded of a to ea, and of b to ib, by Consect. 2. Def. 34. Q.E.D. SCHOLIUM. FRom what we have hitherto Demonstrated, we may not only make an estimate of Magnitudes of the same kind compared together, which is easy to any one who attentively considers them; but also with F. Morgues, deduce a General Rule of expressing the Proportions of any Rectilinear Planes or Solids, contained under Plane Surfaces, by the proportion of one Right Line to another. For since the one may be resolved into Triangles, and the other into Pyramids, having first two Rectilinear Planes given and thus resolved, upon a Right Line I make the ▵ abc (Fig. 49.) Equal to one of the Triangles of either of the Planes e. g. to ABC; than having drawn the Parallel cm, if the ▵ BCD has the same Altitude with the former, you need only join the Base BC to the Base ab. But if the Altitude DS is greater than the Altitude of the other e. g. by ⅕, than you must make b f equal to the Base bc augmented by a fifth part, and the Triangle bcf will = BCD, and the whole acf = to the Rectilinear Figure ABCD. If now therefore I likewise make another Triangle ghi equal to another Rectilinear Figure between the same Parallels, than will the ▵ acf be to the ▵ ghi, that is, the Right Lined Figure ABCD to the Right Lined Figure FGHIK, as of to gh, by Prop. 28. 2. Having 2 Right Lined Solids given, and having resolved them into Triangular Pyramids, they may be transferred between 2 parallel Planes, viz. by augmenting or diminshing their Triangular Bases reciprocally, according to the excess or defect of their Altitudes, as was done above with the Linear Bases; than those Triangular Bases on both sides may be converted into one Triangular Base, and consequently each Solid into a Pyramid equal to itself; which two Pyramids will be one to the other as their Triangular Bases. And because the Proportions of these Bases may be reduced to the Proportion of two Lines each to the other, by Nᵒ 1. of this; therefore also the Reason or Proportion of the two Solids may be expressed by the Proportion of two Lines. Q.E.D. Proposition XXXII. Circle's (β) Eucl. Prop. 2. l. 12. are in the same Proportion to one another as the Squares of their Diameters. Demonstration. Suppose a to be the Diameter of one Circle, and b of another; than by definite. 31. Consect. 1. the Area of the one will be ¼ eaa, and that of the other ¼ ebb. But as aa to bb so is ¼ eaa to ¼ ebb by Consect. 1. Prop. 19 Q E. D. CONSECTARY I THe same will in like manner be manifest of like Sectors' Circles, while for the parts of the Periphery you put and ib, as for the wholes we put ea and ebb: for thus the A● of the one will be ¼ iaa, and of the other ¼ ibb. CONSECTARY II CYlinders whose Altitudes are equal to the Diameters of th● Bases, are in proportion to one another as the Cubes their Diameters; for the Cylinders will be ¼ ea and ¼ ebb, Cubes a and b. CONSECTARY III HEnce also (whatever the Reason of the Sphere is to the Cylinder of the same Diameter and Height; which will hereafter Demonstrate, and which in the mean while will denote by the name of the Reason y) I say, hence Sphe●● which have the same Proportion to one another as these Cyli●ders (viz. as ¼ ea to ¼ ebb, so ¼ yea to ¼ yeb) will also (by C●● sect. 1.) be in the same proportion as the Cubes, a to b is also evident from these Terms themselves. Proposition XXXIII. THE Angle (α) Prop. 18 lib. 12. (β) Eucl. Prop. 20. l. 3. at the Centre of any Circle ACB (Fig. 60) to an Angle at the Circumference which has the same Arch its Base ADB, as 2 to 1. Demonstration. The truth of this has already appeared fro● Scholar definite. 10. Nᵒ 3. but here we will demonstrate it otherwise in its three Cases, after E●clids way. In the first Case DE being conceive Parallel to CB, by Def. 11. Consect. 1 and 2. th● External Angle ACB is = to the Internal A●gle Pag. 95. 61 62 63 64 65 66 67 68 69 70 71 72 ADE, and the Angle BDE, is equal to the alternate Angle BCD, i e. to the other at the Base CDB, by Consect. 2. definite. 13. Therefore BDE is as 1, and CDE, i e. ACB as 2. In the second Case the whole ECB is double of the whole EDB, and the subtracted Angle ECA is double of the subtracted Angle EDA, by Case 1 Therefore the Remainder ACB is also double of the Remainder ADB, by Prop. 26. In the third Case the part ECA is double of the part EDA, and also the part ECB is double of the part EDB, by Case 1 Therefore the whole ACB is double the whole ADB. Q.E.D. CONSECTARYS. I HEnce all Angel's ADB (α) Eucl. prop. 21. 27. 31. lib. 3. in the same Segment are equal, and the Angle ADB (Fig. 61.) in a Semicircle is a Right one; because the Aperture at the Centre answering to it, ACB contains two Right Angles: The Angle in a lesle Segment than a Semicircle EDF, is greater than a Right one; because the Aperture at the Centre EGHFC answering to it, comprehends more than two Right Angles. An Angle, lastly, in a Segment greater than a Semicircle GDH, is lesle than a Right one; because its double at the Centre GCH is lesle than two Right ones. All which we have already otherwise demonstrated in Scholar Def. 10. Nᵒ 6. II All the three Angles (β) prop. 32. l. 1 of any Triangle ABDELLA taken together, are equal to two Right ones; because they are the half of the three at the Centre C, which always make 4 Right ones, by definite. 8. Consect. 2. III Therefore any external Angle JAB, is equal to the two Internal opposite ones at B and D; because that, as well as they with the other contiguous to them BAD, make two Right ones, by Consect. 1. of the same definite. IU. And the greatest Side of a Triangle, because it insists on a (α) Eucl. prop. 21. 27. 31. lib. 3. greater Arch of a Circumscribed Circle, does also necessarily subtend a greater Angle, by virtue of Consect. 1. hereof. Proposition XXXIV. IN Equiangular Triangles (ACB and abc, Fig. 62.) the Sides ●bout the equal Angles are Proportional, viz. as AB to BC, so ab to bc, and as BC to CA so is bc to ca etc. (β) Demonstration. For having described Circles through the Vertex of each Triangle, according to Consect. 6. definite. 8. by reason of the supposed equality of the Angles A and a, B and b, C and c, th● Arches also AB and ab, etc. will necessarily agreed in the number of Degrees and Minutes, by the foregoing 33 Prop, a●● so also the Chords AB and ab, BC and bc, etc. will agreed in th●● number of Parts of the Radius or whole Sine ZA and za, 〈◊〉 Consect. 2. definite. 10. Wherhfore as many such Parts as A●● has, whereof az has also 10000000, so many such also will a● have, whereof az has also 10000000, etc. Therefore AC● to CB as ac to cb, etc. Q. E. D. CONSECTAYS. I WHerefore by the same necessity the Bases of such T●●angles AB and ab, will be proportional to their Altitudes CD and cd, as being Right Sins of the like Arches (〈◊〉 and cb, or rather CE and c e; and so for similar or like Tria●gles (and consequently also Parallelograms) we may rightly suppose that their Bases are as a to ea, and their heights as b 〈◊〉 ebb; though we must not immediately conclude on the contrary, tha● because their Bases and Altitudes are so, therefore they are S●milar. II As also in Similar Parallelepipeds it will be manifest 〈◊〉 any attentive Person, that the Bases are in a duplicate Proportion of the Altitudes. For since the Planes of Similar Solids a●● equal in number, and Similar each to the other, if for A● (Fig. 63.) we put a, and for BC b, AB will = ea and BC = ebb; and so that Basis will be to this as ab to eeab. Moreover having let fall the Perpendiculars EH and EH, the Triangle ●BH and EBH are similar, and by putting c for BE, BE will be ec, putting also d for EH, EH will consequently be ed. But the Reason of the Base ab to the Base eeab, is duplicate of the Reason of d to e d, by Def. 34. Wherhfore in Similar Parallelepipeds we may rightly suppose, that their Bases are as a b to eeab, or as a to eea, and their Altitudes as d to ed. SCHOLIUM I From this Proposition flows first of all the chiefest part of Trigonometry for the Resolution of Right Angled ▵ ▵: For since in any Right Angled Triangle, if one side, e. g. AB (Fig. 64.) be put for the whole Sine, the other BC will be the Tangent of the opposite Angle at A (and in like manner if CB be the whole Sine, BASILIUS will be the Tangent of the Angle C;) but if the Hypothenuse AC be made Radius or whole Sine, than the Side BC will be the Right Sine of the Angle A, or the Arch CD described from the Centre A, and AB the Right Sine of the Angle C, or the Arch A, described from the Centre C, (we will omit mentioning the Secants, because the business may be done without them) which all follow from Def. 10. Wherhfore you may found, I The Angles. 1. From the Sides by inferring As one leg to the other, so the whole Sine to the Tangent of the Angle opposite to the other Leg. 2. From the Hypoth. & one side, by inferring As the Hyp. to the W.S. (whole fine) so the given leg to the S. of the opp. angle II The Sides. 1. From the Hypoth. and Angles: As the W. S. to the Hypoth. so the Sine of the Angle, opposite to the Leg sought, to the Leg itself. 2. From one Leg and the Angles: As the W. S. to the given Leg, so the Tangent of the Angle adjacent to it, to the Leg sought. 3. From the Hypoth. and one of the Sides: Having first found the Angles, it's done by the 2, 1. or by the Pythagorick Theorem. III The Hypothenuse. 1. From the Angles and one of the Legs. As the S. of the Angle, opposite to the given Leg, to that Leg, so the W. S. to the Hypoth. 2. From the Legs given; Having first found the Angles its done by the 1. or by the Pythagorick Theorem. III Inversly also, if two Triangles ABC and ABC (〈◊〉 the Figure of the present Proposition) have one Angle of o●● equal to one Angle of the other (e. g. B and B) and the Sid●● that contain these equal Angles proportional (viz. as AB to B● so AB to BC) than the other Angles (A and A, C and C will be also equal, and the Triangles similar (α) Eucl prop. 6. lib. 6. for to 〈◊〉 like Chords AB and AB, BC and BC, there answer by t●● Hypoth. like or similar Arches, i e. equal in the number 〈◊〉 Degrees and Minutes; and to these also there answer equal Angels both at the Periphery and Centre. IV. (Fig. 65. Nᵒ 1.) If (β) Eucl. 2. lib. 6. the Sides of the Angle BARNES are cut by a Line DE, parallel to the Base BC, the Segments' 〈◊〉 those, Sides will be proportional, viz. A to EC as AD to BD● for by reason of the Parallelism of the Lines BE and BC, th● Triangles ADE and ABC are Equiangular: Therefore as th● whole BASILIUS to the whole AC, so the part AD to the part A● and consequently also the remainder EC to the remainder D● as the part EA to the part AD, by Prop. 26. and alternatively by Prop. 24. EC will be to EA as BD to AD. SCHOLIUM II THere are several useful Geometrical Practices depend 〈◊〉 this Consectary and its Proposition. 1. That (γ) Eucl. l. 9 & 10. l. 6. where we are taught to cut of any part required, e. ● ⅓ from a given Line AB, and so generally to 〈◊〉 or divide any given Line AC, in the same proportion as any other given Line, is supposed 〈◊〉 be divided in D, (and consequently into as ma●● equal parts as you please;) viz. if in the fi●● Case, having drawn any Line OF, you take AD● 1, and make DB 2, and having joined CB, dra●● the Parallel LE: for as AD to DB so is A to AC; that is, as 1 to 2, by this 4th Consect. therefore A is one third of the whole AC, etc. II A Rule (α) Eucl. 11 & 12, l. 6. to found a third Proportional to the 2 Right Lines given AB and BC (N o 2. Fig. 65.) (or a fourth to three given;) if, viz. having drawn OF at pleasure, you make AD equal to BC, and Joining DB, draw the Parallel EC: For as AB to BC, so AD i e. BC) to DE. Now if AD be not equal to BC but to another (viz. a) third Proportional, than by the same Reason DE will be a fourth Proportional. III Another Rule (β) Eucl. 13. lib. 6 & and Eucl. 8. lib. 6. to found a mean Proportional between two Right Lines given AC and CB; which is done by joining both the Lines together, and from the middle of the whole AB describing a Semicircle, and from C erecting the Perpendicular CD: For since the Angle ADB is a Right one, by Consect. 1. of the preceding Proposition, and the two Angles at C are Right ones, and those at A and B common to the whole Triangle ADB, and to the two partial ones ACD and BCD, ●hese two will be Equiangular and Similar to the great one, and consequently to one another: Therefore by the present Proposition, as AC to CD, so CD to CB, Q. E. D. and also as AB to BD so BD to BC, and as AB to AD so AD to AC, etc. IV. The Analytical Praxis of multiplying and dividing Lines ●y Lines, so that the Product or Quotient may be a Line; and also the way of Extracting Roots out of Lines: Which Des Cartes, gives us, p. 2. of his Geom. viz. assuming a certain Line ●or Unity, e. g. AB (in Fig. 65. Nᵒ 2.) if AC is to be multiplied by AD, having joined BD, and drawn the Parallel CE, ●he Product will be A; for it will be as 1 to ●he Multiplier AD, so the Multiplicand AC to ●he Product A; or if A is to be divided ●y AC, having joined EC and drawn the Parallel BD, the Quotient will be AD; (for AC ●he Divisor, will be to A the Dividend, as an ●nit AB to the Quotient AD;) all which are evident from the Nature of Multiplication and Division, and the Precedent Praxes. As also taking CB (in the same Fig. Nᵒ 3) for Unity, if the Square Root is to be extracted out of any other Line AC, this being joined to your Unity in one Line AB▪ and having described thereon a Semicircle, the Perpendicular CD will be the Root sought, as being a Mean Proportional between the two Extremes CB and AC, according to Prop. 17. V A Right Line AGNOSTUS which divides (α) Eucl. 3. lib. 6. any given Angle A into two equal Parts (Fig. 66.) being prolonged, divides the Base BC proportionally to the Legs of the Angle AB and AC For having prolonged CA to E, so that A shall be = to AB● the Angles ABE and AEB will be equal, by Consect. 2. Def. 13. and consequently also equal to each of the halves of the external Angle CAB, by Consect. 3. of the antecedent Proposition Therefore the lines AGNOSTUS and EBB will be parallel, by Cons. 1. Def. 11. Therefore as AC to A, i e. to AB, so GC to GB, by Co●sect. 3. of this Proposition. Q.E.D. VI Hence also there follows further, by conversion of th● last inference, as AC+AB to AC, so GC+GB (i e. BC) 〈◊〉 GC; and inversly GC to BC as AC to AC+AB; and lastly alternatively, GC to AC as BC to AC+AB. N. B. This last Inference follows also immediately from the preceding Consectary. For by reason of the Similitude of the ▵ ▵ ACG and ECB, as GC to AC so BC to CE, i e. to AC+AB. SCHOLIUM III FRom these two last Consectarys there an● these or two or three Practical Rules, t●● first whereof shows, how having the two Legs A● and AC given, and also the Base BC, to found the Se●ments GC and GB, made by the Bisection of the Intercrural Angle● (viz. by this inference, according to Consect. 6: As the Sum 〈◊〉 the Sides to one Side (e. g.) AC;) so the Sum of the Segment of the Base, i e. the whole Base to one of the Segments, vi● that next the said Side GC. 2. It shows on the contrary, how having the Base and one of its Segments given, and moreover the S●● of the Sides, to found separately the Side AC next the known Segment by inferring as the Sum of the Segments, or the Base BC to the Sum of the Sides, so the given Segment GC to the sought AC: or also, 3dly, Having only the Base and Sum of the Sides given, but not the Segment GC, yet to express its Proportion to the next side AC,— viz. in the Quantities of the given Terms, by putting (by Consect. 6.) for GC the value of the Base BC, and for AC the value of the Sum AB+AC; the great use of which last Rule will appear hereafter in the Cyclometry (or Quadrature of the Circle) of Archimedes. VII. In any Triangle ABC (Fig. of the present Proposition) the Sides are to one another as the Sins of their opposite Angles: For they are as the Chords of the double Angles at the Centre, by Prop. 33. therefore they are also one to another as half those Chords, i e. by definite. 10. as the Sins of the half Angles. SCHOLIUM IV. HEnce flow two new Rules of Plane Trigonometry, for Obliqueangled Triangles to found, viz. 1. The other Angles: From 2 given Sides, & an Angle opposite to one of them: by inferring As the Side opposite to the given Angle to the other Side, so is the Sine of the given Angle to the sine of the angle opposite to the other Side; which being given, the third is easily found. II The other Sides: From one side and the angles given, As the Sine of the Angle opposite to the given side, to that side; so is the Sine of the Angle opposite to the side sought to the side sought. So that this way we have reduced all the Cases excepting one of Plane Trigonometry, and consequently all Euthymetry to their original Foundations (for in that Case of having two Sides, and the included Angle given, we may found the rest by the Resolution of the Obliqueangled Triangle into two Right Angled ones; and so it's done by the Rules we have deduced in (Scholar 1.) I say, excepting one, in which from the three sides of an Obliqueangled Triangle given, you are required to found the Angles: the Rule to resolve which we will hereafter deduce in the 2d Consect. of Prop. 45. from that Theorem which Euclid gives us, lib. 2. Prop. 13. VIII. Because in in the Right Angled ▵ BAC (Fig. 67) BC is to CA as CA to CD, by Nᵒ 3. of the 2d Scholar of this Prop. the □ of CA will be = ▭ CE, by Prop. 17. In like manner because as CB to BASILIUS so is BASILIUS to BD; the □ of BASILIUS will be = to ▭ BE: Wherhfore the two Rectangles BE and CE taken together, that is, the □ of the Hypothenuse's BC, will be = to the two □ 's BA and CA taken together: Which is the very Theorem of Pythagoras demonstrated too other ways in Scholar of definite. 13. SCHOLIUM V THis Theorem of Pythagoras as it furnishes us with Rules of adding Squares into one Sum, or subtracting one Square from another; so likewise it helps us to some Foundations whereon, among the rest, the structure of the Tables of Sines relies, etc. Whose use we have already partly shown in Scholar 1 and 4. 1. If several Squares are to be collected into one Sum, having joined the Sides of two of them so as to form a Right Angle, e. g. AB and BC (Fig. 68 Nᵒ 1.) the Hypothenuse AC being drawn, is the Side of a Square equal to them both; and if this Hypothenuse AC be removed from B to D, and the Side of the third Square from B to E, the new Hypothenuse DE will be the Side of a Square equal to the three former taken together. 2. If the Square of the side MN (N o 2.) is to be subtracted from the Square of the side LM. Having described a Semicircle upon LM, and placed the other MN within that Semicircle, than draw the Line LN and that will be the Side of the remaining Square. 3. Having the Right Sine EGLANTINE of any Arch ED given (but how to found the Primary Sines we will show in another place), you may obtain the Sine Compliment CG or OF, by the preceding Numb. viz. by subtracting the □ of the given Sine from the □ of the Radius; and moreover the versed Sine GD by subtracting the Sine Compliment CG from the Radius CD. 4. The Squares of the versed Sine GD, and of the Right sine EGLANTINE being added together, give the □ of the Chord ED of the same Arch, (which all are evident from the Pythagorick Theorem) and half of that EH gives the Right Sine of half that Arch. 5. From the Right Sine EGLANTINE you have the Tangent of that Arch, if you make, as the Sine Compliment CG to the Right Sine GE, so the whole Sine CD to the Tangent G I 6. Lastly, From these Data you may also have the Secants (if required) thus, as the Sine Compliment CG to the W. S. CE, so the W. S. CD to the Secant CI; or as the Right Sine EGLANTINE to the W. S. E.C. so the Tangent ID to the Secant IC; both which are evident by our 34th Proposition. Consect. 9 If the Quadrant of a Circle (CBEG, Fig. 70.) be inclined to another Quadrant (CADG) and two other Perpendicular Quadrants cut both of them, viz. FBAG and FEDG, and the latter do so in the extremities of them both) having let fall Perpendiculars from the common Sections E and B, through the Planes of the Perpendicular Quadrants, and the inclined Quadrant, (viz. on the one side EGLANTINE and BH, as Right Sins of the Segments EC and BC; on the other EI and BK, as Right Sins of the Segments ED and BASILIUS) you'll have 2 Triangles EIG and BKH Right Angled at I and KING, Equiangular at G and H (by reason of the same inclination of the Plane CBEGC) and consequently similar, by our 34th Proposition; wherefore as the Sine EGLANTINE to the Sine EI, so the Sine BH to the sine BK, or as EGLANTINE to BH so EI to BK, and contrariwise. SCHOLIUM VI HEnce you have several Rules of Spherical Trigonometry for resolving Right Angled ▵ ▵ (α) Lansberg. Geom. Triang. lib. 4. Prop. 12. 1. Having given in the Rightangled ▵ ABC there Hypothenuse's BC and the Obliqne Angle ACB, for the Leg AB opposite to this Angle, make: as the sine T (EGLANTINE) to the sine of the Hypoth. (BH) so the sine of the given Angle (EI) to the sine of the Leg sought (BK). 2. Having given the Hypothenuse's BC and the Leg AB for the opposite Angle ACB make as the sine Hypoth. (BH) to S. T. (EC) so the sine of the given Leg (BK) to the sine of the Angle sought (EI) 3. Having given the side AB and the Angle opposite to it ACB, for the Hypothenuse's BC (supposing you know whether it be greater or lesle than a Quadrant) make as the sine of the given Angle EI to the sine T. (EGLANTINE) so the sine of the given Leg (BK) to the sine of the Hypoth. BH). 4. Having given in the Right Angled ▵ EBF (which we take instead of ABC that so we may not be obliged to change the Figure) one Leg EBB and the Hypothenuse's BF for the other Leg OF, you may found its compliment, if you make as the sine Compliment o● the given side (BH) to S. T. (EGLANTINE) so the sine Compliment of the Hypothenuse (BK) to the sine Compl. of the side sought (EI) 5. Having both Legs EBB and OF given, for the Hypothenuse's BF its Compl. BASILIUS may be found thus: as S. T. (EGLANTINE) is to the sine Cpmpl. (BH) of one side EBB, so the sine Compl. (EI) of the other side (OF) to (BK) the sine Compl. of the Hypothenuse. 6. Having given in the same Right Angled Triangle EBF one Leg OF, and the Angle adjacent to it ETB, first prolong into whole Quadrants BA to f, that A f may = BF Hypoth. & BC to e that Ce may = EBB, and AC to d that C d may = D● the measure of the given Angle EFB: secondly from d through e and f let fall a Quadrant through the extremities of the Quadrant B f and Be, that so the ▵ C de may be Right Angled, in which there are given the Hypoth. C d = to the given Angle, and the Angles C = to the Compl. of the given Leg (viz. to the Arch ED) and so, thirdly, there is sought the side de, as the Compliment of the Arch ef, or of the Angle sought ABC, or EBF; viz. by the first case of this, by inferring, as S. T. to the sine Hypoth. cd (i e. of the given Angle EFB;) so the Angle dce (i e. DE the Compl. of the given Leg RF) to the sine de (as the Compl. of the Angle fB e or EBF. 7. Having given, in the same Triangle, the side OF and the opposite Angle EBF (i e. the Arch ef) for the other Angle EFB (that is the Hypoth. cd in the ▵ cde) make by the third of this: As the Sine of the Angle dce (i e. the sine Compl. of the given Leg DE) to the S. T. so the sine of the Leg de (i e. the fine Compl. of the Angle EBF) to the Hypoth. cd (i e. the sine of the Arch DAM or Angle EFB.). 8. Having the Obliqne Angles given to found either of the sides, viz. OF; which may be done thus by the second of this: As the sine of the Hypoth. cd (i e. the sine of the Angle at FLETCHER) to the W. S. so the sine de (i e. the sine Compl. of the Angle at B) to the sine of the Angle dce (i e. the sine Compl. of the side sought EF.) Consect. 10. The same being given as in Consect. 7. if instead of the Right Sins EI and BK, you erect Perpendicularly DL and AM (Fig. 71) because of the similitude of the Triangles DGL and AHM, you'll have, as DG sine T. to DL the Tangent of the Arch DE, so AH the Right Sine of the Arch AC to AM the Tangent of the Arch AB; or as DG to AH, so DL to AM, and contrariwise. SCHOLIUM VII. HEnce flow the other Rules of Spherical Trigonometry for Resolving Right Angled Triangles, viz. 9 Having given the side AC in the ▵ ABC, and the adjacent Angle ACB, for the other side AB, make as the W. S. (DG) to the sine of the given side (AH) so the Tangent of the given Angle (ACB) to the Tangent of the Angle sought (AB.) 10. Having given the side (AB) and the opposite Angle (at C) for the other side (AC, so you know whether it be greater or lesle than a Quadrant) make as the Tangent of the given Angle (DL) to the Tangent of the given Leg (AB) so the whole S. (DG) to the sine of the Leg sought (viz. at AH.) 11. Both sides being given, for the Angles, make, as the sine of one Leg (AH) to the W. S. (DG) so the T. of the other Leg (AM) to the Tangent of the Angle opposite to the same (at C.) 12. Having given moreover in the Right Angled Triangle EBF the Hypothenuse (BF) and the Angle (EFB) for the adjacent side OF, make, as the sine Compl. of the given Angle (AH) to the W. S. so the Tangent Compl. of the Hypoth. (AM) to the Tang. Compl. of the Leg sought (DL) 13. Having given the side (OF) and the adjacent Angle FLETCHER for the Hypoth. BF make; as the W. S. to the sine Compl. of the given Angle (AH) so the Tangent Compliment of the given Leg (DL) to the Tangent Compl. of the Hypoth. (AM.) 14. Having given the Hypoth. (BF) and one side OF for the adjacent Angle (FLETCHER) make as the Tang. Compl. of the given Leg (D. L.) to the W. S. so the Tang. Compl. of the Hypoth. (AM) to the Sine Compl. of the Angle sought (AH). 15. Having given the Hypoth. (BF, i e. the arch Of, or the angle at d) and either of the obliqne angles (at FLETCHER) for the other angle (EBF) make by help of the new Triangle cde, by the 12th of this. As the Sine Compl. of the angle cde (i e. the Sine Compl. of the Hypoth. (AH) to the W. S. so the Tang. Compl. of the Hypoth. cd (i e. Tang. Compl. of the given angle) to the Tang. Compl. of the Side de (i e. to the Tangle of the angle sought ABC or EBF.) 16. Having given the obliqne Angles to found the Hypoth. (BF, or the arch A f, or the angle cde) it is done by the 14 of this Scholar As the Tang. Compl. of the Leg de (i e. the Tangent of the angle ABC or EBF) to the W. S. so the Tangent Compliment of the Hypoth. cd (i e. the Tangent Compliment of the other angle EFB) to the Sine Compl. of the angle cde (i e. the Sine Compl. of the Hypoth. BC sought.) So that now we have with Lansbergius (but much more compendiously) Scientifically Resolved all the Cases of Rigt-angled Triangles; the Resolution of Obliqueangled one's only now remaining. Consect. 11. In Obliqueangled Spherical Triangles, as well as Rightangled ones, the Sins of the angles are directly proportional to the Sins of the opposite Sins. 1. Of the Rightangled one's this is evident from Nᵒ 3 Scholar 6. and from the 9th Consect. For as the Sine of the angle A (Fig. 72.) to the Sine of BD, so the W. S. (i e. of the angle D) to the Sine of AB. 2. The same is immediately evident of an Obliqueangled Triangle ABC, resolved into 2 Rightangled ones. For, The Sine of the angle C is to the Sine of BD as the sine of the gle D to the Sine of AB; and also, The Sine of the angle C to the Sine of BD as the Sine of the angle D to the sine of BC, by the 1. In each Proportionality the means are the Sins of BD and D; therefore the Rectangles of the Extremes of the Sins of AB into the Sine of A, and the Sine of BC into the Sine of C, will be equal among themselves, since the Rectangles of the same Means are equal, by Prop. 18. therefore by Prop. 19 as the Sine of A to the Sine of BC, so the Sine of C to the Sine of AB, Q. E. D. SCHOLIUM VIII. THe latter may appear of Obliqueangled Triangles after this way also; since the Sine of the angle A is to the Sine of BD as the Sine of the angle D to the sine of AB, call the first a, the second ea, the third b, the fourth ebb; and because the Sine of the angle C (which we call c) is likewise to the Sine of BD (i e. to ea) as the Sine of D (i e. b) to the Sine of BC (which will consequently be ) it will be manifest, that the Sine of the angle A is to the sine of BC as the sine of the angle C to the sine of AB, i e. as a to.. , i e ... c to.. ebb. by multiplying the Means and Extremes, whose Rectangles are on both sides eab. Therefore as by the present and precedent Consectary 7, it is universally true, That in any Triangle whether Right Lined or Spherical, Rightangled or Obliqueangled, the Sides or their Sins, are to one another, as the Sins of their opposite Angles (which therefore is commonly called a Common Theorem:) so also hence flow 2 new Rules of Spherical Trigonometry for Obliqueangled Triangles, like those we found in Scholar 4. To found I The other Angles. From 2 sides given of an angle opposite to one of them, by inferring As the sine of the side opposite to the given angle to the sine of the other side, so the sine of the given angle to the sine of the angle sought. II The other Sides. From one side and the angles given, by inferring As the sine of the angle opposite to the given side to the sine of that side, so the sine of the angle opposite to the side sought to the sine of the side sought. And thus we have reduced all the Cases and Rules of Spherical Trigonometry to their original Fountains (for from 2 Sides given and the interjacent Angle, or 2 Angles and their adjacent side, we may found the rest in Obliqueangled Triangles, by resolving them into 2 Rightangled one's; and so by the Rules we have deduced in Scholar 6 and 7) excepting two Cases, viz. when from 3 sides given, the Angles, or from 3 Angles the Sides are sought; to resolve which, we are supplied with Rules from the following Cons. 12. In the given Obliqne angled Spherical Triangle ABC (Fig. 73.) whose Sides are unequal and each lesle than a Quadrant, having produced the sides AB and AC to the Quadrant AD and A, and effected besides what the Figure directs, than will The Arch DE be the Measure of the angle A, OF = AC, and so FB the difference of the Sides AB and AC. BC = BG, and so GF the difference of the third side, and the differences of the rest FB. But now, 1. As EH or DH to CM or FM so will PH be to NM (by reason of the Equiangular Triangles EPH and CNM;) therefore by Prop. 26. so will also DP be to FN. Make therefore DH = a FM = ea, DP = b FN = ebb. AI the R. Sine of the side AB. GM the Sine of the Side AC GL the Sine of GB, or of the side BC. FK the R. Sine of the Arch F B. BY the versed Sine of A B. BLS the vers. Sine of G B or BC. BK the versed Sine of FLETCHER B. KL or NOT the difference of the versed Sins we have now mentioned. EP the Right Sine and DP the versed Sine of the arch DE. CN the R. Sine and FN the versed Sine of the arch FC. Pag. 108. 73 74 75 76 n. 1. 76 n. 2. 77 n. 1. 77 n. 2. 2. By reason of the Equiangular ▵ ▵ FNO and HAI (for FNO is Equiangular to the ▵ FKQ, and that to the ▵ HQM, by reason of the Vertical Angles at QUEEN; and that also to the ▵ HAI by reason of the Common Angle at H) and you have also, As HA', Or DH to AI, so FN to NO. . Wherhfore now, 3. you'll have evidently, the □ DH to FM into AI as DP to NO. . DH NO DP. And Inversly as oea to a so oeb to b. SCHOLIUM IX. SInce therefore the Radius DH or a is known, and also NO the Difference of the versed Sines BL and BK, it is evident, that DP the versed Sine of the angle A will be known also; supposing that the first Quantity oea is likewise known. But this may be had by another Antecedent Inference, if you make, as AH to FM so AI to a fourth oea. a ea oa Hence therefore arises, 1. the Rule: Having given the 3 Sides of an Obliqueangled Triangle, to found any one of the Angles, viz. by inferring, 1. As the Sine of T to the Sine of R, one of the sides comprehending AC; so the sine of the other side AB to a fourth, 2. As this fourth to the sine of T, so the difference of the versed sins of the third side, BC, and the differences of the others to the versed sine of the Angle sought, viz. . But since the sides of a Spherical Triangle may be changed into Angles, and contrariwise the sides being continued [as 〈◊〉 the side AB of the given Triangle ABC (Fig. 47.) be con●nued a Circle, the rest into Semicircles from the Poles b and 〈◊〉 and likewise the Semicircle HI from the Pole A, and the Semicircle FG from the Pole B, and the Semicircle EA from th● Pole C, you'll have a new Triangle a, b, c, the 3 angles o● which will be equal to the 3 sides of the former ABC; as th● angle a or its measure IG, is equal to the side AB, by reaso● each makes a Quadrant joined with the third arch AGNOSTUS; b● the measure of the angle b, is the side AC (viz. in this cas● wherein the side AC is a Quadrant, in the other wherein 〈◊〉 would be greater or lesle than a Quadrant, it would be th● measure of the angle of the Compl. for than the Semicircle H●● described from the Pole A, would not pass through C but beyond or on one side of C. See Pitisc. lib. 1. Prop. 61. p. m. 25.)— the angle c or its measure KL, is equal to the 〈◊〉 BC, because with the third KC they make the Quadrants BK and CL] Therefore, 2. Having given the three Angel's 〈◊〉 the Obliqueangled Triangle abc, you may found any side, e. g. ac, if there be sought the Angle ABC, or rather its Compliment KBF, or its measure FK = ac,— from th● 3 sides given of the ▵ ABC, by the preceding Rule, by inferring, viz. 1. As S. T. to the sine R, of one side comprehending the angle of one side AB (i e. of one angle a adjacent to the side sought) so the sine of the other side BC (i e. of the other angle C) to a fourth. 2. As the fourth to the S. T. so the difference of the versed Sins of the third side AC, and the differences of the others (i e. of the 3d angle b, and the differences of the rest) to the versed Sine of the comprehended angle, or Compliment to a Semicircle (i e. of the side sought ac.) Proposition XXXV. SImilar Plane Figures (α) are to one another in Duplicate Proportion of their Homologous' Sides. Demonstration. For, 1. the Bases of 2 similar Triangles or Parallellograms for which any 2 Homologous Sides, e. g. AB and AB (Fig. 75.) may be taken) and Perpendiculars let fall thereon DE and DE, will be by Consect. 1. Prop. 34, as a to ea, b to ebb. Therefore the Parallellograms and Triangles themselves, will be as ba to eeba, by Consect. 7 and 8. Def. 12. i e. by Def. 34. in duplicate Reason of their Perpendiculars or assumed Sides, which is most conspicuous in Squares, which putting a for the ●ide of one, and ea for the other, are to one another as aa to ●eaa. 2. Like Polygons are resolved into like Tri●ngles, when the Triangles ABC and ABC, (α) Eucl. 19 & 20 lib. 6. and also AED and AED are Equiangular, by Consect. 3. Prop. 34. but GOD and GOD, are also Equiangular, because each of their angles are the remainder of ●qual ones, after equal ones are taken from them. Wherhfore ●he first Triangles are in duplicate Proportion of the sides BC ●nd BC; the second likewise of the sides CD and CD the ●hird are also in the same Proportion of the sides DE and DE, etc. i e. (since by the Hypoth. BC has the same reason to BC ●s CD to CD, and DE to DE) each to each is in duplicate Proportion of the sides BC to BC, or CD to CD, by the first of ●his. Therefore by a Syllepsis, the whole Polygons are in duplicate Proportion of the same Sides: Which is the second thing ●o be demonstrated. 3. Circles and their like Sectors, are as the Squares of their Diameters, by Prop. 32. therefore in duplicate Proportion of them, by the first of this: Which is the third thing: Therefore simi●ar Plane Figures, etc. Q. E. D. CONSECTARYS. THerefore 2 similar Plane Figures are one to another, as the first Homologous' Side, to a third Proportional, by virtue of Definition 34. II Any two Figures described on 4 Proportional Lines (α) Eucl. prop. 22. lib. 6. and similar to 2 others, are likewise Proportional, and contrariwise; for if the simple Reasons or Proportions of Lines be the same, their duplicate Proportions will be the same also, and reciprocally. SCHOLIUM. BUT as this second Consectary confirms Prop. 22 and i● Scholium, so the first teaches us a twofold Geometric Praxis. 1. A Way to express the Proportion of similar Figure's by two Right Lines, viz. by finding a third Proportional to their Homologous' Sides. For as the side of the first 〈◊〉 this third, so will be the first Figure to the second. 2. A wa● to augment or diminish any given Figure in a given Reaso● or Proportion, viz. by finding a mean Proportional between a●● side of the given Figure, and another Line which shall be 〈◊〉 that in a given Proportion, and than by describing thereon 〈◊〉 similar or like Figure. Proposition XXXVI. SImilar or like Solid Figures, are to one another in triplicate Proportion of their Homologous' Sides. Demonstration. For, 1. The similar Bases of two similar Parallelepipedo● (and consequently also of Prisms and Cylinders, by Consect. ●● and 5. definite. 16. and also of Pyramids and Cones, by Consect 3 and 4, of definite. 17.) are, as ab to eeab, by (α) the preceding Proposition, and their Altitudes as c to ec, by Consect. ● Prop. 34. Therefore Parallelepipeds, Cylinders and Prisms (and so the thi●● part of these, Cones and Pyramids) will be as abc to e abc, by Co●sect. 3, 4, 5. definite. 16. i e. they will be, by definite. 34. ●● Consect. 1 and 2. Prop. 34. in Triplicate Proportion of the● Perpendiculars or Homologous Sides. Which is especially Co●●spicuous in Cubes; which, putting a for the Side of one, an● ea for the side, are to one another as a to e a. 2. Polyedross or many sided Figures, may be resolved in● Pyramids of similar Bases and Altitudes; which is evident 〈◊〉 Regular ones, from the Consect. of definite. 21. and cannot b● difficult to understand also of Irregular ones; because the like inclination of their Planes every where similar and equal in number, necessarily require that the whole Altitudes of similar Polyedrous Solids, as well as similar Parallelepipedons, by Consect. 2. Prop. 34. should be in subduplicate Proportion of their Bases, and so these being likewise divided in C and C (Fig. 76. Nᵒ 1.) the parts of their heigths GC and GC, will be in the same Proportion: Whence, e. g. 2 Pyramids standing on simi●ar Bases ABDEF and ABDEF, and having like Altitudes GC and GC, will necessarily be like or similar; and the same ●hing may be likewise judged of others. Or yet, to show it more evidently, the Polyedrous Solids may be resolved into like Triangular Prisms; for, e. g. each of the Triangles of their similar Bases ABDEF and ABDEF (N o 2.) are similar, viz. abf and abf, ABF and ABF, by the preceding Prop. Nᵒ 2. The Planes ABba and A Bba, also AafF and A of FLETCHER, are similar by the Hypoth. and consequently also the Planes BbfF and B bfF (fb is to ba as fb to ba, and also in the one ba to bB, as ba to bB in the other; therefore ex equo as fb to bB so f b to bB, etc.) and so the whole Triangular Prisms will be similar, by definite. 35. and so of others. Therefore similar Polyedrous Solids will be in the same Proportion as similar Pyramids, or Triangular Prisms, i e. by the first of this in Triplicate Proportion of their Sides. (α) Eucl. prop. 12. lib. 12. of Cones and Cylinders. 3. Spheres are as the Cubes of their Diameters (β) Eucl. 18. lib. 12. by Consect. 3. Prop. 32. Therefore they are by the first of this, as a to e a. Therefore similar or like Solids are in Triplicate Proportion of their Homologous' Sides. Q. E. D. CHAP. VI Of the Proportions of Magnitudes of divers sorts compared together. Proposition XXXVII. THE Parallelogram ABCD (Fig. 77. N. 1.) is to the Triangle BCD upon the same base DC, and of the same height as 2 〈◊〉 1. This has been already Demonstrated in Consect. 3. De●●nit. 12. Here we shall give you another Demonstration. Suppose, 1. the whole Base CD divided into four equ●● parts by the transverse Parallel Lines EGLANTINE, HK, LN, than w●● (by reason of the similitude of the ▵ ▵ DGF, DKI, DNS' DCB) GF be 1, KING 2, NM 3, CB 4; and having furthermore continually Bisected the Parts of the Base, the Indivisible or the Portions of the Lines drawn transversly through the Triangle will be 1, 2, 3, 4, 5, 6, 7, 8, etc. ad infinitum, all along in an Arithmetical Progression, beginning from the Poi●● D, as 0; to which the like number of Indivisibles always answer in the Parallelogram equal to the greatest, viz. the Li●● BC. Wherhfore by the 4th Consect. of Prop. 16. all the In●●visibles of the Triangle, to all those of the Parallelogram take●● together, i e. the Triangle itself to the Parallelogram, is as 〈◊〉 to 2. Q. E. D. SCHOLIUM. NOW if any one should doubt whether the Triangle 〈◊〉 Parallelogram may be rightly said to consist of an in●●nite number of Indivisible Lines, he may, with Dr. Wa●●● Pag. 115. 78 79 80 81 82 83 84 85 86 87 88 instead of Lines, conceive infinitely little Parallellograms of the same infinitely little Height, and it will do as well. For having cut the Base (N oh 2) into 4 equal parts by transverse Parallels, there will be circumscribed about the Triangle so many Parallellograms of equal height, being in the same Proportion as their Bases, by Prop. 28. i e. increasing in Arithmetical Progression. In the following Bisection, there will arise 8 such Parallellograms approaching nearer to the Triangle, in the next 16, etc. so that at length infinite such Parallellograms of infinitely lesle height, and ending in the Triangle itself, will constitute or make an infinite Series of Arithmetical Proportionals, beginning not from 0 but 1; to which there will answer in the Parallelogram infinite little Parallellograms of the same height, equal to the greatest. Whence it again follows, by Consect. 9 Prop. 21. that the one Series is to the other, i e. the Triangle to the Parallelogram as 1 to 2; which being here thus once explained, may be the more easily applied to Cases of the like nature hereafter. CONSECTARYS. I SInce in like manner in the Circle (Fig. 79.) the Peripheries at equal intervals from one another, as so many Elements of the Circle, increase in Arithmetical Progression; the Sum of these Elements, i e. the Circle itself will be to the Sum of as many Terms equal to the greatest Periphery, i e. to a Cylindrical Surface, whose Base is the greatest Periphery, and its Altitude the Semidiameter, as 1 to 2. II Hence the Curve Surface of a Cylinder circumscribed about a Sphere, i e. whose Altitude is equal to the Diameter, is Quadruple to its Base. III Also the Sector of the Circle bac, to a Cylindrical Surface, whose Base is the Arch bc, but its Altitude the Semidiameter ac, is as 1 to 2. IU. And because the Surface of the Cone BCD is to its circular Base, as BC to CA, i e. as the √ 2 to 1. by Scholar Prop. 17. the Cylindrical Surface, the Conical Surface and the Circular one we have hitherto made use of will be as 2, √ 2 and 1, and consequently continually Proportional. SCHOLIUM ALICE which may also abundantly appear this way, viz. by putting for the Diameter of the Circle a, for the Semidiameter ½ a and for the Circumference ea, you'll have the Area of the Circle ¼ eaa by Consect. 1 definite. 31. and Multiplying ●he height of the Cylinder AB i e. ½ a by the Periph● ea yo●'l have the Cylindrical Surface ½ eaa by Consect. 6. definite 1● as now is evident also by Consect. 1, 2 and 3. Now if y●● would also have the Surface of the Cone, since its side by the Pythagorick Theorem is and the half of that i. e. (b● Nᵒ 2 of Scholar Prop. 22.) and this half being multiplied by the Periphery of the Base ea, you'll have (by virtue o● Consect. 4. definite. 1.8.) the Su●f●ce of the Cone i. e. (b● the Scholar just now cited) : So that now appears also th● 4th Consect. of this; because the Rectangle of those Extreme●● eaa and ¼ eaa is ⅛ eaa as well as the square of the mean. Proposition XXXVIII. A Parallelepiped (α) Euclid. Prop 7 Coral. lib. 12. BF (Fig. 77. Nᵒ 3) is to a Pyra●●● ABCDE upon the same Base BD and of the same height, 〈◊〉 3 to 1. This was Demonstrated in Consect. 3. definite. 17. bu● here we shall give you another. Demonstration. Suppose 1 the whole Altitude BE divided into 3 equal Pa●● by transverse Plains Parallel to the Base, than will (by reason 〈◊〉 the Similitude of the Pyramids abcd ACBD● and AECDE) the Bases abcd ABCD and ABC● be by Consect. 2. Prop. 34 and Consect. 3. Defi●●● 17 in duplicate Proportion of the Altitudes i e. in duplicate Arithmetical Progression 1, ● 9, moreover 2, bisecting the parts of the Altitude, the quadrangular Sections now double in Number (as the Indivisibles o● Elements of the proposed Pyramid) will be as 1, 4, 9, 16, 2● 36, etc. ad Infinitum, all along in a duplicate Arithmetical Proportion; while in the mean time there answer to them as many Elements in the Parallelepiped equal to the greatest ABCD, wherefore by Consect. 10. Prop. 21. all the Indivisibles of the Pyramid taken together will be to all the Indivisibles of the Parallelepiped also taken together, i e. the Pyramid itself to the Parallelepiped, as 1 to 3. Q. E. D. CONSECTARY. THis Demonstration may be easily accommodated to all other Pyramids and Prisms, and also Cones and Cylinders, (α) Euclid. Prop. 10. lib. 12. since here also (Fig. 78.) the circular Planes ba, BA, and BA are as the squares of the Diameters, and so as 1, 4, 9 and so likewise all the other Elements of the Cone by continual bisection are in duplicate Arithmetical Progression; when in the mean time there answer to them in the Cylinder as many Elements equal to the greatest BASILIUS etc. Proposition THIRTY-NINE. A Cylinder is to a Sphere inscribed in it i e. of the same Base and Altitude as 3 to 2. Demonstration. Suppose 1 (Fig. 80.) the half Altitude GH (for the same proportion which will hold when demonstrated of the half Cylinder AK and Hemisphere AGB, will also hold the same of the whole Cylinder to the whole Sphere) to be divided into 3 equal parts, than will AH, C1, E2, be mean proportionals between the Segments of the Diameter by Prop. 34 Scholar 2 Nᵒ 3, and so by Prop. 17. the Rectangles LHG, L1 G, L2 G equal to the Squares AH, C1, E2, being in order as 9, 8 and 5. and also 2dly, having bisected the former parts of the height, the six Squares cutting the Sphere Across ways will be found to be as 36, 35, 32, 27, 20, 11. etc. in the progression we have shown at large in Consect. 12. Prop 21. Wherhfore since all the Indivisibles of the Hemisphere, viz. the circular Planes answering to the Squares of the said ●aniverse Diameters have the same proportion of Progression, by Pr●● 32. and there answer to them the like number of Elements i● the Cylinder equal to the greatest AH: All these these taken together will be to all the other take● together i e, the whole Cylinder AK to the whol● Hemisphere AGB by virtue of the aforesaid Co●sect. 12. as 3 to 2 (α) A●chim. 32 (al. 31.) lib. I de Sphaer. Q. E. D. SCHOLIUM. I HOnora●us Fabri elegantly deduces this Prop. a priori, in a g●netick Method in his Synopsis Geom. p. 318. (which a● Carotus Renaldinus performs from the same common Foundationi● lib. 1. de Compos. and Resol. p. 301, and the following, but aft●● a more obscure way and from a demonstration further fetched Fabri's is after this Method: The whole Figure (81) ALL being turned round about BZ, the Quadrant ADLBA will describe a Hemisphere, the Square AZ a Cylinder and the triangle BM● a Cone all of the same Base and Altitude. Since therefore Circled are as the squares of their Diameters by Prop. 32. and the Squa● of GE = to the Squares of GD and GF taken together (f● the Square of GF i e. GB× □ GD is = □ BD or BA or G● by the Pythag. Theor.) and so the Circle described by GE ●● be = to 2 Circles described by GD and GF taken together than taking away the Common Circle described by GF there w●● remain the circle described by GF within the Cone equal t● the Annulus or Ring described by DE about the Sphere. A● since this may be demonstrated after the same way in any other case, viz. that a circle described by g f, will be equal to an A●nulus described by de; it will follow, that all Rings or An●● described by the Lines DE or de (i e. all that Solid that conceived to be described by the trilinear Figure ADLM turn round) will be equal to all the Circles described by GF or gf (i● to the Cone generated by the Triangle BLM;) and so as th● Cone i● ⅓ part of the Cylinder generated by ALL, by the C●●sect. of Prop. 38. so also the Solid made by the Triline● ADLM (viz. the Excess of the Cylinder above the Sphere will be ⅓ of the Cylinder, and consequently the Hemisphere Q. E. D. CONSECTAYS. HEnce you have a further Confirmation of Consect. 2. Prop. 32. and Prop. 36. N. 3. II Hence also naturally flows a Confirmation of Consect. 2. definite. 20. and consequently the Dimension of the Sphere both as to its solidity and Surface. For putting a for the Diameter of the Sphere and circumscribed Cylinder, and Ea for the Circumference, the Basis of the greatest Circle will be ¼ eaa, and ●hat multiplied by the Altitude, gives ¼ ea for the Cylinder. Therefore by the present Proposition, ⅙ ea gives the Solidity of the Sphere (by making as 3 to 2 so ¼ to ⅙) This divided by ⅙ a, will give, by virtue of Consect. 1. of the aforesaid Def. 20. and Consect. 3 definite. 17. the Surface of the Sphere eaa. III Therefore the (α) Archim. lib. I de Sph. & Cylind. Prop. 31. (al. 30). Surface of the Sphere eaa, is manifestly Quadruple of the greatest Circle ¼ eaa. IV. The Surface of the Cylinder, without the Bases, made by multiplying the Altitude a by the Circular Periphery of the Base ea, will be eea, equal to the Surface of the Sphere. V Adding therefore the 2 Bases, each whereof is ¼ eaa, the whole Surface of the Cylinder 1 ½ eaa, will be to the Surface of the Sphere eaa as 3 to 2. VI The Square of the Diameter aa to the Area of the Circle ¼ eaa, is as a to ¼ ea, i e. as the Diameter to the 4th part of the circumference. VII. A Cone of the same Base and Altitude with the Sphere and Cylinder, will be by Consect. 2. of this, Prop. and the Consect. of Prop. 38. ea, and of the Cylinders ¼ or ea. Therefore a Cone, Sphere, and Cylinder, of the same height and diameter, are as 1, 2, 3. The Cone therefore is equal to the Excess of the Cylinder above the Sphere; as is otherwise evident in Scholium 1. of this. SCHOLIUM II AND thus we have briefly and directly demonstrated the chief Propositions of Archimedes, in his 1. Book de Sphaer● & Cylind. which he has deduced by a tedious Apparatus, and 〈◊〉 ●●directly. And now if you have a mind to Surveyed the 〈◊〉 and more perplexed way of Archimedes, and compare it with this shorter cut we have given you; take it thus: Archimedes thought it necessary first of all to premise this Lemma; Th●● all the Conical Surfaces of the Conical Body made by Circumvolution of the Polygon, or many angled Figure A, B, C, D, E, etc. (Fig. 82.) inscribed in a Circle, according to definite. 19 I say, those Conical Surfaces taken all together, will be equal to a Circle, whose Radius is a mean Proportional between the Diameter A and a transverse Line BE, drawn from one extremity of the Diameter E to the end of the side AB next to the other extremity. This we will thus demonstrate by the help of specious Arithmetic: Since BN, HN are the Right Sins of equal Arches, CK and CK whole Sines, etc. and the Lines BH, GC, etc. parallel; having drawn obliquely the transverse Lines HC, GD, all the angles at H, C, G, D, etc. will be equal by Consect. 1. definite. 11. and consequently all the Triangles BNA, HNI, ICK, etc. equiangular, both among themselves, and to the ▵ ABE; since the angle at B is a Right one, by Consect. 1 Prop. 33. and the angle at A common with the ▵ BNA. Wherhfore as BN to NA or HN to NI so CK to CI. & GK to KL. and so DM to ML FM to ME so EBB to BASILIUS; and so by making BN, HN, DM, FM = a CK and GK = b, EBB = c, for NA, NI, ML and ME, you may rightly put ea for KING and KL ebb for AB, ec. Which being done you may easily obtain the Conical Surfaces of the inscribed Solid, and the Area of a Circle whose Radius shall be a mean Proportional between A and FB, and it will be evidently manifest, that these two are equal. For, 1 (for Conical Surfaces) the Diameter of the Base BH = 2 a, and the side of the Cone A B = ec: Therefore (making here oh the name of the Reason between the Diameter and Circumference) the circumference will be 2 oa which multiplied by half the side ½ ec, gives the Conical Surface oaec, by Consect. 4. definite. 18. And since the Circumference BH is as before 2 oa, and the circumference CG = 2 ob, half of the sum 2 oa+ + ob, viz. oa + ob is the equated Circumference: which multiplied by the Side BC = ec gives the Surface of the truncated Cone BHGC = oaec − obec, by Consect 5. of the aforesaid Def. since, lastly, the Surface of the truncated Cone DFGC is equal to one, and likewise the conical Surface EDF to the other, by adding you'll have the Sum of all 40 oaec+ + obec. 2. (for the Area of the Circle) the Diameter A is = 4 ea+ + ebb, and BE = c: the Rectangle of these is = 4 eac+ + ebc = (which also is evidently equal to the Rectangle of all the transverse Lines BH, CG, DF into the side A B, as Archimedes proposes in the matter) = to the Square of the Radius in the Circle sought, because the Radius is a mean Proportional between A and BE, and so equal to , so that the whole Diameter is . Therefore, 2. the circumference of this Circle will be , i e. : which multiplied by half the Semidiameter, i e. by gives the Area of the Circle sought . But this Root extracted is , equal to the superior Sum of the conical Surfaces. Q. E. D. Having thus demonstrated the Lemma, we will easily demonstrate with Archimedes (though not after his way) That the Surface of any Sphere, is Quadruple of the greatest Circle in it, which is already evident from the 3d Consect. For since all the conical Surfaces of the inscribed Solid taken together, by the preceding Lemma, are equal to the Area of a Circle whose Radius is a mean Proportional between the Diameter A and the Transverse EBB; and this mean Proportional approaches always so much nearer to the Diameter A, and those Surfaces so much nearer to the Surface of the Sphere, by how many the more sides the inscribed Figure is conceived to have, by Consect. 1 and 2. Def. 18. if you conceive in your mind the Bisection of the Arches AB, BC, etc. to be continued in Infinitum, it will necessarily follow, that all those conical Surfaces will at length end in the Surface of the Sphere itself, and that mean Proportional in the diameter A, and so the Surface of the Sphere will be equal to a Circle, whose Radius is the Diameter A, But that Circle would be Quadruple of the greatest Circle in the present Sphere, by Prop. 35. Therefore the Surface of the Sphere is Quadruple of that Circle also. Q. E. D. Hence also it would be very easy to deduce with Archimedes (though again after another way) that celebrated Proposition, which we have already demonstrated from another Principle in the Prop. of this Scholar viz. That a Cylinder is to a Sphere of the same Diameter and Altitude, as 3 to 2. For by putting a for the Diameter and Altitude, and ea for the Circumference, the Area of the Circle, will be ¼ eaa; and this Area being multiplied by the Altitude a, giveth ¼ ea for the Cylinder, by Consect. 5. definite. 16. and the same Quadruple, i e. eaa multiplied by ⅙ a gives ⅙ ea for the Sphere, by Consect. 1. definite. 20. and Consect. 3. definite. 17. Wherhfore the Cylinder will be to the Sphere as ¼ to ⅙, i e. in the same Denominator as to , i e. as 6 to 4, or 3 to 2. Q.E.D. Whence it is evident, that the Dimension of the Sphere would be every ways absolute if the Proportion of the Diameter to the Circumference were known; which now with Archimedes, we will endeavour to Investigate. Proposition. XL. THE Proportion of the Periphery of a Circle (α) Archim. Cyclom. prop. 2 to the Diameter, is lesle than 3 or to 1. and greater than 3 to 1. Demonstration. The whole force of this Proposition consists in these, that, 1. Any Figure circumscribed about a Circle, has a greater Periphery than the Circle, but any inscribed one a lesle. 2. The Periphery of a circumscribed Figure of 96 sides, has a lesle Proportion to the Diameter, than 3 to 1. To demonstrate this second, we will inquire 3. the Proportion of one side of such a Figure whether circumscribed or inscribed after the following way. For the first part of the Proposition. Suppose the Arch BC (Fig. 84. N. 1.) of 30 degrees, and its Tangent BC making with the Radius AC a Right Angle, to make the Triangle ABC half of an Equilateral one, so that AB shall be to BC in double Reason, viz. as 1000 to 500; which being supposed, AC will be the Root of the Difference of the Squares BC and AB, i e. a little greater than 866, but not quite . Than continually Bisecting the Angles BAC by AGNOSTUS, GAC by AH, HAC by AK, KAC by ALL, BC is half the side of a circumscribed Hexagon, GC the half side of a Dodecagon (or 12 sided Figure) HC of a Polygon of 24 sides, KC of one of 48; lastly, LC of one of 96 sides; and by N. 3. Scholar 3. Prop. 34. GC will be to AC as BC to BA+AC, and also HC to AC as GC to GA+AC, etc. Wherhfore In the first Bisection, of what parts GC is 500, of the same will AC be 1866 and a little more, and AG (which is the Root of the Sum of the □ □ GC and AC) 1931 +. In the second Bisection, of what parts HC is 500, of the same will AC be found to be 3797 +and AH 3830 +. In the third Bisection, of what parts KL is 500 of the same will AC be 7628 + and AK 7644 +. In the 4th Bisection, of what parts LC is 500 of the same will AC be 15272 +. Now therefore LC taken 96 times, will give 48000 the Semiperiphery of the Polygon, which has the same Proportion to the Semi-diameter AC 15272 as the whole Periphery to the whole Diameter. But 48000 contains 15272 , 3 times, and moreover 2181 remaining parts, which are lesle than part of the division, for multiplied by 7 they give only 1526 +. Therefore it is evident, that the Periphery of this Polygon (and much more the Periphery of a lesle Circle than that) will have a lesle Proportion to the Diameter, than 3 to 1. Which is one thing we were to demonstrate. For the 2d Part of the Prop. SUppose the Arch BC to be (N oh 2.) of 60 Degr, that is the Angle BAC at the Periphery of 30 Since the Angle at B is a right one by Consect. 1. Prop. 33. the Triangle ABC will be again half an Equilateral one, and BC the whole side of an Hexagon, and GC of a Dodecagon, etc. So that putting for BC 1000 (as before we put 500 for the side of the Hexagon) let AC be 2000 and AB the Root of the difference of the Squares BC and AC i e. lesle than 1732 viz. 1732 and not quite . Than bisecting continually the Angles BAC, GAC, etc. since the Angles at the Periphery BAG, GAC, GCB, standing on equal Arches BG and GC, are equal by Prop. 33. and the Angle at C, (common to the Triangles GCF and GCA) and the others at H, EDWARD, L are all right ones by Consect. 1. of the aforesaid Prop. these 2 Triangles CGF and CGA are equiangular and consequently by Prop. 34. the Perpendicular GC in the one will be to the Perpendicular GA' in the other as the Hypothenuse CF in the one to the Hypoth. AC in the other i e. (by the foundation we have laid in the former part of the Demonstration of Nᵒ 3, Scholar 3. Prop. 34.) as BC to AB+AC; and in like manner in the following HC will be to HA' as GC to AG+AC, etc. Wherhfore. In the first Bisection, of what parts GC is 1000 of the same AGNOSTUS will be a little lesle than 3732 ; and AC (which is the Root of the Sum of the □ □ AGNOSTUS and GC) will be a little lesle than 3863 . In the second Bifection, of what parts GC is 1000 of the same will AH be a little lesle than 7595 and AC a little lesle than 766 . In the third Bisection, of what parts KC is 1000 of the same will AK be a little lesle than 15257 and AC a little lesle 15290 . In the fourth Bisection, of what parts LC is 1000 of the same will ALL be a little lesle than 30547 ; and AC a little lesle than 30564, and consequently if LC be put 500, AC will be lesle than 15282. Now therefore LC taken 96 times will give 48000 for the Periphery of the inscribed Polygon, and 15282 and a little lesle for the Diameter AC. But 48000 contains 15282 thrice, and moreover a remainder of 2154 parts, which are more than of the Divisor; for of this number makes 215 and so makes 2150 i. e. 2152 . Therefore it is evident that the Periphery of this Inscribed Polygon (and much more the Periphery of a Circle greater than that) will be to its Diameter in a greater proportion than 3 to 1, which is the 2d thing. SCHOLIUM IF any one had rather make use of the small numbers of Archimedes, which he chose for this purpose, by putting in the first part of the Demonstration, for AB 306 and for BC 153, in the second for AC 1560 and for BC 780, by the like process of Demonstration, he may infer the same with Archimedes. We like our Numbers best, though somewhat large, because they may be remembered, and are more proportionate to things, and make also the latter part of our Demonstration like the former. The Proportion in the mean while of the Diameter to the Periphery of the Circle by the Archimedean way is included within such narrow Limits, that they only differ from one another or parts; for Subtracted from leave , as 3 or and 3 or if they are reduced to the same Denomination make on the one hand on the other . Hence it would be easy, having divided the difference into 2 Parts, to express a middle proportion of the Periphery to the Diameter between the 2 Archimedean and Extreme ones as in these Numbers, 1561 to 4970, (or by dividing both sides by 5) as 3123 to 994, or (dividing both sides by 7) as 446 to 142, or (by dividing again by 2) as 223 to 71, etc. While these Numbers become as fit for use as those of Archimedes, which we therefore use before any other, particularly in Dimensions that doubt require an exact Niceness; where they do those may be made use of exhibited by Ptolomey Vieta, Ludolphus a Ceulea, Metius, Snellius, Lansbergius, Hugeus, etc. as if The Diameter be The Circumference will be 10,000,000 31, 416, 666. Ptolomey. 10000,000,000 31, 415, 926, 535. Vieta. 100,000,000,000,000,000000 314, 159, 265, 358, 979, 323, 846 ½, etc. Ludolph. a Ceulen. Proposition XLI. THe Area of a Circle has the same proportion to the Square of it Diameter, as the 4th part of the Circumference to the Diameter. Demonstration. Though we have before Demonstrated this Truth in Consect. 6. Prop. 39 yet here we will give it you again after another way. Since therefore the Circumference is a little lesle than 3 , and ● little more than 3 Diameters, for this excess putting z if the Diameter be 1 we will call the Circumference 3+z therefore the 4th Part of it will be : And the Area of the Circle (by Multiplying the half Semidiameter) i e. ¼ by the Circumference, you'll have both and the square of the Diameter = ● Q. E. D. CONSECTARY. THerefore if the (α) Archim. Prop. 3. Cyclom. Proportion of Archimedes be near enough truth to be made use of, viz. 22 to 7; the Ar●● of the Circle will be to the Square of the Diameter as 11 to 14 because the quarter part of 22, i e. 5 ½ or to the Diana. 7 is in the same Proportion. Proposition XLII. THE Diameter (β) Eucl. last Prop. lib. 10. of a Square AC (Fig. 83.) is incommensurable to the side AB (and consequently also to th● whole Periphery) i e. it bears a Proportion to 〈◊〉 that cannot be exactly expressed by Numbers. Demonstration. For, if for AB you put 1, BC will be als● 1, and by the Pythagorick Theorem AC will be = √ 2. Therefore by Consect. 4. definite. 30. AC is incommensurable to the side AB, etc. Q.E.D. CONSECTARY I IT is notwithstanding Commensurable in Power; for its Square is to the Square of the Side as 2 to 1. CONSECTARY II NOW if the Proportion of the side or whole Periphery ABCDA, to the Diana. AC is to be expressed by Numbers somewhat near, as we have done in the Diameter and Circumference of a Circle; than making the side AB = 100, the Diameter is greater than 141 , lesle than 141 . Proposition XLIII. THE Area of a Circle is Incommensurable to the Square of the Diameter. Demonstration. For dividing the Semidiameter CD (Fig. 85.) into two equal Parts (and consequently the Diameter DF into 4) AC will be 2. viz. √ 4 and AC √ 3 by Scholar 2. Prop. 34. N. 3. the sum √ 4+ √ 3, and the sum of as many equal to the greatest AC 4. Having moreover Bisected the Parts of the Semidiameter, AC will = 4 or the √ 16, ac = √ 15, AC = √ 12, AC = √ 7; the sum ; and the sum of as many equal to the greatest AC is = 16, etc. And thus the last sums will be the Square Numbers increasing in Quadruple Proportion; but the former Sums will be always composed of the Rational Root of every such Square, and of several other irrational Roots of Numbers unevenly decreasing; so that it will be impossible to express those former Sums by any Rational Number, by what we have said in Scholar 2. definite. 30. Wherhfore all the Indivisibles of the Quadrant ADC are to as many of the Square ACDE equal to the greatest, i e. the Quadrant itself ADC to the Square ACDE (and consequently the whole Area of the Circle to this circumscribed Square) will be as a Surd Quantity to a true and truly Square Number, i e. the Area of the Circle will be Incommensurable to the Square of the Diameter, by Consect. 4. of the said definite. Q. E. D. CONSECTARY. AND because the fourth part of the Circumference ha●● the same Proportion to the Diameter, as the Area of the Circle to the Square of the Diameter, by Prop. 41. Therefore also that will be Incommensurable to this, and consequently the whole Circumference will be so to the Diameter. SCHOLIUM. WHerefore it is somewhat Wondered, which G. G. Leibnitius (α) Acta Erudit. Ann. 82. p. 44 & the following. tells us, that the Square of the Diameter being 1, the Area of the Circle will be , etc. ad Infinitum. i e. by adding 1 and , etc.) to , etc. i e. to the Sum of infinite Fractions whose common Numerator is 2. But their Denominators Squares lessened by Unity, and taken out of the Series of the Squares of Natural Numbers by every fourth, omitting the Intermediate one's: Which Sum might seem expressible in Numbers, since all its parts are Fractions reducible to a common Denominaton; while notwithstanding Leibnitius himself confesses, that the Circle is not Commensurable to the Square, nor expressible by any Number. CHAP. VII. Of the Powers of the Sides of Triangles, and other Regular Figures, etc. Proposition XLIV. IN Rightangled Triangles (α) Euc. 47. Lib. 1. (ABC, Fig. 86.) the Square of the Side (BC) that subtends the Right-angle, is equal to the Squares of the other Sides (AB and AC) taken together. Demonstration. Though we have demonstrated this Truth more than once in the foregoing Proposition; yet here we will confirm it again as follows. Having described on each side of the Square BE a Semicircle, which will all necessarily touch one another in one point, and be equal to the Semicircle, BAC, if you conceive as many Triangles inscribed also equal to BAC; it will be evident that the Square BE will contain the said 4 Triangles; and besides the little Square FGHI, whose side FI, u g. is the difference between the greater side of the Triangle CI, and the lesle CF, (for because the lesle side CF = BA, lying in the first Semicircle, if it be continued to I in the second Semicircle makes CI = CA the greater side of the other Triangle, and so in the others. From thence it is evident, That as the Angel's ABC, and ACB together make one right one; so likewise BCF (= CBA) and ECF make also one right one; and consequently ECF is = ACB, and the Arch and the Line EI = to the Arch and the Line AB, etc.) Wherhfore, if the greatest side of the given Triangle BC or BD, etc. be called a, and AC, b and the least AC, or CF, etc. be called c; the 〈◊〉 of the side BC, will be = aa, and the Area of each Triangle ½ bc: and so the 4 Triangles together 2 bc: but the side of the middle little Square will be b − c, and its Square bb + cc = 2 bc: Wherhfore if you add to this the 4 Triangles▪ The Sum of all, i e. the whole Square BE will be bb + cc = aa. Q. E. D. CONSECTARYS. I HEnce having the sides that comprehend the Right-ang●● given, AC = b and AB = c, the Hypothenuse or Barnes that subtends the Right-angle BC will be = √ bb + cc. II But if BC be given = a and AC = b, and you are 〈◊〉 found AB = x; because xx + bb = aa; you'll have (taking awa● from both sides bb) xx = aa − bb: therefore x, i e. AB = √ aa − bb. III If 2 Rightangled Triangles have their Hypothenus● and one Leg equal, the other will also be equal. Proposition XLV. IN Obtuse-angled Triangles (Fig. 87. N. 1. the Square of the Barnes or greatest Side BC that subtends the Obtuse-angle BAC, is eq●● to the Squares of (α) Eucl. prop. 12. lib. 2. the other 2 Sides (AB and AC) taken together and also to 2 Rectangles (GOD) made by one of 〈◊〉 Sides which contain the Obtuse-angle (AC) and 〈◊〉 continuation AD to the Perpendicular BD let fall fr●● the other side. Demonstration. If BC be called a, AB = c, AC = b, AD = x, CD will b● = b + x. Therefore □ BD = cc − xx, by Consect. 2. of the preceding Prop. In like manner if □ CD = be subtracted from the □ BC = aa, you'll have to the same □ BD. Therefore , i e. (adding on both sides xx) . i e. (adding on both sides bb and 2 bx) . Q. E. D. CONSECTARY. IF in this last Equation you subtract from both Sides cc + bb than will and (if you moreover divide both Sides by 2 b) you'll have : Which is the Rule, when you have the Sides of an Obtuse-angled Triangle given, to found the Segment AD, and consequently the Perpendicular BD. Proposition XLVI. IN Acuteangled Triangles (α) Eucl. Prop. 13. l. 2. the Square of any side (e. g. B. C, Fig. 87. N. 2.) subtending any of the Angles, as A is equal to the Squares of the other 2 sides (AB and AC) taken together, lesle 2 Rectangles (GOD) made by one side, containing the Acute-angle (CA) and its Segment AD reaching from the Acute-angle (A) to the Perpendicular (BE) let fall from the other side. Demonstration. Make again BC = a, AC = b, AB = c, AD = x; than will CD = b − x. Therefore cc − xx = □ BD, and (i e. □ BC − □ CD) will also be = □ BD. Therefore . i e. (adding to both sides xx) , i e. (adding on both sides bb, and subtracting 2 bx) . Q. E. D. CONSECTARYS. I IF in the last Equation, except one, you add on both sides bb, and subtract aa, you'll have , and, if moreover you divide both sides by 2 b, you'll have : Which is the Rule, having 3 sides given in an Acuteangled Triangle, to found the Segment AD, and consequently the Perpendicular BD. Knowing therefore the Segments AD and CD, and also the Perpendicular BD in Obliqueangled Triangles, whether Obtuse-angled or Acuteangled, when moreover the sides BC an● AB are likewise given, the Angles of either Rightangled Triangles or Obliqueangled ones, will be known; so that the la●● Case of Plane Trigonometry, which we deferred from Prop. 34 to this place, may hence receive its solution. Proposition XLVII. THE Square of the Tangent of a (α) Eucl. 36. Lib. 3. Circle, is equal to a Rectangle contained under the whole Secant DA, and that part of it whi●● is without the Circle DE, whether the Secant pass through the Centre o● not. Demonstration. For in the first Case, if CB and CE are = b, DE = x, than will CD = b + x, & AD = 2 b + x: therefore □ ADE = 2 . therefore, if from the □ CD you subtract □ CB = bb the remainder will be = □ BD = □ ADE. Q.E.D. In the second Case, the lines remaining as before, make D● = y, FE or FA = Z: therefore the □ ADE will be = , but the □ FC equal to the □ EC − □ FE = i. e. . But the same Square FD is = . Wherhfore taking away from these equal Square● the common one ZZ you'll have 2 i. e. pr. 1st Case = □ BD. Q. E. D. CONSECTARYS. I THerefore the Rectangles of divers secants (as of AD● & ade in Fig. 88, n. 1.) which are equal to the sam● Square. BD, are equal also to one another; which the la●● Equation in our Demonstration (is an ocular proof of. Pag. 133. 89 90 91 92 93 94 95 96 II Therefore by Prop. 19 as a D to AD so is reciprocally DE to D e in the Fig. of the 2 Case. III Tangents to the same Circle from the same Point, as DB & D b (n. 3.) are equal; because the Square of each is equal to the same Rectangle. IU. Nor can there be more Tangents drawn from the same point than two: For if besides DB & D b, D c could also touch the Circle, than it would be equal to them. By Cons. 2. but that is absurd by Cons. 3. Def. 7. SCHOLIUM I HEnce is evident, the Original of the Geometrical Constructions, which Cartes makes use of, p. 6, and 7. in resolving these 3 Equations, , & . For in the First case, since he makes NO or NL (Fig. 89. 11. 1.) or NP ½ a, and the Tangent ZMb, make MNO through the Centre xx will = Z the quantity sought; which thus appears: Making MOTHER = Z, NM will = Z½ a, and it's □ . But the □ OMP (which is by the present Prop. = □ LM, i e. bb) together with the □ NL or NP i e. ¼ aa is = □ NM by the Pathagor. Theor. Therefore . i e. (taking away from both sides ¼ aa) ZZ − a Z = bb i e. (by adding on both sides aZ) ZZ = aZ+ bb: Which is the Equation proposed. In the Second Case, if you make PM = Z (as Cartes makes it) you'll have , and to this again as before you'll have = bb+ + aa. Therefore ZZ+ aZ = bb: Therefore ZZ = − aZ+ bb: Which is the very Equation of the second Case. In the third Case, whether you make the whole Secant RM (N. 2) or that part of it without the Circle QM = Z, the Root sought, there will come out on both sides the same Equation of the third Case; and so it is manifest, that this Equation has those 2 Roots. For if RM be = Z (adding + to the Fig. of Cartes the Line NOT which shall bisect QR, and makes OM = LN) OR or OQ will be = Z − ½ ha', and so the □ OQ = ZZ − aZ+ ¼ aa, and this together with the □ RMQ (which is by virtue of the pres. Prop. = □ LM) = □ NQ. i e. AM, i e. ZZ − , i e. (by adding aZ and taking away ½ aa) ; i. e. (taking away bb) ZZ = aZ − bb: Which is the the very Equation of the third Case. But if QM be made = Z, OQ or OR will = ½ − Z, and it's □ = , as well as the former, and so all the rest. Q. E. D. SCHOLIUM. II NOW if you would immediately deduce these Rules by the present Proposition, without the Pythagorick Theorem. It may (e. g. in the first Case) be done much shorter thus: If MOTHER = Z and NO or NP = ½ a, than will PM = Za: Therefore □ OMP = , or the □ LM, by the pres. Proposition; by adding therefore to both sides aZ, you'll have ZZ = aZ+ bb, which is the very Equation of the first Case. In the second Case, if MR. be Z, QM or PR will = a − Z: Therefore □ RMP = bb, i e. aZ = bb+Z +, i e. aZ − bb = ZZ; but if QM = Z, RM will = a − Z. Therefore □ RMP aZ − ZZ = bb, as before, etc. SCHOLIUM III FRom the 2 Consect. of the present Proposition, flows another Rule for solving the last Case of Plain Trigonometry, which we solved in the Consect. of the foregoing Prop. viz. If you have all the three sides of the Obliqueangled Triangle BCD (Fig. 90) give, if from the Centre C, at the distance of the lesser side CB you describe a Circle, than will, by Consect. 2. of the present Proposition, BD the Base of the Triangle (here we call the greatest side of the Triangle the Base, or in an Equicrural Triangle, one of the greatest) will be to AD, (the sum of the Sides DC+CB) as DE the difference of the Sides, to DF the Segment of the Base without the Ciecle; which being found, if the remainder of the Base within the Circle be divided into two equal parts, you'll have both FG and GB, as also DG; which being given, by help of the Rightangled ▵ ▵ GBC and GDC all the Angles required may be found. Proposition XLVIII. IN any Quadrilateral Figure (α) Ptol. lib. I Almagesh. ABCD (Fig. 91. N. 1.) inscribed in a Circle, the ▭ of the Diagonals AC and BD is equal to the two Rectangles of the opposite sides AB into CD, and AD into BC. Demonstration. Having drawn A so that the Angle BAE shall be equal to the Angle GOD, the Triangles thereby form (for they have the other Angles EBA and ACD in the same Segment equal, by virtue of Consect. 1. Prop. 33.) will be Equiangular one to another and consequently (by Prop. 34) as AC to AD so AB to BE. Wherhfore by making AC = a and CD = ea, and AB = b, BE will be = ebb. In like manner when in the ▵ ▵ BAC and EAD, the respective Angles are equal (viz. adding the common part EAF to BAE and GOD, equal by Constr.) and besides the angles BCA and EDA in the same Segment are also equal; those Triangles will also be Equiangular, and AD will be to DE as AC to CB; wherefore by putting, as before, a for AC, and oa for CB, and c for AD, DE will = oc. Therefore the whole BD = eb + oc. The Rectangle therefore of AC into BD will = eba + oca =; the Rectangle of AB into CD = eba + □ of AD into BC = oac. Q. E. D. SCHOLIUM IN Squares and Rectangles (N. 2.) the thing is self-evident. For in Squares if the side be a, the Diagonals AC and BD will be √ 2 aa, and so their Rectangle = 2 aa will be manifestly equal to the two Rectangles of the opposite sides. In Oblongs, if the two opposite sides are a and the others b. the Diagonals will be , and their Rectangle aa + bb manifestly equal to the two Rectangles of the opposite Sides. Proposition XLIX. THe side (AB) of an Equilateral Triangle (ABC, Fig. 92. N. 1.) inscribed in a (α) Ptol. lib. I Almagesh. Circle, is in Power triple of the Radius (AD) i e. of the □ of AD. Demonstration. MAke AD or FD = a, and so its Square aa. Since Therefore, having drawn DF through the middle of AB, or the middle of the Arch AFB let DE be = ½ a; for the angles at E are right ones, by Consect. 5. definite. 8. and the Hypothenuses AD, OF, are equal, by Scholar of Definition 15. but the side A is common. Therrefore the other Sides FE and ED are equal by, by Consect. 3. Prop. 43.) and the □ of the latter is ¼ aa, which subtracted from aa leaves ¾ aa for the □ of A Therefore the line A is , and consequently AB , i e. , i e. : therefore □ AB = 3 aa. Q E. D. CONSECTARYS. I IF the Radius of a Circle be = a, the side of an Inscribed Regular Triangle will be , e. g. if AD be 10, A B will be ; and if AD be 10,000,000, A B will be , i e. 17320508, and the Perpendicular DE 5000,000. II Hence it is evident, that in the genesis of a Tetraedrum proposed in Def. 22, that the elevation CE (Fig. 44, N. 1.) is to the remaining part of the Diameter of the Sphere CF as 2 to 1; for making the Radius C B = a and its □ aa, the □ of A B or BE will = 3 aa, by the present Proposition. Therefore the □ of CB being subtracted from the □ of BD or BE, there remains the □ of CE = 2 aa. But since CE, C B, CF, are continual Proportionals, by N. 3. Scholar 2. Prop. 34. CE will be to CF as the □ of CE to the Square of C B, by virtue of Prop. 35. i e. as 2 to 1. SCHOLIUM. HEnce you have the Euclidean way of generating (α) Eucl. 12 lib. 12. a Tetraedrum, and inscribing it in a given Sphere, when he bids you divide the Diameter OF of a given Sphere so that EC shall be 2 and CF 1, and than at OF to erect the Perpendicular CA, and by means thereof to describe the Circle A BD, and to inscribe therein an Equilateral Triangle, etc. Proposition L. THE Side (AB) of a Regular Tetragon (or Square) (ABCD, Fig. 92. N. 2.) is double in Power of the Radius (AD). Demonstration. For having drawn the Diameter AC and BD, the Triangle AO B is Rightangled, and consequently, by the Pythagorick Theorem, if the □ of AO and BOY be made equal to aa, than will the □ of A B = 2 aa, Q. E. D. CONSECTARY. THerefore when the Radius of the Circle AO is made = a, the side of the □ A B will = √ 2 aa, e. g. if AO be 10, A B will be √ 200; and if AO be 10,000,000, A B will be √ 200,000,000,000,000, i e. 14142136. Proposition LIVELY THE side AB of a Regular Pentagon (α) Eucl. Prop. 13. lib. 13. (ABCDE) (Fig. 93. N. 1.) is equal in Power to the side of an Hexagon and Decagon inscribed in the same Circle, i e. the □ of AB is equal to the Squares OF an AO taken together. Demonstration. Make AO = a and OF = b, A B = x: We are to demonstrate that xx = aa + bb: which may be done by finding the side A B by the parts BH and HA', (α) Eucl. Prop. 10. lib. 13. after the following way: First of all the angle AO B be 72°, and the others in that Triangle at A and B 54°. But BGG is also 54°, as subtending the Decagonal Arch BF of 36°, and also one half of it FG of 18°. Therefore the ▵ ▵ A BOY and H BOY are Equiangular, and you'll have As A B to BOY so BOY to BH Secondly, in the Triangle BFA the Angles at B and A are equal by N. 3. Consect. 5. Def. 8. and by virtue of the same also the Angles at FLETCHER and A in the ▵ FHA are so too. Wherhfore the ▵ ▵ BFA and FFA are Equiangular, and you'll have As BASILIUS to OF so OF to AH Therefore the whole line A B (because the part AH is found = and BH = ) will be , which was first made = x; so that now is = x, and multiplying both sides by x, aa + bb = xx. Q. E. D. CONSECTARY I THerefore if the Radius of a Circle be (a) the side of a Pentagon A B will be . CONSECTARY II THerefore the □ AI = and □ HEY = i. e. . Therefore HEY = . Which yet may be expressed otherwise, viz. HEY = . Demonstration. Make (α) Eucl. Prop. 1. lib. 14. OA or OF (N. 2.) as before = a, OF = b, and FI now = x; than will HEY = a − x and having drawn the Arch FK at the Interval OF, so that AK may be equal to this, and FI = IK = x; than will the angle IKA = F72°. Therefore the angle AKO = 108; and since KOA is 36°, KAO will be also 36°, and so KA = KO = OF = b. Wherhfore HEY is = b + x, which was above a − x. Therefore 2OI = , i e. a + b. Therefore HEY = . CONSECTARY III THerefore the difference between the Perpendicular of the Triangle DE (Fig. 92 and 93. N. 1.) and the Perpendicular of the Pentagon HEY is = ½ b, by virtue of Consect. 2. of this and of the Demonstrat. of Prop. 49. CONSECTARY IU. HEnce is also evident, by the present Proposition, that which in Fabri's Genesis of an Icosaedr. Def. 22. we said, viz. that (See Fig. 46.) B a is equal to the side of a Pentagon BASILIUS, because, viz. Fa is = to the Semidiameter OB, and BF is the side of a Decagon. Proposition LII. THE side of an Hexagon is in Power equal to the Radius, as being itself equal to it by N. 1. Scholar Def. 15. Proposition LIII. THE side of a Regular Octagon (ABCD, etc. Fig. 94.) is equal in Power to half the side of the Square, and the difference (PB) of that half side from the Radius, taken together. Demonstration. For that the □ of AB is = to the □ of AP+ □ BP, is evident from the Pythag. Theor. But that PO is = PA half the side of the Square, is evident from the equality of the Angles PAO and POA, since each is a half right one or 45°. Wherhfore the side of the Octagon is equal in Power to half the side of the Square, etc. Q. E. D. CONSECTARY. THerefore, if the Radius be = a, AP will be, by virtue of the Pythag. Theor. = and P B = of that □ ½ aa; and of this □ . Therefore the Sum of □ AB = . Therefore the side of the Octagon = . e. g. If AO be 10, AB will be , and if the Radius AO 10000000, AB will be , by virtue of the present Cons. or from the Prop. itself, if AO be 10000000, AP will be virtue of the Cons. of Prop. 50. = 7071068, and consequently BP = 2928932 The Squares of these added together give the □ AB, and the Root thence extracted A B = 7653 668. Proposition LIU. THE side of a Regular Decagon (α) Eucl. 9 lib. 13. Corol. is equal in Power to the greatest part of the side of an Hexagon cut in mean and extreme Reason. Demonstration. Suppose BD (Fig. 95.) divided in mean and extreme Reason in E, and BA to be joined to it long ways = to the side of a Decagon inscribed in the same Circle, whose Radius is BC or AC = BD. Now we are to demonstrate that DE the greatest part of the de of the Hexagon BD divided in mean and extreme Reason, is equal to the side of a Decagon BASILIUS, and the Power of the one equal to the Power of the other. Because the Angle AC B is 36°, A BC and A72°, and consequently C BD 108°, BCD and D will be each 36°, and so the whole ACD 72°. (that so CD may pass precisely through the other end of the side of the Pentagon AF.) Wherhfore the ▵ ▵ ABC and ADC are Equiangular, and AD to AC i e. BD as AC i e. BD to AB. Therefore the whole line AD is divided in mean and extreme Reason. But BD is also divided in the same Reason by Hyp. Wherhfore As AD to D B and D B to BA, So DB to DE and DE to E B. Therefore D B is in the same Proportion to DE as D B to BA. Therefore DE is = BA, and the Power of the one to the Power of the other. Q. E. D. CONSECTARYS. THerefore, if the Radius of the side of the Hexagon is a the side of the Decagon will be , by Scholar 2. Prop. 27. e. g. if the Radius be 10, the side of the Decagon will be , and if the Radius be put 10000000, the side of the Decagon will be = , viz. by adding the Square of the Radius and the Square of half the Radius into one Sum; whence you'll have the side of the Decagon = 6180340; the half whereof 3090170 gives the difference between the Perpendiculars of the Triangle and the Pentagon, by Cons. 3. Prop. 51. II The side therefore of the Pentagon is by Prop. ; for the Square of the Hexagon is aa or aa, the □ of the Decagon : the Root extracted out of the Sum of these is the side of the Pentagon, viz. e. g. if the Radius be 10, the side of the Pentagon will be , and if the Radius be put 10000000, since the side of the Hexagon is equal to it, and the side of the Decagon 6180340, their Squares being added into one Sum, the Root extracted out of that Sum will give the side of the Pentagon, 1755704 nearly; and the sides being collected into one sum, the half of it 8090170 will give the Perpendicular in the Pentagon HEY, by Consect. 2. Prop. 51. SCHOLIUM I TO illustrate what we have deduced in the Consectaries of Prop. 51, you may take the following Notes. If a be put = 10 or , the side of the Decagon will be = , i e. nearly = b; therefore the □ aa = and the □ bb = : therefore aa + bb or the □ AB = , the Perpendicular HEY = divided by 2, that is, . Now the □ of AI is ¼ of the □ A B = the □ HEY = = . Now if you add the □ AI and the □ HEY, the sum will be = □ AO = , i e. , or near 100 Thus likewise, since the Perpendicular above found HEY, in Consect. 1. may be also determined by , since aa is = , and 3 aa , subtracting from it bb = the Remainder will be , and this being divided by 4, you'll have the □ HEY = . and the Root of it extracted nearly; so that those two different quantities in Consect. 1. will rightly express the same Perpendicular HEY. SCHOLIUM II NOW therefore as we have Practical Rules to determine Arithmetically the sides of the Pentagon and Decagon, so also they may be found Geometrically by what we have demonstrated. For if the Semidiameter C B (Fig. 96. N. 1.) be divided into 2 parts, EC will = ½ a; and erecting perpendicularly the Radius CD = aDE will = . Moreover if you cut of OF equal to it, FC will be = = to the side of the Decagon, by Consect. 1. Having therefore drawn DF, which is equal in Power to the Radius or Side of the Hexagon DC, and the side of the Decagon FC together, by the Pythag. Theorem it will be the side of the Pentagon sought. Much to the same purpose is also this other new Construction of the same Problem, wherein BG (Numb. 2.) is the side of the Hexagon B D the side of the Square, to which GF is made equal, so that FC is that side of the Decagon, and DF of the Pentagon; which we thus demonstrate after our way: Having bisected GH the side of an Equil. Triangle, the Square of GE will be , by Prop. 48. which being subtracted from the Square of GF = 2 aa, viz. aa, by Prop. 49. there will remain for the Square of OF aa, and for the line OF , and for FC , which is the side of the Decagon, as DF of the Pentagon, after the same way as before. Proposition LU. THE side of a Quindecagon (or 15 sided Figure) is equal in Power to the half Difference between the side of the Equil. Triangle and the side of the Pentagon, & moreover to the Difference of the Perpendiculars let fall on both sides taken together. Demonstration. For if AB (Fig. 97.) be the side of an Inscribed Triangle, and De the side of a Pentagon parallel to it; AD will be the side of the Quindecagon to be inscribed, by Consect. 4. Def. 15. But this side AD in the little Rightangled Triangle, is equal in Power to the side AH (which is the half Difference between AB and DE) and the side HD (which is the difference between the Perpendiculars CF and CG) taken together by the Pythag. Theor. Q. E. D. CONSECTARY. HEnce if we call the Side A B of the Equil. ▵ c, and mak● the side of the Pentagon DE = d, AH will = HD is = ½ b, by Consect. 3. of Prop. 51. Since therefore the ▵ AH is = and the ▵ HD = the ▵ AD w●●● Therefore the side of the Quindecagon = that is, Collecting the Square of the half difference of the side of the Triangle and Pentagon, and the Square of the difference of the Perpendiculars into one Sum, and than Extracting the Square Root of that Sum, you'll have the side of the Quindecagon sought. E. 9 if the Radius CI be made 10000000 the difference of the sides of the Triangle 17320508, and of the side of the Pentagon 11757704 will be 5564804, and the half of this 2782402; but the difference of the Perpendicular CF from the Perpendicular CG, is 3090170. The Squares therefore of these Two last Numbers being Collected into one Sum, and the Root Extracted will give the side of the Quindecagon 4158234 nearly. SCHOLIUM. HEre we will show the Excellent use of these last Propositions in making the Tables of Signs. For having found above, supposing the Radius of 10000000 parts, the sides of the chief Regular Figures, if they are Bisected, you will have so many Primary Sines; viz. from the side of the Triangle the side of 60 Degrees 8660754, from the Side of the Square, the Sine of 45° 7071068; from the Side of the Pentagon, the Sine of 36° 5877853; from the Side of the Hexagon, the Sine of 30° 5000000; from the Side of the Octagon, the Sine of 22° 30 3826843; from the Side of the Decagon the sine of 18° = 3090170; from the side of the quindecagon lastly the sine of 12° = 2079117. From these seven primary sins you may found afterwards the rest, and consequently all the Tangents and Secants according to the Rule we have deduced, n. 3. Scholar 5. Prop. 34. and which Ph. Lansbergius illustrates in a prolix Example in his Geom. of Triangles Lib. 2. p. 7. and the following. But after what way, having found these greater numbers of sins, Tangents, etc. Logarithms have been of late accommodated to them, remains now to be shown, which in brief is thus; viz. the Logarithms of sins, etc. might immediately be had from the Logarithms of vulgar numbers, if the tables of vulgar numbers were extended so far, as to contain such large numbers; and thus the sine e. g. of oh gr. 34. which is 98900 the Logarithm in the Chiliads of Vlacquus is 49951962916. But because the other sins which are greater than this are not to be found among vulgar numbers (for they ascend not beyond 100000, others only reaching to 10000 or 20000) there is a way found of finding the Logarithms of greater numbers, than what are contained in the Tables. E g. If the Logarithm of the sine of 45° which is 7071068 is to be found, now this whole number is not to be found in any vulgar Tables, yet its first four notes 7071 are to be found in the vulgar Tables of Strauchius with the correspondent Logarithm 3. 8494808, and the five first 70710 in the Tables of Vlacquus with the Log. 4. 8494808372. One of these Logarithms, e g. the latter, is taken out, only by augmenting the Characteristic with so many units, as there remain notes out of the number proposed, which are not found in the Tables, so that the Log. taken thus out will be 6. 8494808372. Than multiply the remaining notes of the proposed number by the difference of this Logarithm from the next following, (which for that purpose is every where added in the Vlacquian Chiliads, and is in this case 61419) and from the Product 4176492 cast away as many notes as adhere to the proposed number beyond the tabular ones, in this case 2; for of the remainder 41764, if they are added to the Logarithm before taken out, there will come the Logarithm required 6. 8494850136, viz. according to the Tables of Vlacq●us, wherein for the Log of 10 you have 10,000,000, 000; but according to those of Strauchius which have for the Logarithm of 10 only 10,000,000, you must cut of the three last notes, that the Logarithm of the given sine may be 6.8494850; as is found in the Strauchian and other tables of sins, except that instead of the Characteristic 6 there precedes the Characteristic 9, whereof we will add this reason: If the Characteristics had been kept, as they were found by the rule just now given, the Logarithm of the whole sine (which is in the Strauchian Tables 10,000,000) would have come out 70,000,000, incongruous enough in Trigonometrical Operations. Wherhfore that Log. of the whole sine might begin from 1, for the easiness of Multiplication and Division they have assumed 100,000,000; the Characteristic being augmented by three, wherewith it was consequently necessary to augment also all the antecedent one's; and hence e. g. the Logarithm of the least sine 2909 gins from the Characteristic 6, which otherwise according to the Tables of vulgar numbers would have been 3. Having found after this way the Logarithms of all the sins (altho' here also if you have found the Logarithms of the signs of 45° and moreover the Logarithm of 30, the Logarithms of all the rest may be compendiously found by addition and substraction from a new principle which now we shall omit) the Logarithms of the Tangents and Secants may easily be found also, only by working, but now Logarithmically, according to the Rules of Scholar 5. Prop. 34. n. 5. and 6. Proposition LVI. THE side of a Tetraëdrum (α) Eucl. 13. lib. 13. or equ● Pyramid is in power to the Diameter of a circumscribed Sphere, as 2 to 3. Demonstration. For because by the genesis of the Tetraëdrum Def. 22 (see its Fig. 44 n. 1.) and Scholar Prop. 49. OC is ⅓ of the semidiameter OB, which we will call a the □ of CB will be = aa by the Pythag. Theor. and so the power (or Sq.) of the side of the Tetraëdrum = aa by Prop. 49. but the power of the Diana. 2 a or a is aa. Therefore the power of the side of the Tetraëdrum is to the power of the Diana. as 24 to 36. i e. (dividing each side by 12) as 2 to 3. Q. E. D. Or more short. The □ of CB is = 2 by Schol of Prop. 49. and the □ of EC = 4. Therefore the □ EBB = 6. But the □ OF is = 9 Therefore the □ of EBB is to the □ of OF as 6 to 9, i e. as 2 to 3. Q. E. D. CONSECTARY. THerefore if the Diana. OF be made = a, the side EBB will be = . Proposition LVII. THE side of the Octaëdrum (α) Eucl. 14. lib. 13. is in power one half of the Diameter of the circumscribed Sphere. Demonstration. For since by the genesis of the Octaëdrum Def 22. (see Fig. 44. n. 2.) CA, CB, CF, etc. are so many radii of great circles, if for Radius you put a, the square of OF will be = 2 aa by the Pythag. Theor. But the square of the Diana. FG = 2 a is 4 aa. Therefore the power of the side is to the power of the Diana as 2 to 4, i e. as 1 to 2. Q. E. D. Moore short. Because OF is also the side of a square inscribed in the greatest circle by the gen. of the Octaëdrum; the □ of OF will be by Prop. 50. to the □ of FC as 2 to 1: Therefore to the square of FG as 2 to 4, by Prop. 35. Q. E D. CONSECTARY. THerefore if the Diana. of the sphere be made (a) the side of the Octaëdrum OF will be . Proposition LVIII. THE side of the Hexaëdrum or Cube (α) Eucl. 15. lib. 13. 〈◊〉 in Power subtriple of the diameter of the circumscribed sphere. Demonstration. Making a the side of the inscribed cube GF or FE (Fig. 98.) the square of the diagonal GE of the base of the cube will be 2 aa by the Pythag. Theor. and by the same Reason the square of the diam. of the cube and the circumscribed sphere GD will be = to the square of GE+ □ DE = 3 aa. Q. E. D. CONSECTARYS. I THerefore if the diameter of the sphere be made = a, the side of the cube AB will be . II The diameter of the sphere is equal in power to the side of the Tetraëdrum and cube taken together. For if the power of the diam. of the sphere be made aa the power of the side of the Tetraëdrum will be ⅔ aa by Consect. Prop. 56. and the power of the side of the cube ⅓ aa by the pres. Consect. 1. Wherhfore these two powers jointly make aa. Q. E. D. Proposition LIX. THE side of the Dodecaëdrum (α) Eucl. Consect. 1. Prop. 17. lib. 13. is equal in power to the greater part of the side of the cube divided in mean and extreme reason. Pag. 150. 97 98 99 100 101 102 103 Demonstration. For if to the side of the cube AB (vid. Fig. 45. n. 1.) and to its upper base ABCD you conceive to be accommodated or fitted a regular Pentagon according to the genesis of a Dodecaëdrum laid down in Def. 22, and at the interval B e you make the arch ef, the ▵ ▵ AB e and A ef will be equiangular; (for the angles at A and B being 36°, and A ebb 108, having drawn ef, the angles B of and B fe are each 72°; therefore A of the remaining angle will be 36°) wherefore as AB to B e (i e. B f) so A e (i e. B e or B f) to A f. Therefore the side of the cube AB is divided in mean and extreme reason in f, and B e the side of the Dodecaëdrum is = to the greater part B f. Q. E. D. SCHOLIUM. HEnce would arise a new method of dividing a given line in mean and extreme reason, viz. if you apply to the given line a part of the equilateral Pentagon by means of the angles A and B 36°, and at the interval B e you cut off B f. This angle may be had geometrically, if another regular Pentagon be inscribed in a circle, and having drawn also a like subtense, if the angles at the subtense are made at A and B equal, by Eucl. 23. lib. 1. Proposition LX. THE side of an Icosaëdrum (α) Eucl. Prop. 16. Coral. lib. 13. is equal in power to the side of a Pentagon in a circle containing only five sides of the Icosaëdrum; and the semi diameter of this circle is in power subquintuple of the Diana. of the sphere of the circumscribed Icosaëdrum. Demonstration. Both these are evident from the genesis of the Icosaëdrum in Def. 22. The first immediately hence, because all the other sides of the triangles (Fig. 99) A a, B a, etc. are made equal to the side of the Pentagon AB by Consect. 4. Prop. 51. The latter from this inference; if for OA the radius of the circle you put a (since the side of the Pentagon, which here is also the side of the Icosaëdrum, it will be equal in power to the radius and side of the Decagon taken together by the aforesaid Prop.) the altitude OG will be the side of the Decagon = by Consect. 1. Prop. 54. to which the equal inferior part o being added, and the intermediate altitude OH oh = a, you'll have the whole diameter of the circumscribed sphere GH = i. e. i e. i. e. and so the square of the diameter of the sphere will be 5 aa: Therefore the square of the diameter of the sphere is to the square of the semidiam. of the circle containing the five sides of the Icosaëdrum as 5 to 1. Q. E. D. SCHOLIUM. IT is also evident that a sphere described on the diameter GH will pass through the other angles of this Icosaëdrum; for assuming the centre between OH and oh the radius FG will be = . But FAVORINA is also = ; for the □ of FOYES is = ¼ aa, and the □ AO = aa: Therefore the sum is = aa = □ FA. Q. E. D. CONSECTARY I THerefore, if the radius of the circle ABCDE remain a, you'll have the altitude OG , and the side of the Icosaëdrum , by Cons. 1. and 2. Prop. 54. and the diam of the circumscribed Sphere 2 , as is evident from the Demonstration. CONSECTARY II Being a general one of the five last Propositions. IF AB (Fig. 100L) be the diameter of a sphere (α) Eucl. Prop. 18. and los●, lib. 13. divided in D so that AD shall be ⅓ AB, than (having erected the perpendicular DF) BF will be the side of the Tetraëdrum by Prop. 56. and OF the side of the Hexaëdrum by Prop. 58. Cons. 2. and BE or A (erecting from the centre the perpendicular CE) will be the side of the Octaëdrum by Prop. 57 Now if OF be cut in mean and extreme reason in OH, you'll have AO the side of the Dodecaëdrum by Prop. 59 Lastly, if you erect BG double of CB, HI will be double of CI, and the □ of HI = 4 □ of CI; consequently the □ CHANGED or CB = 5 □ CI. Therefore the □ of AB (double of CH) is also = to 5 □ of HI (which is double of CI) therefore HI is the radius of the circle circumscribing the Pentagon of the Icosaëdrum, and IB the side of the Decagon inscribed in the same circle, and HB the side of the Pentagon, and also the side of the Icosaëdrum, by Prop. 60. The End of the first Book. THE SECOND BOOK. SECTION I Containing DEFINITIONS. Definition I IF a Cone ABC (Fig. 101.) be conceived to be cut by a plane at right angles to the side of the cone, e. g. BA; the Plane EFGHE arising by this section, and terminated on the external surface of the cone by the curve line HEG, etc. was anciently by Euclid, Archimedes, etc. called the Conic Section; and if the sides of the cone AB and BC made a right angle at B, as n. 1. the section was particularly called the Section of a rightangled Cone; but if it made an obtuse angle, as n. 2. it was called the Section of an obtuse-angled Cone; if, lastly, it made an acute one, as num. 3. it was called (3) the Section of an acuteangled Cone. Definition II BUT afterwards their Successors, particularly Apollonius Pergaeus, found from the properties of these Curves, which their Predecessors had happily discovered, that the same (all of them) might be generated in one and the same cone whether rightangled, obtuse-angled, or acuteangled, if the section OF (Fig. 102.) is made in the first case parallel to the opposite side BC; in the second case, if it meet that side produced upwards; for the third, when it meets downwards with the diameter of the base AC produced to D. And to give new names (for the old ones would do not more now) to these Sections, to distinguish them one from another, nominating them from their Properties hereafter demonstrated, they called the first a Parabola, the second an Hyperbola, the third an Ellipsis. Definition III BUT it is evident, that only the plane making the section of the second case, being according to the line FED produced or carried on, (Fig. 103.) will fall upon the vertical Cone aBc comprehended under the sides AB, CB, etc. continued onwards, and there produce another Section opposite to the former; whence these, viz. GEHG and gehg are called opposite Sections. Definition IU. BEsides these names of the sections, there are several others made use of to denote various lines drawn and considered both within and without those sections, the chief whereof we shall here explain. And first of all, in general the line OF so let fall through the middle of the section from its top E (which is called the Vertex of the section) to the diameter of the base of the Cone AC (produced if occasion be) that it shall bisect the line GH and all others parallel to it, is called the Diameter of that Section; and particularly it is called the Axis of the Section if it cuts them at right angles or perpendicularly; as also they name those lines GH, KN, etc. which are cut indifferently by the diameters, but at right angles by the Axis, those, I say, they call Ordinates', or Ordinate Applicates, and their halves, FG, IK, etc. Semiordinates', (or some also call the latter Ordinates', and the former double Ordinates') and the parts of the Axe or Diameter OF, EI, etc. are called Abscissa's (by some intercepted axes and diameters) Definition V PArticularly in the hyperbola they call the continuation of the axe or diameter ED till it meet the opposite side cB, i e. to the vertex of the opposite section, the Latus transversum of the Hyperbola, to which there answers in the Ellipsis the axis or longest diameter, and so by latter Authors is called by the same name, but by Apollonius the transverse Axe or the transverse Diameter, as also the shortest axe or diameter is called the Conjugate, and its middle point OH is called the Centre of the Section or of the opposite Sections. Definition VI. THey called also a certain line EL (Fig. 101.) by the name of Latus Rectum, which is particularly to be found in all the sections, as we shall hereafter show: And because this Latus Rectum is a sort of a Rule or Measure, according to which the squares or powers of the ordinates' used to be estimated or valued (as we will show in its proper place) therefore the Ancients used to call it by a Periphrasis Linea secundum quam possunt Ordinatim applicatae, or the measure of the powers of the Ordinates'; by some latter Authors it is called the Parameter. Now a mean proportional PQ found between this Latus Rectum and the Latus Transversum (Fig. 104 n. 1. and 2) (see also hereafter Consect. 2. Prop. 8.) and drawn through the centre OH parallel to the Ordinates' is called the second Axis or Diameter, or the Conjugate of the Hyperbola. Definition. VII. NOW if we conceive the diameter or conjugate axe PQ brought down to the Hyperbola so that its middle point OH shall touch its vertex in E, and from the centre OH you draw the right lines OR, OS, through the ends of this tangent line lordship and cue, these are the lines which Apollonius, Prop. 1. lib. 2. demonstrates, that though by being continued, they always approach nearer and nearer to the curve GEH, and come so Pag. 156. 104 105 106 107 much the nearer by how much the farther they are continued, yet they will never concur with it or touch it, for which reason they are called Asymptotes or non-coincident lines; and by some the Intactae. Which non-coincidency appears most manifestly where the hyperbolical section of the cone is made parallel to the triangular section through the axis of the cone ABC (n. 3.) along the line e. f. parallel to the axe BF. For if we conceive the hyperbola geh to move forwards always parallel to itself, according to the direction of the equal and parallel lines gG, fF, and hH, till it stands in the position GEH; it is manifest that the curve line GEH is distant on both sides from the right lines BC, BA, the length of the versed sine of the equal arches hC and ga' in the circumference of the circular base, while in the mean time it is evident that they approach nearer and nearer to them. So that hence there flow the following CONSECTARIES. I IN this case the sides of the cone are the Asymptotes of the hyperbola, while it is manifest, that the point B is its centre, and EBB half the transverse diameter; which appears from n 1. and 2. of the pres. Fig. for the section of being made parallel to the the axe of the cone DF by def. 5. de (which in the case n. 3. would coincide with dq) is the transverse diameter, but the triangles dpq and O pE are equiangular, and consequently as pE is to ½ pq, so is OE to ½ dq. II The lines AGNOSTUS and HG (num. 3.) are equal, as being the versed sins of equal arches; and in like manner (n. 1. and 2.) RG and HIS are equal, since FR and FS as well as E lordship and E cue are so also (for as OE to E p, − E cue so OF to FR − FS and the semiordinates' FG and FH are also equal. III Consequently □ □ of RG into GS and of HIS into HR are equal, etc. all which hereafter we will more universally demonstrate. Definition VIII. IF a Parabolic plane (α) Archim. de Conoid. & Spher. Def. 1. HEGFH (Fig. 105. n. 1.) together with a triangle HEG inscribed in it, and a rectangle HL circumscribed about it, be conceived to be moved round about the axe OF on the point F; it will be evident that by the triangle there will be generated a cone, by the rectangle a cylinder, and by the parabola a parabolic solid, which with the comprehended cone, and the comprehending cylinder, will have the common base HIGK and the same altitude OF, and was by Archimedes named a Parabolic Conoid. Definition IX. IF moreover an (α) Archim. de Conoid. & Spher. Def. 3. hyperbolic plane HEGFH (n. 2.) with the inscribed triangle HEG, and another circumscribed one ROSALURA made by the Asymptotes OR, OS, be conceived to be turned about the common axe OEF on the point F; it will be evident that there will be described by the inscribed triangle a cone comprehended within side, and by the hyperbola an Hyperbolical Conoid upon the same base HIGK and of the same height OF; and by the ▵ ROS another cone which Archimedes calls the comprehending cone, whose base is RTSV, and its altitude composed of the axis of the hyperbola OF and half the transverse axe OE (which Archimedes called the additament of the axe of the hyperbola) and which we may commodiously divide into two parts, viz into the cone OPMQL, whose base has for its diameter the conjugate axe PQ, and its altitude equal to half the transverse axe; and into a Curticone or detruncated cone terminated by the two bases PMQL and RTSV, but answering in altitude to the conoid and inscribed cone: From which, as comprehending it, if you take away the included conoid, there will remain the hollow curticone terminated below by the Annulus or ring RGIUHSKT and above by the circular base PMQL, and generated in the circumvolution by the intermediate lines EP, GRACCUS, etc. or the mixtilinear plane EGRP. Now if we suppose instead of this comprehending cone a circumscribed cylinder on the same base and of the same altitude with the conoid and included cone, you'll have every thing like as in Def. 8. CONSECTARYS. NOW if we suppose the 1. case of Def. 9 to be expressed by the Fig. of Def. 7. n. 3. and conceive the present figure brought thence to be turned round about the axe BEF (Fig. 106.) we may deduce these following things in the room of Consectaries for the foundation of our future demonstrations. I The lines EQ, RS, HC, etc. of the mixtilinear space comprehended between the hyperbola and the Asymptotes (viz. the excess of the ordinates') altho' they are unequal, and by descending always grow lesle, yet in this circumvolution they will describe equal circular spaces, viz. EQ a whole circle, (or circular plane) but RS and HC, etc. circular Annuli or rings all of the same bigness; which will thus appear to any one who compares this figure with the former: Since the spaces generated by the lines EQ, FC, etc. are as the squares of those lines, and the □ of F h or FC exceeds the square of fh or FH by the excess of the square F f or E e or EQ, consequently the quantity generated by FC will exceed that generated by FH the excess of that generated by EQ; and also that generated by FH by the excess of that generated by HC; it is manifest that the circle generated by EQ will be equal to the annular space generated by HC; and the same will in like manner be evident of any spaces produced by RS. II Therefore the hollow detruncated cone generated by the space EHCQ according to Def. 9 will be equal to a cylinder generated by the rectangle FIQE; for all the indivisibles of the one, are equal to all the indivisibles of the other by Consect. 1. Definition X. IF, lastly, (α) Archim. de Conoid. & Spher. Def. 6. an elliptical plane be turned about either of the axes, viz. either the longest DE (Fig. 107. n. 1) or the shorter AB (n. 2.) there will be thence form an elliptical solid, called by Archimedes a spheroid; which in the first case will be an oblong or erect one, in the other a flat or depressed one: And it is self evident, that if before this circumvolution of the ellipsis, there be inscribed in one of its halves a triangle, and also a rectangle circumscribed about it, having the same altitudes and bases with the semi ellipse, there will afterwards in the circumvolution be described by the triangle a cone, by the rectangle a cylinder, to which afterwards we will also compare the half spheroid; as also both the conoids with their inscribed and circumscribed cones and cylinders. Definition XI. IF upon the right line A (Fig. 18.) you conceive a wheel or circle to roll, until its point A, with which it touches the said line, come to touch it again in E; the circle will measure out the line A equal to its periphery; but the point A by its motion will describe the curve line AFE, which is called a Trochoid or Cycloid; and the area which this curve with the subtense A comprehends, is named the Cycloidal Space; and the circle by whose motion they are determined is called the generating Circle. CONSECTARY. IT is evident from the genesis of this curve that the describing point a will always be as much distant in the circle from the point of contact d or c, as the point A in the right line passed over A, is from the same point of contact, i e. if the point d is distant from A the fourth part of the line A, the arch da will also be the fourth part of the circle considered in this second station; and the point c being distant from A half of the interval A, the arch ca will be also half of the Pag. 160. 108 109 110 111 circle, and so the point a coincide in the curve with FLETCHER: And when the point e is distant from E only an eighth part of the whole line A, the arch ea will also be the eighth part of the whole circle. Definition XII. IF the right line BASILIUS (num. 1. 109.) one end at B remaining fixed, be moved round at the other end with an equal motion from A through C, D, E to A back again, and in the mean while, there be conceived another movable point in it to move with an equal motion along the line BASILIUS from B to A, so that in the same moment wherein the movable point A absolves one revolution, the other movable point shall also have passed through its right lined way, coinciding with the point A returned to its first situation; this extremity A by its revolution will describe the circle ACDEA, and that other movable point another curve B, 1, 2, 3, 4, etc. which with Archimedes we will call a Helix or spiral Line, and the plane space comprehended under this spiral line and the right line BASILIUS in the first station is called a spiral space. Now if we suppose, e. g. the right lined motion of the point moving along BASILIUS to be twice slower than in the former case, so that (see num. 2.) in the same time that the point A makes one whole Revolution, the other movable point shall come to FLETCHER, making half the way BASILIUS, and than at length shall concur or meet with the extremity, when that shall have performed the other revolution; and so there will be described a double spiral line, whose parts with Archimedes we will so distinguish, that as he calls the part of the right line BF, passed over in the first revolution, simply the first line, and the circle made by the right line BF the first Circle; so we will call that part of the curve which is described in that time or revolution B 136912 the first Helix or the first Spiral, and the area comprehended under it the first spiral space: And, as the other part of the right line FAVORINA passed over in the other revolution is called the second line, and the circle marked out by the whole line BASILIUS the second Circle; so the curve described in the mean while 12, 15, 18, 24, may be called the second spiral line, and the space comprehended under it the second spiral space, and so onwards From these Definitions there flow the following CONSECTARIES. I THE lines B 12, B 11, B 10, etc. drawn out, and making equal angles to the first or second spiral (and after the same manner (α) Archim. Prop. 2. of Spirals. B 12, B 10, B 8, etc. or B 12, B 9, B 6, etc.) are arithmetically proportional, as is evident. II The lines drawn out to the first spiral as B 7, B 10, etc. are one among another as the arches of the circles intercepted between BASILIUS and the said lines (β) Archim. Prop. 4. B 7, B 10, etc. which is also evident to any one who considers what we did suppose; for in the same time as the end A passes over seven parts of the circle, the other movable point will also run over seven parts of the right line BASILIUS, etc. III Lastly, The right lines drawn from the initial point (γ) Archim. Prop. 15. B to the second spiral e. g. B 19 and B 22 (num. 2.) will be one to another as the aforesaid arches together with the whole periphery added to both sides: for at the same time the extremity A runs through the whole circle or 12 parts and moreover 7 parts (i e. in all 19 parts) in the same time the other movable point passes through 12 parts of the right line BASILIUS (in this case divided into 24 parts) and moreover 7 parts, that is, in all 19; and so in the others. Definition XIII. IF a right line BAF be conceived so to move within the right angles ADC, CDE, that on the one hand a certain point C fixed in one leg of the Norma or ruler may always glide along, and on the other hand a certain movable point A may always run along the other side of the Norma (which complicated motion of the describing rule BAF, after what way it may be organically procured, may be seen by Pag. 163. 112 113 114 115 116 117 118 the 110th. Figure;) by the extreme point of the movable line B there will be described a curve called by its Inventor Nicomedes the first Conchoid, whereof this is a property, that the right lines CB, C b, drawn from its centre C to its Ambitus or curvity are not themselves, as in the circle, equal, but yet have all the parts, AB, ab, intercepted between the curve, and the directrix A are equal; as is evident from its genesis. Definition FOURTEEN. IF, the diameters of a circle being AB and CD (Fig. 111. n. 1.) cutting one another at right angles, you take BE or B e and BF or B f equal arches, and from E or e you draw the perpendiculars EGLANTINE or Eglantine, and through these from D to FLETCHER or f you draw the transverse lines DF or D f; the several points of intersection H, h, etc. decently connected together will exhibit the curve line D h, H hB (which may be continued also without the circle if you please, and) which is commonly attributed to Diocles, and called a Cissoid. Definition XU. IF, having divided the right line A (Fig. 112.) into the equal parts AB, BC, etc. from the points of the division A, B, C, etc. you draw the parallel lines A a, B b, C c, etc. in geometrical progression (as e. g. let A a be 1, B b 4, C c 16, D d, 64, etc. or B b 10, C c 100L, D d 1000, E e 10000, etc.) and further bisecting AB, BC, &c, FLETCHER, G, H, I, you let fall mean proportionals between the next Collaterals' Ff, Gg, Hh, Two, etc. and continued to do so till the parallels are brought near to one another; the curve line drawn through the extremities of these lines a f, b g, c h, d ie, will be the logarithmical line of the moderns, whose properties and uses are very excellent. SCHOLIUM. AMong those uses, that is none of the lest, from which this curve borrows its name, viz. in showing the nature and invention of Logarithms. For, e g. 1. If this line was accurately delineated in a large space, the parts AB, BC, etc. being taken so big, that they might be subdivided not only into 100 or 1000 but into 10000 or 100000 parts; making AB 100000 (and so A 00000) AC would be 200000, AD 300000, etc. while in the mean while there answer to these as primary Logarithms in arithmetical progression the geometrical proportional numbers, A a 1, B b 10, C c 100L, D d 1000, E e 10000, etc. Whence, 2. It's Logarithm may be assigned to any given intermediate number, e. g. to the number 982, for having cut of this number from Dd by a geometrical scale on the line DM, if you draw Mn parallel to AD, and nN parallel to DM, it will give AN on the same scale, viz. the Logarithm sought, and reciprocally. But if, 3. it seem difficult to delineate a Figure so large, yet at lest the clear conception of such a delineation evidently shows the arithmetical method, which those ingenious Men have made use of, who have made the tables of Logarithms with a great expense of Labour and pains, viz. by finding continual mean proportionals, arithmetical ones between any two Logarithms already known, and geometrical ones between two vulgar numbers answering to them, etc. by comparing what we have noted in Scholar 2. Prop. 20. Lib. 1. And we will note, 4 out of Pardies, that, since the Logarithms of numbers distant from one another by a decuple proportion, differ by the number 100000, having found the Logarithms of all the numbers from 1000 to 10000 you will at the same time have all the Logarithms of all the other numbers that are between 100 and 1000, between 10 and 100, and between 1 and 10, only changing the characteristic, and lessening it in the first case by unity, in the second by 2, in the third by 3; as e. g. if of the number 9, 900 you had found the Logarithm 399, 563, the Logarithm of the subdecuple number 990 would be (viz. substracting from the former 100000) 299, 563. and the Logarithm again of this subdecuple of this 99 would be 199, 563, etc. Thus in the Chiliads of Briggs to the number 99000 Answ. 4,99563,51946 9900 — 3,99563,51946 990 — 2,99563,51946 99 lastly — 1,99563,51946. But there will not arise such advantage for making Logarithms by this observation as it may at first sight seem to promise', because there are 9000 numbers between 1000 and 10000 whose Logarithms must be found also, and but 900 between 100 and 1000, and but 90 between 10 and 100, and but 9 between 1 and 10, and so in all 999, which is not the ninth part of the former. Definition XVI. IF the radius AD (Fig. 113.) be conceived to move equally about the point A through the periphery of the quadrant DB, while in the mean time the side of the square DC remaining always parallel to itself, descends also with an equal motion through DAM, so that in the same moment the radius AD and the aforesaid side DC shall fall upon the base AB; or (if any one should think that this way the proportion of a right line to a circular one is supposed by a sort of Petitio Principii or begging the question) the right line DAM as well as the quadrant DB being divided into as many equal parts as you please (e. g. here both of them into 8) and drawing through these from the centre A so many Radii and through them parallel lines; the points of intersection being orderly connected together will exhibit a curve line, whose invention is attributed to Dinostratus and Nicomedes in the fourth Book of Pappus Alexandrinus, and which from its use is called a Quadratrix, it having among the rest this property, that from AB it cuts of a part A, which is a third proportional to the quadrant DB and its radius DA; which hereafter we will demonstrate. In the mean while from this description of it, you have these CONSECTARIES. I IF through any point H assumed in the Quadratrix you draw the radius AHI, and from the same point the perpendiculars H h and H e, the whole arch DB will always be to the part IB cut of, as the whole line DAM to the part ha' cut of, or H e equal to it. II Consequently therefore any given arch or angle of the quadrant e. g. IB or JAB may by help of the quadratrix be divided into three equal parts or as many as you please, or in what proportion soever you will; while having drawn the radius AI, the perpendicular H a let fall from the point of the quadratrix H, may be divided into three or as many equal parts as you please, or in any proportion whatsoever, and through these sections radius drawn to divide the arch. BOOK II SECTION II CHAP. I Of the chief Properties of the Conic Sections. Proposition I IN the Parabola (GKEH Fig. 114) the (α) I Property of the Parab. Apoll. Prop. 11. Lib. 1. square of the semiordinate (IK) is equal to the rectangle IL made by the Latus Rectum EL and the abscissa EI. Demonstration. MAke the sides of the cone that is supposed to be cut, AB = a, BC = b, and moreover EBB = oa, and EI = ebb, and AC = c; therefore NI will be = ec, by reason of the similitude of the ▵ ▵ BCA and EIN; and EP or IO = oc, by reason of the similitude of the ▵ ▵ ABC and EBP. Therefore ▭ NIO = oecc = □ IK by the Scholar of Prop. 34 (n. 3) and Prop 17. Lib. 1. Now if a line be sought which with the abscissa EI shall make the ▭ IL = □ IK you will have it by dividing the said square by the Abscissa EI. viz. i. e. = EL. And this is called the Latus Rectum, viz. in relation to the Abscissa EI with which it makes that rectangle, which, it's evident, is = □ IK, and from this equality the section has the name of Parabola, in Apollonius. CONSECTARYS. I THis Latus Rectum, expressed by the quantity , may be found out after a shorter way, if you make as b to c (the side of the cone parallel to the section BC at the Diameter of the base AC) so oc (the side EP called by some the Latus Primarium) to a fourth. II But if any one, with Apollonius, had rather express this by mere data in the cone itself as cut (because oc or that Latus Primarium EP is not a line belonging to the cone itself) he may easily perceive, if the quantity of the Latus Rectum found above, be multiplied by the other side of the cone a, there will be produced the equivalent which instead of the proportion above will furnish us with this other, as ab − too cc − so oa □ of AB into BC − □ AC − EBB to a fourth; which is the very proportion of Apollonius in Prop. 9 Lib. 1. and confirms our former. SCHOLIUM I HEnce you have an easy and plain way of describing a Parabola, having the top of the axe and the Latus Rectum given, viz. by drawing several semiordinates' whose extreme points connected together will exhibit the curvity of the Parabola. But you may found as many semiordinates' as you please, if having cut of as many parts of the Axe as you please, you found as many mean proportionals between the Latus Rectum and each of those parts or Abscissa's. See n. 2. and 3. Fig. 47. Introduct. to Specious Analysis. SCHOLIUM II HEnce also we have a new genesis of the parabola in Plan● from the spculations of De Witte, viz. if the rectilinear angle HBG (Fig. 115.) conceived to be movable about the fixed point B be conceived so to move out of its first situation with its other leg BH along the immovable rule OF, that it may at the same time move also the ruler HG, from its first situation DK, all along parallel to itself, and with the other leg BG let it all along cut the said ruler HG, and with this point of its intersection continually moving from B towards G it will describe a curve. That this curve will be the parabola of the ancients is hence manifest, because it will have this same first property of the parabola. For, 1. if the angle HBG (n. 1.) be supposed to be a right one, and BD or HI = a, BY or KG = b (viz. in that station of the angle and rule HG by which they denote the point G in the intersection) you'll have by reason of the right angle at B, BY, i e. b a mean proportional Between HI i e. a, and IG or BK, and so this as an abscissa = . Wherhfore if BK i e. be multiplied by BD = a, the rectangle DBK will be = bb = □ KG; which is the first property of the parabola: So that it follows, since the same inference may be made of any other point in this curve, that this curve will be the parabola, BD or HI its Latus Rectum, KG a semiordinate, and BK its axis, etc. 2. If the angle HBG be an obliqne one (num. 2) it may be easily shown from what we have supposed that the ▵ ▵ DBH and BKG will be equiangular: Therefore as BD (i e. a) to DH sc. BY (i. e. b) so KG sc BY (i. e. b) to BK (i e. ) Therefore again the □ DBK = bb □ KG. QED. Consect. 3. It is also evident in this second case, that BK drawn parallel to the axe, but not through the middle of the parabola, will be a diameter which will have for its vertex B, its Latus Rectum BD, and semiordinate GK, etc. Consect. 4. Therefore you may found the Latus Rectum in a given parabola geometrically, if you draw any semiordinate whatsoever IK (Fig. 116.) and make the abscissa OF equal to it, and from FLETCHER draw a parallel to the semiordinate IK, and from E draw the right line EKE through KING cutting of FH the Latus Rectum sought; since as EI to IK so is OF (i. e. IK) too FH by Prop 34. lib. 1. wherefore having the abscissa and semiordinate given arithmetically, the Latus Rectum will be a third proportional. Consect. 5. Since therefore the Latus Rectum found above is , if you conceive it to be applied to the parabola in LM, so that N shall be that point which is called the Focus, LN will be and it's square and this divided by the Latus Rectum will give occ for the abscissa EN: So that the distance of the Focus from the Vertex will be ¼ of the Latus Rectum. Consect. 6. Since therefore EN is = if for OF you put ib, NF will be = , whose square will be found to be , to which if there be added □ GF = oicc, by Prop. 1. the square of NG will be = whose root (as the extraction of it and, without that, the analogy of the square NF with the square NG manifestly shows) will be ; so that a right line drawn from the Focus to the end of the ordinate, will always be equal to the abscissa OF +EN i e. (if EO be made equal to EN) to the compounded line FOYES. SCHOLIUM III HEnce you have an easier way of describing the parabola in Plano from the given Focus and Vertex, viz. (Fig: 117.) the axis being prolonged through the vertex E to OH, so that EO shall = EN, if a ruler HI be so moved by the hand G, according to PQ, from OF to HI, that putting in a style or pin, it shall always keep the part of the Thread NGI (which must be of the same length with the rule HI) as fast as if it were glued to it (which perhaps might also be done with the Compasses by an artifice which we will hereafter also accommodate to the hyperbola) and at the same time it will describe in Plano the part of the line EGR. That this will be a parabola is evident from the foregoing Consect. because as the whole thread is always = to the ruler IH; so the part GN is always necessarily equal to the part GH, i e. to the line FOYES. Moreover from the same sixth Consect. and Fig. 116. may be drawn another easy way of describing the parabola in Plano from the Focus and Vertex given through innumerable points G to be found after the same way: viz. If from any assumed point in the axe F you draw to the axe a perpendicular, and at the interval FOYES from the Focus N you make an intersection in G. Which innumerable points G will be determined with the same facility, having given only, or assumed the axis and Latus Rectum, by virtue of the present Proposition. For if, having assumed at pleasure the point F in the axis, you found a mean proportional between the Latus Rectum and the abscissa OF, a semiordinate FG made equal to it, will denote or mark the point G in the parabola sought. Proposition II IN the hyperbola (GKEH Fig 118) (α) I Property of the Hyperb. Apoll. Prop. 12. Lib. 1. the square of the semiordinate (IK) is equal to the rectangle (ILL) made of the Latus Rectum (EL) and the abscissa (EI) together with another rectangle LS of the said abscissa (EI or LR) and RS a fourth proportional to DE the Latus Transversum, (EL) the Latus Rectum, and EI the Abscissa. Demonstration. Suppose the side of the cone AB here also = a, and BM parallel to the section = b, and the intercepted line AM = c, and EI = ebb; all according to the analogy we have observed in the parabola; and NI will be as there = ec. Making moreover MC = d and the Latus Transversum DE = ob, so that DIEGO shall be = ob + ebb; than will (by reason of the similitude of the ▵ ▵ BMC, DEP, and DIO) EP be = odd, and IO = od + ed, and so QO = ed. Therefore □ NIO will be = oecd + eecd = □ IK But this square divided by the Abscissa EI = ebb gives or for the line IS which with the abscissa would make the rectangle ES = to the said square ●K. Now therefore, if here also we call a Line the Latus Rectum found after the same way as in the parabola, viz by making as b − to c − so odd to a fourth (as a line parallel to the section— to the intercepted diam. so the Latus Primarium, but that the other part will be a fourth proportional to bc and ed or to ebb, ec▪ and ed, or (to speak with Apollonius as we have done in the Prop.) to ob, and ebb (for in these three cases you I have the same fourth ) Wherhfore now it is evident that the square of the semiordinate oecd + eecd is equal to the rectangle IL (made by the Latus Rectum into the abscissa ebb = oecd) together with ▭ LS of this fourth proportional into the same abscissa Ebb, which is = eecd. Which was to be found and demonstrated. CONSECTARYS. I HEnce you have in the first place the reason why Apollonius called this Section an Hyperbola; viz. because the square of the ordinate IK exceeds or is greater than the rectangle of the Latus Rectum and the Abscissa. II Since therefore the Latus Rectum here also as well as in the parabola is found by making as b to c so odd to (i e. as the parallel to the section BM is to the intercepted Diana. AM so is the Latus Primarium EP to a fourth EL.) If any one had rather express this Latus Rectum after Apollonius' way, he will easily perceive, this quantity being found and multiplied both Numerator and Denominator by b the parallel to the section, there will come out the equivalent quantity which gives us instead of the former proportion this other, as bb − too cd − so ob to a fourth; □ BM − ▭ AMC— Latus Transversum to a fourth; which is that of Apollonius in Prop. 12. Lib. 1. and consequently herein confirms our former. III You may also have this Latus Rectum geometrically, by finding a third proportional (as we have done in the parabola Consect. 4. Prop. 1.) to the abscissa EI (Fig. 119.) and the semiordinate IK (= OF;) and than found a fourth proportional EL to DIEGO (the sum of the Latus Transversum and abscissa) and FH already found, or IS equal to it, and DE (the Latus Transversum) and that will be the Latus Rectum sought. Pag. 172. 119 120 121 122 123 124 125 126 126 127 128 SHCOLIUM. FRom this third Consectary, we may reciprocally from the Latus Rectum and transverse given, found out and apply as many semiordinates' to the axe as you please, and so describe the hyperbola through their (ends or) infinite points: viz. if assuming any part of the abscissa EI, you make as DE to EL so DIEGO to IS; and than found a mean proportional IK between IS and the abscissa EI, and that will be the semiordinate sought: And both this praxis and the Consect. may be abundantly proved by setting it down in, and making use of, the literal Calculus. Proposition III IN the Ellipsis (KDEK, Fig. 120.) the (α) I Propert. of the Ellipse of Apoll. Prop. 13. Lib. 1. square of the semiordinate (IK) is equal to the rectangle (ILL) of the Latus Rectum (EL) and the abscissa (EI) (lesle or) taking first out another rectangle (LS) of the same abscissa (EI or LR) and RS a fourth proportional to (DE) the Latus Transversum (EL) the Latus Rectum and (EI) the abscissa. Demonstration. Suppose the side of the cone to be AB here also = a and BM parallel to the section = b and the intercepted AM = c, and EI = ebb; and NI will be again = ec, all as in the hyperbola. And making also here as in the hyperbola MC = d, and the Latus Transversum DE = ob, so that DIEGO will be ob − ebb; than will (by reason of the similitude of the ▵ ▵ BMC, DEP and DIO) EP be = odd, and IO = odd − ed. Therefore ▭ of NIO will be = oecd − eecd = □ IK. But this square divided by the abscissa EI = ebb gives or for that line IS which with the abscissa would make the rectangle ES = to the said square IK. Now therefore if we call the Latus Rectum a right line found after the same way as in the parabola, by making according to Cons. 1. Prop. 1. as b to c − so odd − to a fourth i. e. as the line parallel to the section— to the intercepted diameter— so the Latus Primarium, etc. It is manifest that the Latus Rectum is one part of the line just now found; and the other part is a fourth proportional to b, c and ed, or (to speak with Apollonius as we have done in the Prop.) to and ebb (for there will come out the same quantity ) wherefore now it is evident that the □ of the semiordinate IK is equal to the ▭ IL (of the Latus Rectum and the abscissa ebb = oecd) having first taken out thence the ▭ LS, or eecd out of that fourth proportional by the same abscissa ebb; which was to be found and demonstrated. CONSECTARYS. I HEnce you have first of all the reason of the name of the Ellipse, which Apollonius gave to this section; viz. because the square of the semiordinate IK is defective of, or lesle than the rectangle of the Latus Rectum and the abscissa. II Since therefore the Latus Rectum here also as well as in the parabola and hyperbola, is found by making as b to c so odd to (i e. as BM parallel to the section is to the intercept. diam. AM so the Latus Primarium EP to a fourth EL) now if any one had rather express this Latus Rectum after Apollonius' way, he will easily see that the quantity above found being multiplied both Numerator and Denominator by b, that there will come out an equivalent one , which instead of the former proportion will give this other, as bb − too cd − so ob to a fourth; BM − ▭ AMC— Latus Transvers. to a fourth; which is the same with that we have also found in the hyperbola, and which also Apollonius has Prop. 13. Lib. 1. III This Latus Rectum may also be had geometrically, if you found, 1. in the hyperbola a third proportional FH to the abscissa EI (Fig. 121.) and semiordinate IK (= EF.) 2. But EL a fourth proportional to DIEGO (the difference of the Latus Transversum and the abscissa) and the found FH, or IS equal to it, and the Latus Transversum DE, is the Latus Rectum sought. SCHOLIUM. FRom this third Consect. we may reciprocally, having the Latus Rectum and Transversum given, apply as many semiordinates' to the axe as you please, and so draw the ellipsis through as many points given as you please, viz. if, taking any abscissa EI, you make as DE to EL so DIEGO to a fourth IS; than between this IS and the abscissa EI found a mean proportional IK, and that will be the semiordinate sought: And this Praxis also and the third Consect. may be abundantly proved by making use of a literal Calculus. For e. g. here a fourth proportional to ob, and ob − ebb will be and a mean proportional between this fourth and ebb will be , etc. Proposition IU. IN a Parabola (α) 2. Property of the ●arab. 20. Prop. Apoll. Lib. 1. Con. the squares of the ordinates' are to one another as the abscissa's. Demonstration. For if OF (Fig. 122.) be called ib, as above EI was called ebb, since the Latus Rectum is the square of FG will be = oicc. Therefore □ IK will be to □ FG as oecc— oicc e to i or ebb to ib. Q. E. D. CONSECTARY. HEence having drawn LO parallel to the axe or diameter OF, if it be cut by the transverse line EGLANTINE in M and by the curve of the parabola in KING; than will OL, ML, and KL be continual proportionals. For OF is to EN as FG to NM or IK, by reason of the similitude of the ▵ ▵ EFG and ENM. But the squares FG and IK are in duplicate proportion of OF to EN by Prop. 35. Lib. 1. and are also in the same proportion as the abscissa's OF and EI by the pres. Therefore OF to EI is also in duplicate proportion of OF to EN i e. OF is to EN as EN to EI Q. E. D. OL is to ML as ML to KL Q. E. D. Proposition V IN the hyperbola and Ellipsis (α) Property of the Hyperbola and Ellips'. Apol. 21. Lib. I the squares of the Ordinates' are as the rectangles contained under the lines which are intercepted between them, and the Vertex's of the Latus Transversum's. Demonstration. For, if OF (Fig. 118. and 120.) be called ib, as EI was above called ebb, than will according to Prop. 2. and the 3d. deduction. GF = oicd + eicd in the Hyperb. oicd − eicd in the Ellips'. and the ▭ DFE = oibb + iibb in the Hyperb. oibb − iibb in the Ellips'. Therefore the □ KING is to the square GF as oecd± eecd to oicd ± iicd i e. as oe± ee to oi± two. and ▭ DIE is to the ▭ DFE, as oebb± eebb to oibb± iibb i e. in like manner as oe± ee to oi± two. Q. E. D. CONSECTARY I IN the Ellipsis this may be more commodiously expressed apart thus; the squares of the ordinates' (KING and GF) are as the rectangles contained under the segments of the Diameter (viz. DIE and DFE) in which sense this property is also common to the circle, as in which the squares of the ordinates' are always equal to the rectangles of the segments. CONSECTARY II THerefore, if the Latus Rectum be conceived to be applied in the hyperbola, so that N shall be the Focus; (see Fig. 123.) than will LN = , and its square be . But as the □ KING to the square LN, so is the ▭ DIE to the ▭ DNE i e. oecd + eecd to so is oebb + eebb to . But now the ▭ of the whole DE and the part added EN into the part added EN, i e. ▭ DNE (= together with the square of half CE (= ) is = □ compounded of half and the part added CN = by Prop. 9 lib. 1. Wherhfore CN the distance of the Focus from the centre is = . But is the fourth part of the ▭ of the Latus Transversum ob and the Latus Rectum (or the fourth part of the figure, as Apollonius calls it) and is the □ of i. e. of half the Latus Transversum. Wherhfore we have found the following Rule of determining the Focus in an hyperbola: If a fourth part of the figure (or the rectangle of the Latus Rectum into the Transversum) be added to the square of half the Latus Transversum, and from the sum you extract the square root; that will be the distance of the Focus from the centre CN: And hence substracting half the Latus Transversum CE, you will have distance of the Focus from the Vertex EN. CONSECTARY III IN like manner in the Ellipsis having drawn the ordinate LM (Fig. 124.) that the Focus may be in N, the □ LN would be as above, and by a like inference □ DNE = . But now □ DNE together with the square of the difference CN is equal to the □ of half CE by Propos. 8. lib. 1. and consequently the □ CN is = □ CE − □ DNE, that is, . Wherhfore CN the distance of the Focus from the centre is = . Wherhfore we have found the following Rule to determine the Focus in the Ellipse. If the fourth part of the figure (or the rectangle of the Latus Rectum into the Latus Transversum) be substracted from the square of half the Latus Transversum, and from the remainder you subtract the square root; that will be the distance of the Focus from the Centre CN: And taking hence half the Latus Transversum CE, you'll have the distance of the Focus from the Vertex EN. SCHOLIUM I BOth the Rules are easy in the practice, for since is nothing but the square of CE, and nothing but the rectangle of ¼ DE into LM; if between LM and ¼ DE or MORE (Fig. 125.) you found a mean proportional MN, (and so whose □ is equal to that □) and in the hyperbola join to it at right angles MC = CE, the hypothenusa CN will be the distance sought of the Focus from the centre: And the same may be had in the Ellipsis, if (n. 2.) having described a semicircle upon CM = CE you draw or apply the mean found MN, and draw CN. SCHOLIUM II HEnce also we have (α) De Wit Elem. Curv. Lib. 1. Cap. 3. Prop. 13. a new genesis of the Ellipse in Plano about the diameters given, from the speculations of Monsieur de Witt; viz. If about the rectilinear angle DCB (Fig. 126. n. 1. and 2.) considered as immovable, the rule NLK (which all of it will equal the greater semidiameter CB, and with the prominent part LK the lesser CD) be so moved that N going from C to D, and L from B to C may perpetually glide along the sides of the angle, the extreme point in the KING in the mean while describing the curve BKE (and in a like application the other quadrants) and that this curve thus described will be the ellipsis of the ancients is hence manifest, because it has the second property of the ellipse just now described. For, 1. if the angle DCB or NCB be supposed to be a right one (as in Fig. 126. num. 1.) and the rule KN in the same station, it marked out the point KING, and having applied the semiordinate KING, and drawn the perpendicular LM, from the square KL and the square CE (as being equal) subtract mentally the equal squares LM and CI, and there will remain by virtue of the Pythagorick Theor. on the one hand □ KM and on the other by Prop. 8. lib. 1. □ DIE equal among themselves. But now the square of KING is to the square of KM (i e. to the □ DIE) as the square of KN to the square of KL (i e. as the square of CB to the square of CE) by reason of the similitude of the ▵ ▵ KLM and KNI; and since the same may be demonstrated after the same manner of any other semiordinate KING i viz. that its □ KING i is to the □ D iE as the square CB to the square CE. It also follows, that the □ KING is to the □ DIE as □ EDWARD i to the □ D iE, and alternatively, the square KING will be to the square KING i as the □ DIE to the □ D iE; which is the second property of the ellipse. 2. If the angle NCE be not a right one (as in Fig. 126. n. 2. and the like cases) having drawn NO and KP parallel to the rule nlB in the first station, [in which station the angle NCE, to which the flexible ruler is to be made, is determined, viz. by letting fall the perpendicular B l from the extremity of one diameter upon the other, and moreover by adding or substracting the difference of the semidiameters In] having also drawn the Ordinate KIM, and PI parallel to CN; which being done the ▵ ▵ CB l and IKF, and also CB n and IKP will be similar. Wherhfore having joined NP, from the parallelism of the lines IP and NC and the similitude of the aforesoid ▵ ▵, as also of NCO and nC l, it will be easy to conclude that NCIP is a parallelogram. Wherhfore since KN is = CE and □ KN = □ CE, having substracted the squares of the equal lines NP and CI, there will remain on the one hand □ KP on the other the □ DIE equal among themselves as above. Therefore the square of KING will be to the square of KP (i e. to the □ DIE) as the square of BC to the square of B n (i. e. to the square of CE) as in the former case: And since here also the same may be demonstrated after the same manner of any other semiordinate KING i; we may infer as above, that the □ □ KING and KING i are to one another as the rectangles DIE and D iE, etc. But after what way the same ellipses may be described by these right lined angles without any of these rulers through infinite points given, will be be manifest from the same figures to any attentive Person. For having once determined the angle NCE or nCD (num. 2. e. g.) if NL or nl be applied where you please by help of a pair of compasses, and continued to KING, so that LK or lk shall be equal to lB, you will have every where the point KING, etc. CONSECTARY IU. SInce in the hyperbola (Fig. 123.) the □ CN − □ CE = □ DNE, and in the ellipsis (Fig. 124.) □ CE − □ CN = □ DNE, by virtue of Prop. 9 and 8. lib. 1 if for CN you put on both sides for brevity's sake m, than will the □ DNE in the hyperbola be rightly expressed in these terms , and in the ellipsis in these . Proposition VI. IN the parabola (α) 3. Property of the Parab. the Latus Rectum is to the sum of two semiordinates' (e. g. IK+FG i e. HO in Fig. 122.) as their difference (OG) to the difference of the abscissa's (IF or (KO.) Demonstration. For if the greater abscissa OF be made = ib, and the lesle EI = ebb, the semiordinates' answering to them FG and IK will be √ ooic and √ oecc as is deduced from Prop. 1. Wherhfore if you set in the same series 1 2 3 The Latus R. − sum of the semiord. − their diff. − √ oicc + √ oecc − √ oicc − √ oecc 4 − diff. of the absciss. − ib − ebb and multiply the extremes and means, you'll have on both sides the same product oicc − oecc, which will prove by virtue of Prop. 19 lib. 1. the proportionality of the said quantities. Q. E. D. SCHOLIUM. THis is that property of the parabola, whereon the Clavis Geometrica Catholica of Mr. Thomas Baker is founded, which as unknown to the ancients, nor yet taken notice of by Des Cartes, he thinks was the reason why that incomparable Wit could not hit upon those universal rules for solving all Equations howsoever affected. Concerning which we shall speak further in its place. We will only further here note, that Baker was not the first Inventor of this property, but had it, as he himself ingeniously confesses, out of a Manuscript communicated to him by Tho. Strode of Maperton, Esquire. Proposition VII. IN the hyperbola and ellipsis (α) The 3. Property of the Hyperb. and Ellips'. Apollon. lib. I Prop. 21. pars prior. the Latus Rectum is to the Latus Transversum, 〈◊〉 the square of any semiordinate (e. g. IK in Fig. 118▪ and 120.) to the rectangle (DIE) contained under the lines intercepted between it and the Vertex's of the Latus Transversum. Demonstration. For the Latus Rectum is on both sides , the Latus Transversum ob, etc. Wherhfore if you make in the same series as the Latus R. to the Lat. Transv. so the □ IK to ▭ DIE in hyperb. in ellips. The rectangles of the extremes and means will both be ooebcd± oeebcd, and so will prove the proportionality of the said quantities, by Prop. 19 lib. 1. Q. E. D. CONSECTARY I HEnce having given in the ellipsis (see Fig. 124) the Latus Rectum and the transverse axe, you may easily obtain the second axe or diameter, if you make as the Lat. Transv. to the Lat. Rect. so the □ DCE to □ AC . CONSECTARY II THerefore the □ of the whole AB will be = oocd = □ of the Latus Rectum into the Lat. Transv. (which Apllonius calls the Figure) so that the second Axe (and any second Diameter) will be a mean proportional between the Latus Rectum and the Latus Transversum. Hence in the hyperbola also the second or conjugate diameter may be called a mean proportional between the Latus Rectum and Transversum, i e. √ oocd or a line which is equal in power to the Figure, as Apollonius speaks. SCHOLIUM I HEnce may be derived another and more simple way of delineating organically the ellipsis in Plano about the given axes AB, DE (Fig 127.) which Schooten has given us▪ viz. by the help of two equal rulers CG and GK movable about the points G and G: If, viz. the portions CF and HK are equal to half the lesser axe AC, but taken with both the augments (viz. CF+FG+GH) may = ½ the greater axe CD or CI; and the point EDWARD moving along the produced line DE the point H may describe the curve EHAD. That this will be an ellipsis will be evident by virtue of this seventh Prop. from a property that agrees to this curve in all its points H. For having drawn circles about each diameter, and the lines IHN, FOE perpendicular to CE; and having made the Latus Rectum EL, which is a third proportional to DE and AB by the second Consect. of this Prop. etc. by reason of the similitude of the triangles CFO, CIN, FOYES will be to FC as IN to IC, and alternatively FOE to IN as FC to IC i e. as AC to CE or AB to DE. Therefore also the square of FOYES (or HN) will be to the square of IN, as the square of AB to the square of DE, by Prop. 22. lib. 1. i e. as EL the Latus Rectum to ED the Latus Transversum, by Prop. 35. lib. 1. But the □ IN is = DNE from the proporty of the circle. Therefore □ FOYES (or of the semiordinate HN) is to the ▭ DNE as EL the Latus Rectum to ED the Latus Transv. therefore by virtue of the pres. Prop. the point H is in the Ellipsis, and so any other, etc. Q. E. D. CONSECTARY III NOW if in the ellipsis the □ of AC the second Axe (= by Consect. 1.) and □ CN the distance of the Focus from the centre (= by Censect 3. Prop. 5. the figure whereof you may see n. 124.) be joined in one sum; the □ AN will be = , and so the line AN = i. e. to half the Latus Transversum: So that hence having the axes given you may found the Foci, if from A at the interval CD you cut the transverse axe in N and N. CONSECTARY IU. NOW if, on the contrary, in an hyperbola (Fig. 123.) the □ AC or OF = be substracted from the □ CF or CN = by virtue of Consect. 2. Prop. 5. there will remain and its root , i e. half the Latus Transversum CD: So that here also, the axes being given, you may found the Focus's, if from the vertex E you make OF a perpendicular to the axe = to the second Axe AC, and at the interval CF from the centre C you cut the Latus Transversum continued in N and N. SCHOLIUM II BUT now, that the right lines KN and KN drawn from any other point (e. g. KING) to the Foci, when taken together in the ellipsis, but when substracted the one from the other in the hyperbola, are equal to the Latus Transversum DE, we will a little after demonstrate more universally, and also show an easy and plain Praxis of delineating the ellipsis and hyperbola in Plano, having the axes and consequently the Foci given. CONSECTARY V SInce we have before demonstrated Consect. 2. and 3. Prop. 5. that the □ DNE in the hyperbola and also in the ellipsis is = ; and here in Consect. 1. the □ of the second semi-diameter AC is also = ; it is evident that this □ AC is equal to the □ DNE. CONSECTARY VI. IT is hence moreover evident, if the square of half the transverse diameter GE = be compared with the square of half the second diameter AC or OF = , multiplying both sides by 4 and dividing by oh; they will be to one another as obb to ocd i e. further dividing both sides by b, as ob to the Latus Transversum to the Latus Rectum. CONSECTARY VII. BUT since also the □ DIE is to the □ IK as the Latus Transversum to the Latus Rectum, by virtue of the present 7. Prop. the square of CE the transverse semidiam. will be to the square of AC the second semidiam. (or by the 5th. Consect. of this, to the □ DNE) as the □ DIE to the square of IK. CONSECTARY VIII. YOU may also now have the □ IK (which otherwise in the hyperb. is eocd + eecd, in the ellipse oecd − eecd, by virtue of Prop. 2. and 3.) in other terms, if you make as □ CE to the □ DNE so the □ DIE to a fourth; i e. as to by virtue of Consect. 4. Prop. 5. (so oebb + eebb in the hyperb. as to so oebb − eebb in the ellipsis. For hence by the Golden Rule the square IK may be inferred as a fourth proportional. In the hyperbola ; In the ellipsis : The use of which quantities will presently appear. Proposition VIII. THE Aggregate in the ellipse Difference in the hyperb. of the right lines (α) Apollon. Lib. 3. Prop. 51. and 52. KN and Kn (Fig. 128.) drawn from the same point KING to both the Focus' is equal to the transverse axe DE. An Ocular Demonstration. WHich consists wholly in this to found the lines KN and KING n by help of the rightangled triangles IKN and IK n (sc. the hypothenuses having the sides given) and afterwards see if the sum of both in the ellipse, and difference in the hyperbola be = ob, i e. to the transverse axe DE. I In the Ellipsis. Putting for CN (which above Prop. 5. Cons. 3. was found to be ) I say putting for it m, you'll have IN = CI+CN = ½ ob − eb + m I n = C n − CI = m − ½ ob + ebb Therefore □ IN = □ I n = Add to each □ IK which was found in Prop. 7. Consect. 8. in the ellipsis = and you'll have □ KN = and by extracting the roots (which is easy) you'll have and ; Sum ob. Q. E. D. II In the Hyperbola. Putting again m for CN (which above Cons. 2. Prop. 5. was found to be ) and you'll have IN = CI+CN = ½ ob + eb + m I n = CI − C n = ½ ob + ebb − m. Therefore □ IN = and □ I n = Add to both the □ IK which was found in Prop. 7. Consect. 8. in the hyperbola and you'll have □ KN = □ KING n = and extracting the roots out of these (which is easy) you'll have KN = . KING n = (which is a false or impossible root, for it would be CE − CN and moreover— another quantity. Or KING n = ; which is a true and possible root. The difference therefore of the true roots is = ob. Q. E. D. SCHOLIUM I WE first of all tried to make a literal Demonstration by using the quantity of the square IK as you have it expressed Prop. 2. and 3. and the quantity IN as it was compounded of CI = ob + eb+C + = , etc. but we found it very tedious in making only the squares of IN and I n. Than for the furred quantity CN we substituted another, viz. m, and we produced the squares of IN and I n as above, but we added the square of IK in its first value; and thus we obtained the squares KN and KING n, but in such terms, that the exact roots could not be extracted, but must be exhibited as furred quantities, and consequently we must make use of the rules belonging to them to found their sum or difference, which we laid down Cons. 3. Prop. 7. and Consect. Prop. 10. Lib. 1. which though it would succeed, yet would be full of trouble and tediousness. Therefore at length when we came to use those other terms which express the square IK, the business succeeded as easy as we could wish, and that in a plain and easy way and not lesle pleasant, which I doubt not but will also be the opinion of the Reader, who shall compare this with other demonstrations of the same thing, which only lead indirectly to this truth, or with them, which the Wit has given us in Elem▪ Curvar. lin. p. m. 293. and 302. and which he thinks easy and short enough in respect of others both of the ancients and moderns, and which we have reduced into this yet more distinct form, and accommodated to our schemes. Preparation for the Hyperbola. Make as CD to CN EE − CN so CI to CM so that the □ of DCM MCE will be = □ NCI nCI. Because therefore it will be by Consect. 7. Prop. 7. as □ CD to □ DNE, so the □ DIE to the □ IK. And also by composition, as □ CD to □ CD+ □ DNE i e. □ CN per 9 lib I so DIE to DIE + □ IK. Therefore by a Syllepsis, as □ CD to the □ CN so □ CD + DIE i e. □ CI to □ CN + DIE + □ IK. But also by the Hypothesis. as the □ CD to the □ CN so □ CI to □ CM. Therefore □ CM is = □ CN + DIE + □ IK. Demonstration. Since therefore it is certain that the difference between DM and EM is the transverse axe DE; if it be demonstrated that DM is = KN and EM = EDWARD n, the business will be done, because the difference between KN and KING n is also the transverse axe DE. Resolve the □ KN. It is certain that NI cue + IK cue = KN q. Substitute for NI cue, by the 7. Lib. 1. CI q+C + q+2NC +. Preparation for the Ellipsis. Make as CD to CN so CI to CM. So that the □ DCM is = □ NCI. Because therefore by Consect. 7. Prop. 7. as □ CD to □ DNE so □ DIE to the □ IK; Than also by dividing, as □ CD to □ CD − DNE i e. □ CN, by 8. l. 1. so DIE to DIE − □ IK. Therefore by a Dialepsis, as □ CD to □ CN, so □ CD − □ DIE i e. CI □ by 8. cit. to □ CN − DIE+ □ IK. But also by the Hypothesis, as □ CD to □ CN, so □ CI to □ CM: Therefore □ CM is = □ CN − DIE + □ IK. Demonstration. Since therefore it is certain that the sum of DM and EM is the transverse axe DE; if it be demonstrated that DM is = KN and EM = EDWARD n, the business will be done, because the sum of KN and KING n is also equal to the transverse axe DE. Resolve the □ KN. It is certain that NI q+● + cue = KN q. Substitute for NI cue, by the 7. lib. 1. CI q+C + q+2NC + Than will CI q+C + q+2NC●+I + cue = KN q. Substitute for C● cue, by the 9 lib. 1. CD q+DI +; than will CD q+DIE+C + q+2NC●+ + cue = KN q. Resolve also □ DM. It is certain that CM q+C + q + 2DCM 2NCI = DM, by the 7. lib. 1. Substitute for CM cue its value by the Preparation, and you'll have CN cue − D●E+●K q+C + q+2NC + = DM cue: Which were before = KN q. Therefore KN = DM; which is one. In like manner resolve □ Kn. It is certain that nl q+● + cue = KING nq. Substitute for nI cue, by Consect. 1. Prop. 10. Lib. 1. C● q+C + cue − 2 nCI, and you'll have CI q+C + cue − 2 nCI+IK cue = KING nq. Substitute for CI cue, by the 9 lib. 1. CD q+D● +, and you'll have = KING nq. Resolve also the □ EM. It is certain that 2CD q+2C + cue − DM cue = EM cue per 13. lib. 1. Than will CI q+C + q+2NCI+I + cue = KN q. Substitute for CI cue by the 8. lib. 1. CD cue − DIE; than will . Resolve also the □ DM. It is certain that CM cue 2DCM 2NCI = DM cue per 7. lib. 1. Substitute for CM cue its value from the preparation, and you'll have CN cue − DIE+IK q+C + q+2NC + = DM cue: Which before were = KN q. Therefore KN = DM; which is one. In like manner resolve the □ Kn. It is certain that nI q+I + cue = KING nq. Substitute for nI cue by Consect. 1. Prop. 10. lib. 1. CI q+C + cue − 2NCI, and you'll have CI q+C + cue − 2NCI+IK cue = KING nq. Substitute for CI cue per 8. lib. 1. CD cue − DIE, and you'll have CD cue − DIE + CN cue − 2NCI+IK cue = KING nq. Resolve also □ EM. It is certain that 2CD q+2C + cue − DM cue = EM cue per 13.1. Substitute the value of DM cue first found above, and you'll have CD q+C + cue − 2 nCI = EM q. Substiute for CM cue the value as in the preparation, and you'll have CD q+C + q+DI + − 2 nCI+IK cue = EM cue: Which were before = KING nq. Therefore KING n = EM; which is the other. SHCOLIUM II Pag. 192. 129 130 131 132 133 134 As to the hyperbola, there is a mechanic method of drawing that also, not unlike the others, from a like property in that, communicated by the same Van Schooten, viz. If having found the Focus' N and n (Fig. 129. n. 3.) you tie a thread NFO in the Focus N and at the end of the ruler not of the length of the transverse axe DE; than putting in a pen or the movable leg of a pair of compasses (nor would it be difficult to accommodate the practice we before made use of to this also) draw or move it within the thread NFO from OH to E, so that the part of the thread NOT may always keep close to the ruler as if it were glued to it. For if we call the length of the thread X, and the transverse axe ob as above, the ruler not will be, by the Hypoth. = X + ob. Make now the part of the thread OF = ½ X, the remainder or other part will be NF = ½ X and nF = ½ X + ob, and the difference between FN and FLETCHER n, = ob. Make OF = ¾ X, than will FN be ½ X and FLETCHER n ¼ X + ob, the difference still remaining ob and so ad infinitum. In short, since the difference of the whole thread and of the whole ruler is ob, and in drawing them, the same OF is taken from both, there will always be the same difference of the remainders. Hence also assuming at pleasure the points N and n you may describe hyperbolas so, the thread NFO be shorter than the ruler nFO: For if it were equal there would be described a right line perpendicular to N n, through the middle point C. There yet remains one method of describing hyperbolas and ellipses in Plano, by finding the several points without the help or Apparatus of any threads or instruments, viz. in the ellipsis, having given or assumed the transverse axis DE and the Foci N and n (Fig. 130. n. 1.) if from N at any arbitrary distance, but not greater than half the transverse axe NF, you make an arch, and keeping the same opening of the compasses you cut of, from the transverse axe, EGLANTINE, and than, taking the remaining interval GD, from n you make another arch● cutting the former in FLETCHER, and so you will have one point of the ellipse, and after the same way you may have innumerable others, f, f, f, etc. In like manner to delineate the hyperbola, having given o● assumed the transverse axe DE and the Focus' N and n (n. 2.) if from N at any arbitrary distance NF you strike an arch, and keeping the same aperture of the compasses from the diameter continued, you cut of EGLANTINE, and than at the interval GD from n make another arch cutting the former in FLETCHER, you will have one point of the hyperbola, and after the same way innumerable others, f, f, etc. Proposition IX. IF the secondary axe, or conjugate diameter of the hyperbola AB (Fig. 131.) be applied parallel to the vertex E, so that it may touch the hyperbola, and OE, EP are equal like BC and AC, and from the centre C you draw through OH and P right lines running on ad insinitum, and lastly QR parallel to the Tangent OPEN; you'll have the following CONSECTARIES. I THE parts QG and HR intercepted between the curve and those right lines CQ, CR will be equal; for by reason of the similitude of the ▵ ▵ CEP and CFR as also CEO, CFQ as CE is to EO (and EP) so will CF be to FQ and FR, and consequently these will be equal; and so taking away the semiordinates' FG and FH which are also equal, the remainders GQ and HR will be also equal, and consequently the □ □ QGR, GRH, etc. all equal among themselves: Which we had already deduced before in Consect. 2. and 3. Def. 7. II The rectangle QGR will be = □ EO or EP = i. e. (as Apollonius speaks) to the fourth part of the figure: For by reason of the similitude of the ▵ ▵ CEO, CFQ, CE will be to EO as CF to FQ: i e. as the □ CE to the □ EO i e. as (by Cons 2. 7.) the Lat. Transv. to the Lat. Rect. so the □ CF to th' □ FQ i e. as (by the 7. Prop.) as the □ DFE to the □ FG But now if from the □ CF you take the □ DFE, there will remain the □ CE, by Prop. 9 lib. 1. and if from the □ FQ you take the □ FG there will remain the □ QGR, by Prop. 8. lib. 1. wherefore that remaining □ CE to this remaining □ QGR will be, as was the whole square CF to the whole square FQ, by Prop. 26. lib. 1. i e. as was the □ CE to the □ EO; consequently the □ QGR and the square EO (to which the same square CE bears the same proportion) will be equal among themselves. III Since this is also after the same manner certain of any other rectangle ggr or grh, etc. it follows that all such rectangles are equal among themselves. IV. Wherhfore it is most evident, since the lines FR, fr, etc. and so GRACCUS and gr grow so much the longer, by how much the more remote they are from the vertex E; that on the contrary the lines QG and qg must necessarily so much the more decrease and grow shorter, and consequently the right line CQ approach so much nearer and nearer to the curve EGLANTINE. V But that they can never meet or coincide altho' produced ad infinitum will thus appear; if it were possible there could be any concourse or meeting, so that the point G and Q or g and cue could any where coincide, it would follow from Consect. 2. that as the □ DFE to the square FG so the square CF to the square FQ i e▪ to the same square FG; and so that the □ DFE would be = □ CF; which is absurd by Prop. 9 lib. 1. so that now it is evident that the lines COQ and CPR drawn according to Consect. 1. are really Asymptotes (i e. they will never (α) Apoll. Prop. 1. lib. 2. coincide (viz. with the curve of the hyperbola) as Apollonius has named them. VI Having drawn the right lines from G and g parallel to both the asymptotes, viz. GS and gs and likewise GT and gt, the rectangles TGS and tgs will be (α) Apollon. Prop. 12. lib. 2. equal among themselves. For by reason of the similitude of the ▵ ▵ TQG and tqg, first, TG will be to QG as tg to qg; and, by reason of the equality of the □ □ QGR and qgr, secondly, QG will be to gr reciprocally as qg to GRACCUS, by Prop 19▪ lib. 1. and by reason of the similitude of the ▵ ▵ SGR and sgr, thirdly, as gr to gs so GRACCUS to GS, wherefore (since in two series 1. 2. 3. as TG to QG to gr to gs so tg to qg to GRACCUS to GS) you'll have ex aequo or by proportion of equality as TG to gs so tg to GS, by Prop. 24 lib. 1. Therefore, by Prop. 17. of the same, the □ of TG into GS = □ of tg into gs. Q. E. D. SCHOLIUM. HEnce, lastly, we have a new genesis of the hyperbola in Plano about its given diameters from the speculations of (β) De Wit Elem. Curv. lib. I cap. 2. prop. 3. De Wit; if, viz. having drawn the lines AB and OF cross one another at pleasure (Fig. 132) to the angle BCF you conform the movable angle BCD (and being to be delineated in the opposite hyperbola equal to the contiguous ACD) one of whose legs is conceived to be indefinitely extended, but the other CD of any arbitrary length; and to the end of it D apply the slit of a movable ruler GD about the point G at any arbitrary interval GD (but yet parallel to the leg CB in this first station) and so carrying together along with it the movable angle BCD about the line ECF, but so that the leg CD may always remain fast to it, and the other CB be intersected in its progress by the ruler GDH, e. g. in b or β This point of intersection, thus continually moved on, will describe the curve bGβ, which we thus prove to be an hyperbola: Because the ruler GDH turning about the pole G, and carried from D e. g. to d or δ cuts the leg of the movable angle CB brought to the situation cb or γβ, and in the mean while remaining always parallel to itself; and having drawn from the points of intersection b and β and G the lines GI', bK and βη parallel to the ruler CF, because e. g. in the second station, having taken the common quantity cD from the equal ones CD and cd, the remainders C c and D d are equal, and by reason of the similitude of the ▵ ▵ dcb and dDG, as D d i e. C c or bK to DG, so dc i e. DC or GI' to cb; the rectangle of KING b into bc will be = □ of DG into GI', by Prop. 18. lib. 1. and in like manner, when in the third station having added the common line Dγ to the equal ones CD and γδ, the whole lines Dδ and Cγ are equal, and, by reason of the similitude of the ▵ ▵ βγδ and GDδ as Dδ i. e. Cγ or βη is to DG so is γδ i e. DC or GI' to γβ; the □ of ηβ into βγ = □ of DG into GI', by the same 18. Prop. Wherhfore the three points b, G, β, (and so all the others that may be determined the same way) are in the hyperbola, whose asymptotes are CB and CF and its centre C, etc. by the present Prop. Consect. 6. Q. E. D. You may also determine innumerable points of this curve separately without the motion we have now prescribed, viz. as the point a in the opposite hyperbola, if through any assumed point c in the asymptote CE you draw a parallel to the other asymptote CA, and having made cd equal to CD, from G through d draw G da, and so in others. CHAP. II Of Parabolical, Hyperbolical and Elliptical Spaces. Proposition X. THE (α) Archim. de Quadratur. Parab. Prop. 17. and 24. Parabolic Space (i e. in Fig. 133. that comprehended under the right line GH and the parabola GEH) is to a circumscribing Parallelogram GK, as 4 to 6 (or 2 to 3) but to an inscribed ▵ GEH as 4 to 3. Demonstration. Suppose FH divided first into two than into four equal parts, and draw parallel to the axe OF the lines ef, ef, etc. dividing also OF into four parts, the first fg will be 3, the second 2, the third 1, by Prop. 34. lib. 1. but as ef is to ge so is ge to he, by Consect. 1. Prop. 4. Therefore he in the diameter OF is = oh, in the first ef it is = ¼ (for as ef, 4, to goe, 1, so ge, 1, to he, ¼) in the second of a portion of he is = , in the third to , and so the portions eh in the trilinear figure E hHK make a series in a duplicate arithmetical progression, viz. 1, 4, 9, 16: After the same manner, if the parts F f, etc. are bisected, you'll found the portions eh in the external trilinear figure to make this series of numbers ⅛, , , , , , , , and so onwards. Wherhfore since the portions eh or the indivisibles of the trilinear space circumscribed about the parabola are always in a duplicate arithmetical progression: the sum of them all will be to the sum of as many indivisibles of the parallelogram FK, ●qual to the line KH, i e. the trilinear space itself to this parallelogram as 1 to 3, by Consect. 10. Prop. 21. lib. 1. Wherhfore the semi-parabola ●E hH will be as 2, and the ▵ FEH as 1 ½; therefore the whole parabola as 4, and the whole ▵ GEH as 3, and the whole parallelogram GK as 6. Q E. D. CONSECTARY I IT is evident (α) Archim. Prop. 19 with the Coral. that in the first division, the second line fh (i e. that drawn from the middle of the base FH) is three such parts whereof FE is 4; for eh is i. e. 1, therefore fh is 3. CONSECTARY II IT is also evident, that this demonstration will hold of any parabolic segment. Proposition XI. THE Elliptical Space (α) Archim. lib. de Conoid. etc. Prop. 5. comprehended by the Ellipsis DAEB (Fig. 127.) is to a circle described on the transverse axe DE, as the Axis Rectus or conjugate diameter AB to the transverse axe DE. Demonstration. THis is in the first place evident from the genesis of the ellipse we deduced in Scholar 1. Prop. 7. for in that deduction we shown that FOE, i e. HN was to NI as AB to DE: Which since it is true of all the other indivisibles or ordinates' HN and IN ad infinitum; it is manifest that the planes themselves constituted of these indivisibles will have the same reason among themselves, as the Axis Rectus AB to the transverse DE. Q. E. D. CONSECTARY I THerefore the quadrature of the ellipse will be evident, if that of the circle be demonstrated. CONSECTARY II SInce a circle described on the lest diameter AB will be to one described on the greater diameter DE, as AB to a third proportional by Prop. 35. lib. 1. it follows by virtue of the present Prop. that the ellipse is a mean proportional between the greater and lesser circle, i e. as the ellipse is to the greater circle so is the lesser circle to it, viz. as AB to DE. CONSECTARY III HEnce you may have a double method of determining the area of an ellipse. 1. If having found the area of the greater circle, you should infer, as the greater diameter of the ellipsis to the lesle, so the area of the circle found to the area of the ellipse sought. 2. If having also found the area of the lesser circle, you found a mean proportional between that and the area of the greater. SCHOLIUM. WE may also show the last part of the second Consect. thus, 1. If having described the circle E adbE (Fig. 134.) about the least axis of the ellipse we conceive a regular hexagon to be inscribed, and an ellipse coinciding with one end E of its transverse axe, and with the other or opposite one D to be so elevated, that with the point d it may perpendicularly hung over the circle, and further from all the angles of the figure inscribed in the circle you erect▪ the perpendiculars gG, bB, etc. it is certain that the sides ED and E d of the triangle DE d will be cut by the parallel planes FG gf, etc. into proportional parts, and that those by reason of the similitude of the ▵ ▵ FDG and fdg, and so also the other rectangles will be among themselves as the intercepted parts of the lines ID and id, CI and ci, and in infinitum, (viz. of how many sides soever the inscribed figure consists:) Wherhfore also all the parts of the ellipse taken together will be to all the parts of the circle taken together, i e. the whole ellipse to the whole Pag. 201. 135 136 137 138 139 140 141 circle as all the parts of the diameter ED or ab, i e. as DE itself to AB. Q. E D. CONSECTARY IU. IT is also evident that both these demonstrations of the present Prop. will be also the same in any segments of the ellipsis or circle. Proposition XII. ANY Hyperbolical space GEHG (Fig. 135) is to any Hyperbolic figure of equal height gEhg [whose Latus Rectum and Transversum are equal (as in the circle) and also equal to the Latus Transversum of the former DE, as the Axis Rectus (or conjugate) AB is to the Latus Transversum DE (as in the ellipsis.) Demonstration. By the Hypoth. and Prop. 7. and its second Consect. the □ F g is = □ DFE. Wherhfore this □ DFE i e. the □ F g is to the □ FG as the Latus Transversam to the Latus Rectum of the hyperbola GEHG, by the same seventh Prop. i. e. (by Consect. 2 of the same) as the square of the Latus Transvers. DE to the square of the conjugate AB: Therefore the roots of these squares will be also proportional, viz. F g to FG as DE to AB; and consequently (since the same is true of any other ordinates' ad infinitum) the whole hyperbola goe hg will be to the whole one GEHG as DE to AB. Q. E. D. CONSECTARY I THerefore having found the quadrature of such an hyperbola, whose Latus Rectum and Transversum are equal, you may have also the quadrature of any other hyperbola. CONSECTARY II IT is evident that the same demonstration will hold in any other hyperbolas. Proposition XIII. ANY Parabolic segments upon the same base, and hyperbolical and elliptical ones described about the same conjugate (one whereof shall be a right one, the other a scalene) and constitued between the same parallels, are equal. Demonstration. I It is evident of Parabola's; for both the right one GEHG, and the scalene one G EHG (Fig. 136. n. 1.) (for the demonstration of Prop. 10. will hold in both) is to a ▵ inscribed in them as 4 to 3. But the triangles GEH and G EH are equal, by Consect. 5. Def. 12. or Prop. 28. lib. 1. Therefore the Parabola's also. Or thus, in the right parabola GEHG every thing is the same as in 1. and 4. Prop. of this Book, viz. EI = ebb, OF = ib, the square IK = oecc, the □ FG = oicc. And because therefore in the scalene Parabola also the square FG remains = oicc, make F E = n, and found both the abscissa EI, and the □ answering to it IK. 1. For the abscissa; as FE to EI so FLETCHER E to EI, per Consect. 4 Prop. 34 lib. 1. 2. For the □ IK; as FLETCHER E to EI so □ FG to □ IK, . per Prop. 4. of this. Therefore the □ IK = □ IK and IK = IK, and this in any case ad infinitum: Therefore the one parabola is = to the other. Q. E. D. II The business is much after the same way evident of ellipses and hyperbolas. For making all things in the ellipsis and right hyperbola (n. 2. and 3. Fig. 136.) as in Prop. 2, 3, 5, 7. viz. the □ IK oecd − eecd in the ellipsis, oecd + eecd in the hyperbola, the □ AB oocd by Consect. 2. Prop. 7. EI = ebb, DE = ob, etc. if in obliqne ones for the Latus Transversum DE you put n, and seek the Latus Rectum and abscissa EI, you may by means of these also have the square IK, by Prop. 2. and 3. 1. For the Latus Rectum. As n to √ oocd so √ oocd to by Cons. 2.7▪ 2. For EI the abscissa. As ob to ebb so n to = EI. 3. For the side RS □ deficient or exceeding, from Prop. 2. and 3. As n to so to = RS. Now the abscissa multiplied by the Lat. Rect. The abscissa multiplied by RS. by gives □ oecd. by gives □ eecd. The sum of these □ □ oecd + eecd in the hyperb. = □ IK by Prop. 2. evidently = □ IK The difference of these □ □ oecd − eecd gives in the ellipsis □ IK by Prop. 3. evidently = □ IK. Wherhfore the lines IK and IK, and the whole KL and KL will be equal; and since the same thing is evident after the same way of all other lines of this kind ad infinitum, the elliptical and hyperbolical segments will be so also. Q. E. D. CHAP. III Of Conoids and Spheroids. Proposition FOURTEEN. A Parablick Conoid (α) Archimed. Prop 23. and 24. (al. 26. and 27.) is subduple of a Cylinder, and in sesquialteran reason (or as 1 ½) of a cone of the same base and altitude. Demonstration. Because in the parabola the □ AD (Fig. 137.) is to the □ SHOULD, as BD to BH, i e. as 3 to 1, and so to the □ TI as BD to BY, i e. as 3 to 2, by Prop. 4. of this; it is evident that these squares of SHOULD and TI and AD and consequently of the whole lines also SAINT h, T i, AC, and the circles answering to them will be in arithmetical progression, 1, 2, 3; and moreover if there are new Bisections in infinitum, as the abscissa's so also the squares and circles of the ordinates', by virtue of the aforesaid fourth Prop. will always be in arithmetical Progression 1, 2, 3, 4, 5, 6, etc. It is evident that an infinite series of circles in the conoid, considered as its indivisibles, will be to a series of as many circles equal to the greatest AC, i e. the conoid to the cylinder OF as 1 to 2, or as 1 ½ to 3, by Consect. 9 Prop. 21. or Consect. 4. Prop. 16. lib. 1. but to the same cylinder OF the inscribed cone ABC is as 1 to 3, by Prop. 38. lib. 1. therefore the cylinder, conoid and cone are as 3, 1 and 1. Q.E.D. Proposition XU. THE half of (α) Archim. 29. and 30. (al. 32. and 33.) any Spheroid, or any other segment of it is in subsesquialteran proportion to the cylinder, and double of the cone having the same base and altitude. Demonstration. Having divided the altitude BD (Fig. 138.) e. g. into three equal parts, because in the ellipse as well as in the circle the square of AD is to the square of SHOULD as the □ GDB to the □ GHB, i e. as 9 to 5, and so to the square TI as 9 to 8, by Consect. 1. Prop. 5. of this; and in like manner if you make new bisections, the squares (and consequently the circles) of the ordinates' go on or decrease by a progression of odd numbers, as 36, 35, 32, 27, 20, 11, and so ad infinitum, the bisections being continued on; as we have shown in the sphere and circumscribed cylinder Prop. 39 lib. 1. and it will necessarily follow here also (by virtue of Consect. 12. Prop. 21.) that the whole cylinder will be to the inscribed segment of the spheroid, as 3 to 2; and since the same cylinder is to the cone ABC as 3 to 1, also the segment of the spheroid will be to the cone as 2 to 1. Q. E. D. Proposition XVI. AN hyperbolical Conoid (α) Archi●. Prop. 27. and 28. (al. 30. and 31.) is to a cone of the same base and altitude, as the aggregate of the axe of the hyperbola that forms it and half the Latus Transversum, to the aggregate of the said axis and Latus Transversum. Demonstration, containing also the Invention of this Proportion. Make (in Fig. 139.) CE = a, OF = b, OE = c; than will CF = a + b. Since therefore as CE to OE so CF to FQ the □ EO will = cc and □ FQ = . But as these squares so also are the circles of the lines EO and FQ to one another, by Prop. 32. lib. 1. and so the cone COP will be as , and the cone CQR as (viz. by multiplying the third part of the altitude CF by the base FQ:) Having therefore substracted the cone COP from the cone CQR, there will remain the truncated cone QOPR , and from this solid truncated cone having further substracted the hollow truncated cone, which the space EHRP produced in the genesis of the conoid (and which according to Consect. 2. definite. 9 is as bcc) there will remain the hyperbolical conoid i. e. (by substituting now the values of the axe or abscissa OF, and of half the Lat. Transv. EC, and of the conjugate diam. OPEN, etc. found in the demonstrations of the preceding Chapter, viz. for a, ebb for b, and √ oocd for c or oocd for cc) the hyperbolical conoid will come out i. e. . But the cone GEH (multiplying the third part of OF into the □ GH, i e. ⅓ ebb into 4 oecd+ + eecd) is as . Therefore the conoid is to the cone as 6 eebocd+ + e bcd to 4 eebocd+ + e bcd, i e. (dividing on both sides by 4 eecd) as ½ ob + ebb to ob + ebb. Which was to be found and demonstrated. SCHOLIUM. IF any one had rather proceed herein by indivisibles, as in the precedent Prop. having divided the axe OF (Fig. 140.) again into three equal parts, and assuming the values of the lines determined in the hyperbola, viz. ebb for the abscissa OF, ob for the transverse axe, for the Latus Rectum, oecd + eecd for the square of the semiordinate FG, etc. the lowest and greatest circle of the diameter HG will be as oecd + eecd, and, if you make as the Latus Transv. to the Latus Rectum, so the □ D fE ob − made of ob+ + ebb into ⅔ ebb (i e. ⅔ oebb + eebb) to a fourth; there will come out ⅔ oecd + eecd for the second circle of the diam. hg; and by the same inference (as ob to so ob+ + ebb into ⅓ ebb to a fourth) for the third circle of the diameter HG ⅓ oecd + eecd; so that these indivisibles [for which here and in the precedent also the partial circumscribed cylinders may be assumed] proceed in a double series of numbers, the first in a simple arithmetical progression 3, 2, 1, the latter in a duplicate Arithmetical progression of squares 9, 4, 1; and the same if you make further new bisections, will necessarily hap ad Infinitum, (the former numbers e. g. in the first bisection will be ⅚ oecd the latter eecd, &c) it is manifest from the consectaries of Prop. 21. lib. 1. that the whole cylinder HK will in like manner be expressed by a double series of parts answering, in numbers to the indivisibles of the conoid made by any bisection, but in magnitude to the greatest of them all, and in the sum of its first series of parts will be to the sum of the first in the conoid, both being infinite, as 2 to 1 or 3 to 1 ½ oecd, by Consect. 9 of the said Prop. 21. and the sum of its latter to the sum of the former in the conoid will be as 3 to 1 eecd and so the whole cylinder to the whole conoid as 3 oecd+ + eecd to 1 ½ oecd + eecd i e. (dividing by ecd) as 3 o+ + e to 1 ½ o + e i e. multiplying both sides by b) as 3 ob+ + ebb to 1 ½ ob + ebb; and consequently the cone (which is ⅓ of the cylinder) to the conoid as ob + ebb to 1 ½ ob + ebb. Q.E.D. CONSECTARY. HEnce also appears the proportion of the hyperbolic conoid to a cylinder of the same base and altitude, which we did not express in the Prop. viz. as the aggregate of the axe and half the Latus Transversum to triple the aggregate of the said axe and Latus Transversum. CHAP. IU. Of Spiral Lines and Spaces. Proposition XVII. THE (α) Archim. Prop. 24. de Spiral. first spiral space is subtriple of the first circle, i e. as 1 to 3. Demonstration. Having divided the circumference of the circle into (Fig. 141. n. 1.) three equal parts by lines drawn from the initial point, beginning from the first line BASILIUS, the line BC will be as 1, BD as 2, BA as 3, by Consect. 1. Def. 12. of this book, and consequently the sectors circumscribed about the spiral will be CB c as 1, DB d as 4, AB a as 9, by Prop. 32. lib. 1. and in like manner, if you make new bisections, the lines drawn from the point B to the spiral, will be 1, 2, 3, 4, 5, 6; but the circumscribed sectors, 1, 4, 9, 16, 25, 36; and so the circumscribed partial sectors ad infinitum will proceed in an order of squares, there being always as many sectors in the circle equal to the greatest of them. Therefore all the sectors that can be circumscribed ad infinitum about the spiral space, i e. the spiral space itself (in which at last they end) to so many equal to the greatest, i e. to the circle, is as 1 to 3, by Consect. 10. Prop. 21. lib. 1. Q. E. D. CONSECTARY I SInce the first circle is to the second as 1 to 4 (i e. as 3 to 12) by Def. 12. of this, and Prop. 31. lib. 1. and the first spiral space to the first circle as 1 to 3 by the present Prop. the same spiral space will be to the second circle as 1 to 12; and to the third by a like inference as 1 to 27, to the fourth as 1 to 48, etc. CONSECTARY II THE first spiral line is equal to half the circumference of the first circle. For the lines or radii of the sectors, and consequently their peripheries or arches proceed in a simple arithmetical reason, as 1, 2, 3, 4, 5, 6, etc. while in the mean time the whole periphery of the circle contains so many arches equal to the greatest. Therefore the whole periphery of the circle is to an infinite series of circumscribed arches, i e. to the spiral line itself, as 2 to 1, by Consect. 9 Prop. 21. lib. 1. Proposition XVIII. THE whole spiral (α) Archim. Prop. 25. space comprehended under the second right line EA and the second spiral EGIA (see Fig. 141. n. 2.) is to the second circle as 7. to 12. Demonstration. For having divided the circumference of the circle first into three equal parts, there will be drawn to the second spiral four right lines BE, BG, BY and BASILIUS being as 3, 4, 5, 6, and but only three sectors circumscribed, viz GB g, IB i and AB a, which proceed according to the squares of the three latter lines, viz. 16, 25, 36, so that the sum is 77, while the sum of three equal to the greatest is 108, and so the one to the other (dividing both sides by 9) as 12 to 8 ▪ Having moreover bisected the arches and parts of the line BE, so that that shall be 6, the second BF will be 7, and so the other five 8, 9, 10, 11, 12; and the sectors answering to them (excepting the first) 49, 64, 81, 100, 121, 144, so that their sum shall be 559, while the sum of six equal to the greatest, i e. the whole circle is 864, and so one to the other (dividing both by 72) as 12 to 7 . In the other bisection of the arches and the parts of the line BE, so that the one shall be 12, the second 13, etc. to the thirteenth BASILIUS which will be 24, the sum of twelve sectors will be found to be 4250; and the sum of as many equal to the greatest 6912, and so the one to the other (dividing both sides by 576) as 12 to 7 s. . Therefore the proportion will be I In the first case 12 to 7+1+½+ viz. . II In the second case 12 to 7+½+¼+ viz. . III In the third case 12 to 7+¼+⅛+ , etc. The first and second fractions thus decreasing by ½ the latter by ¼. Wherhfore the proportion of the second circle to the second spiral space will be as 12 to 7+1+½+ − ½ − ¼ − − ¼ etc. − ⅛ etc. − etc. By virtue of Consect. 3. and 8. = 0 = 0 Prop. 21. lib. 1. i e. as 12 to 7. Q. E. D. CONSECTARY I BEcause the second circle is to the first spiral space as 12 to 1, by Consect. 1. of the preceding Prop. and to the second spiral space as 12 to 7, by the present. it will be to the second space without the first (viz. BCDEAIGE) as 12 to 6 i e. as 2 to 1. CONSECTARY II THerefore the second space separately to the first is as 6 to 1. CONSECTARY III SInce in the trisection of both these circles, first and second, there arise six lines, and as many sectors, viz. three lines BC, BD, BE, i e. 1, 2, 3, to which there answer three arches in the same progression within the second circle, and also as many equal to its greatest; therefore the sum of all the unequal arches will be 21, but the sum of the equal ones of both circles (each of which in the first are equivalent to 3, in the second to 6) will be 27. Wherhfore the sum of both the Peripheries to the sum of all the circumscribed arches will be as 27 to 21, i e. (dividing both sides by 9) as 3 to 2 ⅓. Moreover bisecting the arches of the circles and the parts of the line BASILIUS, there will arise six circumscribed unequal arches within the first circle, which are as 1, 2, 3, 4, 5, 6, and as many within the second 7, 8, 9, 10, 11, 12; the sum of all which is 78, while the sum of as many equal ones on both sides is 108. Wherhfore the one will be to the other, i e. the sum of both the peripheries to twelve circumscribed arches taken together, is now as 108 to 78, i e. (dividing both sides by 36) as 3 to 2 ⅙. And making yet another bisection, the proportion will be found to be as 3 to 2 , etc. and hence at length may be evidently inferred; that the sum of both the peripheries will be to the sum of all the arches circumscribible ad infinitum, i e. to the whole helix as 3 to 2+⅓ − ⅙ − etc. = 0. that is, as 3 to 2. Q. E. D. CONSECTARY IU. THerefore, since the periphery of the second circle is double of the first, that alone will be equal to the whole spiral. CONSECTARY V THerefore, if the periphery of the second circle be 2, the periphery of the first will be 1, and the first spiral line ½ by Consect. 2. of the anteced. Prop. wherefore the second spiral alone will be 1 ½, and so the periphery of the second circle alone will be to the second spiral alone as 2 to 1 ½ i e. as 4 to 3; and to the first alone as 4 to 1. SCHOLIUM I BUT as Consect. 4. may be also deduced after another way, viz. by comparing only the arches of the second circle with the correspondent circumscripts, but considering them as taken twice (because that circle is twice turned round while the whole helix or spiral is described) and finding in the first trisection the proportion of double the second periphery to all the circumscripts as 12 to 7; and in the succeeding bisection as 12 to 6 ½; in the second bisection as 12 to 6 ¼, etc. and at length by inferring, that the second periphery is double of all the arches circumscribible about the whole helix ad infinitum, that is to the helix itself. as 12 to 6+1 − ½ − ¼ etc. = 0. i e. as 12 to 6; and consequently the simple second periphery will be to the whole helix as 6 to 6: Thus the 5. Consect. may be separately had after the same manner, if instead of the first trisection, you only bisect; (vid. Fig. 141. n. 3.) for so in the first bisection the arches circumscribed about the second spiral line would be separately two semicircles D d, 3 and A a, 4, (for as the line BC is one, BE, 2, BD, 3, BA, 4; so the arch described by the radius BD is 3 and described by the radius BA = 4,) and their sum 7; while the sum of two equal to the greatest is 8. In the second bisection (when BE is 4) BF and its arch is made 5, the arch BD 6, the arch BG 7, the arch BA 8, the sum 26; while the sum of so many quadrants equal to the greatest is 32. Thus in the third bisection the sum of eight Octants circumscribed about the second helix will be found to be 100, the sum of so many = to the greatest 128, etc. Wherhfore the periphery of the second circle in the first case will be to the arches circumscribed about the second helix as 4 to 3+½; in the second as 4 to 3+¼; in the third as 4 to 3+⅛, etc. and so to all the arches circumscribible in infinitum, i e. to the second helix itself as 4 to 3+½ − ¼ − ⅛ etc. = 0. i e. as 4 to 3. Q. E. D. By the same method you may easily found the proportion of the third circle to the third spiral space, and of that periphery either to the whole spiral, or separately to the third, as will be evident to any one who tries. I For the third spiral space. (Fig. 142) BC 1 BF 4 BY 7 49 BD 2 BG 5 BK 8 64 BE 3 BH 6 BASILIUS 9 81 are the three first sectors circumscribed about the parts of the third helix. The sum of these three sectors is 194; and the sum of so many equal to the greatest 243. Therefore the first proportion of the one sum to the other will be as 243 to 194, i e. (dividing both sides by 9) as 27 to 21 . In the first bisection there will be seven lines: BH 12, BL 13 169 Sectors circumscribed about the parts of the third helix. BY 14 196 BM 15 225 BK 16 256 BN 17 289 BASILIUS 18 324 Sum 1459; while in the mean time the sum of as many equal to the greatest is 944, and so the second proportion as 1944 to 1459 i e. (dividing both sides by 72) as 27 to 20 . In the second Bisection there will be thirteen lines, viz. BH 24, the rest 25, 26, etc. but the sum of the sectors, i e. of the square numbers answering to the twelve latter will be found to be 11306; while in the mean time the sum of as many equal to the greatest will be 15552, so that you will have the third proportion of this sum to the other, viz. as 15552 to 11306, i e. (dividing both sides by 576) as 27 to 19 . Therefore the I proportion will be as 27 to 19+2+ i. e. to 19+2+½+ II— as 27 to 19+1+ i. e. to 19+1+¼+ III— as 27 to 19+ — i e. to 19+½+⅛+ . Therefore the proportion of the third circle to the third spiral space will be as 27 to 19+2+½+ − 1 − ¼ − i. e. as 27 to 19 − ½ etc. − ⅛ etc. − etc. = 0. = 0. = 0. Q. E. D. II For the third spiral line. If instead of the first trisection (as lesle commodious for the end proposed) you make use here also, as before, of bisection in the same figure, there will come out six lines from the point B to the helix, viz. B m, 1, BE, 2, B n, 3, BH, 4, B oh, 5, BA, 6; to which there answer as many semicircular arches in the same progression, and to the greatest of the two as many equal to 2, 4, 6; so that the sum of the unequal ones is 21, and of the equal ones 24, and so the proportion of three peripheries together to all the circumscripts together will be as 24 to 21 (and dividing both by 6) as 4 to 3 ½. In the second bisection the twelve unequal lines and arches make the sum 78, and as many equal to the greatest of the four will give the sum 96; so that the second proportion will be 96 to 78, i e. (dividing both sides by 24) 4 to 3 ½. In the third bisection the proportion will come out as 384 to 300, i e. (dividing both sides by 96) as 4 to 3 ½, etc. Therefore the proportion of the three circles together to the whole Helix will be as 4 to 3+½ − ¼ − ⅛ etc. = 0. i e. as 4 to 3 or 12 to 9 Q. E. D. CONSECTARY VI. NOW, if the periphery of the first circle be made 2, the second will be 4, the th●rd 6, and consequently the sum 12; it will be manifest that the third periphery separately will be to the whole helix as 6 to 9, i e. as 2 to 3. CONSECTARY VII. AND because the second periphery (which is 4) is equal to the first and second helix together, by the above Consect. 4. the remaining third spiral will be 5, and so the proportion of the third periphery to it as 6 to 5. CONSECTARY VIII. WHerefore the proportions of each of the peripheries to their correspondent spirals will be in a progression of ordinal numbers, viz. so that the latter of every two will denote the periphery of a circle, and the former an inscribed spiral; and consequently the spiral lines will be in an arithmetical progression of odd numbers, and the peripheries of the circles in a progression of even ones. 1— The first Spiral, 2— The first Periphery, 3— The second Spiral, 4— The second Periphery, 5— The third Spiral, 6 etc.— The third Periphery, etc. SHCOLIUM II THE seventh Consectary may also be easily deduced separately this way: In the first bisection the line BASILIUS and its periphery is 6, the line B o and its periphery 5, the sum of the circumscribed Peripheries 11; the sum of as many equal to the greatest 12. Therefore the periphery of the third circle will be to the two circumscripts as 12 to 11, i e. as 6 to 5 ½. In the second bisection the four circumscribed quadrants will be 12, 11, 10, 9, their sum 42; and the sum of four equal to the greatest, i e. the periphery of the third circle 48. Therefore the proportion is now as 48 to 42, i e. (dividing both sides by 8) as 6 to 5 ¼. Thus you will have the third proportion as 192 to 164, i e. (dividing both sides by 32) as 6 to 5 ⅛. Wherhfore the proportion of the third periphery to the third helix or spiral is as 6 to 5+½ − ¼ − ⅛ etc. = 0. i e. as 6 to 5. Q. E. D. CONSECTARY IX. AS Consect. 8. supplies us with a rule to determine the proportion of every spiral of every order to the periphery of the correspondent circle, viz. if the number of the order be doubled for the periphery of the circle, and the next antecedent odd number be taken for the spiral line; so what we have hitherto demonstrated supplies also another rule, to define the proportion of the spiral space in any order to its circle. For since the circles are in a progression of Squares 1, 4, 9, 16, etc. but the first circle is to the first space as 3 to 1 (i e. a, 1 to ⅓) by Prop. 17. and the second to the second as 12 to 7 (i e. as 4 to 2 ⅓) by Prop. 18. the third to the third as 27 to 19 (i e. as 9 to 6 ⅓) by Scholar 1. of this. And contemplating both these series one by another, Of the circles, 1, 4, 9 Of the spaces, ⅓, 2 ⅓, 6 ⅓. We see the numbers of the spaces are produced, if from the spuare numbers of the circles you subtract their roots, and add to the remainder ⅓. Wherhfore, if, e. g. we were to determine the proportion of the fourth circle to the fourth spiral space; the square of 4 viz. 16 would give the circle; hence substracting the root 4, there will remain 12, and adding ⅓ you would have the fourth spiral space 12 ⅓; and in like manner the spiral space 20 ⅓ would answer to the circle 25, etc. And that this is certain is hence evident, that if we multiply these numbers 16 and 12 ⅓, also 25 and 20 ⅓ by 3, that we may have those proportions in whole numbers, 48 and 37, 75 and 61, these are those very numbers Archimedes had hinted at in the Coral. of Prop. 25. Pag. 217. 142 143 144 145 146 CONSECTARY X. NAY what we have now said, is that very Coral. comprehending also that 25th. Proposition, viz. that a spiral space of any order is to its correspondent circle, as the rectangle of the semidiameters of this and the preceding circle together with a third part of the square of the difference between both semidiameters to the square of the greatest semidiameter. For, if e. g. the proportion of the third spiral space to the third circle be required, since the semidiameter of this third circle is as 3, and the semidiameter of the second precedent one is 2, and so the difference 1; the rectangle of 2 into 3 i e. 6, together with ⅓ of the square of the difference will define the third spiral space 6 ⅓; since the third circle may be defined by the square of the semidiameter of the greater, viz. by 9, and so in the rest; as the numbers we have found show, or further that may be found according to given Rules which may be here seen in the following Table. Orders. I TWO III IV FIVE VI VII VIII XI X Circles. 1 4 9 16 25 36 49 64 81 100 The whole spaces, the preced. ones being included. ⅓ 2 ⅓ 6 ⅓ 12 ⅓ 20 ⅓ 30 ⅓ 42 ⅓ 56 ⅓ 72 ⅓ 90 ⅓ Separate spaces the preced. ones being excluded. ⅓ 2 4 6 8 10 12 14 16 18 CONSECTARY XI. OUT of which table it is obvious to sight, that the second space excluding the first is sextuple of the first, as we have already deduced in Consect. 2. Prop. 18. and the third separate space double of the second, and the fourth triple of the same second, and the fifth quadruple, and so onwards. SCHOLIUM III AND this shall suffice for spirals, which comprehends not only the chief Theorems of Archimedes of spiral spaces, but also the chief of spiral lines (whereof Archimedes has left nothing.) If any should have a mind to carry on our method further, he may easily demonstrate after the same way what remains in Archimedes, and what Dr. Wallis in his Arithmetic of Infinites from Prop. 5. to the 38, and what others have done on this Argument. CHAP. V Of the Conchoid, Cissoid, Cycloid, Quadratrix, etc. Proposition XIX. THE first conchoid of Nicomedes Bbb (Fig. 110) on both sides of the perpendicular cDb approaches nearer always to the directrix or horizontal line A, and yet will never coincide with it, altho' it be conceived to be produced on both sides ad infinitum. Demonstration. For since only D b is perpendicular to A, and all the rest ab are so much the more inclined to it by how much the more remote they are from the middle one D b, and all in the mean while are equal both to it and to one another, by Def. 13. it is evident that the points b and B will come so much the nearer to A, by how much the farther they recede from the middle line D b. And yet because the lines BAC and bac are all right ones, whose points A, a, are in a right line A it is equally as impossible that the point b or B, which is always in the conchoid should ever touch this right line, as it is impossible that the point C should be in it, by virtue of the aforecited Def. Q. E. D. Proposition XX. YET no other right line can be drawn between the directrix A and the conchoid, but what will cut it if produced. Demonstration. For if such a right line be made parallel to A, as GH, and you make, as DI to IC so D b to a fourth, which will be greater than IC, as D b is greater than DIEGO, and consequently, if making that an interval you draw the circular arch from C, it will necessarily cut the line GH e. g. in G. Drawing therefore C aG. you'll have as DIEGO to IC so aG to GC, i e. to that fourth proportional before found, by virtue of Prop. 34. lib. 1. but as DI to IC, so was also D b to the same fourth by Construc. Therefore aG and D b, which have both the same proportion to the same quantity, are equal; and consequently the point G is in the conchoid by virtue of Def. 13. and consequently the right line GH being produced will cut that produced also, on both sides, by the same reason. Much more will it cut it on either side if it be not parallel to the directrix A, which is very obvious. Therefore no right line can be drawn between the conchoid, etc. Q. E. D. CONSECTARY. HEnce, besides orher Problems, that may be very easily solved, which requires, having any rectilinear angle given ABC (Fig 143.) and a point without it, from that point to draw a right line DEF, so that part of it OF, which is intercepted between the legs of the angle, shall be equal to a given line Z. For if you draw the perpendicular DGH from the given point D through the nearest leg of the angle BC, and make GH equal to the given line Z, and from the centre C at the interval GH describe the conchoid IHK, which will be necessarily cut by the other leg of the angle by virtue of the present Prop. e. g. in FLETCHER, the line DF being drawn will give the intercepted part = GH by the nature of the conchoid, and consequently = to the given line Z. SCHOLIUM. BY means of this Consectary Nicomedes solves that noble Problem of finding two mean proportionals, after this way, which we will here show from Eutocius, but drawn into a compendium, and somewhat changed as to the order. Let two given lines AB and BC (Fig. 144.) between which you are to found two mean proportionals, be joined together at right angles, and divide both into two parts in D and E, and having completed the rectangle ABCL, from L through D draw LG to BC prolonged; so that after this way GB may become = ALL or BC: Having let fall a perpendicular from E cut of from C at the interval CF = AD the part OF, and having drawn FG make CH parallel to it; and lastly through the legs of the angle KCH draw the right line FHK, so that the part HK shall be equal to the line CF, by the preceding Consect. and also draw the right line KM from KING through L to the continued line BASILIUS: All which being done, CK and AM will be two mean proportionals between AB and BC; which after our way we thus demonstrate: By reason of the similitude of the ▵ ▵ MAL and LCK MALFORT is to LC or AB as ALL or BC to CK b − ebb − c − ec and moreover, as MALFORT to AD so GC to CK i e. FH to HK b − ½ ebb − 2 c − ec by reason of GF and CHANGED being parallel, by Consect. 4. Prop. 34. lib. 1. therefore since HK is = AD = ½ ebb, FH will be = A = b, and consequently MD = FK, viz. both b+ + ebb, and the square of both = bb + eb+ + eebb = □ EF+EK by virtue of the Pythag. Theor. Now if to these equal quantities you add the equal □ □ DX and EC = ¼ cc, their sum, viz. □ MD+ □ DX i e. □ MX will be bb + ebb+ + eebb+ + cc, equal to the sum of these, viz. □ EF+ □ EC i e. □ CF (by the Pythag. Theor. or EX by Construct.) + □ KX; whence these two things now follow: 1. That the lines MX and KX are equal. 2. If from those equal sums you take away the common quantities ¼ eebb+ + cc, the remainders will be equal, viz. bb + ebb = ecc + eecc; and (since the part taken away, viz. bb is manifestly to the other part taken away, viz. ecc as the remainder ebb to the remainder eecc, and the whole with the parts taken away and the remainders are in the same proportion by Prop. 26. lib. 1.) separately also bb will = ecc and ebb = eecc. But from the latter equation it follows that as ebb to ec so ec to b by virtue of the 19 Prop. lib. 1. AB to CK so CK to MA and by the same reason it follows from the former Equation as ec to b so b to c CK to MA so MA to BC i e. CK and MALFORT are two mean Proportionals between AB and BC. Q. E. D. From which deduction you have also manifest the foundation of that mechanical way, which Hiero Alexandrinus makes use of in Eutocius, lib. 2. of the Sphere and Cylinder, and which Swenterus has put into his practical Geometry lib. 1. Tract. 1. Prop. 23. when, viz having joined in the form of a rectangle the given right lines AB and BC (Fig. 145.) and continued them at the other ends, he so long moves the ruler in L, having a movable centre, backwards and forwards, till XK and XM by help of a pair of compasses are found equal. To which, another way of Philo's is not unlike, and flows from the same fountain, wherein, having made on AC a semicircle, the movable ruler in L is so long moved backwarks and forwards, until LM and NK are found equal: Which seems to Eutocius to be more accommodated to practice, and easier to be performed by help of a ruler divided into small equal particles. Proposition XXI. IF from any point of the other diameter in the generating circle e. g. from G (Fig. 111. n. 1.) you draw a perpendicular GE through the cissoid of Diocles. the lines CG, GE, GD, and GH will be continual proportionals. Demonstration. For since GE and IF, as right sins, and also GD and IC as versed sins of equal arches by the Hypoth. are equal; you'll have as ID too IF (i e. CG to GE) so IF to IC (i e. GE to GD) per n. 3. Scholar 2. Prop. 34. lib. 1. But GD is to GH as ID too IF (i e. as GE to GD) by the forecited Prop. 34. lib. 1. Therefore CG to GE, GE to GD, and GD to GH, are all in the same continual proportion. Q. E. D. CONSECTARY. HEnce it was easy for Diocles to found two mean proportionals x and y between two given right lines FIVE and Z; (Fig. 146) for he made (having first described his curve DHB) as FIVE to Z so CL to LK, and having drawn CKH to the curve, and through H the perpendicular GE, he had between CG and GH two mean proportionals GE and GD by virtue of the present Prop. when in the mean while CG the first would be to GH the last, as CL to LK, i e. as the first FIVE to the last Z given by virtue of the Constr. Therefore nothing remained but to make, 1. as CG to GE so FIVE to X; and lastly, as GE to GD so x to y. SCHOLIUM. IT may not be amiss to mention here another way of finding two mean proportionals between any two given lines by the help of two Parabola's, which Menechmus formerly made use of, viz. by joining at right angles the given lines AB and BC (Fig. 147.) and prolonging them as occasion shall require through E and D; and than describing a Parabola about BE as its axis, so made that BC shall be its Latus Rectum, and in like manner describing another Parabola about BD as its axe, that shall have AB for its Latus Rectum, and that shall cut the former in FLETCHER: Which being done, the semiordinate FE (or BD which is equal to it) being drawn to the point of Intersection FLETCHER, will be the two mean proportionals sought. For by virtue of the fourth Consectary of Prop. 1. of this Book, DF or BE is a mean proportional between AB and BD, and in like manner OF or BD is a mean proportional between BE and BC, and consequently as AB to BE so BE to BD, and as BE to BD so BD to BC; Q. E. D. To this way of Menechmus that of Des Cartes is not unlike, which he gives us p. m. 91. except only that instead of two Parabola's, he makes use only of one and a circle in room of the other: In imitation of whom Renatus Franciscus Slusius has since shown infinite methods of doing the same thing by help of a circle, and either infinite Ellipses or Hyperbola's, in his ingenious Treatise which he thence names his Mesolabium. Proposition XXII. ANY semiordinate of the Cycloid as BF (Fig. 148.) or bf is equal to its corresponding Sine in the generating circle as BD, bd, together with the arch of that sine AD or Ad. Demonstration. For the motion of the point A describing the semi-cycloid AFE, by Def. 11. is compounded of the motion of the orb (or wheel) B along the semicircle ADC, and of the motion of the centre along the right line BC equal to CE, and consequently to the semi circle itself, or motion of the orb, Therefore as the point A moving to E by the motion of the orb (or wheel) moved or was carried from the diameter AC through the whole semi circle ADC till it came to AC again, and by the motion of the Centre passes through the whole space BG or CE, which is equal to the semicircular arch; thus the same A, when come to FLETCHER, will have described the quadrant AD, by moving from the diameter AC the quantity of the sine BD, and moreover by the motion of its centre (which is equal to the motion of the Orb) moves from AC the space of DF: And so the semiordinate BF will be equal to the arch AD and to its sine BD taken together; and in like manner the semiordinate bf will be equal to the arch A d and its sine bd, etc. Q E. D. CONSECTARY I HEnce may be easily assigned by help of the cycloid a right line equal to the semiperiphery or any given arch AD or A d; viz. CE double, or taken twice for the whole circumference, and single for the semiperiphery or half circumference, DF for the quadrant A d, df for the arch AD, etc. CONSECTARY II WHerefore the quadrature of the circle may be geometrically obtained according to Consect. 2. of Def. 15. lib. 1. CONSECTARY III IF you take B e, be, double of the sins BD, bd, etc. so that all the indivisibles bd taken together, may be to all the indivisibles be taken together as BD to B e, a curve described through the points e will be an ellipsis by Prop. 11. and the curvilinear space ADC eA will be equal to the semicircle ACDA. CONSECTARY IU. AND since DF (i e. D e + of) is equal to the quadrant DAM, by virtue of the present Prop. BD+FG will be also equal to the quadrant (because the whole BG or CE is = to a semicircle) and consequently of and FG will be equal; in like manner since df both above and below is equal to the arch damn, below bd + fg will be = to the remaining Arch dC: and above bd + ef (i e. df) will be equal to the equal arch damn. Therefore of above and fg below are equal, and (since the same may be shown of all the indivisibles of the same sort throughout) the trilinear figure FGE will be equal to the trilinear eFA. Proposition XXIII. THE cycloidal space is triple of the generating circle i e. the semi-cycloidal space AECA is triple of the semicircle ADCA. Demonstration. Since the parallelogram BCEG is equal to the whole circle by Consect. 2. of Def. 15. lib. 1. i e. to the semi-ellipse A eCA, by the present construction the Trapezium C egg will be equal to the quadrant of the ellipse or the semicircle. But the trilinear space FEG is = to the trilinear space FA e, by the fourth Consect. of the preced. therefore also the trilinear space A eCEA is equal to the semicircle. Therefore the whole cycloidal space is equal to the three semicircles. Q. E. D. Or thus. Since the whole parallelogram A is equal to two circles and the semi ellipsis A eCA to one; the remaining space A eCEC to one circle, and its half A eGC to a semicircle. But the trilinear space A of is equal to the trilinear space FGE by Consect. 4. of the preced. Therefore the one being substituted in the other's place the trilinear space AFEC will be equal to the semicircle: Therefore the remainder of the Parallelogram, i e. the cycloidal space AFECA will be equal to three semicircles. Q. E. D. SCHOLIUM. TO these short Demonstrations, which we confess we own for the most part to Hon. Faber, we will subjoin another somewhat more prolix, but yet not unpleasant, which we found in Carolus Renaldinus, lib. 1. de Resol. & Compos. Math. p. 299. But here we will give it the Reader more plain, and free from all Scruples, and likewise much easier. It is performed in these inferences, 1. That the right lined Parallelogram A baB (Fig. 149 n. 1) is equal to the curvilinear space A bdaBDA. 2. That as that is divided into two equal parts by its right lined diagonal A a, so likewise is this by the semi-cycloid A aa, so that the right lined triangle A aB is equal to the curvilinear space A aaBDA. 3. Therefore the one as well as the other is equal to the generating circle; and consequently, 4 If to this curvilinear space there be added the semicircle ADBA the semi-cycloidal space A aaBA will be equal to three semicircles. The first is evident, while if you take from the right lined parallelogram the semicircle ADBA on the one side, and on the other add the semicircle abda, there will arise the curvilinear Parallelogram. The third is evident from Consect. 2. Def. 15. lib. 1. Because the line B a is equal to the semiperiphery, which multiplied by the diameter BC gives the area of the circle. The fourth is self evident; and so there remains only the second to be demonstrated, viz. That the curvilinear parallelogram is divided into two equal parts by the cycloid, i e. that the external trilinear Figure A adbA is eqaal to the internal one A aaBDA; which may be thus shown: Having divided the base B a into three equal parts, and drawn through them three semicircles, and moreover the transverse right lines D d and E e through the intersections of the semicircles and the cycloid; it is certain from the genesis of the cycloid, by virtue of the Cons. of Def. 11. that as the right line a1 is a third part of the whole aB, so the arch 1 a is a third part of the generating periphery, and by the same reason the arch 2 a two thirds, and so the remaining arch a11s also ⅓; insomuch that the first arch 1 a, and the last a11, and consequently their right sins of, ag, and likewise their versed one's f1, g11s, are equal, and so the curvilined partial Parallellograms, both above and below, all upon equal bases, and of the same height (viz. the two linear ones ae21 and da 11. 1.) are equal among themselves, and so likewise the two pricked or pointed ones D a 2B and aeb1. Wherhfore if now the base B a (n. 2) be conceived to be divided into six equal parts, and having drawn semicircles, and transverse lines through their intersections with the cycloid, the arches and will be and and of a semicircle, Pag. 227. 147 148 149 150 and so the versed sins of each, i e. the altitudes of the corresponding parallellograms will be equal, and consequently the parallellograms of the internal and external trilinear space that are alike noted or signed will be equal to each other. Now this inscription of curvilinear parallellograms always respectively equal both in number and magnitude, since it may be continued in both the trilinear figures ad infinitum; it will evidently follow, that the trilinear figures themselves, whose infinite inscripts are always equal, will be likewise equal to one another. Proposition XXIV. THE base of the quadratrix A (Fig. 150) and the semidiameter of the generating quadrant AD and the quadrant itself BD are in continual Proportion. Demonstration. For the quadrant DB is to the radius DA as the arch IB to the perpendicular H e by Consect. 1. Def. 16 and I b is to H e as A b to A e by Prop. 34. lib. 1. But the arch IB (if it be conceived to be lesle and lesle ad infinitum) will at length coincide with I b, as ending in the same moment in the point B, wherein H e will end in the point E, and so A e will end in A and A b in AB. Therefore at length DB will be to DAM as IB (i e. I b) to H e, i e, as A b to A e, i e. as AB (or DA) to A Q. E. D. SCHOLIUM I CLavius about the end of the sixth Book of Euclid, and others, demonstrate this indirectly by a deduction ad Absurdum, or concluding the opposite much after this manner: If DAM or AB is not to A as DB to DA, suppose it to be so to the greater A f or the lesle A e. In the first case therefore, because AB is to A f as DB to DA per Hypoth. i e. as KING f to A f the quadrant KING f and the radius AB or DA will be equal. But as BD is to IB so is KING f to H f by reason of the similitude of the arches; and as BD to IB so also DA (or = KING f) to the sine H e, by Consect. 1. Def 16. Therefore the sine H e and the arch H f (to which the same KING f bears the same proportion) will be equal; which is absurd. In the latter case, because AB would be to A e as DB to DA by the hypoth. i e. as L e to A e, the quadrant L e and the radius AB or DA would be again equal. But as BD is to IB so is L e to M e by reason of the similitude of the arches; and as BD to IB so also is DAM (ie = L e) to H e by Consect. 1. Def. 16. Therefore the tangent H e and the arch M e (to which the same L e bears the same proportion) will be equal, which is again absurd. Wherhfore BD is to DA, not as DAM to a greater A f or a lesle A e; therefore as DAM to A Q. E. D. CONSECTARY I WHerefore it is evident from what we have deduced, if by means of the base of the quadratrix A you draw a quadrant, the side of the quadratrix DA will be equal to it, and consequently double of the semi periphery, and quadruple of the whole periphery. CONSECTARY II IT is evident also that you may obtain a right line equal to the quadrant DB of any given circle, if, having described a quadratrix, you make as A to AD so AD to a third equal to the quadrant DB: Which third proportional taken four times will be equal to the whole periphery. CONSECTARY III YOU may also obtain a right line equal to any lesle arch, if you make, as DA to H e so a third proportional found (i e. the quadrant DB) to a fourth, by virtue of Cons. 1. Def. 16. CONSECTARY IU. THE quadrature of the circle therefore, by virtue of Cons. 2. Def. 15. lib. 1. as likewise the trisection of an angle, by virtue of Consect. 2. Def. 16. lib 2. may be Geometrically obtained, if the Quadratrix might be numbered among Geometrical Curves. SHCOLIUM II CLavius was also of this Opinion in the book afore mentioned, who thought that if the quadratrix be excluded out of the number of geometrical curves, by the same reason you may also exclude the ellipse, parabola, and hyperbola, since they as well as this are commonly described through innumerable points. But by that great Man's leave, we may deny this consequence, by the same reason as Des Cartes has denied the converse of it in his Geom. lordship 18. and 19 by virtue of which he suspects the ancients took the conic sections, etc. for mechanic or non-geometrick lines, because they did the spiral, quadratrix, etc. for such. But this is the difference between the description of the quadratrix and the conic sections through points, that all and every of the points of the conic sections, relating to any given point of the axis, may be geometrically determined; but all the points of the quadratrix promiscuously related to any point of the generating quadrant, cannot be geometrically determined, but only those which respect some certain point, from which the quadrant may be divided into two arches of known proportion. For if, e g. in the quadrant BD the point X be given at pleasure, it will be impossible by Clavius' Rule to define a point of the quadratrix answering to it, because the proportion of the arches DX and BX is unknown, and consequently neither can a proportional section of the right line AD be made: Not to mention that the last point E (which is the primary and most necessary one to the quadrature) even by Clavius' own confession cannot be geometrically defined. We may pass the like judgement on Archimedes' spiral and such like curves, which are conceived to be described by two motions independent on one another; as will be manifest to any one who compares the genesis of the spiral with that of the quadratrix and what we have hitherto said. Whence neither will Monantholius' trisection of a given angle (which he essays) by means of a spiral be enough geometrical; which in his Book de Puncto, Cap. 7. p. 24. he attempts to perform thus: To the centre of a described spiral and its first helical or spiral line BASILIUS (Fig. 151.) he applies the angle ABC equal to the given one abc; than having drawn circles through FLETCHER and A where the legs of the angle cut the spiral, he divides the intermediate space DAM into three equal parts in 1, and KING: And than through these points he draws circle's cutting the helix in L and M; and lastly having drawn BLN, BMO, he easily demonstrates from the genesis of the spiral that the arches AO, ON, NC are equal. And so after the same manner not only any angle or arch, but the whole periphery may be geometrically divided into as many parts as you please; only supposing that this spiral line may be numbered among geometrical ones; as we have heretofore hinted that the cycloid, conchoid, cissoid, and logarithmical curve, etc. might be; and we have above sixteen Years ago declared our opinion for it in our Germane Edition of Archimedes; and now are therein confirm d by those celebrated Mathematicians Leibnitz, Craige, etc. who number lines of this kind, altho' they cannot be expressed by our common equations, among geometrical ones, notwithstanding the contrary opinion of Des Cartes, etc. because they admit of equations of an indefinite or transcendent degree, and are capable of a Calculus as well as others, though it be of a nature and kind different from that commonly used. See the Acta Erud. Lips. ann. 84. p. 234. and ann. 86. p. 292. and 294. CHAP. VI The Conclusion, or Epilogue of the whole Work. NOW we may at length understand what Honoratus Fabri delivers concerning the distribution of figurate magnitudes into certain Classes, in his Synopsis Geom. p. 57 and the following. 1. The first Class contains elementary figures, or equal indivisibles, such as, 1. All Parallellograms, as the Square, Oblong, Rhombus and Rhomboid, the elements whereof are equal right lines, as in Def. 12. lib. 1. 2. Convex or concave Surfaces, the elements whereof are curve lines moved through right lines by a parallel motion; among which are chief reckoned cylindrical surfaces, whereof see Def. 16. lib. 1. about the end. 3. Parallelepipeds, and among them the cube, whose indivisibles are squares, or other Parallellograms. 4. Prisms made by the motion of a Triangle, Trapezium, or any Polygonous Body, along a right line, all the indivisibles whereof are consequently similar and equal to the generating plane. 2. The second Class contains Figures whose Elements decrease in a simple arithmetical Progression; such are, 1. Triangles, as is evident from Prop. 37. lib. 1. 2. The circle, and its Sectors, as resolvible into concentric Peripheries according to Cons. 1. and 3. of the aforecited Prop. 3. The Cylinder as resolvible into concentric cylindric Surfaces, as its indivisibles. 4. The Surface of a Cone, whose elements are circular Peripheries, and also of the Pyramid whose indivisibles are similar angular Peripheries every where increasing in arithmetical Progression. 5. The Parabolic Conoid, whose indivisibles are Circles according to the proportion of the abscissa's in arithmetical Progression, by virtue of Prop. 14. lib. 2. etc. 3. The third Class contains elementary Figures increasing in duplicate arithmetical Progression; such are, 1. The Pyramid and Cone, the first whereof may be resolved into angular Planes, the second into circular ones increasing according to a series of square numbers; as is evident from Prop. 38. lib. 1. and its Consectary. 2. The trilinear parabolic Space, as defined Prop. 10. lib. 2. by the letters E h HK. 3. The Sphere, as far as it may be resolved into spherical concentric Surfaces, every one whereof may be considered as a base, taking the semidiameter for the altitude. 4. The Cone, as resolvible into parallel conical Surfaces described by the parallel indivisibles of the Triangle. 5. The remainder of a Cylinder after an Hemisphere of the same base and altitude is taken out, according to Scholar 1. of Prop. 39 lib. 1. 4. The fourth Class would comprehend all magnitudes resolvible into elements or indivisibles increasing in triplicate, quadruplicate, etc. Arithmetical Progression; such we have not treated of, but may be found among Planes terminated by Curves of superior Genders; see Fabri's Synopsis, p. m. 67. 5. The fifth Class is of those Magnitudes, whose indivisibles decrease, proceeding from a square number by odd numbers, as, 36, 35, 32, 27, 20, 11, etc. such are, first, an Hemisphere, as is evident from Prop. 39 lib. 1. 2 An Hemispheroid, as in Prop. 15 lib 2. 3. A Semi-parabola, as may be gathered from the demonstration of Prop. ●0 lib. 2. For since the indivisibles of the circumscribed trilinear figure ebb are found in a duplicate arithmetical Progression, ¼, , , , the indivisibles of the semi-parabola will necessarily be , , , , etc. 6. We may make a sixth Class of those Magnitudes whose indivisibles decrease in a like Progression, not of the numbers themselves descending by odd steps from a given square, but of their roots, which are for the most part furred ones; such as is first, the Semicircle, as is evident from Prop. 43. lib. 1. and by virtue of Prop. 5. lib. 2. and also the semi-ellipse, etc. 7. The seventh Class comprehends those Magnitudes, whose Elements are in a Progression of a double series of numbers, as in the Parabolick Conoid, as may be seen in the Scholium of Prop. 16. lib. 2. But, to omit the other Classes of Magnitudes of a superior Gender, the consideration whereof these Elements either have not touched on, or only by the by; (which any one who pleases may see in Faber s Synopsis, especially those which he comprehends under the sixth and seventh Classes, p. 70 and the following) about those we have here particularly noted, there remain only two things to be taken notice of. 1. That since in the first Class we place Parallellograms and Cylinders, in the second Triangles, in the third Pyramids and Cones, in the fifth Hemispheres, in the sixth semicircles, etc. We may with Hon. Faber call the first Class, that of Cylindrical or Parallelogrammatick Figures; the second, the Class of Triangular Figures; the third, of Pyramidals; the fifth, of Hemispherical Figures; the sixth of semicircular ones, etc. 2. That having ranged or reduced after this manner homogeneous Figures, or those of like condition, to a few Classes, their dimension, and consequently almost the whole business of measuring may be very compendiously reduced to a few Rules; whereof we will here give the Reader a short Specimen, in the following CONSECTARIES. I THE dimension of Parallelogrammatick Figures, i e. of those of the first Class, may be had, by multiplying the whole base by the whole altitude: See lib. 1. Def. 12. Cons. 7. Def. 18. Cons. 6. Def. 16. Cons 3. and 4. II The dimension of Triangular Figures, i e. of those of the second Class, may be had by the multiplication of the whole Base by half the Altitude, or of half the Base by the whole altitude; [see lib. 1. Def. 12. Consect. 8. Def. 15. Consect. 2. Def. 18. Consect. 4. lib. 2. Prop. 14.] and their Proportion is to their respective circumscribing Parallellograms, as 1 to 2; [see besides the Prop. already cited, lib. 1. Prop. 37. and its first Consect.] III The dimension of Pyramidals, i e. of magnitudes of the third Classis, may be obtained by the multiplication of the Base by the third part of the of Altitude; [see lib. 1. Def. 17. Consect. 3. and 4. and Def. 20. Consect. 1. etc.] and their Proportion to the corresponding Figures of the Class of the same Base and Altitude is as 1 to 3. [see besides the Prop. already cited Prop. 38. lib. 1. and its Cons. Prop. 39 and Scholar 1. lib. 2. Prop. 10. etc. IV. The Proportion of Hemispherical Magnitudes, i e. of the fifth Classis to corresponding one's of the first Class of the same Base and Altitude is as 2 to 3; [see lib. 1. Prop. 39 lib. 2. Prop. 10. and 15.] and so their dimension may be had by multiplying their Base by ⅔ of their Altitude. V The Proportion of semicircular Magnitudes, i e. of those of the sixth Class to so many corresponding one's of the first Class of the same Base and Altitude cannot be expressed by whole numbers or by a small fraction [see lib. 1. Prop. 43. and lib. 2. Prop. 11.] and consequently their exact numeral dimension cannot be had. AN INTRODUCTION TO SPECIOUS ANALYSIS, OR, The New Geometry, chief according to the Method of Des Cartes, But much facilitated by later Inventions, etc. By J. CHRIST. STURMIUS. THE PREFACE TO THE READER. SINCE those ingenious Mathematicians of this present Age, which is now drawing to a Conclusion, Vieta, Ougthred, Harriot, Cartes, Schooten, Beaune, Van Hudde, Heuraet, de Witte, and Slusius, and several other Famous Men coeval with them, have by their Endeavours improved the Algebra of the Ancients, raised it to vastly an higher pitch, and brought it from Numbers to universal Symbols, and not only found the excellent uses of it in Geometry themselves, but also communicated them to others; almost all Countries have furnished us with some excellent People, who treading in the footsteps of their Predecessors, have endeavoured to advance it yet further: And even our Times are not without those of the highest rank, as Wallis, Baker, Renaldinus, Mengolius, Huggeas, Malbranch, Leibnitz, Craan, and several others, who endeavour to promote this Science, deservedly reputed the very Apex of Human Reason, and carry it more and more towards its utmost Perfection by daily augmenting it with new and curious Inventions. But in the mean time, while these ingenious Men wholly busy themselves in promoting it, there are few found who condescend to explain the first Principles of it, and show a ready way to young Beginners to arrive at the knowledge of those Inventions. It is not long since a certain Friend of mine, who for some time has publicly and successfully taught these Sciences, complained to me by his Letters, of the want of a good Guide to Specious Analysis, whereby he might instill the Principles of that admirable Art into his Auditors. To whose Desires being not just than at leisure to satisfy, I immediately after projected this Introduction, which at length you see finished, in the Form we now present you with it, wherein we have all along consulted to suit the Endeavours of young Beginners, as far as possible; as we thought ourselves engaged by the Duties of our Professorship to do, & have comprised the Precepts of the Art in six or seven Pages, and accommodated Examples of every kind to illustrate them. ●herein if I have but indifferently accomplished my Design▪ I shall not think my Labour lost. We here add it to our Mathesis Enucleata, both as being properly a part of it, and more especially, because this Introduction presupposes the Reader to be acquainted with the first Principles of Specious Computation, which we have therein laid down. And so we commit our Endeavours to the Perusal and Censure of the Candid Reader. INTRODUCTION TO SPECIOUS ANALYSIS. THE Analytick Art, or Specious Analysis, is solely subvervient to finding of Theorems, and resolving Problems, by leading us from certain Data or given Quantities, into the knowledge of unknown and sought ones, by a Chain of certain and infallible Consequences: This admirable Artifice may be reduced to four Primary Heads, viz. Denomination, Reduction, Equation, and Effection (if the Problem be a Geometrical one) or Construction. I DENOMINATION. BY Denomination is understood a preparatory imposition of Names peculiar to each Quantity, whereby every one of the Quantities given or sought, are denoted by one or more peculiar Letters of the Alphabet at pleasure, but with this (arbitrarious) difference, that known or given quantities are marked by the former Letters of the Alphabet, a, b, c, etc. and the unknown or sought one's by the latter, z, y, x, etc. But although this imposition of Names, is, as we have said, altogether arbitrarious, yet there often happens not a little facility to the Solution itself, by its being chosen as accommodate as possible to the conditions of the quantities given and sought; which any one w●ll learn better by Use than Precepts: As we found that both Theorems may be demonstrated, and Problems resolved e. g. by an extraordinary Compendium, if we denote any reason of two given Homogeneous quantities by a and e a, b and i b, d and oh d, etc. (v z. by expressing the Names of the Reasons by e, and i, and oh, etc.) and continued proportionality by a, ea, e a, e a, etc. and discontinued or discrete by b i b, c i c, d i d, or after the like manner, as we have done in our Math. Enucl. Lib. 1. Cap. 2, 3, 4, 7. and Lib. 2. Cap. 1. etc. II EQVATION. HAving thus given each quantity its Name, and making no further distinction between the quantities given and those sought, but treating them all promiscuously, and as already known, you must carefully search into and discuss all the Circumstances of the Question, and making various Comparisons of the quantities, by adding, substracting, multiplying, and dividing them, etc. till at length, which is the chief aim and design of it, you can express one and the same quantity two ways, which is that we call an Equation: And you must found as many of these Equations, or Equalities of literal quantities, (as expressing the same thing) as there are several unknown quantities in the Question, independent on each other, and consequently denominated by so many different Letters, z, y, x, etc. But if so many Equations cannot be found, after having exhausted all the Circumstances of the Questions by one or two Equations; that is a sign the other unknown quantities may be assumed at pleasure: Which the Examples we shall hereafter bring will more fully show. But as here also (as likewise in all this Art) Ingenuity and Use do more than Rules and Precepts; yet we will here show the principal Fountains, for the sake of young Beginners, whence Equations, according to circumstances obvious in the Question, are usually had. These are partly Axioms self evident, E. g. That the whole is equal to all its parts taken together. That those quantities which are equal to one third, are equal among themselves. That the Products or Rectangles under the Parts or Segments, are equal to the Product of the whole. Partly some universal Theorems that are certain and already demonstrated, as, Three (α) Eucl. 6.17. continual Proportionals being proposed, the Rectangle of the Extremes is equal to the Square of the mean. (β) id. 6.16. Four being proposed, whether in continued or discontinued Proportion, the Product or Rectangle of the Extremes is equal to that of the Means. And several others such like, which we have demonstrated in Cap. 2, 3, and 4. Lib. 1. of our Mathesis Enucleat. partly in the last place, some particular Geometrical Theorems already demonstrated, as e. g. that common Pythagorick one. That in rightangled Triangles (γ) id. 1.47. the Square of the Hypothenusa is equal to the two Squares of the sides. That the Square of the Tangent of a (δ) id. 3.36. Circle is equal to the Rectangle of the Secant and that Segment of it that falls without the Circle; the first whereof, we have demonstrated, Lib▪ 1. Math. Enuc. Def. 13. Scholar and also Prop. 34. Consect. 8. also Prop 44. after various ways; to which may be numbered Prop. 34. with Scholar 11. n. 3. Prop. 37. and following, Prop. 45. and 46. also the 48. and several others in Lib. 1. Math. Enucl. and likewise Lib. 2 Prop. 1, 2, 3, and several following. And as for Examples both of Denomination, and Equations found after various ways, you may see them hereafter follow, and some we will here give you by way of Anticipation. III REDUCTION. An Equation thus found must be reduced, i e. those two equal quantities, which for the most part are very much compounded of the quantities given and sought together, must be reduced to such a form, by adding or substracting something to or from each part, or multiplying and dividing by the same, etc. that the unknown or sought quantity alone, or its Square or Cube or Biquadrate, etc. may be found on the one side, and on the other the quantity expressed by mere given or known Letters, or affected with the unknown and sought one's; such are these Forms which follow, distinguished by their Names prefixed to them, A simple Equation, z = b, or y = . A pure Quadratick, . A pure Cubick, . An affected Quadratick, . An affected Cubick, etc. A Biquadratick, etc. To one of which, or some other like them, when yo● reduced your Equation first found, there are Rules in rea● whereby the Value of the unknown or sought quantity z or x, may be either expressed in Numbers, if the Quest● an Arithmetical one, or geometrically determined if it Geometrical one: Which is that we call Effection or Coaction. Thus therefore the whole, or at lest the chief business Analytics, is conversant in finding a convenient or fit Eq●on: For Reduction is very easy, and consisting only in Operations and mere Axioms, as e. g. If to equal quantities you add or subtract equal ones Aggregates or Remainders will be equal; Pag. 5 Fig. I Fig. II Fig. III Fig. IU. Fig. V VI VII VIII IX X XI XII XIII FOURTEEN XV XVI XVII IV. EFFECTION, or CONSTRUCTION. 1. In simple Equations. suppose z = b, the quantity b is sought. If z be = .... or or x = .... or or make as c to b so a to z, as b to a so a to x, as h − 1 to b + g so f to y, as h+ + to b − g so f to z, etc. every where according to n. 2. Scholar 2. Prop. 34. Lib. 1. Math. Enuc. If z be = , the Resolution of it into Proportionals 〈◊〉 be more difficult, because neither of the Letters are found ●ce in the Numerator. That therefore you may have e. g. twice, you must make as k to n so m to a fourth Proportio●l which call lordship; than will, by virtue of Prop. 18. Lib. 1. ● = mn, and the proposed Equation be changed into this , to be now constructed from the 2 d. Case. Or if , found a mean Proportional between k ● l, which call lordship; and between m and n, which call cue, ●ording to n. 3. of the aforecited Scholar; and the proposed Equation, by virtue of Prop. 17. will be in this form: . Make therefore in the rightangled ▵ (Fig. 1.) B = p and BC = cue; and the □ AC by virtue of the Py●gorick Theorem, = pp + qq: Which since it must be di●ed by razors − s, make further, by Prop. 18 as r − s to the , so is to y, according to the afore●ed. 4 In like manner if x be = , make 1st. as b to m so n to a fourth which call k; and so putting bk for mn, the Equation will be reduced to the second Case under this form: Or thus: Found a mean Proportional between b and g, which call lordship, and between m and n, which call cue; and the proposed Equation will be in this Form: Make therefore (in Fig. 2.) AB = p, and having on this described a Semicircle, apply BC = cue; than will, by virtue of Scholar 5. Prop. 34. □ AC = pp − qq: Which since it must be divided by c + d, make farther, as c + d to so to x; all from the same Foundations, whence you have the Construction of the third Case. 5. If z be = ; make first as f to a, so a to a third Proportional m, and you'll have (putting fm for aa) , i e. . Make secondly as f to m, so b to a fourth n, and by putting fn for mb you'll have z = , i e. , wherefore thirdly you'll have as g to n so c to z. 6. If y be = make first as m to n, so pomell to a fourth, which call n, and by putting now mn for hl, you'll have i. e. = y. Therefore you'll now have secondly, as m to n, so pomell to y by Case 2: So that the Construction of the fifth and sixth Cases is nothing but reiterations of the Rule of three, according to what we have often inculcated, N. 2. and 3. Scholar 2. Prop. 34. 2. In simple Quadratick Equations. 1. IF xx = ab or y = 1 c or z = ¾ dd you'll have that is to a mean proportional between a and b 1 and c ¾ d and d and so the Construction will be had from n. 3. Scholar 2. Prop. 34. (see Fig. 3) 2. If y = fg + kl or x = fg − kl you'll have make therefore on the one side the Rightangled Triangle ABC (Fig. 4.) whose side AB is = to a mean Proportional between f and g, BC is = to a mean Proportional between k and l; On the other a Rightangled ▵ (Fig. 5.) whose side AB is = to a mean Proportional between f and g, and the side BC = to a mean Proportional between k and l; and on the one hand the Hypothenusa, on the other the side AC will be the value of y AC will be the value of x sought. And all by virtue of the Pythag. Theor. and according to Scholar 5. Prop. 34. or the Consectaries of Prop. 44. See Fig 4. and 5. 3. If z = i. e. extracting the Roots on both sides, z will = , and so be the second case of simple Equations. 4. If y be = , make first as l to f, so g to a fourth which call n, and by putting In for fg, you'll have y = i. e. . Make Secondly, as m to n, so h to a fourth, which call lordship; and by putting mp for nh, you'll have y = i. e. pk, and so the first case of the present Equations. 5. If in the first place the Rectangles fg and lm being turned into Squares, and collected into one Sum, make them = nn. Than (since cc and cd are multiplied by qb + bd) in like manner qb and bd added make pp; and you'll have . Thirdly (since pp is already multiplied by cc + cd) having added cc + cd into one Sum that they may e. g. make rr; x will = , and so be the third case of the present Equations. 3. In affected quadratick Equations. 1. IF zz be = az + bb, than will ; which may be thus in short demonstrated a priori; Since z − az = bb per Hypoth. and that first quantity if it be added to ¼ aa, it becomes an exact Square, the root whereof is z − ½ ha'; therefore , and consequently ; and lastly ; which last Root is a false one and lesle than nothing, but yet gives you the proposed Equation back again as well as the former; as will be evident to any one who tries, viz. having transferred ½ a on the other side, and so the two equal quantities z − ½ ha' and being squared. For here will come out ¼ aa + bb as well as if the radical Sign were affected with the Sign + because − by − giveth +. Therefore , and taking away on both sides ¼ aa, zz − az = bb i e. zz = az + bb. The value therefore of this Root will be had geometrically, by making (in Fig. 6.) CD = ½ a and DE = b, that the Hypothenusa CE may be ; and moreover, drawing out on both sides CD, and at the interval CE describing a Semicircle AEB: This being done, AD will be the value sought of the true Root z, and DB of the false one. 2. If y be = − ay + bb, than will ; which again may thus appear: Since y + aye is = bb per Hypoth. adding to both sides ¼ aa, the first quantity will be an exact Square, and . Therefore the Roots will be also equal, viz. , and consequently ; which is a false Root. The value of these Roots may be had geometrically, viz. of the true Root DB in Fig. 6. or BE in Fig. 7. and of the false one in the first AD, in the second A 3. If xx is = axe − bb, you'll have or . Which may be demonstrated after the same way a priori, as the former Cases, viz. Since x − axe is = − bb, adding on both sides ¼ aa, the former quantity will be an exact Square, viz. . Therefore the Root of the one x − ½ a = to the Root of the other, viz. , and adding on both sides ½ a, which is one of the true Roots. Or which in this case is also a true one. But the value of each may be obtained by making (Fig. 8.) CB = ½ a and by erecting BD perpendicularly = b, and making the Semicircle BEA, and drawing DE parallel to CB, and letting fall the Perpendicular OF: For thus CF will be and consequently OF, , and FB . Or, with Cartes, making (Fig. 9) CB = ½ a and BD = b, drawing DF parallel to CB, that FD may be one root and ED the other; as is manifest from the precedent Construction, and its Rule. See also another Deduction from Cartes' Constructions, Scholar 1 Prop. 47. Lib. 1. Math. Enucl. NB. 1 The ingenious Schooten has before showed this Method of demonstrating, and also of finding out these Rules in his Comment on the Geometry of Des Cartes, p. m. 163. and moreover deduces another ingenious Method for all the three Cases of these Equations, by taking away the second term in the Equation, p. 290, and the following, where we may make only this Remark concerning the third Case; that perhaps the Rule might be better deduced, if we make x = ½ a − z rather than x = z − ½ a. NB. 2. If any one has a mind to see the new Constructions of affected quadratick Equations of the Abbot Catelan, he may found them in Acta Erud. Lips. Ann. 1682. p. 86. and in the 27th Journal des Scavans. 1 Dec. 1681. IV. For Cubick and Biquadratick Equations both simple and affected, and also for all before mentioned, and consequently universally for all not exceeding the fourth Dimension. THE value of the unknown Quantity or Root may for any Case be determined by one general Rule, found out by Mr. Thomas Baker an Englishman, occasioned by what Des Cartes had taught concerning this matter, Lib. 3. Geom. p. 85, and the following; but now very much perfected by this Rule, and made more simple. Now that this Rule may be the better comprehended by Learners, we will premise these following things. 1. That all Equations occurring under those Forms which we have before shown in the Article of Reduction, or the like, must always for this purpose be so changed as to have all the terms or parts of the Equation both known and unknown, affected and not affected, brought over to one side promiscuously, and so on the other there will stand oh or naught, as e. g. let , or ; or ; or ; or etc. which also was usual to Des Cartes in Lib. 3. 2. In all Equations the known quantity or Co-efficient of the second Term we will generally denote by the Letter p, that of the third Term by the Letter cue, of the fourth by r, and the fifth (or absolute Number) by SAINT; according to Cartes, but with some little alteration: So that hence the Equations we have before been treating of, and all others like them (every where denoting the unknown quantity by x) may all be reduced to these forms: , etc. etc. 3. These and the like Equations may either occur whole, or with all their Terms, as here, or deprived of one or more of them, as the following Examples will show, where we will always put an Asterisk in the place of the deficient Term. 4. The unknown quantity in any Equation has † N. This is an error, as has been proved by Dr. Wallis. as many divers Roots or Values, as the Equation has Dimensions; which Des Cartes shows, Lib. 3. Geom. p. 69. at the same time evidently demonstrating this, viz. that some of those Roots may be false ones, i e. lesle than nothing: Which from him we here suppose. When therefore Des Cartes in his Construction of Cubick and biquadratick Equations, p. 85, and the following, requires as a necessary Condition, the ejection of the second Term in the given Equation, unless it were already wanting, and so was obliged to show a way to eject it, with several other Preparations; and afterwards, when by help of his Rule delivered p. 91. he had found a way of finding two mean Proportionals, and dividing any given Angle into three equal parts, than he use● it for solving other solid Problems, or finding two mean Proportionals, or trisecting an Angle. But the general Rule of Baker has no need of these methods or helps, neither of the Ejection of the second Term, nor any other Preparation, but immediately shows us a way, by the help of a Circle and Parabola to found all the Roots of any given Equation, both true and false, whether the Equation want any term or not, and howsoever affected, after the way we will now, and perhaps a little more distinctly, show. 1. It supposes with Cartes a Parabola NAM to be already described, (See Fig. 10. and 11.) whose Latus Rectum shall be L or 1, and its Axe aye; which Des Cartes only making use of, and never thinking of the other Diameters, was forced to take away the second Term of the Equation, etc. Baker therefore (strangely perfecting the Cartesian Geometry by this one thought) if the quantity lordship or second Term be in the Equation applies to the Axe aye (or draws an ordinate to it) BA = i. e. he erects at top of the Axe a on the right hand the perpendicular aE = and from E draws EA y parallel to the Axe aye; whereby he obtains the Diameter A y sought. 2. Having made this Preparation, the whole business depends on this, to found the Centre of the Circle to be described through the Parabola, which (by virtue of some arbitrary suppositions in the beginning) he always seeks on the left side of the Axe or Diameter, by help of two Lines or b, and DH or d; viz. by placing the former upon the Axe from a to D, if lordship be wanting in the Equation, or upon the Diameter A y from A to D if lordship be there; and letting fall from the point D the latter to add or AD perpendicularly towards the left hand. 3. He shows how to found the quantity of either of these Line (which is here very requisite) in any given Equation, by ● certain general Rule (which he calls the Central Rule, because it alone helps to found the Centre H) comprehended in these terms: Directions for the Bookbinder. The Eight half-sheet Plates that are to be folded in, and the single Leaf marked Page 89, are to be placed in those Pages of the Introduction to Specious Analysis which the figures at the top of them direct to. A SYNOPSIS of Mr. Baker's CLAVIS, to be annexed to Page 13, of the Introduction to the Specious Analysis. Of Aequations. Of Aequations. Of Central Rules. Class. I 1. . N B. The Sign ∽ denotes a dubious Case, viz. That either the Antecedent must be Substracted from the Consequent, or the Consequent from the Antecedent, according as the matter will bear. This Rule as it stands here whole, only answers to those ovations wherein are all the Terms p, cue, and razors; and in the ●●●n time may also be easily accommodated to all other Cases, 〈◊〉 observing these things. 1. Whatever Term, or Quantities ●●r, be wanting in the proposed Equation, that must also be ●●ctively omitted, or put out of the general Central Rule, that 〈◊〉 remaining quantities may determine the special or particular … ral Rule. 2. As for what belongs to the Signs, viz. whether 〈◊〉 ∽ (which latter Sign denotes a dubious Case, either that the 〈◊〉 must be substracted from the latter, or contrariwise, as 〈◊〉 matter will bear) must be put in the Central Rule, he 〈◊〉 (α) that in the Rule you'll always have , unless 〈◊〉 in the proposed Equation p and r are affected with divers … s: (β) By what Sign soever in the proposed Equation it … ens that the quantity cue is marked with, it must be noted 〈◊〉 the contrary one (altho' involved with other quantities) … e Rule; as may be seen in the application of the Rule to … ecial Cases done by the Author himself for the sake of Be … ers', and is exhibited in the Synopsis hereunto adjoining, … h yet we have thought fit to give at the end of this Trea … much more contract as to the Central Rules, in a short … pendium by way of Appendix. By these Rules therefore, the quantities of the Lines add … D and DH will be so determined, that the parts in the 〈◊〉 marked with the Sign + (taken either aggregately or 〈◊〉) will be put downwards from a to A towards y, and on … ft hand of D; but the negative Parts, or those affected … the Sign—, will be cut of, on the one part above, on … her on the right hand: Which being done the Centre H … e found. From the Centre H through the Vertex of the Axe a (if the … D is found in the Axe) or in the other Case through the 〈◊〉 of the Diameter A, you must draw a Circle which by … g or touching the Parabola will deermine the Roots sought, if the Equation be not a Biquadratick i e. has not the quantity S; otherwise another Point L or Z must be found, (vid. Fig. 12. and 13.) and a Circle described on the Radius HL or HZ, according to Des Cartes p. 86, and following, of his Geometry. 7. Viz. If you have − S, you must take on the Line H a or HA' produced, on the one side AI = L or 1, and on the other AK = , and describing a Semicircle on IK, draw ALL perpendicular to AH, to obtain the point L. (see Fig. 12.) But if you have +SS, than in another Semicircle described on AH, apply the Line AZ = to ALL found, thereby to obtain Point Z, (see Fig. 13.) 8. A Circle therefore described from H through a or A, if SAINT be wanting, but through L if there be − S, and through Z if +SS, may touch or cut the Parabola either in 1, 2, 3 or 4 Points; from which if you let fall Perpendiculars to the Axe or Diameter, you will obtain all the Roots of your Equation both true and false. 9 And, 1. If in the Equation p be wanting and − r be there, the true Roots will be on the left side of the Axe, as NOT, and the false ones as MOTHER on the right side. 2. But if there be in the Equation p and − p, the true Roots will fall on the left side of the Diameter, and the false ones on the right; but if + p, on the contrary the true will be on the right hand and the false on the left. 10. But if the Circle neither touches nor cuts the Parabola in any point, it is a sign that the Equation is impossible, and has no Root either true or false, but only imaginary ones. All which, how they may be found out, and that they are undoubtedly true, are demonstrated a posteriori, in an easy and plain way by the Author, wherefore we shall not give the Demonstrations of them here; but remit the Reader, after he has made a little progress in this Art, to the Author himself. 11. Wherhfore now, (omitting also in this place the Doctrine of the Composition of the plain and solid Geometrical Loci, or Places, which would serve for a Compliment of the Analytick Art) we will show the Practice of these Rules already delivered, premonishing only this from Mr. Baker, if the Latus Rectum be made Unity, that L in the Central Rules and all its Powers may be omitted, and so the Rules exhibited more compendiously, as we have already done in our Synopsis, and may be seen from the form of a general Central Rule hereunto annexed. To which Premonition of Baker we may also add this, if any given Line in the Problem itself be taken for Unity, which may be often very commodiously done [as a in the former Problem p. 91. Geom. Cartes, and the Line NOT in the latter, and a again in the Equation p. 83. the last line] and than the same Line also may be taken for the Latus Rectum of the Parabola to be described, if we have a mind to make use of this Compendium for abbreviating the Central Rules. For otherwise if we would construct all Problems, as Baker rightly asserts we may, by only one Parabola, we shall fall often into very tedious Prolixities. SOME EXAMPLES OF SPECIOUS ANALYSIS, In each kind of Equations. I In Simple Equations. PROBLEM I HAving the sum of any two sides given for forming a Triangle ABC, to found each of the sides, and form the Triangle. suppose e. g. three Lines given in Fig. 14. the first = AB+AC in the Triangle sought, the second = AB+BC, the third = BC+AC, to found each of the sides e. g. to found AB, which being known, the rest will be so also. SOLUTION. 1. Denomination. Make AC+AB = a; AB+BC = b; BC+AC = c; AB = x; than will AC = a − x, and BC = b − x, and so the Denomination be complete. 2. Equation. Now if the values of the two last Lines BC and AC be added into one Sum, which we had before given; you'll have this Equation a + b − 2 x = c. 3. Reduction. By adding on both sides 2 x, you'll have ; and substracting from both sides ; and dividing both sides by . 4. The Effection or Geometrical Construction, which the Equation thus reduced will help us to Join A = a and ED = b in one Line AD, and from this backwards cut of DF = c; and divide OF which remains into two equal parts in B, and you'll have AB the first side of the Triangle to be form; and BE will give the other side AC, which substracted from ED, will leave GD = to the third side BC; of which you may now form the Triangle ABC. 5. A general Rule for Arithmetical Cases. Add the two former Sums, and from the Aggregate subtract the third Sum; half the Remainder will give the side AB common to the two former Sums. For an Example take this Question: There are three Towns of ancient Hetruria, viz. Forum Cassii (which the Letter A denotes in ▵ ABC) Sudertum (B) and Volsinii (C) which are at this distance one from another; if you go from Volsinii to Forum Cassii and thence to Sudertum, you mst go 330 Furlongs; from Forum Cassii to Sudertum and thence to Volsinii there are 306 Furlongs; lastly, from Sudertum to Volsinii and thence to Forum Cassii 272 Furlongs. How far is each Town distant from each other. PROBLEM II IN a rightangled Triangle ABC, having given the Base AB, and the difference of the Perpendicular AC and the Hypothenusa BC to found the Perpendicular and Hypothenusa, and form the Triangle. Make e. g. the Base AB (Fig. 15.) and the difference of the Perpendicular and Hypothenusa BD, to found the Perpendicular AC; which being known, the Hypothenusa AC will be known also, if the given difference be added to the found Perpendicular. SOLUTION. 1. Denomination. Make AB = a, BD = b, AC = x; than will BC = x + b. 2. Equation by the Pythagorick Theorem, , viz. the two Squares of the Sides to the Square of the Hypothenusa. 3. Reduction. Substracting from both sides xx, you'll have ; and moreover by substracting also ; and dividing by . 4. Effection or Geometrical Construction. Having described upon the given Base AB a Semicircle, apply therein the given difference BD, and draw AD, whose Square is = aa − bb. Since this must be divided by 2 b, make, as A = 2 b to AD = , so AD = to AC the Perpendicular sought. To which if you add CF = BD, you will have OF = to the Hypothenusa sought BC; which will come of course together with the whole Triangle sought, if the found Perpendicular AC be erected at right Angles on the given Base AB. 5. The Rule for Arithmetical Cases. From the square of the given Base subtract the square of the given difference, and divide the Remainder by the double difference; and you'll have the Perpendicular sought. E. g. suppose the Base = 20 foot, and the difference between the Perpendicular and Hypothenusa 10. PROBLEM III IN the right angled Triangle ABC, having given the side AC and the sum of the other side AB and the Hypothenusa BC, to found the other side and the Hypothenusa separately, and form the Triangle. Suppose the given side (that is to be) AC (Fig. 16.) and the sum of the other sides AD, to found the side AB, which being known the Hypothenusa BC will be known also. SOLUTION. 1. Denomination. Make AC = a, AD = b, AB = x, than will BC = b − x. 2. Equation. and substracting xx. 3. Reduction. ; and adding ; and substracting ; and dividing by . 4. The Effection or Construction is like the former, and so will be manifest only by inspecting that Scheme. 5. The Arithmetical Rule. From the square of the given sum subtract the square of the given side, and divide the Remainder by double the given sum; and you'll have the other side, and substracting that from the given sum, you have the Hypothenusa also. E. g. let one side be 15, and the sum of the other two 45. PROBLEM IU. HAving given the Perpendiculars and sum of the Bases of two rightangled Triangles having equal Hypothenuses, to found the Bases separately, and form the Triangles. Suppose e. g. to form the Triangle ABC (see Fig. 17.) you have given the Perpendicular AB, and for the other ▵ ADC, the Perpendicular CD, and the given sum of the Bases BE, to found the Bases singly, viz. the lesle for the greatest Perpendicular, and the greater AD for the lesle Perpendicular. SOLUTION. 1. Denomination. Make AB = a, CD = b, the sum BE = c; make the lesser Base BC = x; the greater AD will = c − x. 2. Equation. Since the Hypothenuses of the two Triangles are supposed equal, the two □ □ AB+BC, i e. xx + aa will be = to the two □ □ AD+CD, i e. . 3. Reduction. By taking away therefore xx and adding 2 cx, aa+ + cx will = bb + cc; and further taking away from both sides ; and dividing both sides by . 4. Geometrical Construction. Join the Lines b and c CD and BE at right Angles, (n. 2.) and the square of Hypothenusa DE will = bb + cc. Upon this Hypothen● having described a Semicircle, apply therein the Line AD, ● the square of A will = bb + cc − aa. Which, since it m● be further divided by 2 c, make (n. 3.) as BF = 2 c to ● = , so BG to BC, the lesser Base sought. 5. The Arithmetical Rule. From the sum of the squa● of the lesser Perpendicular and the sum of the Bases subtract 〈◊〉 square of the greater Perpendicular, and the Remainder di●ded by the double sum of the Bases, will give the lesser Barnes E. g. make AB 76, CD 57, and BE 114. PROBLEM V HAving given the Perpendiculars of two rightangled T●angles standing on the same given Base, to found the Se●ments of the Hypothenuses. E. g. suppose the common give Base be AB (Fig. 18.) and the Perpendicular of one Triang● AD of the other BC; to found the segments of the Hypothen●ses, cutting one another geometrically. SOLUTION. If a Geometrical Solution be required there is no need 〈◊〉 any Analysis; for having erected perpendicularly on the common given Base AB the given Perpendiculars AD and B● the Hypothenuses AC, BD being drawn immediately, exhibe their Segments EA, EBB, EC, ED. But if it be to be do arithmetically by a general Rule, than an Analysis will be necessary. 1. Denomination. Make the common Basis AB = a, B● = b, AD = c; and, having found OF, all the rest may had (for as AB to BC so OF to FE; which being given, y● have also GD and HC, and consequently also DE, CE, & ● make OF = x, than will BF or HE = a − x. XVIII XIX XX XXI XXII XXIII XXIII XXIV XXV XXVI XXVII XXVIII As BASILIUS to AD so BF to FE. . Therefore . Reduction. Multiplying both sides by a you'll have bx, c − cx; and adding on both sides cx, bx + cx = ac dividing both sides by . The Arithmetical Rule. Multiply the commom base ●●e least Perpendicular, and divide the Product by the sum ●e Perpendiculars; and you'll have the lesser segment of the 〈◊〉 which being given you'll have all the rest. E. g. sup● AB = 10, BC = 9, AD = 6. PROBLEM VI. OH inscribe a Rhombus in a given Oblong, i e. having the sides of the Oblong AB and BC given, (Fig. 19) to found Segment BF or DE, which being cut of, the remainder FC ●E will be the side of the Rhomb sought. SOLUTION. Make AB = a, BC = b, BF = x: FC or FAVORINA will be = x (so far the Denomination.) Therefore the square of FAVORINA, ●ch is bb − 2 bx + xx will be = aa + xx, viz. to the two ●res of AB and BF (so far the Equation;) and substracting both sides ; and , and ; and . (so far the Reduction.) The Geometrical Construction. Having described a semi●le on BC (n 2.) apply CD or AB, and the □ BD will be bb − aa. Which since it must be divided by 2 b, make, ●E = 2 b to BD = , so BD to BF sought, and ●e cut of the side of the Oblong BC (n. 1.) The Arithmetical Rule. From the square of the greater side subtract the square of the lesser, and divide the remainder by double the greater side, and the quotient will give the Segment BF sought. E. g. suppose AB = 4, and BC = 8. PROBLEM VII. TO inscribe the greatest square possible in a given Triangle, i e. having given the height of the Triangle CD (Fig. 20) and the Base AB, to found a portion of the altitude CE, which being cut of there shall remain ED = FG. SOLUTION. Make the base AB = a; the altitude CD = b, CE = x; than will ED or FG = b − x. By reason of the similitude of the Triangles ABC and FGC you'll have as AB to CD so FG to CE. Therefore the Rectangles of the means and extremes will be equal, i e. axe = bb − bx; and adding on both sides bx, ax + bx = bb, and dividing by . Construction. Upon the side of the Triangle CB produced, make CHANGED = b, and HI = a, so that the whole Line shall be a + b. And having joined ID and parallel to it HE drawn from H, the part CE will be cut of, which is that sought. For as CI to CD so CHANGED to CE, according to the second case of simple Effections. Arithmetical Rule. Square the given height of the Triangle, and divide the Product by the sum of the base and altitude; and the quote is the part to be cut of CE. E. g. suppose CD = 10, and AB = 15. PROBLEM VIII. IN an acuteangled Triangle having all the sides given, to found the Perpendicular that shall fall from the Vertex on the Base, i e. having given AB, AC, BC (Fig. 21.) to found AD or BD (for having found the one you may easily found the other) Coral. Prop. 13. Lib. 2. Eucl. SOLUTION. If there be only required a Geometrical Construction of this Problem, there will be no need of any Analysis; for having form a Triangle ABC of the three given sides, you need only let fall the perpendicular AD from the Vertex A, which would determine the Segment BD. To found the general Arithmetical Rule, which is the general Corollary of Euclid, or if any one for exercise sake had rather determine the Perpendicular DAM by the segment of the Base BD, than the latter by the former, the Analysis will proceed thus: Make AB = a, BC = b, AC = c, BD = x; than will CD be = b − x. Wherhfore by the Pythagorick Theorem □ AD = aa − xx, and by the same reason the same □ AD = . Therefore ; and by adding on both sides ; and by transferring , and dividing by . The Arithmetical Rule. Subtract the square of the lesser side from the Sum of the □ □ of the base and greater side, and the remainder divided by double the base will give its greater segment: If the □ of the greatest side be substracted from the sum of the other squares, etc. you will have the lesle segment CD. Geometrical Construction Having described a semi circle upon AB (n. 2.) apply therein AC, and the □ BC will = aa − ccx, and continuing AC to B, till C B be = CB (n. 1.) the □ of BB will = aa − cc + bb; which since it is to be divided by 2 b make as BE = 2 b to BB = , so BF = BB to BD the segment sought = BD n 1. PROBLEM IX. IN an obtuse angled Triangle having the thr●e sides given to found the Perpendicular let fall from the Vertex to the Base being continued: i e. Having given AB, BC, AC (Fig. 22. n. 1.) to found AD or CD (for the one being found the other will readily be so also) Coral Prop. 12. Lib. 2. Eucl. SOLUTION. What we premonished about the former Problem, we understand to be premonished here also. For the rest make here also AB = a, BC = b, AC = c, CD = x; than will BD = b + x: Wherhfore by the Pythagorick Theorem the □ AD will = cc − xx, and by the same Theorem the same □ AD = . Therefore ; and adding to both sides ; and transposing cc and ; and dividing by . The Arithmetical Rule. Subtract from the square of the greater side the sum of the squares of the base and lesser side; and the Remainder divided by double the base will give its continuation to the Perpendicular. Geometrical Construction from the Equation reduced: Having described a semicircle upon AB (n. 2.) apply therein AC, and the □ of CB drawn will ; which since it must be divided by 2 b make, as CF = 2 b to CE , so CE to CD the segment sought, n 1. PROBLEM X. Commonly ascribed to Archimedes. THE Diameter AB of a given semicircle (Fig. 22 n. 1.) being any how divided in L, and from L erecting a Perpendicular LX, and upon the segments LA and LB having described two other semicircles, whose semidiameters are also given as well as that of the great Circle CB; to found the Radii FM and Vy of the little Circles that are to be so described, that they shall touch the Perpendicular LX, the Cavity of the greater semicircle, and the Convexities of the lesle. SOLUTION. I For the Radius FM. 1. Denomination. Make CB = a, EBB = b than will CE = a − b; for which for brevity's sake put c. And let FM or FN or FK = x: Therefore OF will be = b + x, and CF (substracting FK from CK) = a − x. Wherhfore now you'll have at lest the names of the three sides in the ▵ CFE, so that according to Poblem 8. the Segment of the base GE may be determined (which indeed is determined already, as being = LE − LG or MF i e. b − x) for which in the mean time we will put y; and now will CG = c − y. 2. For the Equation. If the □ GE = y be substracted from the □ OF = , you'll have the square of the Perpendicular FG = ; and, if □ CG = be substracted from the □ CF = , you'll have the same □ of the Perpendicular FG = . Therefore . 3. Reduction. And taking from both sides the quantities xx and yy, ; and adding 2 axe and cc, but taking away aa from both sides, ; and adding 2 axe and cc, and taking away from each side ; and dividing by . but the same y or EGLANTINE is = EL − MF i e. b − x. Therefore ; which is a new and more principal Equation: And multiplying both sides by 2 c (you have a new Reduction) ; and adding 2 cx, and transposing the others, ; and dividing by . The Geometrical Construction of this first Case. Add the determinate (n 2.) quantity 2 bc into one sum with the quantity aa, as in the beginning, n. 3. Than from this sum subtract successively the quantities bb and cc, and there will come out (the same n 3) FH, whose □ : Which since it must be divided by , make (the same n. 3) as FI = to FH = , so FH to FM the Radius sought of the little Circle to be described. This quantity FM being thus found, place it from L to G (n. 1.) and from G erect a Perpendicular, which being cut of at the interval CF (which may be had, if from CB or CK you cut of FK = FM) or from E at the interval OF (which is composed of the Radii EN and FN) gives the Centre of the little Circle to be described. The Arithmetical Rule. Add twice the □ CEB to the square of the greatest semi-diameter CB, and from the sum subtract the Aggregate of the □ □ CE and EBB; divide the remainder by the sum of all the three Diameters, (AB, ALL and LB) i e. by double the greatest AB; and you'll have the Radius FM, etc. For Example sake let a be = 12, b = 4; c will be = 8, and x will be produced = 2 ⅔. II For the Radius Vy by help of the obtuse-angled ▵ DVC. 1. Denomination. CA = a as above, DA or DL = b, and putting x again for the sought FIVE y or VK, CV will be = a − x, DL or DOCTOR = b, and consequently DV = b + x, and DC = a − b, for which for brevity's sake we will put c. Now you'll have at lest in Denomination in the ▵ CUD the three sides, so that aecording to Problem 9 the segment CW may be determined, for which in the mean while we will put y; than will DW = c + y, which is the same as DL − WL or FIVE y i e. b − x. 2. For the Equation. If the □ CW = yy subtract it from the □ CV = , and you'll have the □ of the Perpendicular VW, ; and if the □ DW = , subtract it from the □ DV = , and you'll have the same □ of the Perpendicular VW = Therefore . 3. Reduction. Therefore taking from both sides xx and yy, ; and adding 2 cy and 2 axe, ; and substracting , and dividing by 2 c, But if you add to the same y or CW DC = c, you'll have DW = i. e. reducing this c to the same Denomination, = DW. But the same DW = DL − WL = b − x. Therefore , and multiplying , and adding 2 cx, and transposing the rest, , and dividing by just as above in the first Case. 4. The Geometrical Construction therefore will be the same as there. See Fig. 23. n. 4. and 5. 5. The Arithmetical Rule is also the same, but the given quantities in this Example, which the figure of the Problem will show, thus vary, while a remains 12, b will be 8, and c 4, from which data (or given quantities) there will notwithstanding come out again, for x or the Radius V y 2 ⅔. II Some Examples of simple or pure Quadratick Equations. PROBLEM I TO make a Square equal to a given Rectangle; i e. having given the sides of the Rectangle, to found the side of an equal Square, Eucl. Prop. 14. Lib. 2. Suppose e. g. the given sides of the Oblong to be AB and BC (Fig. 24.) to found the Line BD whose square shall be equal to that Rectangle. SOLUTION. Make AB = a and BC = b, and the side of the square sought = x, and the Equation will be ab = xx; and extracting the root on both sides . Geometrical Construction. Join AB and BC in one right line, and describing a semicircle upon the whole AC, from the common juncture B erect the Perpendicular BD which will be the side of the square sought, according to Case 1. of the Effection of pure quadraticks. Arithmetical Rule. Multiply the given sides of the Oblong by one another, and the square root extracted out of the Product will be the side of the square sought. PROBLEM II THE square of the Hypothenusa in a rightangled ▵ being given, as also the difference of the other two squares to found the sides. E. g. If the Hypothenusa be BC (Fig. 25.) and the difference of the squares of both the legs, and consequently its Leg also BE given (for the squares being given the sides are also given geometrically) to found the sides of the rightangled ▵ which shall have these conditions; or more plainly, to found one side e. g. the lesser which being found, the other, or the greater, will be found also. SOLUTION. Let the □ of the given Hypothenusa = aa, and the square by which the two other differ = bb. Let the lesle side = x, and it's □ = xx. Wherhfore the greater will be xx + bb. And since the sum of these is = to the □ of the Hypothenusa, you'll have ; and substracting ; and dividing by . Therefore . Geometrical Construction. Having described a semicircle on BC, and applied therein BE, the □ EC will = aa − bb; and having described another semi circle upon EC divided into two Quadrants the □ DC will be , and so DC = or the side sought; which being also transferred upon the other semicircle described on BC, viz. from C to A gives the other side AB and the whole ▵ sought. The Arithmetical Rule. From the square of the Hypothenusa subtract the given difference, and the square root extracted out of half the remainder gives the lesser side of the ▵ sought. PROBLEM III HAving an equilateral ▵ ABC given (Fig. 26. n. 1.) to found the Centre and Semi diameter of a Circle that shall circumscribe it. i e. Found the BD the side of an Hexagon that may be inscribed in it. For if we consider the thing as already done; it will be manifest that the side of the Hexagon BD will fall perpendicularly on the side of the ▵ AB, as making an angle in a semicircle, so having bisected the Hypothenusa DA you'll have E the Centre sought. SOLUTION. Make the side of the Triangle AB = a, BD = x, than will AD = 2 x. Since therefore the square BD i. e. xx being substracted out of the square AD i e. 4 xx, there remains the square AB 3 xx, you'll have the Equation ; and dividing by 3 ; therefore The Geometrical Construction. Having produced AB (n. 2.) to FLETCHER a third part of it, the square of a mean proportional BD between BF and BASILIUS will be ⅓ aa or , and so the Line BD = . Therefore the Hypothenusa DAM being divided in two in E, or at the interval BD, making the intersection from B and A, you'll have the Centre sought. The Arithmetical Rule. Divide the square of the given side into three equal parts, and the square Root of a third part will give the semi-diameter A or BE sought, by the intersection of two of which you'll have the Centre. PROBLEM IU. HAving given, in a rightangled Parallelogram, the Diagonal, or for a rightangled ▵, the Hypothenusa and the proportion of the sides, to found the sides separately and construct the Parallelogram or ▵. Suppose e. g. the given Diagonal to be AB (Fig. 27. n 1.) and the given reason of the sides as AD to DE, to found the sides. SOLUTION. Make AB = a, the reason of AD to DE as b to c; make the lesser side = x, than will the greater be . For the Equation, the □ □ of the sides are , □ AB; and multiplying both sides by bb, ; and dividing by bb + cc, . Therefore i. e. Extracting the roots as far as possible Another Solution. Call the name of the given reason e, so that assuming any line for unity, the value of e may also be expressed by a right line, which shall be equal e. g. to DE above. Wherhfore because we make the lesle side x, the greater will be ex, and so i. e. dividing by i. e. The Geometrical Construction. The last Equation above being reduced to this proportion as the to b so a to x, make (n. 2) AD and DE at right angles, and A will be = , and continuing A and AD make as A to AD so AB to AC the lesser side sought. Having therefore drawn BC which determines the lesser side AC, the greater side and so the ▵ ABC will be already form, and may be easily completed into a Rectangle. In the other Solution the last Equation agrees with the precedent (for it gives us this proportion as to 1 so a to x in which 1 is = b, and ee = cc by what we have supposed) and so the Construction will be the same. The Arithmetical Rule may be more commodiously expresed by this last Equation under the last form but one, after this way, divide the □ of the Diagonal by the □ of the name of the Reason lessened by unity, and the root extracted out of the Remainder is the lesser side sought. PROBLEM V. (Which is in Pappus Alexandrinus, and in Cartes' Geometry, p. 83. in a Biquadratick affected Equation, and p. 84. he gives us thereon a very remarkable Note.) HAving given the Square AD (Fig. 28) and a right line BN, you are to produce the side AC to E, so that OF drawn from E towards B shall be equal to BN. It will be evident, if you imagine a semicircle to pass through the points B and E, that the most commodious way will be to found the line DG, that you may have the Diameter BG; upon which having afterwards described a semi circle, there will be need of no other operation to satisfy the question, than to produce the side AC till it occur to the prescribed Periphery. SOLUTION. (As found by Van Schooten, p. 316. in his Comment on Cartes' Geometry, which we will here give somewhat more distinct.) 1. Denomination. Make BD or DC = a, BN or FE = c, BF = y, and DG = x; the Perpendicular EH will be = a, and EGLANTINE = BF, viz y (because the ▵ EHG is similar to ▵ BDF, by n. 3 Schol 2 Prop. 34. Lib. 1. Math●s. Enucl. and BD in the one = to EH in the other) and BG = a + x, BE = y + c; and BH will have its Denomination, if you m●ke (by reason of the similarity of the ▵ ▵ BFD and ●EH) as BF to BD so BE to BH . and you'll have also HG = , i e. having reduced them all to the same Denomination, , i e. . Having therefore named all the lines you have occasion for, you must found two Equations, because there are assumed two unknown quantities, viz. x and y. 2. For the first Equation and its Reduction. By reason of the similiarity of the ▵ ▵ BGE and BEH, as BG to GE so BE to EH : Therefore the Rectangle of the Extremes will be = to the Rectangle of the means, i e. ; and taking from both sides . 3. For the second Equation and its Reduction. Since BH, HE and HG are continual proportionals, the Rectangle of the extremes are equal to the square of the mean, i e. ; and multiplying both sides by yy, and dividing by , and taking away ayy and transposing the rest, ; and dividing by ; i. e. dividing actually as far as may be by . 4. The comparison of these two Equations thus reduced▪ gives a third new one, in which there will be only one unknown quantity, viz. ; and adding to both sides cy, ; and multiplying by ; i. e. and dividing both sides by ; and adding . Therefore . 5. The Geometrical Construction, which is the same Pap● prescribes in Cartes, viz. having prolonged the side of 〈◊〉 square BASILIUS to N, so that BN shall be = to a given right li● since BASILIUS is = a, and BN = c, the Hypothenusa DN will . Having therefore made DG = DN, a● described a semicircle upon the whole line BG, if AC be prolonged until it occur to the Periphery in E, you'll have do that which was required. PROBLEM VI (Which Van Schooten has in his Comment, p. m. 150, and following.) HAving given a right line AB, from the ends of it A an● B (Fig. 29.) to inflect two right lines AC and B● which shall contain an angle ACB = to the given one D, an● whose squares shall be in a given proportion to the Triang● ACB, viz. as 4 d to a. Viz. You must determine the point C, which the two rig● lines AH and HC or EH and HC will do, assuming the middle point E in the line AB. Wherhfore here will be two unknown quantities HE and HC, and consequently two Equ●tions to be found in the Solution; one whereof the giv● proportion in the Question supplies us with, and the other 〈◊〉 have from the similar Triangles AIC and GFD, which represent equal angles. SOLUTION. XXIX XXX XXXI XXXII XXXIII XXXIV XXXV XXXVI XXXVII And the ▵ ACB will be = aye: And since the ▵ ▵ ● and AIC are similar, and the sides of the former FD and arbitrary, so that for FD we may put b and for FC, c; ●he sides of the latter are determined by the similitude of ▵ ▵ ABI and HDB, as being rightangled ones, and ha● the common angle B; they will be obtained by making the Hypothen. BC to the Hypoth. AB so the base HB to ●ase BY, i e. , whence substracting BC = e, there remains CI = . For the first Equation, by virtue of the Problem as to aye. And the Rectangle of the ●mes is = to the Rectangle of the means, i e. . For the other Equation, since as DF to FG so CI to AI the Rectangle of the extremes will again be = to the Re●gle of the means, i e. ; multiplying both sides by ; ●h is the second Equation. The Reduction of both Equations. 〈◊〉 first was . ●refore dividing by . substracting . ●he latter Equation was , i e. ●ituting again the value ee, which was ; and by transposition, ; and dividing by , or , or (putting 2 f for ) . Wherhfore we have the value of xx twice expressed, but by quantities partly unknown, because that is found on both sides. Wherhfore now we must make a new comparison of their values, whence you'll have this new 5. Third Equation, in which there is only one of the unknown quantities: ; and adding on both sides both yy and aa, ; or dividing by 2, ; and transposing ; and dividing by ; which is the value of the quantity y in known terms. But this value in one of the precedent Equations, viz. in this , being substituted for y and its square for yy, will give ; i. e. all being reduced to the same denomination, ; and . 6. The Geometrical Construction, which Schooten gives us p. 153. Having made the angle KAB (n. 2. Fig. 29.) equal to the given one D, erect from A, ALL perpendicular to KA, meeting the Perpendicular EM in L; and from the Centre L, at the interval of the given right line d, describe a Circle that shall cut KA and EL produced to KING and M. Than assuming EN = KA, join MALFORT, and from N draw NH parallel to it, which shall meet AB in H. Afterwards, having described from L, at the interval LA, the segment of a Circle ACB, draw from H, HC perpendicular to AB meeting the circumference in C, and join AC, CB. NB. The reason of this elegant Construction, which the Author concealed, for the sake of Learners we will here show. 1. Therefore, he reduced the last Equation (extracting the root, as well as it could bear, both of Numerator and Denominator) to this: multipl. by , so that after this way the Construction would be reduced to this proportion, as d + f to a so . 2. He made the angle KAE = to the given one D, and the angle KAL a right one, so that having described the segment of a Circle from L the inscribed angle will also be made equal to the given one, according to the 33. Lib. 3. Eucl. 3 By doing this, EL expresses the quantity f, since by reason of the similarity of the ▵ ▵ KOA s. GFD, n. 1. and AEL (for the angel's LAE and AKO are equal, because each makes a right one with the same third Angle KAO) you have as KO to OA so A to EL i e. 4. Making now LM and LK = d you had EM = d + f, and AK = (for the □ ALL is = to aa + ff, which being substracted from □ LK = .) 5. Wherhfore there now remains nothing to construct the last Equation above, but to make EN = AK, and to draw HN parallel too AM; for thus was the whole proportion as EM to EA so EN to EH to x. Q. e. f. For the point H being determined, a perpendicular HC thence erected in the segment already described defines the Point C, which answers the Question. PROBLEM VII. HAving given the four sides of a Quadrangle to be inscribed in a Circle, to found the Diagonals and their Segments, and so to construct the Quadrangle, and inscricbe it in the Circle. As e. g. suppose the given sides are AB, BC, CD, DA (Fig. 30 n. 1.) which now we suppose to be joined in●o a quadrangle inscribed in the Circle, the Diagonals also AC and BD being drawn (n. 3.) to found first the segments of the diagonals A e, B e, etc. which being had, the Construction is ready. SOLUTION. Denomination. Make AB = a, AC = b, CD = c, DAMN = d, A e = x [for this segment alone being found, the rest will be found also, as will be evident from the process.] Since therefore the vertical angles at e are equal, and likewise the angles in the same segment BCA, BDA, also DAC, DBC, etc. are equal, the Triangles A eD and B eC, also A ebb and C eD are similar: wherefore it will follow that, 1. As DA to A e so CB to B e 2. As AB to B e so CD to C e 3. As AB to A e so CD to D e Therefore the whole Diagonal AC will be = and BD = . 2. The Equation. But now by Prop. 48. Lib. 1. Math. Enucl. the Rectangle of the Diagonals is equal to the two Rectangles of the opposite sides. Diagoras. AC, Diagoras. BD, Therefore the □ of the Diagonals . 3. Reduction, i e. taking (a) for unity ; i. e. the quantities on the left hand being reduced to the same denomination. ; and multiplying both sides by dd, ; and dividing both sides by d, ; and than dividing both sides by ; i. e. in the present case, where b by chance happens to be = a, Therefore or in our case 4. The Geometrical Construction, which, by supposing a (and in the present case also b) to be unity, aught to determine, 1. The quantities cd, , cc, and their aggregate with unity. 2. The aggregate of cd and dd. 3. To divide the one by the other. And, 4. To extract the root out of the quotient, or also to extract the roots first out of each quantity, and divide them by one another; which may all of them be separately done in so many separate Diagrams, but more elegantly connected together after the following or some such like way. 1. Join AD and DC (n. 2.) into one line, and having described a semicircle thereupon, erect the Perpendicular DE; and the line A drawn will = . 2. Making the angle CAG at pleasure, make OF = AB, and draw CG parallel to the line DF; so FG will be = . Now if, 3. in the vertical angle you make AH = CD, the line HI drawn parallel to DF will cut of AI = cd. 4. In AK erected = to AB, if you take ALL = AH or CD, and draw LM parallel to KH, you'll have AM = cc. 5. Having prolonged AGNOSTUS to N and AH to OH, so that GN shall be = AI+AM and AO = AB or AK, and having described a semicircle upon the whole line NOT, a perpendicular erected AP will be = and so, 6. if AQ be made = A and ARE = OF or AB, and you draw a line RS from R parallel to PQ; AS will be = x, i e. the segment sought A e of the Diagonal AC; which being given, by force of the first Inference premised in the Denomination above, by drawing DS and, having made DT = BC, TV parallel to it; you'll have also the other segment B e = SV and by their Intersection on the line AB (n. 3.) the point e, through which the Diagonals must be drawn which will be terminated by the other given sides, and thence you'll have the quadrilateral figure ABCD sought, to be circumscribed about the Circle, according to Consect. 6 Defin. 8. Mathes. Enucl. NB. Unless we had here consulted the Learner's ease, the artifice of this Construction might be proposed after a more short and occult way, thus: Make DE a mean proportional between AD and DC, and draw A Than having made any angle CAG, make OF = AB, and at this Interval describe the circle FROK, and draw CG parallel to DF. Moreover in the opposite vertical angle, having made AH = CD, draw HI parallel to DF, and having erected the perpendicular AK, and thence the abscissa ALL = AH, make LM parallel to HK, and thence having prolonged AH to OH, and GN being made equal to AI+AM, make AP a meam proportional between AO and AN, cutting the hidden circle in R; and lastly having made AQ = A, if RS be drawn parallel to QP, you'll have AS the value of x sought, etc. III Some Examples of Affected Quadratick Equations. PROBLEM I HAving given, to make a right angled Triangle ABC, the differences of the lesser and greater side, and of the greater, and the Hypothenusa, to found the sides separately and form the Triangle. E. g. Having given the right line DB (Fig. 31.) for the difference of the perpendicular and base, and CE for the difference of the base and Hypothenusa, to found the perpendicular AC, which being found, you'll have also, by what we have supposed, the base AB, and the hypothenusa BC. SOLUTION. Make the difference DB = a, CE = b; put x for the perpendicular; the base, which is greater than that will be x + a and the Hypothenusa x + a + b. Therefore by virtue of the Pythagorick Theorem, ; and substracting from both sides . Wherhfore by the first case of affected quadratick Equations . Construction. Found a mean proportional AK between AH = 2 b and AI = a (n. 2.) (Fig. 31.) and having made both OF and AG = b, place the Hypothenusa KF from ALL, and cut of GC equal to the hypothenusa GL; thus you'll have AC the perpendicular of the Triangle sought, and adding DB you'll also have the base AB, and from thence having drawn the hypothenusa BC, it Will be found to differ by the excess required CE. The Arithmetical Rule. Join twice the product of the differences multiplied by one another, to twice the square of the difference of the base and the hypothenusa; and if the square root of this sum being extracted be added to the aforesaid difference, you'll have the perpendicular sought. Suppose e. g. both the differences of CE and DB = 10. PROBLEM II IN a rightangled ▵ having given the Hypothenusa and sum of the sides, to found the sides. E. g. If the Hypothenusa BC be given (Fig. 32.) and the sum of the sides CAB, to found the sides AB and AC separately, to form the Triangle. SOLUTION. Make the Hypothenusa BC = a, the sum of the sides = b. Make one side e. g. AB = x, than will the other side AC be = b − x. Therefore ; and adding 2 bx, and taking away ; and dividing by . Therefore according to Case 1. of affected Quadraticks. i. e. or . The Geometrical Construction. Having described a semicircle upon BD = BC so a apply therein the equal lines BE and DE, and having described another semicircle on BE apply therein BF = ½ b, to be prolonged farther out. Lastly, if another little semi circle be described at the ●nterval OF, the whole line AB will be the true root or the side sought, and GB the false root, etc. The Arithmetical Rule. From half the square of the Hypothenusa subtract the fourth part of the square of the given sum, and the root extracted out of the remainder, if it be added to half the sum, will give one side of the Triangle; and substracted from the given sum, will give also the other Suppose e. g. BC to be 20, and the sum of the sides 28. PROBLEM III HAving given again in the same ▵ the Hypothenusa, as above, and the difference of the sides DB (Fig. 33.) to found the sides. SOLUTION. Make the lesle side x, the difference of the sides = b; the greater side will be x + b. Let the Hypothenusa be = a. Therefore, ; and taking away ; and dividing by . Therefore by case 2, i. e. . The Geometrical Construction. Having described a semicircle upon BD = BC or a, apply therein the equal lines BC and DC, and having described another semicircle on DC apply in it DF = ½ b, and if at the same interval you cut of FAVORINA from FC, the remainder AC will be the lesser side sought, etc. The Arithmetical Rule. From half the square of the Hypothenusa subtract the square of half the difference, and if you take half the difference from the root extracted out of the remainder, you'll have the lesser side of the Triangle required, and by adding to it the given difference you'll have also the greater. E. g. Let the Hypothenusa be 20, and the difference of the sides 4. PROBLEM IU. HAving given the Area of a rightangled Parallelogram, and the difference of the sides to found the sides. E. g. If the Area is = to the square of the given line DF, and the difference of the sides ED (Fig. 34.) to found the sides of the rectangle. SOLUTION. Make the given Area = aa, the difference of the sides = b, the lesser side x; than the greater will be x + b. Therefore the Area xx + bx = aa; and substracting bx xx = − bx + aa. Therefore according to case 2, . The Geometrical Construction. Join at right angles AG = a, and GH = ½ b, and having drawn AH and prolonged it, describe the little Circle at the interval GH: so you'll have A the lesser side, and AD the greater of the Rectangle sought, etc. The Arithmetical Rule. Add the given Area and the square of half the difference, and having the sum, subtract and add the difference from or to the root extracted, and so you'll have the greater and lesle sides of the rectangle. PROBLEM V HAving given for a rightangled Triangle the difference of both the Legs from the Hypothenusa, to found the sides and so the whole Triangle. E. g. Suppose the difference of the lesle side to be BD and of the greater DE (n. 1. Fig. 35.) to found the sides themselves, and so make the Triangle. SOLUTION. For BD put a, for DE, b. Let the greater side be x; the Hypothenusa will be x + b; therefore the lesser side will be x + b − a. Now the □ □ of the sides are = to the □ of the Hypothenusa, i e. 2 xx − 2 ax+ + bx + bb − 2 ab + aa = xx+ + bx + bb; and taking away aa, ; and adding 2 axe and 2 ab, and taking away . Therefore i. e. . The Geometrical Construction. Between the given differences BD and DE▪ found (n 2.) a mean proportional DF▪ and join to it at right angles the equal line FG, and cut of DH equal to DG; and so you'll have BH the greater side of the triangle sought. This being prolonged to C, so that HC shall be = b, having described a semicircle upon the whole line BC apply therein BA = BH; and having drawn AC, the Triangle sought ABC, will be form. The Arithmetical Rule. If the square root extracted from the double rectangle of the differences be added to the greater difference, you'll have the greater side sought, etc. PROBLEM VI. HAving given, to make two unequal Rectangles, but of equal height, the sum of their Bases with the Area of … her (viz. the greater,) and the proportion of the sides of the … her (viz the lest,) to found the sides separately. E. g. Let the sum of the bases be AB (n. 1. Fig. 36) and the square of the line BC = to the Area of the greater rectangle; and let the sides of the lesser rectangle be to one another as CD to DE: To found the sides of both the rectangles; i e. to found the common altitude, which being found the other sides will be easily obtained from the Data; or to found the base of the greater which, with the same ease, will discover the rest. SOLUTION. Make AB = a, and the Area of the greater rectangle = bb; ●●d the proportion of the altitude to the base in the lesser, as c ●d; to found e. g. the greater base which call x. Therefore 〈◊〉 common altitude will be = , and the base of the lesser rectangle = a − x. Wherhfore you'll have for the Equation, as c to d so to a − x. Therefore ac − cx = ; and multipl. by x, acx − cxx = bbd; and adding cxx and taking away bbd, acx − bbd = cxx. Now that you may conveniently divide both sides by c, make first as c to b so d to a fourth which call f, and than put cf for bd, and you'll have ; and dividing by ; and so according to case 3 or . The Geometrical Construction. Found first the quantity f● (num. 2. Fig. 36.) according to the following proportion, as c to b so d to f; and a mean proportional between b and f will be = . Than having at the interval ½ a described a semi circle (n. 3.) upon the given line AB, and erected BD = , and having made OF = to it, CF will be ▪ To which AC being added will give x for one value and FB for the other. And for the common altitude, which we called , make as x to b, so b to a fourth, i e. as OF to FH so FH to FG; which will be the altitude of both rectangles A g and B g which may now easily be constructed. The Arithmetical Rule might easily be had from this Equation reduced; but you may have it more commodiously from this other SOLUTION. Let the Denomination remain the same as above, only her● put x for the common altitude, and express the reason of th● lesser base of the rectangle to this altitude by e, and that bas● will be = ex: Therefore the base of the greater Rectangle will be = a − ex. Having now multiplied the com●mon altitude by each base, the area of the greater rectangle will be axe − exx, and hence you'll have the Equation axe − exx = bb; and adding exx, and taking away bb, axe − bb = exx; and dividing by . Therefore by case 3 . Wherhfore now this will be the Arithmetical Rule. If from the fourth part of the square of the sum of the bases divided by the □ of the name of the reason you subtract the given area divided by the same name of the reason, and if the root extracted ●ut of the remainder be added to or substracted from half the sum of the bases divided by the same name of the reason; this sum or remainder will give the altitude of the given Rectangles, and ●hat multiplied by the name of the reason one of the bases: And that being substracted from the given sum of the bases ●ill give the other base. For Example, let the sum of the ba● be 16, the area of one of the rectangles 30, the ●ame of the reason which the common altitude has to the base ●f the other rectangle = 2. There will come out the common altitude, on the one side 5, on the other 3, etc. PROBLEM VII. HAving given the Perpendicular of a rightangled Triangle let fall from the right angle, and its Base, to found the ●ments of the Base, and so to form the Triangle. E g. If the base of the rightangled Triangle you are to ●m be AB (Fig. 37.) and the length of the perpendicular ●● or BF; to found the segments of the base, and so the point 〈◊〉, from which you are to make the perpendicular CD, to form ●e Tiangle ABG. SOLUTION. Let the given base be = a; and the given perpendicular ●b: Than will one of the segments of the base be = x; ●d the other = a − x, and b a mean proportional between ●e said segments, i e. x to b as b to a − x; ●herefore axe − xx = bb; and by adding xx and taking a●y bb, axe − bb = xx. Therefore by case 3 . The Geometrical Construction. Having described a semicircle upon the given line AB, if you erect the perpendicular BE, and from the point G (which is determined by EGLANTINE parallel to AB) let fall GD equal to it, you will have the two segments sought, AD = and DB = , which Construction, it cannot be denied, but it may be evident to any attentive person even without the Analysis. But that case may by the by be taken notice of wherein the given perpendicular would not be BE but BF. For in this case the perpendicular BF being erected upon AB, the parallel FG would not cut the semicircle; which is an infallible sign that the Problem in this case is impossible, where the perpendicular is supposed to be greater than half the base; which is inconsistent with a right angle. The Arithmetical Rule. From the square of half the base take the square of the given perpendicular, and add or subtract the square root extracted out of the remainder, to or from half the base; and on the one hand the sum will give the greater segment, and on the other the difference will give the lesle. PROBLEM VIII. HAving given the perpendicular of a rightangled Triangle that is to be let fall from the right angle, and the difference of the segments of the base, to found the segments, and describe the Triangle. E. g. If the perpendicular is, as above, BE, and the difference of the segments AH (n. 1. Fig. 38.) to found the segments AD and DB, from whose common term you are to erect a perpendicular DG or DC to form the Triangle. XXXVIII THIRTY-NINE XL XL XLI XLI XLII SOLUTION. Make the lesser segment = x, and the difference of the seg●ts = a, the greater segment will be x + a Make the gi● perpendicular as before = b: Therefore you'll have as x + a to b so b to x; and consequently, xx + axe = bb; and substracting axe, xx = bb − axe. Wherhfore according to case 2. . The Geometrical Construction. Make HD = ½ a, DG equal perpendicular to b; HG will be = = HB or ●, viz having drawn a semicircle from H through G. herefore DB is the lesle segment, and AD the greater; and ●ing drawn AGNOSTUS and BG, or on the other side, (making the ●pendicular DC = DG) having drawn AC and BC, the liangle will be constructed. Or with Cartes, make (n. 2.) ● = ½ a and EBB = b, and having described a Circle from ●hro ' E draw BHA; and so you'll have the two segments ●ght AB the greater and DB the lesser. The Arithmetical Rule. Join the squares of the half diffe●ce and perpendicular into one sum, and than having extras the root subtract half the difference from it; and the re●nder will be the lesser segment sought; and having added ● difference you'll have also the greater. PROBLEM IX. HAving given for a rigbt-angled Triangle one segment of the base and the side adjacent to the other segment, to found rest and construct the Triangle. As if the lesser segment of the base DB be given (Fig. 39.1.) and the side AC adjacent to the other segment; to found greater segment of the base, which being found the rest easily obtained, and consequently the whole Triangle. SOLUTION. Make the greater segment = b, the given side = c, the segment sought = x. Now if we suppose the triangle ABC to be already found, it is evident, 1. If from the square of AC you subtract the □ AD, you'll have the □ CD = cc − xx. 2. The same □ CD may also be otherwise hence obtained, because, the angle at C being a right one, CD is a mean proportional between BD and DA, i e. between b and x; whence the rectangle of the extremes bx is = □ of the mean CD. Wherhfore now it follows, 3. that cc − xx = bx; and adding xx, cc = bx + xx; and substracting bx, − bx + cc = xx. Therefore according to case 2. x = − ½ b + √ ¼ bb + cc. Geometrical Construction. Join OF = ½ b (n. 2. Fig. 39) and FAVORINA = c at right angles, and having described a Circle from E through FLETCHER draw AEB; so you'll have DAM the greater segment and DB the lesle; having erected therefore a perpendicular from D, and described a semicircle upon AB, you'll have C the vertex of the triangle sought, whence you are to draw the sides AC and BC. The Arithmetical Rule. Join the □ of half the given segment, and the □ of the given side into one sum; and having extracted the root of it, if you thence take half the given segment, you'll have the segment sought. PROBLEM X. HAving given in an obliqne angled Triangle the perpendicular height, and the difference of the segments of the base, and the difference of the other sides, to found the sides and form the triangle. As, if the altitude CD be given (n. 1. Fig. 40.) and also the difference of the segments of the base EBB, and the difference of the sides FB (as is evident from the triangle ABC (n. 2.) conceived to be so form beforehand) and you are to determine the base itself and both sides, etc. SOLUTION. Make the given perpendicular CD = a (see n. 2. Fig. 40.) EBB = b, FB = c: For the lesser segment of the base AD put x, and the greater will be x + b. It is now evident that you may obtain the □ CB by the addition of the □ □ DC and BD, , and the □ AC by the addition of the □ □ AD and DC, viz. aa + xx: So that the side AC will be = , and the side BC = . But since also this same side BC may be obtained by adding the difference c to the side AC, so that it shall be = : you'll have this Equation, ; and squaring both sides, ; and substracting from both sides ; and again squaring it, ; and substracting from both sides 4 ccxx (because c is lesle than b) and transposing the rest, and dividing by i. e. dividing the affected quantities by 4 both above and underneath, and actually dividing the former part by . Therefore according to the second case, ; or reducing ¼ bb to the same denomination with the rest ; and leaving out those quantities that destroy one another . The Arithmetical Rule. Multiply four times the square of FB by the square of the perpendicular CD, and add to it the product of the square of EBB into the □ FB, and from the sum subtract the biquadrate of FB; and the remainder will be the first thing found. Than subtract four times the square of FB from four times the square of EBB; and the remainder will be the second thing found. Lastly, divide the first thing found by the second, and from the quotient take half after having extracted the root: Thus you'll have the lesser segment of the base sought, etc. E. g. In numbers you may put 2 for FB, 4 for EBB, 12 for CD. As for the Geometrical Construction, the quantity of the last Equation contained under the radical sign will help us to this proportion, as so cc to a fourth; or dividing all by 4, as so ¼ cc to a fourth, which is ¼ of the quantity under the radical sign. Assuming therefore the quantity c for unity, make (n. 3.) IK = c IN and KL = b; NOT will be = bb, and substracting OPEN = cc (i e. to unity) there will remain NP = bb − cc. In like manner IS and KM = a, ST will = aa; to which if you add SX = ¼NO, and take thence XV = ¼ unity; TV will be = . Wherhfore if you make NR equal to this TV, and PQ = ¼ of unity or cc, since NP is = bb − cc; by the rule of proportion there will come out DR¼ of that quantity, which is under the radical sign. Therefore this being taken four times will give DZ for the whole quantity; to which if you join D y = to unity, and, having described a semicircle upon the whole line YZ, erect the perpendicular DE; this will be the root of the said quantity, and taking hence moreover OF = ½ b, you'll have DE or DA the lesle segment of the base sought. Therefore adding GB = b to DG, DB will be the greater segment, and, having let fall the perpendicular DC = a, BC and AC will be the sides sought. Q. E. F. IV. Some Examples of Affected Biquadratick Equations, but like Affected Quadratick ones. PROBLEM I TO found a square ABCD (such as in the mean while we'll suppose n. 1. to be in Fig. 41.) from which having taken away another square AEFG, which shall be half the former, ●here will be left the Rectangle GC whose Area is given. E. g. Suppose the given area equal to the square of the given line LM, to found the true sides of the squares AB and A, answering to these supposed ones, n. 1. SOLUTION. Make the area of the rectangle that is to remain = bb, and GB = x; BC or AB will be = , and substracting hence GB, the remaining side of the lesser square AGNOSTUS = − x, 〈◊〉. . Since therefore the square of this is supposed ●o be half of the square of AB, this will be the Equation: ; ●nd multiplying by xx, ●nd multiplying by 2, ; and substracting 2 b, and adding ; and dividing by 2, . NB. The same Equation may be obtained, if, putting x for GB or FH, and having found the □ of AGNOSTUS or GF as above, you infer . This last Equation, though it be a biquadratick, yet may be rightly esteemed only a quadratick one, because there is neither x nor single x in it, and so you may substitute this for it, , viz. by supposing y = xx. Whence according to the third case of affected quadraticks, y will = i. e. or = . Therefore . Geometrical Construction. Now if the given line b be assumed for unity, bb and b will be = = to the same line Therefore, if between LM as unity, and MN = ½ b viz you found a mean proportional MORE (n. 2. Fig. 41.) that wil● be = , which being substracted from LM, and added to it, will give the two values of the quantity y. Moreover therefore by extracting its roots, i e. by finding other mean proportionals LR and LS between the quantities found LP and LQ and unity (n. 3.) they will be the two values of the quantity x sought; the first whereof LR will satisfy the question, and the other LS be impossible. Wherhfore to form the square itself, since its side will be = ; by making (n 4.) as x to b so b to a fourth, it will be obtained: And this may be further proved, if finding a mean proportional BK between BY = LR and the side of the □ BC, it be equal to the given quantity LM. Arithmetical Rule. From the given area or the square of the given line LM subtract the root of half the biquadrate of the same line; thus you will have the value of the □ FC, viz. xx: Therefore extracting further the square root of ●●is, it will be the value of x sought. PROBLEM II TO found another square ABCD (Fig. 42. n. 1.) out of the middle whereof if you take another square EFGH, which ●all be a fourth part of the former, the area of the rectangle ●K intercepted between BC and FG prolonged, shall be equal 〈◊〉 the square of a given line LM; i e. having these given to ●nd the segment BY, and consequently also the side BC or ●B. SOLUTION. Make the area of the given rectangle, or the square of LM 〈◊〉 to bb, and the side sought of the rectangle BY = x; the o●er side BC will be = , and having substracted out of it 〈◊〉 and GK (i e. 2 x) the side of the lesser square FG will ● − 2 x, i e. ; whose square since it is the ●orth part of the greater square by the Hypothesis, you'll have ; 〈◊〉 multiplying both sides by ; 〈◊〉 taking away 4 b, and adding ; and dividing by 4, . Therefore according to the third case of affected quadratick ●quations, i. e. . Therefore . Geometrical Construction. If the given line b be taken 〈◊〉 unity, b and bb will be equal to it. Therefore if between LM as unity, and MN = 3 ⅓ b, you found a mean proportional MORE (n. 2. Fig. 42.) 'twill be ; which subst●cted from MQ = 2 b, or added to it, will give two values the quantity xx, viz. PQ and IQ; the first whereof will 〈◊〉 only a true one, and of use here. Now therefore a mea● proportional QR found between PQ and unity will express the quantity sought x. Therefore for forming the square itself, since its side AB = , you may proceed as in the former Constructon, (vi● n. 3) PROBLEM III HAving given the base of a rightangled Triangle, and mean proportional between the Hypothenusa and Perpendicular, to found the Triangle. As if the given base be A (Fig 43.) and the mean proportional between AC and B● be CD; to found the perpendicular BC, and Hypothenu● AC. SOLUTION. Make the given base = a, and the mean proportional = the perpendicular BC = x, than will the Hypothenusa be 〈◊〉 the Hypoth, − . Pag. 56. XLIII XLIV XLV XLVI XLVII XLVIII herefore ; ●d multiplying both sides by ; ●d substracting . herefore by the second case of affected quadraticks , and . Or thus. Make the Hypothenusa AC = x, than will the perpendi●ar be BC = . Therefore ; ●d multiplying by . ●●erefore by case 1. and . Geometrical Construction; the first for the latter Equation. 〈◊〉 be put for unity, the line AB will be also = aa, and king, as a to b so b to a third, i e. as LM to MN so LO OPEN, and you'll have bb. Having erected the perpendicu● AQ = OP upon AM, and drawn MQ or M n equal to , and consequently A n will be = , 〈◊〉 the value of xx. Moreover a mean proportional AC ●nd between A n and ARE unity will be the value of x, i e. Hypothenusa sought; which being found, you may easily implete the Triangle ABC. 2. In the case of the former Equation, making every thing before AK would be the value of the quantity xx, i e. . Therefore a mean proportional RT ●nd between RS = AK and ARE unity will be the value of x, i e. the perpendicular sought, and so AT the Hypothenusa of the Triangle sought. Arithmetical Rule. In the first Solution add the biquadrate of the given mean proportional to the biquadrate of half the given base; and having extracted the square root of the sum, take from it half the square of the given base; the root of the remainder will give the perpendicular of the triangle sought, and the root of the sum will give the hypothenusa of it. PROBLEM IU. HAving the Hypothenusa of a rightangled Triangle given, and a mean proportional between the sides to found the Triangle. As if the hypothenusa be AC (Fig. 44.) and a mean proportional between the sides BD, to found the sides AB and BC. SOLUTION. Make the given Hypothenusa = a, and the mean proportional = b, and the perpendicular BC = x; the basis AB by the hypoth. will be . Therefore ; and multiplying by xx, ; and substracting b, . Therefore by the third case, and . Geometrical Construction. If a be put for unity, AC will be also = aa, and by making as AC to CG (a to b) so OF to GH (b to a third) this third will be GH = bb. Assuming therefore OC = ½ aa = OB the radius of a semicircle, and having erected CD = bb = BE parallel to it, EO will be , and consquently EC = , and EA , viz. the double value of the quantity xx. Therefore for the double value of x, you must extract the roots out of them, i e. you must found the mean proportionals ALL and AM between unity AC and AI = EC on the one side, and AK = A on the other; Although these last may be more compendiously had, and the triangle itself immediately constructed, if having found EC and EA, you draw CB and AB: For these will be those two last mean proportionals = = ALL and AM; for by reason of the ▵ ▵ ABC, AEB, and BEC, BC is a mean proportional between AC and CE, and AB a mean proportional between the same AC and A by the 8. Lib. 6. Eucl. which is Consect. 3. Scholar 2. Prop. 34. Lib. 1. Math. Enucl. PROBLEM V HAving given the Area and Diagonal of a rightangled Parallelogram, to found the sides and so the Parallelogram. As if the given Area be = to the square of a given line BD (Fig. 45.) and the Diagonal AC, to found the sides AB and BC. SOLUTION. If for the given Area, or square of the line BD you put bb, and make the Diagonal AC = a, and put for the lesser side BC, x; the other side will be . Therefore ; and multiplying by xx, ; and substracting b . Which Equation, since it is the same with that of the preceding Problem (which is no wonder, since this fifth perfectly coincides with the fourth; for the Diagonal AC is the hypothenusa, and BD, whose square is = to the given area of the rectangle, is a mean proportional between the sides AB and BC) and so will have the same Construction, (see Fig. 45.) and the same Arithmetical Rule, which may be easily form from the last Equation of the preceding Problem. PROBLEM VI. HAving given the first of three proportional lines, and another whose square is equal to both the squares of the other two, to found those two proportionals. As if AC (Fig. 46.) be the first of the three proportionals, and another line ED given, whose square equals the two squares of the others taken together; to found those two as second and third proportionals. SOLUTION. If for AC you put a, and make the given line ED = c, and the second proportional = x, the third will be . Wherhfore the squares of the two last will be + xx = cc, □ ED by the hypoth. and multiplying both sides by ; and substracting . Therefore and . Geometrical Construction. If a be put for unity, AC will also = aa and a, and making as AC to CD (a to c) so OF to DE (c to a third) DE will be = cc. Now having made AK = DE, i e. cc, a mean proportional AI found between AC and AK will be . Therefore taking AO = ½AC, viz. ½ aa, the hypothenusa HEY will be = . And OA = ½ a being substracted from HEY or O equal to it will leave AH the value of xx; and the root of that being extracted, i e. finding another mean proportional AGNOSTUS between AC and AH, it will be the value of x, i e. the second of the proportionals sought, and since AC is the first given, AH will be the third. Q. E. F. NB. This Construction may be abbreviated, and the first Operation, by which you found DE, to which afterwards AK is made equal, may be omitted. For since you make use of a mean proportional between CA and DE sought, which is = to the given line ED, and afterwards AI a mean proportional between AC and AK is sought; it is evident that AI will be equal to ED given, and consequently that they may be immediately joined at right angles at the beginning of the given line AC, and the rest may than be done as before. Some Examples of Cubick and Biquadratick Equations, both simple and affected, whether reducible or not. PROBLEM I BEtween two given right angles to found two mean proportionals. E. g. Suppose given AB the first and CD the fourth, (Fig. 47. n. 1.) between these to found two mean proportionals. SOLUTION. Make the first of the given quantities AB = a, the other CD = cue, the first mean = x, than will the latter be , and consequently = cue; and multiplying by aa, x = ** aaq. The Central Rule will b = AD = DH. i e. according to a supposition we shall by and by make, ½ a = AD, and ½ cue = DH. Geometrical Construction. If AB or a be made unity, and also the Latus Rectum of your Parabola, and you describe, by means of this Latus Rectum, the Parabola, according to Scholar 1. Prop. 1. Lib. 2. Math. Enucl. [see n. 2. and 3. Fig 47.] ●n which AB is the Latus Rectum; A1, A2, etc. the Abscis●a's; AI, AH, etc. the semiordinates'; make moreover (n. 4) AD = ½ a, and having from D erected a perpendicular = ½ cue, describe a circle at the interval AH, cutting the parabola in N: Which being done, a perpendicular to the Axe will be the root sought or the value of x, i e. the first of the means, and consequently OA the other; since NO by the first property of the Parabola (see Prop. 1. Lib. 2. Mathes. Enucl.) is a mean proportional between the Latus Rectum AB, and the abscissa AO. And by this means there will come out, by Baker's Central Rule the very construction of Des Cartes, Geom. p. m. 91. The Arithmetical Rule. Multiply the square of the first by the fourth given, and the cube root extracted out of the product, will express the first of the means sought. PROBLEM II HAving given the solid Contents of a solid or an hollow Parallelepiped, and the proportion of the sides, to found the sides. As, if the given capacity or solid contents be = to the cube of a certain given line IK (Fig. 48. n. 1.) and the proportion of the height to the length be as AB to BC, and to the latitude as the same AB to BD; to found first the altitude, which being had, the other Dimensions will also be known, by the given proportions. SOLUTION. Make IK = a, AB = b, BC = c, and BD = d; and lastly the height sought = x, than as b to c, so x to the length required ; and as b to d, so x to the latitude sought . Multiplying therefore these three dimensions of the Parallelepiped together, you'll have its capacity or solid contents = a; and multiplying by bb, cdx = a bb; and dividing by cd, i e. . Therefore the Central Rule will be the samc as above, = AD and = DH, i e. according to the supposition which will by and by follow, = AD and = DH. Geometrical Construction. If IK or a be made unity, and at the same time Latus Rectum, and by means of it you describe a Parabola, after the way we have shown, Fig. 47. n. 2. and 3. and shall always hereafter make use of; and than to prepare the quantity (which in the Central Rule is the the quantity ) make (n. 2.) IK − IN = BC − KL = BD − MN as a to c so d to e; so that for cd you may put ae, and afterwards divide by a both above and underneath; you'll have the quantity . Therefore by further inferring as 2 e to bb, so aa to a fourth IO − IP = IT − IK − IQ, and you'll have the quantity DH, which will determine the centre, after AD is made equal . Having therefore described from that centre a circle through the vertex of the Parabola A (n. 3.) a semiordinate NOT drawn from the intersection will be the altitude sought, which will easily give you the length and breadth by the reasons above shown. Another Solution. Which will be more accommodated to the Arithmetical Rule. Let the rest of our Positions or Data remain as above, but the name of the proportion which the altitude has to the length be = e, and of that which it has to the latitude = i, the length will be = to ex, and the latitude to ix. Wheref● multiplying the sides together, you'll have the whole solidity ; and dividing by ei, . Therefore Hence The Arithmetical Rule. Multiply together the given na● of the reasons, and divide the given cube by the produ● which done, the cubick root extracted out of the quotient 〈◊〉 be the altitude of the solid sought. Another Geometrical Construction. Now if we would a● construct this Equation x = geometrically, putting AB = ● for unity, BC and BD will be the names of the reasons = ● and i. Making therefore first IK − IN − KL − MN as a to e so i to a fourth f (Fig. 49. n 1.) af ● be = ei, and the proposed Equation will have this form: i. e. aa/ f. Therefore 2. making as f to a so a to a third IQ, that will be the value of 〈◊〉 But hence 3 by extracting the cube root, i e. by finding t● mean proportionals between unity b, viz. AB and the 〈◊〉 found ●Q; the first of them will give the root sought. NB. The same Central Rule would come out according Baker's ●s Central Rule, which would have the same form of Equation, as the precedent Example x ** − aa = oh, taking b for the Lat. Rect. and unity, = AD and = DH, = AD and i. e. ½IQ = DH (see Fig. 49. n. 2.) Pag. 64. XLIX L LII LIII PROBLEM III HAving given the solid contents of a solid or hollow Parallelepiped, and the difference of the sides, to found the sides. 〈◊〉 if the given capacity be equal to the cube of any given ●e LM (Fig. 50. n. 1.) and the difference whereby the ●gth exceeds the breadth = NOT, and the difference by which 〈◊〉 breadth exceeds the altitude or depth = PQ, to found the ●gth, breadth and depth. SOLUTION. Make the side of the given cube = a, the excess of the ●gth above the breadth NOT = b, and of the breadth above 〈◊〉 depth PQ = c, make the depth = x, the breadth = x + c, ●d the length = x + b + c. Multiplying therefore these ●ree dimensions together. ● according to the forms of Baker and Cartes Wherhfore the Central Rule contracted by the supposition ●hich will hereafter follow will be this, = AD and = DH. ● e. by virtue of the supposition just now mentioned (which ●kes LM viz. a for unity and also for Lat. Rectum) = AD. And = DH. Or more short; i. e. = AD; and (∽ i. e. = DH. Geometrical Construction. If LM or a be taken for unity or Latus Rectum of the Parabola to be described, that being described (Fig. 50. n. 3.) you are first of all to determine two quantities AD and DH; which may be done two ways; either by Baker's form of his Central Rule, or by ours immediately divided by the quantities of the last Equation. 1. For AD by our form, = AD, you must make (Fig. 50. n. 1.) as a to b so b to a third (AB to AC so BE to CF) which will be bb, and D the eighth part of this CF must be added to A the half of AB. And by Baker's form you must make, 1. as AB (= a, n. 2.) to AC (= ½ p i e. ½ b + c) so BE (= ¼ p i e. ¼ b+ + c) to a fourth CF (which will be = ▪) 2. Make moreover as AB to AGNOSTUS (a to b) so BH to GI' (c to bc) and, as AB to AK (a to c) so BH to KL (c too cc) and the two quantities found GI' and KL (bc and cc) being added into one sum will give the quantity cue = MN, the half whereof MORE will express the quantity in the Rule, and to be substracted from the former . Actually therefore to determine the quantity AD not on the Axe, but on another Diameter of the described parabola n. 3. (because the quantity lordship is in the Equation) having made a perpendicular to the Axe aE = i. e. to the line BE n. 2. and from E having drawn EA parallel to the axe, according to our form AD n. 1. transfer it only on the diameter of the parabola n. 3. from A to D, either by parts i. e. ½LM from A to c, and i. e. ⅛CF n. 1. from c to D: But according to Baker's form, first you must put = ½LM n. 1. from A to 1. Secondly you must put from 1 to 2 the quantity = CF n. 2. Thirdly you must put from 2 to 3 backwards the quantity to be substracted = MOTHER n. 2. which being done, the point D will be determined. [It is evident by comparing these two ways of Construction, that we may join our forms not incommodiously to Baker 's; because by ours the quantity AD was obtained more compendiously than by Baker 's, which will also often hap hereafter. And where this Compendium cannot be had, there is another not inconsiderable one, that, if the quantities AD and DH determined according to both ways shall coincide, (which happens in the present case) we may be so much the more sure of the truth] 2. As for DH by our form, you must put it from D to e on a perpendicular erected from D on the left hand, the quantity = BE n. 2. falling here upon the Ax. Than for the quantity make n. 4. as a to ½ bb (LM to LN) so ⅛ b to (MOTHER to NP) and this quantity must be put from e to f in a perpendicular to the Diana. Thirdly, for the quantity you must farther make n. 4. as a to ½ bb (DM to LN) so ¼ c to a fourth (MQ to NR) which must be put from f to g. Lastly the quantity (which is = MOTHER n. 2.) must be put backwards (because to be substracted) from g to H, which is the centre sought. In like manner by Baker's form, first = BE is put from D to 1 even to the Ax. Secondly, for the quantity make, n. 4. as a to (LM to LS = CF n. 2.) so to a fourth (MT = AC n. 2. to SV) and this SV is further put (n. 3.) from 1 to 2. Thirdly, make in the same Fig. n. 4. for the quantity , as a to (LM to MT) so to a fourth (LX to XZ) and this XZ is put backwads (n. 3.) from 2 to 3, which precisely coincides with the point g. Lastly the remaining quantity (= MOTHER n. 2. and so by what we have said above, precisely coinciding with the interval g H) is put backwards from g to H the Centre sought. [Hence it appears again that Baker 's form is more laborious than ours; though both accurately agreed, and hereafter, for the most part, we shall use them both together, though in ' the work itself, rather in Figures, than in that tedious process of words, which we have here for once made use of, that it might be as an Example for the following Constructions.] Having therefore found by one or both ways the Centre H, and thence described a circle through the vertex of the parabola A, the intersection N will give the perpendicular to the diameter NOT, the value of the quantity x sought. PROBLEM IU. TO divide a given angle NOP, or a given arch NQTP (Fig. 51. n. 1.) into three equal parts; i e. having given or assumed at pleasure the Radius NO, and consequently also the Chord of the arch NP, to found NQ the subtense of the third part of the given arch. SOLUTION. If NO be made unity, NP = cue, and NQ be supposed = z; having drawn QS parallel to TO, you'll have three similar triangles NOQ, QNR and RQS. For since the angle QOP is double of the angle QON, and the same (as being at the Centre) double also of the angle at the Periphery QNR; it will be equal to the angle QON. But the angle at Q is common to both triangles: Therefore the whole are equi-angular, and consequently the legs NQ and NR equal, as also NO and QO; and by the like reason also PY and PT. Wherhfore if RS should be added to RY, the line NP by this addition would be triple of the line NQ; and so would give the Equation, if RS was determined; which may be done by means of the ▵ QRS, similar to the two former NOQ and QNR; for the angle RQS is equal to the alternate one QOP = QNR, and the angle at R common to the triangles QNR and RQS, etc. Wherhfore as NO to NQ so NQ to QR CITIZEN − z − z − zz and as NQ to QR so QR to RS z − zz − zz − Therefore according to what we have above said ; and substracting ; or Therefore the Central Rule will be (supposing also unity NOT for the Latus Rectum,) = AD i e. by our ½+ i. e. 2NO = AD form, and = DH. and = DH. Geometrical Construction. Having described your parabola (Fig. 51. n. 2.) take on its Axe (because the quantity lordship is wanting in the Equation) AD = 2NO, and from D having erected a perpendicular = NP to H; that will be the Centre, from which a circle described through A, by cutting the parabola in three places, will give the three roots of the Equation, viz. NOT and no true ones, the first whereof will express the quantity sought NQ (n. 1.) the latter the line NV, being the subtense of the third part of the compl. of the arch; and MORE will express the false root, which is equal to the former taken together: All the same as in Cartes p. 91. but here somewhat plainer and easier. PROBLEM V HAving three sides given of a quadilateral Figure to be inscribed in a circle, AB, BC, CD, (Fig. 52. n. 1.) to found the fourth side, which shall be the Diameter of the Circle. SOLUTION. If we consider the business as already done, and make AB = a, BC = b, CD = c, and AD = y; we shall have first in the rightangled ▵ ABDELLA, □ BD = yy − aa, and (since in the obtuse angled ▵ BCD the □ BD is = □ □ BC+CD+2 ▭ of BC into CE) if those two □ □ BC+CD (i e. bb + cc) be substracted from □ BD (yy − aa) you'll have 2. yy − aa − bb − cc = to the two said rectangles of BC into CE. But these two rectangles may also be otherwise obtained, 3, if the segment CE be otherwise determined; which may be done by help of the similar ▵ ▵ ABDELLA and CED (for the angles at B and E are right ones, and ECD and BAD equal, because each with the same third BCD makes two right ones; the one ECD by reason of its contiguity, the other at A by the 22. Eucl. Lib. 3.) viz. by inferring as DAM to AB so DC to CE y − a − c − for now multiplying CE = by BC = b, the ▭ of BC into CE will be = and two such . Now therefore yy − aa − bb − cc = ; and multipl. by y, y * − aa y = 2 abc; i e. by Baker's and Cartes' forms. − bb y = 2 abc; i e. by Baker's and Cartes' forms. − cc y = 2 abc; i e. by Baker's and Cartes' forms. y * − aa y − 2 abc = o. − bb y − 2 abc = o. − cc y − 2 abc = o. Therefore the Central Rule will be (supposing the same quantity e. g. a for unity and Lat. Rect.) = AD and = DH, i e. according to our form, i. e. = AD, and Geometrical Construction, which, without any circumlocution, from our form is founded on the foll. in Fig. 52. n. 2. LM = a n. 3. A 1. = LM from n. 2. MN and LP = b 1, 2 = ½PQ PQ = bb 2, 2 = ½ST LS and MOTHER = c DH = PR PR = bc MOTHER and mother two false roots ST = cc NO the true root; upon which having described a semicircle the quadrilateral will be easily made. According to Baker's form, for AD there would be n. 3. first A c = ½LM, than cD = = VX, half the line US, which is compounded of LM, PQ and N-ab; but DH is = PR as above. PROBLEM VI. HAving given to form a rightangled Triangle the lest side BASILIUS (Fig. 53. n. 1.) and the difference of the segments of the base, to found the difference of the sides, and so form the Triangle. If we represent the business as already done, having given AB and EC to found FC. SOLUTION. Make AB = a, and EC = b, and FC = x; than will BC = a + x: Therefore the □ AC = 2 aa+ + ax + xx and the line AC and . Now therefore ACE i e. multiplied by b or √ bb i e. is = GCF [but GC is = 2 a + x] i e. 2 ax + xx by Cons. 1. Prop. 47. Lib. 1. Mathes. Enucl. and squaring both sides ; and transposing all, . − bb lordship cue (r) (s) Therefore (taking a for 1 and the Latus Rectum of the parabola) the Central Rule will be, = AD and = DH i e. according to our form and Reduction, Geometrical Construction. First from our form, the compendiousness whereof will here appear, for it requires only one preparation n. 2. in which LM = a, MN and LO = b, OPEN = bb, which being premised, and the diameter A y (because the quantity lordship is in the Equation) being drawn (n. 3.) make AI = ½LM, and 1, 2 or ●D = ½OP, and DH = LM. The rest therefore being also performed, which the quantity SAINT occurring in the present Equation requires, according to the last precepts of our introduction, you'll have NO the value of x sought; whence (n. 4.) at the interval AB having described a Circle, and made a right angle at B, if FC be made = to the found NOT, you'll have the ▵ ABC required, and EC will be found at the beginning of the prescribed magnitude. Now if you were to found the centre H by Baker's form without our Reduction, 1. you must put of (n. 3.) ½AB from A to c. 2. For the quantity make as 1 to so to a fourth, which would be = 2AB, viz. 2DM, and to be set from c to d. 3. The quantity (to obtain which you must subtract OPEN (n. 2.) from the quadruple of LM, and divide the remainder by 2) must be set of backwards from d to e, by thus determining the point D. 4. The quantity p, which here is precisely = LM, must be transferred from D to f on a perpendicular erected from D. 5. For the quantity make as 1 to (= 2AB) so (also = ) to a fourth quadruple of LM; and this must further be produced from f to the point g (which here the paper will not permit) 6. For the quantity make as 1 to so to a fourth, (which would be = to the quadruple of LM, but taking away OPEN) which must be set backwards from g to h. 7. Lastly the quantity (= OPEN) set of from h backwards or towards the right hand to i will at length give the point H required. Another Solution of the same Problem. This Problem may be more easily solved, and will give a far more simple Equation, if you are to found not FD but A Make therefore (n. 1. Fig. 53.) = x, and the rest as above; AC will be = x + b, and it's □ xx+ + bx + bb; therefore the □ BC = xx+ + bx + bb − aa; therefore the □ of the tangent HC = xx+ + bbx + bb − 2 aa. But the rectangle ACE will be bb + bx. Therefore by Prop. 47. Lib. 1. Math. Enucl. ; and turning all over to the right hand, or . Therefore by case 2 of affected quadraticks, . The Geometrical Construction may be performed according to the rules of quadraticks Fig. 54. n. 1. as will be evident to any attentive Reader. Having therefore described a circle at the interval BASILIUS, whether it be done from any arbitrary centre (see n. 4. Fig. 53.) or upon A found in the pres. Fig. making an intersection at the said interval in B; and applying A, and producing it until EC become equal to the given quantity cue, and at length having drawn BC you'll have the triangle rightangled at B, and also the difference of the sides FC. But to make it more short and elegant; having determined A by a little circle (Fig 54 n. 3.) prolong it to the opposite part of the circumference in C, and draw AB and CB; for as the radius of the little circle is ½ b, so EC is = b. Now if you would construct the Equation above by Baker's rule (that its universality may also be confirmed by an Example in quadraticks) viz. ; LIV LV LVI LVII The Construction therefore will consist in these: LM (n. 2. ●g. 54) = a, MOTHER = ¼ b, LP = ½ b, therefore PR = : ●N = PR viz. , therefore QUEEN'S = . In Fig. 53. n. 3. ● = ¼ b, A1 = ½LM, 1, 2, or 1 D = PR; D1 = LP MOTHER, 1, 2, or 1 H = PQ Having drawn a circle from 〈◊〉 through A you'll have the true root RO = to A sought; ●d MORE the false root. NB. Hence it is evident that one Problem may have seve●l Solutions and Constructions, some more easy and simple, ●hers more compound and laborious; viz. according as the ●●known quantity is assumed more or lesle commodious to the purpose: which may not be amiss here to note for the sake 〈◊〉 Learners. PROBLEM VII. SUppose a right line BD (Fig. 55.) any how divided in A, to divide it again in C, so that the square of BASILIUS shall ●e to the square of AC as AC to CD. SOLUTION. Since the first segments BASILIUS and AD are given, call the first 〈◊〉 and the other b, and call AC the quantity AC x; and CD ●ill be = b − x. Now therefore since we suppose as the □ of AB to the □ AC so AC to CD aab − aax will = x, and transferring the quantities on ●he left hand to the right, x * aax − aab = o. Wherhfore the Central Rule (taking a for 1 and the L. R.) will be ½ − = AD and = DH. i e. by our form i. e. oh = AD, that D may fall on the vertex of the parabola; and i. e. = DH. The Construction therefore will be very simple, as is evident from the fifty fifth Figure. PROBLEM VIII. THere is given AB the capital line of a horn work (which we represent (though rudely) n. 1. Fig. 56) and the Gorge AD, also part of the line of defence OF, to found the face BE, the flank DE, the curtain (or the chord) DF, also the angle of the Bastion ABE, etc. and so the whole delineation of the horn work. It is evident if you have the flank DE or the curtain DF, the rest will be had also. Suppose therefore, the capital line AB and the gorge AD, and part of the line of defence OF to be of the magnitudes denoted by the Letters a, b, c, on the right hand. SOLUTION. Make AB = a, AD = b, and OF = c, and DF = x; than will OF = x + b, and by reason of the similitude of the ▵ ▵ BAF and EDE and ECB, as FAVORINA to AB so FD to DE But now □ □ DF+DE are = = □ OF i e. ; or giving the same denomination to all the quantities on the left hand, and multiplying both sides by and translating all the quantities on the right hand by the contrary signs to the left, Wherhfore (putting a for 1 and L. Rectum) the Central Rule will be, (because the quantity cue is negative in the Equation, for cc is greater than aa + bb) = AD; and = DH. or according to our Reduction, = AD and i. e. i. e. = DH. Wherhfore the Geometrical Construction requires no other preparatory determination by our form than of the quantities for AD, for DH at the centre, and bbcc to determine the radius of the circle; which are exhibited by n. 2. Fig. 56. viz NP is = cc, RS' = bcc, RV = bbcc, which are found by means of LM = a, LR = b, LN and MOTHER = c, MQ = NP and MT = RS. Having therefore described a parabola, n. 3. and drawn its diameter, transfer AD = ½NP upon it (because the quantity lordship is in the Equation) and also ½RS from D upon H perpendicularly, and on the right hand, (because DH = − ;) and so you'll have the centre H; through which having drawn KAI so that AK shall be = to the quantity bbcc or SAINT, i e. RV, etc. a circle described at the interval HL will cut the parabola in M and N, and applying the magnitude NO it will be that of the Curtain sought; upon which, n. 4. having laid down the circumference of the horned work by help of the given lines AB and AD, you'll have the line OF, of the magnitude which was above supposed. Now if any one has a mind to do the same thing by Baker's way; by laying down first the interval AB = and than making bc = , and lastly, putting cd for the quantity ; he will fall upon the same point D, and in like manner may express the other parts of the Central Rule by the intervals D e, ef, fg, and setting back the last gh, he will fall upon the same centre H: But this is done with a great deal more trouble and labour to determine so many quantities, and also is in more danger of erring by cutting of so many parts separately, as experience will show; and thus we have by a new argument shown the advantage of our Reduction. Another Solution of the same Problem. Things remaining as before (only assuming the given lines AB, AD and OF, n. 1. Fig. 56. one half lesle, that the Scheme may take up lesle room) make BE = x, as the first or chief unknown quantity; than will BF = x + c, and it's □ xx+ + cx + cc: And since as BE to BC = AD so BF to OF a fourth, which will be and it's square = . Wherhfore if this square be substracted from the square of BF, there will remain the square of BASILIUS, i e. ; i. e. all being reduced to the same denomination, ; or according to the forms of Cartes and Baker, Therefore the Central Rule (putting again a for 1 and the L. R.) will be = AD and = DH; or by our Reduction, i. e. = AD; and i. e. = DH. Geometrical Construction. Having therefore described a parabola (Fig. 56. n. 5.) and drawn the diameter A y, make A1 = a and 1, 2 = , so you'll have the point D; make moreover D or 2, 3 = c, and backwards 3, 4 = (we here omit to express the Geometrical determination of these quantities and as being very easy) and you'll have the point H, etc. and there will come out the quantity sought NO; which since it is equal to half BE n. 4. the business will be done; which Baker's form will also give, exactly the same, but after a more tedious process. PROBLEM IX. IN any Triangle ABC (the scheme whereof see n. 1. Fig. 57) suppose given the Perpendicular AD, and the differences of the lest side from the two others EC and FC to found all the three sides. i e. Chief the lest side AB which being found, the others will be so also. SOLUTION. Make AD = a, EC = b, and FC = c, AB = x; than will BC = x + b and its square be xx+ + bx + bb, and AC = x + c and its square be xx+ + cx + cc; and BD will be , and DC . But the □ BC may also be obtained otherwise, and the Equation also, if □ □ BD+DC+2 ▭ of BD by DC be added into one sum according to Prop. 4. Lib. 2. Eucl. viz. , multiplied by BD √ xx − aa, gives the rectangle of the segments and this doubled i e. multiplied by √ 4, giveth the quantity which is contained under the radical sign in the Equation] Therefore turning all over on the left hand which are before the sign √ to the right hand, prefixing to them the contrary signs, you'll have and taking away the Vinculum on the left hand, and squaring on the right and adding and substracting on both sides, as much as can be, and transferring all to the left, and dividing all by 3, Note, I sought this Equation also after two other ways; 1. By a comparison of the □ AC with the two squares AB+BC, after 2 □ □ CBD thence substracted, according to Prop. 13. Lib. 2. Eucl which is the 46. Lib. 1. Math. Enucl. and I formed the same with the present. 2. By putting at the beginning y for x + b and z for x + c, and going on after the former methods, till you have this Equation, in which▪ when afterwards I substituted the values answering the quantities yy and zz, etc. This same last Equation came out a little easier, but (NB) with all the contrary signs. Now to form the Central Rule, and thence make the Geometrical Construction, we must determine first each of the quantities p, cue, razors and S, that we may know whether they are negative or positive; and you'll found (n. 2. Fig. 57) p = G2 positive, cue = H negative, and KING = S also negative; and that by help of the quantities LM = b or 1, MN and LO = a, OPEN = aa, MQ and LR = c, RS = cc, and also LT = cc, TV and LX = c, XY = c. Wherhfore the form of the last Equation will be like this, x + px³ − qxx − rx − S = oh, and so the Central Rule (taking here b for 1 and the L. R.) = AD and = DH. Wherhfore, having now described a parabola (as may be seen n. 4.) having found the diameter A y transfer upon it first A b = ½LM (n. 2) and than bc = DP (n. 3.) i e. ; and thirdly cD = i. e. ½H (n. 2.) moreover from D to e put of MB (n. 3. = , and from e to f put of DOCTOR = , and from f to g put of CF = ; and from g backwards to h put of half the quantity r, or I5 (n. 2.) and having done the rest as usual, you'll have NOT, the side required of the Triangle to be described; the description whereof will be now easy (n. 5.) having all the three sides known. This may serve for an Examen, if having described a semicircle AGB upon AB = NOT, you apply the given line AD, and from B through D draw indefinitely BDC: Than at the interval AB having described the Arches A and BF, add the given line EC to BE, for thus having joined the points A and C, FC aught to be equal to FC before given. SCHOLIUM. WE have here omitted our Reduction, because it would be too tedious, and would express the quantities AD and DH (especially the latter) in terms too prolix. For AD would be = (viz. because is found = and taken in itself = ; but here [where by virtue of the Central Rule it is taken positively, when it is in itself negative] under contrary signs it is = or or or yet more contractedly (because b is unity) AD = ; which parts may be expressed without any great difficulty on the Diameter A y, by its portions A1S, 1, 2; 2, 3; 3D: But the other quantity DH, or the definition of the Centre H, would also have some tediousness, as because If you take away out of the quantities and (since this latter is to be substracted, and so left, as it is, under the sign −; but the other, also negative in itself, but here positively expressed in the Central Rule, must have all the contrary signs) I say, if you take out of these quantities those which destroy one another, and add the rest with the two former quantities, they will be = = DH. or a little more contracted (because b is 1) = DH. But now if any one has a mind to illustrate this by a numeral Example and try the truth, etc. of the quantities found; they may make e. g. a = 12, b = 1, and c = 2; and they will easily found that in the last Equation the quantity p will be 4, and cue − 190, razors − 388, S − 195: Secondly in the Central Rule of Baker they'll found = ½, = 2, and = 95, and so the whole line AD = 97 ½; and further = 1, = 4, = 190, and , and so the whole line DH = 195 − 194 i e. = 1. Thirdly, likewise in our Reduction (if we proceed by each part corresponding to Baker's) , and . The sum for AD 97 ½; but further ; and to be substracted; and so the sum for DH = 195 − 194 = 1. Which same quantities will fourthly come out, if the quantities AD and DH contracted, as they are expressed in letters above, be resolved into numbers. PROBLEM X. YOU are to build a Fort on the given Polygons EAF (see Fig. 58. n. 1.) whose capital line AB shall equal the aggregate of the gorge and flank, and the squares of these added together shall be equal to the square of the given line GH, and the solid made by the multiplication of the square of the flank by the gorge, shall be equal to the cube of the given line IK. LVIII LIX LX SOLUTION. Make the Gorge AC = z, whose square zz substracted from bb the square of the given line b, will leave the square of the flank DC = bb − zz. Now this square being multiplied by the gorge AC or z will give bbz − z = g, the cube of the given line IK; and adding to both sides z, and substracting bbz, oh = z * − bbz + g. Therefore if we take g or IK for 1 and L, g will be the line g, and The Central Rule: = AD. and = DH. i e. according to our Reduction, = AD and DH. Geometrical Construction. Having described a Parabola (n. 2. Fig. 58.) make on its axe A1 = ½IK and 1, 2, s. viz. 1D = ½MN (from n. 1.) and DH = ½IK. Than having described a circle from H, and found the true root NOT upon the given angle EAF (n 3.) make AC = NOT, and having erected the perpendicular CD divide it by AD = GH (n. 1.) and make AB = AC+CD; and the Fort will be drawn. PROBLEM XI. IN a rightangled Triangle ABC (which we denote by n. 1. Fig. 59) having given the greater side AC, and made the lesle side AB = to the segment CE, which shall cut of from the base BC a perpendicular let fall from the right angle A; to found these lines AB or CE, and consequently the whole triangle. SOLUTION. Make AC = a, CE or AB = x. Therefore, 1. you'll have aa − xx = □ A And because the ▵ ▵ BEAUMONT and CAESAR are similar, you'll have as AC to CE so BA to A And so, 2. □ A = . Therefore = aa − xx; and multipl. by aa, x = a − aaxx; or, according to the form of Cartes and Baker, transposing all to the left, x * + aaxx* − a = oh i e. x * + qxx* − S = o. Therefore (taking a for 1 and L. R.) the Central Rule will be i. e. oh = AD and i. e. oh = DH; that H may fall on the vertex A. Geometrical Construction. Since a is assumed for unity and L, the quantity SAINT also and Latus Rectum i e. AI and AK and consequently the mean proportional ALL and the radius HL will be = = to the given side AC, and consequently at that interval having described a circle, through the Parabola rightly delineated, you'll have NO the value of the quantity x, i e. of the lesser side AB. Having drawn therefore NA, which is = to AC by Construction, if you draw to it the perpendicular AB cutting NOT produced to B, you'll have the Triangle sought ABC, and AB (which will be a sign of a true Solution) will be found = NOT or CE. Another Construction. Since in the Equation above found there is neither x nor x, it may be looked upon as a quadratick, and constructed after the same way, as several other like it among the Examples n. 4. viz. because ; according to case 2 of affected quadraticks, , and i.e. Wherhfore (n. 3. Fig. 59) if AC be made = a and CD a, the mean proportional CG will be = , and taking hence GF = ½ a, there will remain FC = : And now between this or CH equal to it, and unity AC, having found another mean proportional CE it will be the value of the quantity x sought = NOT (n. 2.) PROBLEM XII. IN a rightangled Triangle ABC (Fig. 60. n. 1.) there is given the Perpendicular BASILIUS, a segment of the Hypothenusa BD, and a segment of the Base EC, from C to the perpendicular DE let fall from the end of the segment BD; to found A DC, and consequently the Base AC and the Hypothenusa BC, and so the whole Triangle. SOLUTION. Make AB = a, BD = b, and EC = c, and DC = x; which being given, the rest cannot be wanting: Therefore xx − cc = □ DE. But the same □ DE may be had, if you infer as BC to BASILIUS so DC to DE And than square the quantity DE, the square will be ; and multiplying both sides by and substracting also Pag. 89. LXI so PV = to it, that PX may be ; and lastly PEA y = ½ cue, that PZ may be . These being thus prepared, if (n. 4) A b be made = ½AB (n. 1.) bc = RT, and cd backwards = ½ PQ we shall light on the point D; and, if we make D e = ½BD, ef = PX (n. 5. which interval was too big to be represented in the Paper) and from f you put backwards fg = PZ, and on from g beyond to the right hand gh = 2RS; we shall light on the point H, etc. Which of these two methods is the shortest and fittest for practice, any one, never so little experienced, may here see; and first Learners may take notice if they would construct by Baker's form, in the Diagrams n. 3. n. 5. and the like, they must take care to make the angel's MISTOOK, SPT pretty large; which we have here represented the lesle to save charges in cutting on Copper. PROBLEM XIII. HAving given the Diameter of a Circle CD (n. 1. Fig. 61.) and the line BG, which falls on it perpendicularly, (which we have here only rudely delineated) to found the point A. from which a right line AC being drawn shall so cut the line BG in FLETCHER, that OF, FG, GD shall be three continual Proportionals. SOLUTION. If CF be found, the point A will be also had, and the section of the line BG will be made. Make therefore CF = x, and (because the perpendicular BG is given, there will also be given the segments of the diameter CG and GD) make GD = b, and CG = c; than will BG = √ bc▪ and CD = b + c. Since therefore the ▵ ▵ GOD and CGF are rightangled, and have the angle at C common, they will be similar. Therefore, as CF to CG so CD to CA Therefore OF = and FG . But by the Hypothesis, as OF to FG so FG to GD . Therefore the rectangle of the extremes will be equal to that of the means, ; and multiplying by ; and transposing all to the left, ; i. e. by the Cartesian form, . Therefore (taking b for 1 and L.) the Central Rule will be, = AD and = DH. or according to our Reduction, i. e. = AD and i. e. = DH. Geometrical Construction. Having described upon the given line CD (n 2.) a semicircle, and applied in it the given perpendicular BG, as the figure shows, you'll have the segments of the Diameter GD = b, and to the quantity p in Baker's form, and CG = c, which (n. 3. where LM = b, LO and MN = c) will give OPEN = cc and to the quantity q. Wherhfore having described a Parabola (n. 4.) and the line US = 2 ½ b, having cut of the fourth part of XZ, and the eighth of YZ (whereof the one will be = , viz. , and the other to ) if you transfer A1S = XZ upon the diameter of the Parabola A y, and moreover 1, 2 or 1 D = to half OPEN (n. 3.) and transversly D = YZ and backwards 3, 4 = ¼ OPEN, as also 4, 5 = ½CG (n. 2.) you'll have the centre H, and having described a Circle at the interval HA', the root NOT must be transferred from (n. 2.) C to FLETCHER, and continued to A the point sought. In Baker's Form (because the quantity lordship is = b or 1) is = and = , and the quantity cue or cc = OPEN, (n. 3.) make therefore in the Diameter of the Parabola A b = ½GD, and bc = ⅛GD, (n. 2.) and lastly cd = ½OP (n. 3.) and you'll have the point D the same as before. Make moreover D e = ¼GD and cf = CG, and than backwards fg = ¼OP, and lastly gh = ½GD, and you'll have the same centre H, and the coincidence of the parts in both forms will be pleasant to observe; which otherwise seldom happens. Other Solutions of the same Problem. Carolus Renaldinus, from whose Treatise de Resol. & Compos. Math. Lib. 2. we have the present Problem, proceeds to solve it in another way, changing it plainly into another Problem: viz. he observes, 1. That the angles FAD (see n. 1. of our 61. Fig.) and FGD, since both are right ones on the same common base FD, are in circle. Hence he infers, 2. (by virtue of the Coral. of the 26. Prop. 3. Eucl.) that the □ □ DCG and ACF are equal, and consequently CD, CA, FC and CG are four continued proportionals. Than he observes, 3. That GD is the excess of the first of these proportionals above the fourth CG, and OF is the excess of the second AC above the third CF; and so, since 4. the rectangle of OF and GD is = to the square of the mean proportional FG (for OF, FG, GD, are supposed to be continual proportionals) and this □ FG is the excess, by which the square of the third CF exceeds the square of the fourth CG; now the present Problem will be 5. reduced to this other: Having two right lines (CD and CG) given to found two such mean proportionals (AC and FC) that the ▭ of the excess of the first above the fourth (viz. of FAVORINA into GD) shall be equal to the excess, by which the square of the third (FC) exceeds the square of the fourth (CG) viz. by the square FG. Wherhfore instead of the former he solves this latter Problem, putting for CG, b, for GD, c, so that the first of the given lines CD shall be = b + c, and the other GD = b; calling the first mean proportional AC, x; and thence denominating the latter (viz. multiplying the fourth by the first, and dividing the product by the second) and moreover he finds the excess of the first (b + c) above the fourth (b) to be c, and the excess of the second (x) above the third to be i. e. ; so that the □ □ of these two excesses is , and because the □ of the third FC is = , having substracted bb = □ GC, there is given the □ FG = = ▭ of the excesses we just now found. So that now you'll have the Equation etc. We also endeavoured to found another Solution, by finding an Equation from the line BD (n. 1. Fig. 61.) as which might be twice obtained by means of the two rightangled Triangles FAD and FGD, since it is the hypothenusa of both. But here, besides the former denominations of our Solution we must first give a denomination to the line AD, by making as CF to FG so CD to AD , etc. But whosoever shall prosecute this Solution of ours, or that of Renaldinus to the end, will found much more labour and difficulty in either, than in the first we have given. APPENDIX. THE Invention of the special Central Rule for the case of Problem 1. Of Cubic● Equations, etc. which may serve for an Example for all the other special ones which belong to our Synopsis, p. 354. (and from these special ones) to found a general one. In Fig. 47. n. 4. make AD = b, DH = d; and so we shall have the □ of the radius HA' = bb + dd. But this □ HA' or HN, may be also had otherwise, by putting, 2. For the quantity NOT, as sought, the letter x, and by inferring from the known Property of the Parabola, as L to NO so NOT to AO and substracting AD = b from AO, you'll have DO or PN = ; whose □ is . But you also have PH = DH − DP or NOT, i e. d − x; whose □ is therefore = dd − 2 dx + xx. Wherhfore adding the □ □ PN and PH, you'll have the □ HN = . Wherhfore the Equation will now be readily had: (□ HN)— (□ HA') ; and taking from both sides . and multiplying all by L and also dividing by . +L x 3. And now farther comparing this Equation with a form like ours in Probl. 1. viz. with this, ; It is manifest, since in ours also the third term, viz. cue is wanting, that the correspondent one to it in the former − 2 bl+l is equivalent to 0; and adding to both sides 2 bls, L will be made = 2 bls; and dividing by 2L, will = b or AD. In like manner, since − razors in our form answers to the quantity 2L d in the former; 2L d will = r, and, dividing by 2L, DH or d will = : Which is the other Member of the Central Rule to be found. NB. The Analytick Art has this particularly to be admired in in, that it finds its own Rules by an Analysis: Whereof we have here an evident Example, and several others in the Resolution of affected Quadraticks, p. 345. and the following. II The Invention of the Central Rule, in the Case of Fig. 11. and the like. 1. □ HA' = bb + dd as above. 2 Putting x for NOT, as sought, we may infer from a new Property of the Parabola, which we have demonstrated Prop. 6. lib. 2 as L to NO so OR to AO, i e. (putting a for BASILIUS or FOYES given; that NO − OF i e. NF or OR shall be = x − a) as L to x so x − a to = AO. Therefore having substracted AD i e. b from AO, you'll have DO or HP = ; whose □ is . But PN also is = NOT − PO or HD, = x − d is = . Therefore having added the □ □ PH and PN, there will come out □ HN , = □ HA' i i. bb + dd; and taking away from both sides ; and multiplying every where by L and diving by . 3. And now comparing this Equation with another form, which shall be like an Equation arising from the Solution of some Problem, e. g. with this ; to this you'll have = this other, . 4. Wherhfore, because in these equal forms, first, 2 a is = p, a will be = i. e. the line BASILIUS. Secondly, Because aa (or ) ; Therefore , and dividing by = AD. Thirdly, Because i. e. ; therefore dividing by 2L, d will be = i. e. i. e. resolving into equivalent Terms expressed by lordship and cue, ; which is the other member of the Central Rule to be found. viz. a is = Therefore ab will be = Therefore . Dr. Wallis is of Opinion this is nothing but an Improvement of the Ancients Method of Exhaustions. THE INDEX OF THE FIRST BOOK. SECTION I CHAP. I COntaining the Definitions or Explications of the Terms which relate to the Object of the Mathematics. Page 1 CHAP. II Containing the Explication of those Terms which relate to the Affections of the Objects of the Mathematics. p. 35 SECTION II CHAP. I Of the Composition and Division of Quantities. p. 55 CHAP. II Of the Powers of Quantities. p. 59 CHAP. III Of Progression or Arithmetical Proportionals. p. 66 CHAP. IV. Of Geometrical Proportion in general. p. 69 CHAP▪ V Of the Proportions or Reasons of Magnitudes of the same kind in particular. p. 90 CHAP. VI Of the Proportions of Magnitudes of divers sorts compared together. p. 114 CHAP. VII. Of the Powers of the sides of Triangles and regular Figures, etc. p. 12 BOOK II SECTION I COntaining Definitions. p. 154 SECTION II CHAP. I Of the Chief Properties of the Conic Sections. p. 166 CHAP. II Of Parabolical, and Hyperbolical, and Elliptical Spaces. p. 198 CHAP. III Of Conoids and Spheroids. p. 204 CHAP. IV. Of Spiral Lines and Spaces. p. 208 CHAP. V Of the Conchoid, Cissoid, Cycloid, and Quadratrix, etc. p. 218 CHAP. VI The Epilogue of the whole Work. p: 231 An Introduction to Specious Analysis; or, The New Geometry of Des Cartes, etc. pag. 1 Some Examples of Specious Analysis in each kind of Equations. p. 16 1. In Simple Equations. ibid. 2. Some Examples of pure Quadratick Equations. p. 28 3. Some Examples of Quadratick affected Equations. p. 41 4. Some Examples of affected Biquadratick Equations, but like affected Quadratick ones. p. 53 5. Some Examples of Cubick and Biquadratick Equations, both simple and affected, whether reducible or not. p. 61 Appendix. p. 93 FINIS.