<--º-º-º-º-ºr-wºrk...a…, x-aa-ººr " zºº.º.º.º.º.º. -----º--- .*.* * * * * * ***** * * * * * * * * * * * * * ~ *º-ºº-ººººººº sºººººº…º.º. 4 ºf . . . a zºº sº.º.º.º.º.º.º.º. - .*.*...*. & ..., " . . . z-z-z-z-z-rz-z-z-z-z-z-z-z-z--------- • *- : 3...? --, .. m: - sº-z-z-z-z-z-z-r- - - - - →w : | 1,113*ARY of TJ11. # isitºf Mi(Hiſ |UnivElo" ºutlºw || - } -- º | – s ... f* --→ -- ". . . . ` - 1 J H. H º | - - x * ‘, * $º, tº ; \ * \ *..." ^^ . . - - '--> \\ w - w *~ *. * ºš. º w 1–3 º | F itſ) M THE K.IHRARY' () F' f º PROFESSOR W. W. BEMAN - Sºl *|| º A.H. (87), A.M. l'S7: ; º ºf TEACHERUFMATHEMATICs | 1922 : "...T. tº | 87 l - r lºcºm Lºſ t---~~~ * *… !--→ * – - : - * ........S. # * * * …º.º.º.” ºr QA º 72 2. TREATISE G E o M ETR Y, CONTAINING THE FIRST SIX BOOKS OF EUCLID'S ELEMENTS, METHo Di CALLY ARRANG ED AND CONCISELY DEMONSTRATED. TOGETHER WITH The Gºlements of $olio &contetry. —sº- By D. gr.ESSWELL, M. A. * FELLOW OF TRINITY COLLEGE, CAMBRIDGE. ; - g *-*-as CAMBRIDGE : * Printed by J. Smith, Printer to the University; for J. DEighton & Sons, cAMBRIDGE ; AND For G. & w. B. whittaken, AVE MARIA LANE, LoNDON. —sº- 1819 sº E. R. R. A.T A. Pref. Page xiv.ºl. 4, from the bottom for acquired read required. Page 60. l. 7. from the top for being read be. 170. I. 3. from the bottom for a Part read an Aliquot Part. 208. l. 13. from the top for if read of. 216. l. 3. from the bottom after (Th. 80.) for B read q B. 228. i. 8. from the top for GB read GP. 275. l. 7... . . . . . . . . . . at the beginning insert angles. 284. l. 3... . . . . . . . . . . insert Book I. 295. l. 3 . . . . . . . insert Book I. 304. 1. S. . . . . . . . . . . for acute read obtuse. 321. I. 12.... . . . . . . . . for XII. read XI. 376. l. 15. . . . . . . . . . . . after sides insert or else of the n diagonals. 420. l. 12. . . . . . . . . . . . for Th. 157. read Th. 158. 439. l. 2. from the bottom, for AC read AK. 456. l. 4. from the bottom for Th. 154. read Th. 157. 472. last line for Th. 32. read Th. 33. 496. l. 5. . . . . and diagram, below the letter I, for h read k. —-Q- NOTE on Th, 28. and Th. 171. The last case of Th. 28. might be proved by drawing from C (see fig. p. 79.) a parallel to the diameter of the parallelogram ABCD, and, again, a parallel to the straight line which joins B, and the point in which the former parallel meets AF, and so on, until a straight line be drawn which meets AF either in E, or in some point between E and F. For thus will there be a series of parallelograms of which ABCD is the first, and EBCF the last, and which, by the two former cases of Th. 28, and Def. 11. Cor. 1, are equal to one another. Also, Th. 171. might be divided into three cases, and proved, in a similar manner; the two first cases, as the first case in p. 469, which is one of those two, and the last case of it, by supposing c, B and c, A, (see fig. p. 454.) to be joined, and a plane to be drawn through ab parallel to the plane of the A chA, and cutting the planes ACda, BCdb, and so on ; for thus will there be a series of trian- gular prisms, of which ABC — abo is the first, and ABD — abd the last, and which, by the two former cases of the theorem, and Def. 1 1. Cor. 1. are equal to one another. 427660 W A N EXPLANATION OF THE SYMBOLS employed in this Treatise as Abbreviations. —sº- F . . . . . denotes. . . . . . is equal to or equal to. *** - > . . . . . . . . . . . . . . . . is greater than. < . . . . . . . . . . . . . . . . is less than. +. . . . . . . . . . . . . . . . together with. A. . . . . . . . . . . . . . . . . . angle. 4. . . . . . . . . . . . . . . . . . angles. AB, or A.B. . . . . . . . ... a straight line, of which the points denoted by A and B are the extreme, or any two different, points. e- A.B. . . . . . . . . . . . . . . . a circular arch, of which the points denoted by A and B are the extremities. AB x CD. . . . . . . . . . . . a rectangle, of which AB and CD are adjacent sides. AB for AB X AB i.e...a square, having AB for one of its sides. & tº º e º dº dº º te e º e º e º tº . . a triangle. /*- : . . . . . . . . . . . . . . ... triangles. D. . . . . . . . . . . . . . . . . a parallelogram. [T]. . . . . . . . . . . . . . . . . parallelograms. A : B. . . . . . . . . . . . . . . the ratio of A to B. A : B :: C : D. . . . . . ... the ratio of A to B is equivalent to the ratio of C to D. AP = < B if c > = < D. i. e. if c > D then A- B, if C = D then A= B, if C3 D then A & B. ‘. . . . . . . . . . . . . . . . . . . . therefore. JP R. E. F. A. C. E. The publication of a treatise on Geometry, in an University, where Euclid's Elements have long been used as the text book in that science, might in itself be considered as an indirect attack upon the credit of the Grecian Geometrician. Of the writer of such a treatise it may, therefore, justly be expected, that he should state explicitly what it is which he thinks amiss, in a work of the highest authority in its kind. It will become him, indeed, to offer this statement in the form of an apology. Nor will it be enough that he is able to point out some faults in the compilation of Euclid. That it should contain some errors, and be chargeable with some defects, is nothing more than what is common to all human performances; and its want of absolute perfection has not escaped the notice either of ancient, or of modern, commentators. But the question really is, whether it be so faulty, as to justify an attempt, which aims at new modelling 0. ii P. R. E. F. A. C. E. j almost the whole system of elementary Geometry. This, then, is a question, the discussion of which, although it cannot be very short, can hardly be avoided, on an occasion like the present. For although a zealous advocate for the advancement of science may, perhaps, be inclined, without scruple, to en- courage labours, from which, even when they fail of attaining their proposed object, some incidental ad- vantage is commonly found to result; yet others, from an habitual veneration for the sources whence they derived their knowledge, will be disposed to discountenance, rather than to approve, any con- siderable innovation in the doctrine of Geometry. We think ourselves obliged, therefore, to proceed, with as much brevity as the nature of the subject will allow, to point out the principal objections, which may be urged against the accuracy of Euclid's Ele- ments; acknowledging, at the same time, with the rest of the literary world, many well known claims of high merit in that book, and leaving it entirely to the reader's judgment to determine, whether something still better might not be produced in the present day. I. In the first place, then, it may be asserted, that Euclid's definitions are not all of them the best that might have been given; nor are they disposed in their proper places; nor is every thing defined by them, which ought to have been defined. ** PRE FA C E. iii Nothing is of greater importance to the perfection of a system of Geometry, than right definitions. They not only serve to explain and to fix the meaning of words, but they are, in reality, the basis upon which the whole superstructure of Geometrical science is built: they are the very principles, from which all our knowledge of the properties of magnitudes is derived; in a tacit, if not in an expressed, reference to them, all our reasonings terminate; and it is truly surprising to observe, what a multitude, and what a diversity, of consequences, are necessarily included in many single definitions ; in that of a triangle, for example, or in that of a circle. What is it, we may ask, which is the subject of any demonstration, in the doctrine of Geometry not any matter of fact, not any object of sense, not the diagram, which is drawn, which is always an imperfect representation, and which serves only as an aid to the memory and the imagination; but a certain hypothesis, a certain set of ideas, conveyed to the mind by certain definitions, the consistency, or the inconsistency, of which, with some other hypothesis, or with some other set of ideas, is attempted to be proved. Upon the clearness, therefore, and the precision of the definitions, in Geometry, the evidency of the conclusions must mainly depend. It must be confessed, however, that these geometrical definitions, are not each of them iv. PR E FA C Ee what is commonly understood by the word definition; namely, an enumeration of the simple ideas, which constitute a complex idea, together with an ex- planation of the order, and the manner, in which they are combined. Of several of them the object is merely to place the mind, as it were, in such a point of view, as that it cannot fail distinctly to apprehend some particular idea, which, strictly speaking, cannot be defined. And this is the case with some of the most fundamental and important of all the definitions in Geometry. In no other way can we attempt to explain, what is meant by a surface, a line, a point, a plane surface, and a straight line. These are purely mathematical terms; they indicate things which have only an ideal existence; and which cannot, therefore, be made the objects of apprehension to any of our senses. They form, however, the constituent parts of other definitions properly so called. Their meaning ought, therefore, to be fixt, with as much accuracy as is possible. But in order to put the mind of the reader completely in possession of the notions, which these terms are intended to convey, there is only one method which it is safe to pursue. It is plain, that we must not begin with what can neither be defined, nor exhibited. We must first direct the attention of the learner to ideas, with which he cannot but be familiar; or which may easily, and without any PR E FA C E. v ambiguity, be imparted to him; and which, in their modifications, or by being considered in a particular manner, will suggest all that we intend him to perceive. If, for example, he reflect on the idea which he has of an orange, or of a block of stone hewn from a quarry, or of any other mass of matter, he will perceive, that it occupies a portion of space, to the exclusion of all other bodies; and thus he will acquire the notion of solid space. He will then easily be made to comprehend the divisibility of solid space into two contiguous parts, by a common boundary, which he may be taught to call a surface. Further, the two contiguous parts into which a solid may thus be supposed to be divisible, may either, as he will readily imagine, be both of them exactly and entirely of the same form at that boundary, and then the boundary is called a plane surface; or the two contiguous parts may be of different forms at their common boundary, the one being prominent, where the part in contact with it is hollow. The notions of a line, as the boundary which divides a surface into two contiguous parts; of a straight line, as that which divides a plane surface into two contiguous parts, which have the same form at their common boundary; and of a point, as the extremity of a line, or as the boundary common to two contiguous parts of a line, will next follow, and will be apprehended without difficulty. vi PR E FA C E. Now the method of Euclid is the contrary of this. He begins by defining a point. It would, however, hardly be possible for any one to conceive what is meant by a point, from his negative, and, we might almost say, his aenigmatical description of it—anuetov éariv of wépos ov6év. Again, to know what a line is, from Euclid's account of it, it is necessary in the first place for the mind to have a clear conception of uñkos drxares, of length unaccompanied with breadth. A straight line he defines to be that its éétarov Tots ép eavrns a muetois keirai. It will not be found easy, perhaps, to give an exact translation of this passage, nor to explain, quite satisfactorily, its grammatical construction. It is evident, however, that the word evenly, by which ééta'ov is commonly rendered, requires explication as much as the word straight. Another definition of a straight line is, indeed, men- tioned by Proclus, -ótt travra arms ºrd uépm Taaw duows éqapuéel, which is much less liable to ob- jection; although, without explanation, it might seem to be applicable also to the arches of equal circles. But is it not still better, to guide the mind through a train of thought, at the end of which it will unavoidably arrive at the idea of a straight line To do this, is to teach, as it were, the generation of a straight line, much after the manner in which Euclid himself describes what a sphere is ; namely, by calling our attention to an intelligible hypo- PR E F A C E. vii thesis, involving in it, as is perfectly evident, no absurdity. The meaning of the word angle Euclid has en- deavoured to explain by the word inclination. But even if this latter term were the more easily in- telligible of the two, which it is not, it would still be injudiciously chosen. To make use of it is to in- troduce into Geometry a new species of quantity, which is neither linear, nor superficial, nor solid. The use of this term is, besides, accompanied with some degree of obscurity and of difficulty. Even in the simple case of adding two angles together, or of subtracting one angle from another, it cannot, perhaps, be adhered to, with strict propriety: and, in the hands of Euclid, it is found to vitiate the demonstra- tion of a most important proposition, the last of his sixth book. For if an angle be an inclination, it is not very easy to see, as hath been already suggested, how a multiple may be taken of a mere inclination. Indeed, of a right angle, or of an obtuse angle, no multiple whatever can, properly speaking, be called an inclination. And in the proposition referred to, if one of the two given angles be half of a right angle, and the other a right angle, then, the one being supposed to be quadrupled and the other to be doubled, it is manifest that these multiples cannot in any sense, be called inclinations; because the two straight lines, which bound each of them, will be in viii PR E FA C E . one and the same straight line, and will, therefore, have no inclination to one another: so that the multiple itself, and that of which it is the multiple, are things of a different kind. If, now, an angle be defined to be a portion of an unlimited plane lying between two indefinite straight lines in the plane, which meet one another, and which are not in the same straight line, no such absurdity will follow. For, although the double of a right angle will not, ac- cording to this latter definition, be an angle, it will still be a magnitude of the same kind as an angle; in the same manner, as when the bases of two equal equilateral triangles in the same plane, are made wholly to coincide, the aggregate although it is a multiple of one of them, is no longer a triangle: it continues however to be a plane rectilineal figure. But, if we extend our definition of an angle, and call it a portion of an unlimited plane, bounded by two straight lines which meet one another, without the restriction, that the two straight lines are not to be in one and the same straight line, the double of a right angle will then be itself an angle. To give to the word angle so extended a sense, is, undoubtedly at variance with common usage: it is, however, very convenient, thus far at least, to generalize our notion of what is meant by that term, whether we retain the name Or not. Whatever weight may be allowed to the foregoing PR EFA C E. ix remarks, it is certain that many of Euclid's de- finitions are misplaced. For no hypothesis, the con- sistency of which is not manifest in itself, ought to be proposed to the learner, before its consistency has been actually demonstrated to him. He ought not to be told, as he is in the Elements, of parallels, before it has been proved that there may be straight lines, which can no where meet; nor of a square, before he knows that there may be a quadrilateral rectilineal figure having all its angles right angles, and all its sides equal to one another; nor of a tangent to a circle, before he has been convinced, that a straight line may meet a circle in one point, and yet, when it is produced, not cut the circle. To call his attention to these things before the possibility of their existence has been shewn, is only to create needless perplexity in his mind, and to waste his time. The proper place to introduce a definition of this questionable kind, is next after the particular theorem in which the foundation for it has been laid. It is a defect of great moment, that Euclid has omitted to define what is meant by geometrical equality. We may, indeed, venture to assert that, by the want of this definition, the very ground-work of his Geometry is rendered unsound. Wherever he has inferred the equality of two figures, whether b X PR E FA C E. plane or solid, without having shewn their possible congruity; in a word, wherever, in reasoning about this kind of relation, in respect of such figures, he has used the axiom, “ that if equals be taken from equals the remainders are equal,” his proofs are de- fective. A term of such extensive use, as is that of equality, which is applied to lines, to angles, to limited spaces, to numbers, to portions of time, to velocities, and to forces, and which always indicates a comparison of two or more objects of thought, must, it is evident, require a definition. And after the precise sense in which it is used by the geometrician has thus been fixt, the axiom, above cited, which has probably been borrowed from the science of arith- metic, will not be found to be self-evident, when it is asserted of geometrical magnitudes; excepting in the case of straight lines, that of plane rectilineal angles", and that in which, from two figures, that can be made perfectly to coincide, without at all altering the form of either, there are taken two parts, which may not only be made to coincide, without any alteration of their form, but which are also similarly * Although we have defined an angle to be a space indefinitely extended one way, yet the definitions, which we have afterwards given of the meanings of the terms equal, greater, and less, will be found to be as applicable to angles as they are to any finite magnitudes. P R. E. F. A. C. E. xi posited : as, for example, when from equal circles there are taken equal circles, supposed to be, re- spectively, concentric with them. But this subject has been so fully and so ably treated of by Barrow, in the third and fourth of his Mathematical Lectures for the year 1665, that it is needless to enlarge upon it here. We shall only further remark upon it, therefore, that if Euclid had defined equality, he would have been led to introduce some other defini- tions, which are also wanting in his treatise; namely, those which belong to the terms homogeneous, greater, and less. ! * II. To proceed, then, to another class of ob- jections. It is well known, that under the same head of axioms, or koival évvotal, Euclid has ranged pro- positions, which differ widely in their nature one from another. Some of them are either pure definitions, or, so far as they are self-evidently true, they are the immediate consequences of definitions; one of them is a theorem, requiring, and easily admitting of, demonstration; and the last of them belongs to that peculiar kind of propositions, which Archimedes has very appropriately called Xaugaváueva; that is, as- sumptions. These assumptions may, in truth, be considered as marks of the present imperfection of the science of Geometry. They ought to be regarded as having been originally introduced, only in order xii PR E FA C E. that the progress of the science might not be im- peded. Whenever, therefore, it is perceived, that they can be readily deduced from some plainer hy- pothesis, they ought no longer to retain their form; and even whilst they are suffered to remain, some proof of them, although it may be tedious, ought, perhaps, to be given, separately, for the satisfaction of the scrupulous. III. But it is not so much the name, given by Euclid to the proposition which constitutes his last axiom, nor the place which he has assigned to it, that we are disposed to censure, as the choice which he has made of the proposition itself. Indeed, almost all the editors of his Elements, whether they be Greek, or Arabian, or of modern Europe, have ob- jected to that assumption, on which he has founded the doctrine of parallel lines: nor can it be denied, that he has therein assumed what is far from being sufficiently evident in itself, to warrant; its being so proposed. The learner, also, experiences some diffi- culty, in clearly comprehending what is meant by the enunciation of theproposition ; in consequence of his being required by it to compare the aggregate of two separate angles with that of two right angles. Euclid's doctrine of parallels does, therefore, really need correction. It must be acknowledged, however, that it is not easy to remove this imperfection al- PREFA C E. * xiii together from elementary Geometry. Nor are we able, in the present state of the science, to do any thing better, than to substitute, for the axiom in question, another assumption, which is more simple, and more readily admissible; and which has, also, been employed by Euclid himself, although he has no where stated it in formal terms. IV. The blemishes which occur in Euclid's de- monstrations, notwithstanding their reputation for correctness, have not escaped animadversion: such are the occasional use of arithmetical principles, with- out necessity, as in Prop. 20. Book III. and in Prop. 16. Book IV; the omission of cases included under the general enunciation of a proposition, and to which the proof given is not applicable, as in Prop. 35. Book I, and in Prop. 26. Book III; the contrary error of giving separate proofs of problems or theorems belonging to one general proposition, as in Prop. 1. and Prop. 22. Book I, in Prop. 47. Book I, and Prop. 31. Book VI, and in Prop. 14. Book II, and Prop. 25. Book VI; and the enunciation of pro- positions as generally true, which are not true, without a restriction as in Prop. 32. Book VI: these, and various instances, also, of paralogism, have been sufficiently made known, by his most ancient as well as by his latest editors. It is unnecessary, therefore, to dwell upon them in this place. XIV PR E FA C E. V. What we are next to notice is, that Euclid's treatise does not appear to be perfectly well adapted to the present advanced state of mathematical learn- ing. A treatise on the Elements of Geometry is, at this day, to be regarded as subsidiary to the study of the sciences of Natural Philosophy, and Universal Arithmetic. Now the latter of these sciences enables us, on the one hand, most easily to investigate several properties of solids of which Euclid's geometrical demonstrations are complicated and tedious : and, on the other hand, it presupposes a knowledge of certain theorems, which are not to be found in Euclid. So that his Elements, if they are considered as in- troductory to the doctrine of fluxions, may be said to contain both more than enough, and less than enough. The same remark may, also, be applied to them, when we take into account the use which is made of Geometry in the developement of the different branches of Natural Philosophy. - If, for example, it were proposed to the student, to find the place of the centre of gravity of a given pyramid, by the method of fluxions, that method would furnish him with a general formula, for the acquired computation; but he would look, in vain, in Euclid's treatise, for the particular geometrical theorem, which enables him to apply the formula to the case in question. PR E FA C E. XV VI. Neither are the materials of Euclid's book, if they were otherwise unobjectionable, judiciously arranged. It will hardly be disputed, that the most convenient system of arrangement, in the teaching of any science, is that which begins with the most simple things; which, through them, advances to what is more complex and difficult; and which always places things of the same kind under the same class. But is this the system which has been adopted by Euclid 2 In his first book the properties of straight lines, angles, triangles, and parallelograms, are mixt together without any regular and determinate classifi- cation. His second book, the reading of which is always found an irksome task to the learner, contains only particular cases of a set of more general pro- positions, which admit of very short and easy de- monstrations in their appropriate places. And in the remaining parts of his work, if we except the fourth and the fifth books, a similar want of subdivision and methodical arrangement is also observable. We speak from ample experience, when we assert, that this want of order in Euclid's Elements renders what is therein taught difficult to be recollected; that it hinders the mind of the learner from obtaining a clear and comprehensive view of the whole structure of Geometry; and that it serves to hide from him the relations, which the several parts of the science xvi PR E FA C E. bear to one another : so that even after a careful perusal, he is at a loss to enumerate, for example, and to describe, the properties of triangles, with which he has been made acquainted. The other objections, which have been mentioned, might, it is evident, be removed by corrections and additions: and an edition of Euclid's Elements, con- taining a more faithful English translation of the text, than has hitherto been given, and exhibiting such emendations, 3S appear to be necessary, always apart from the body of the work itself, would undoubtedly be a valuable accession to our general stock of ma- thematical books. But the last fault, which we have pointed out, if it really exist, cannot be cured, without taking to pieces, as it were, the whole fabric of Euclid's Geometry, and rebuilding it after another model. And this will not only render some alteration necessary in the form of the detached fragments, but will also require the addition of new materials. In the following treatise, the whole of Elementary Geometry is distributed into two parts; the first of which contains the Elements of Plane, the second those of Solid, Geometry : and in each of them such all arrangement has been adopted, as seems most likely to facilitate the study of the subject, and to give the student the clearest, and the most complete comprehension of it. In both the parts, the de- PREF Ack. xvii finitions are introduced, as they become necessary 3. and not one of them is laid down, unless its con- sistency be self-evident, without a reference to some previous demonstration, which may prove that no absurdity is \nvolved in it. The doctrine of Plane Geometry is comprised in two books, of which the first, consisting wholly of Theorems, includes the general doctrine of Proportionality. The introduction of the subject last mentioned, occasions a kind of subdivision of the elementary theorems, which make up the first book; the object of investigation, in those which precede it, besides the circumstance of position, is simply whether the one of two compared magnitudes be equal to the other, or less, or greater than it; and the properties, therein demonstrated, may be called, for the sake of distinction, the primary re- lations of geometrical magnitudes. In the theorems, which follow the chapter on Proportionality, the enquiry is not confined to the circumstance of position, and to the relation of mere equality or inequality, but serves to determine whether the first of four assigned magnitudes contains, or is contained, by, the second, as often as the third contains, or is contained by, the fourth; or rather whether they be proportionals in a more general sense, which includes that particular case, whenever the magnitudes are actually com- mensurable. The doctrine of Ratios, as it is not C xviii PR E FA C E. founded upon any principles that are peculiar to Geometry, might have been made preliminary to the study of that science. And thus, we should have had no need to distinguish from the-rest, those relations, which, for want of a better term, we have called primary; all the properties of triangles might have been demonstrated in an uninterrupted series; and so might the properties of every other class of māgni- tudes. This would, doubtless, have been a more perfect form of arrangement; but it would have been attended with the preponderating disadvantage, of first presenting to the learner, that which is the more difficult, instead of that which is the more easy. After a long and careful consideration, the ancient doctrine of ratios, as it is delivered in Euclid's Elements, has been adopted, without much alteration, in our chapter on proportional magnitudes. But as the subject is in itself somewhat abstruse and difficult, and as it has been observed to be seldom thoroughly understood from definitions and theorems alone, some pains have been taken fully to explain the true nature and the grounds of the doctrine of proportio- mality, in a scholium, which the learner will do well to read with much attention. This scholium may, perhaps, deserve to be recommended to the perusal of those, who continue to study Geometry in Euclid's Elements. - PR E FA C E. xix. The second book, of our first Part, consists entirely of problems. But although the problems are thus given separately, no constructions are supposed to be made in the proofs of the theorems, but such as are either manifestly possible, or are actually shewn to be possible. We readily perceive, for example, when we consider the nature of continued quantity, that, in every given finite line, there is a point which divides it into two equal parts; and it follows, from the definition of a circle, that every given circle has a centre : so that in reasoning about lines and circles, we may always, at once, without any impropriety, assume these points. In developing the theory of Geometry, we may, also, shew the possible earistence of a point, of a line, or of a surface, which shall have some particular position, or, in the case of a line or a surface, some assigned magnitude : this is, indeed, the most proper mode of proceeding, whilst we are pursuing the course of a science of pure reasoning. And if we were occupied with theory alone, no ad- vantage, excepting that of abridged enunciations, would result from giving to any proposition the form of a problem. But there is a practical Geometry, to which the form of a problem, proposing, as it does, to determine some required point, line, angle, or figure, and proceeding in the way of precept, is peculiarly well adapted. It is true, that the postulates, XX PR E FA C E. which are laid down, in order, if they be granted, to render the solutions of the problems rigorously logical, suppose the possibility of performing certain opera- tions, which are really impossible; and that, therefore, our practical Geometry falls short of complete ex- actitude. It is, nevertheless, an useful branch of human knowledge. It is a resource, almost indis- pensible, to the mechanist, the land-surveyor, the architect, and the geographer. To these, and several other artists, it is important to know the method, by which a particular point, line, angle, or figure, may be determined, subject only to the error arising from the want of an absolutely true plane to operate upon, and from the imperfection of the rule and the compass, which must be employed in the operation. It is useful, also, to the teacher of Geometry, to be enabled rightly to construct the diagrams which he makes use of in his demonstrations. With a view; therefore, principally to this practical application of the science, a considerable portion of Geometry is not improperly cast into the form of problems; and, with the same view, it is most convenient to keep this class of pro- positions distinct from the rest. Those, however, who object to such a separation, may refer to the problems, as they find them to be wanted, in reading the demonstrations of the several theorems. For the solutions of the problems are not made to depend PR E FA C E. xxi upon those theorems, in the proofs of which they themselves might be quoted. But it is evident, that nothing will have been gained, with respect to absolute exactness, by thus changing the order of the pro- positions; because the problems suppose the practica- bility of what is actually impracticable. On the other hand, when the possible existence of a point, of a line, or of a surface, is shewn to be consistent with the definitions previously laid down, there is no defect whatever in any part of the chain of reasoning. * To our second Part, which treats of Solid Geo- metry, no problems have been subjoined; and what has been already said may serve to account for this omission. The problems which are denominated solid problems, are, for the most part, of no practical utility, inasmuch as they cannot be constructed by means of the rule and compass. Af & Throughout the following work, certain symbols are used, for the sake of conciseness, many of which are borrowed from the language of universal Arith- metic. They are merely abbreviated representations of the words, and the phrases, for which they are severally put. Without at all altering the nature of the demonstrations, they render the force of them more easily and more quickly perceived. The same plan had been long since partially adopted by Barrow, who took it from Herigone, and it has been followed, xxii PREF Ac E. also, by most of the foreign mathematicians, who have lately written upon the subject of Geometry. The marks employed by Herigone to express an angle, a triangle, and a circle, are in common use at this day. Other parts of his notation do not appear to have been thought worthy of being copied. Having denoted the parallelism of straight lines by the symbol = , he signified equality by 2 | 2, majority by 3 2, and minority by 2 || 3. Instead of the common mark for subtraction, he substitutes - , and takes—AB, to express a straight line, the extremities of which are denoted by A and B. Such a line is, however, more meatly expressed by AB; a circular arch, of which A and B mark the extremities, being also expressed by TB. This notation has, therefore, been adopted in the following treatise: and there is, evidently, a great convenience in the use of it, in all those propositions which relate to arches of circles and their chords, and also in Solid Geometry, where a plane, as well as a straight line, is often indicated by two letters only. Wherever, then, the mark AB occurs, it is to be understood as an abridgment of the phrase, “the straight line of which two different points are denoted by the letters A and B:” and when such a line is first mentioned, it ought either to be expressed by that mark, or else to be described in words: in speaking of the same line afterwards, it PR E FA C E. xxiii may be denoted simply by AB, without repeating the words “the straight line,” or the mark above AB, which is put for them; unless in that particular case in which the points designated by A and B, are, at the same time, the extremities of an arch, and of a straight line, in the diagram. Any deviations from these rules for expressing a straight line, which may occur in the following pages, are to be considered as errors of the press. These, and other such mistakes, which will doubtless be found there, more frequently than they have been marked in the list of errata, will not, it may be presumed, create much perplexity to the reader. And with respect to faults of a weightier kind, which he may easily perhaps discover in the treatise now offered for his perusal, it is hoped that he will look with some degree of indulgence, upon the imperfections incident to a method of arrange- ment, which, although it is not altogether new, has not had the benefit of the reiterated labours and corrections of men eminent for their ingenuity and their mathematical knowledge. Trinity College, December 7, 1819. THE ELEMENTS OF GEOMETRY. PRELIMINARY DEFINITIONS, AND PRINCIPLES DER IV E D FROM THE M. —O— DEFINITION I. GEOMETRY is a science, which has for it's object the determination of the properties of magnitudes in respect of quantity and position, and the measure- ment of extension. f II. The space occupied by a body is it's Bulk, or it's Solid Content; and such a portion of space is called a Solid. III. A Surface is the boundary which divides a solid from the surrounding space; or which divides a solid into two contiguous parts. CoR. A surface cannot have any thickness. A 2 Elements of Geometry. IV. A Line is the boundary of a surface; or it is the boundary which divides a surface into two contiguous parts. CoR. A line cannot have either breadth or thickness. V. The extremity of a finite line, as also the boundary which divides any line into two contiguous parts, is called a Point. * CoR. A point has neither length, nor breadth, nor thickness; that is, it has no parts; but it has position: it may, therefore, be considered as an indivisible mark in linear magnitude. VI. Any portion of space, solid or superficial, which is enclosed by one or more boundaries, is called a Figure. "º VII. A Plane Surface is the boundary which divides solid space, or a portion of solid space, into two parts, which, at that common boundary, have both of them the same form; whether the whole be compared with the whole, or any part of the one with the part con- tiguous to it of the other. Preliminary Definitions. 3 CoR. Two plane, surfaces cannot have a part that is common to both of them. For if they had a common part, it is manifest that they would not both of them divide the sur- rounding space into two portions having, at their common boundary, the same form. VIII. A Straight Line is the boundary that divides a plane surface into two parts, which, at that common boundary, have both of them the same form, whether | _T | they be compared the whole with the whole, or any part of the one with the part contiguous to it of the other. # CoR. 1. Two straight lines in a plane eannot have a common part or segment; because, then, it is evident, the two straight lines would not both of them divide the plane into two portions having, at their commonsboundary, the same form: for the same reason, there cannot lie two straight lines in a plane, between two given points in that plane, which do not coincide with one another: wherefore if two straight lines in a plane have any two points common 4. Elements of Geometry. to them both, they shall wholly coincide with one another. CoR. 2. The common section of two planes, which cut one another, is a straight line. For it cannot (Def. 7. Cor.) be a plane surface; it is, therefore, a line: and, if it be not a straight line, it will turn it's convexity toward one of the parts, into which one of the two planes divides unlimited space, and it's concavity toward the other; and thus one of the two planes will not divide solid space into two parts, which at that common boundary have the same form; which is contrary to the definition: wherefore, the common section of two planes, which cut one another, cannot be any other than a straight line. CoR. 3. A straight line, which has any two given points of it in a given plane, has all it's points in that plane. - For, by the definition, the straight line is in Preliminary Definitions. 5 some one plane; and if this and the given plane be not in the same plane, their common section is (Cor. 2.) a straight line, which necessarily passes through the two given points; and which, being the common section, is also in the cutting plane; unless, there- fore, the part of this common section, between the two given points, coincides with the straight line first proposed, two straight lines in a plane can inclose a space; which, (Cor. 1.) is impossible: wherefore, the proposed straight line coincides with the common section of the two planes, between the two given points, and therefore (Cor. 1.) wholly coincides with it: the proposed straight line has, therefore, all it's points in the given plane. CoR. 4. There cannot be more than one straight line which joins two given points in solid space. For a plane which passes through one of the given points may be supposed to be turned so as to pass through the other; and (Cor. 3.) any straight line which passes through the two given points lies wholly in that plane; more than one straight line cannot, therefore, pass through the two given points; otherwise two straight lines in a plane would enclose a superficial space, which (Cor. 1.) is impossible. , CoR. 5. Any straight line may be always sup- posed to be so placed, as that it shall coincide with another given straight line. For it is manifest that a point of the one may be supposed to be placed on a point of the other; and there is no absurdity in supposing the former line to be turned about that common point, so that some f 6 Elements of Geometry. other point of it shall also fall upon the latter line; and then the two straight lines will be (Cor. 3.) in the same plane, and will, therefore, (Cor. 1.) coincide with one another. - CoR. 6. Any two straight lines, which meet one another, may be supposed to be in the same plane. For the plane, in which either of them is, may be supposed to be turned so as that it shall pass through a point in the other line, besides the point in which the two lines meet, and then (Cor. 3.) the two straight lines will be in the same plane. CoR. 7. If three points a, b, c, of a plane (X) which are not in a straight line, be also in another plane (P.) then every point of the one of these two planes is in the other; that is, the two planes wholly coincide with one another. For (Cor. 3.) the straight lines passing through a, b, and b, c, and c, a, are common to both the planes: but let p be any point of the plane (F) • * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * = which is not in any one of those three straight lines; and, since every point, but c, of the straightline passing Preliminary Definitions. 7 through b, c, is (Cor. 1.) out of the straight line passing through a, c, there is manifestly some point q of the straight line passing through b, c, such that p and q are on contrary sides of the straight line passing through a and c; wherefore the straight line joining the two points p and q of the plane (Y) will necessarily cut the straight line passing through a and c, which is also in the plane (P); let på cut that line in r: therefore, q and r being points that are common to the planes (X) and (P), the straight line q rp is, (Cor. 3.) common to both those planes; wherefore, the point p, the extremity of Trp, is common to both the planes: and, in the same manner, it may be shewn, that any other point in either of the two planes, is common to them both ; therefore, any point of the one of the two planes is in the other. CoR. 8. It is manifest, therefore, that any plane may always be supposed to be so placed, as that it shall coincide with another given plane. CoR. 9. It is also manifest from Cor. 7, that any two straight lines which meet one another, and which (Cor. 6.) are, therefore, in some plane, cannot be in two different planes, which do not coincide with one another. IX. If two straight lines in an unlimited plane be ter- minated one way by a point, in which they meet, but are otherwise indefinite, either of the two portions, into which they divide the plane, is called a Plane Rectilineal Angle: the two straight lines, which bound 8 t Elements of Geometry, the angle, are said to contain the angle, and the point in which they meet is called its Summit; and, of the angles made by two indefinite straight lines which cut one another, any two adjacent angles, that are both wholly on the same side of either of the two lines, are called the Supplements each of the other. * ScholIUM. If two indefinite straight lines AB, CD, cut one another in E, it is manifest that between either of the segments of the one and either of the segments of the other, there are two portions of the plane, each of which answers to the definition of an angle: thus, the straight lines EA, ED, terminated in the point E, but indefinite towards A and D, divide the unlimited plane, in which they lie, into two portions, either of which may be called an angle. The same may be said of the two portions into which an unlimited plane is divided, by two indefinite straight lines that meet in a point, but are in one and the same straight line. So much latitude has been left to the sense of the Preliminary Definitions. : 9 word angle, on account of the use which is made of that term in the doctrine of Trigonometry. The angles which, principally, come under consideration in treat- ing of Geometry, are such, as if either of their con- taining straight lines be supposed to be produced, the whole angle will lie on the same side of that line; and in this sense, unless the contrary be specified, is the word angle always to be understood in the following treatise. Angles which have not that property, have been called Reverse Angles, and Re-entering Angles. X. Magnitudes are said to be of the same kind, when they admit of being put together so as to form one whole. XI. Two magnitudes are said to be equal, if the one, or all the parts into which it has been divided, can be applied to the other, so as wholly to coincide with it: two magnitudes are said to be unequal, if the one be a part, or equal to a part of the other; and of two unequal magnitudes, that which is a part, or equal to a part, of the other, is called the less, and the other the greater. - CoR. I. Magnitudes which are each of them equal to the same magnitude, or which each of them contain the same magnitude, or equal magnitudes, the same number of times, are equal to one another. For if the first and second of three magnitudes be º B 10 Elements of Geometry. each of them equal to the third, when the first is applied to the third so as to coincide with it, let the second be supposed to be applied to the third so as to coincide with it; then, the first and the second, both of them coinciding with the third, will necessa- rily coincide with one another : and, hence, the rest of the corollary is manifestly true. CoR. 2. If to equal magnitudes there be added equal magnitudes, of the same kind with them, the wholes are equal. CoR. 3. If two plane rectilineal angles be placed so as that the summit of the one shall coincide with the summit of the other, and either of the containing straight lines of the one angle with either of the con- taining straight lines of the other, and so as that the planes of the two angles shall coincide, and that the two other containing straight lines shall be on the same side of those which coincide with one another, then, if the plane rectilineal angles be equal, it is manifest that the two other containing straight lines must also coincide: and the same supposition being made, as to the summits and to two of the containing straightlines, if the two other containing straight lines can be placed so as to coincide, then (Def. 8. Cor. 7.) the planes of the two angles will coincide, and the angles will be equal to one another. CoR. 4. If from equal straight lines there be taken equal straight lines, or if from equal plane rectilineal angles there be taken equal plane rectilineal angles, the remainders shall be equal. CoR. 5. If to unequal straight lines equal straight Preliminary Definitions. 11 * lines be added, and if to unequal plane rectilineal angles equal plane rectilineal angles be added, the wholes shall be unequal ; also, if from unequal straight lines equal straight lines be taken, and from equal plane rectilineal angles equal plane rectilineal angles be taken, the remainders shall be unequal. ,- CoR. 6. Either half of the one of two equal straight lines is equal to either half of the other: likewise, either half of the one of two equal plane rectilineal angles is equal to either half of the other, XII. That part of the science of Geometry which treats of the measurement and position of plane figures, of lines that are in plane surfaces, and of plane recti- lineal angles, is called Plane Geometry; and the remaining part, which treats of the measurement and position of solids, of surfaces that are not plane, and of solid spaces lying between planes which meet one another, but which are otherwise indefinite, is called Solid Geometry. § THE ELEMENTs of GEOMETRY. . PART I. THE ELEMENTS OF PLANE GEom ETRY. THE £itmentº of #lant (ſºtomittty. —O— Book I. ELEMENTARY THEOREMS OF P L A N E G E O M ET. R. Y. -sº- C H AIP. I. ON STRAIGHT LINES AND PLANE RECTILINEAL ANGLES. —O— THEoREM I. Any two angles, which are the supplements each of the other, are, together, equal to any other two angles, which are also the supplements each of the other. Let the / AEL, LEB, be the supplements each C R ``s,” __ T. H G. JK A. -* TE of the other; likewise let the // HGF, FGK, be the I6 Elementary Theorems of Plane Geometry, supplements each of the other; and, first, let the containing straight lines of these several angles not be in the same two straight lines: The / AEL, LEB are together equal to the // HGF, FGK. For the straight line GH may (Def. 8. Cor. 5.) be supposed to be so placed, as that the point G shall coincide with the point E, and GH with the straight line EA ; wherefore (Def. 8. Cor. 1.) GK will then likewise coincide with EB; also, the plane in which are HGK and GF may then (Def. 8. Cor. 7.) be supposed to be so placed, as that it shall coincide with the plane in which are MEB and EL, and that GFand EL shall be on the same side of AEB; if, therefore, GF coincide with EL, it is manifest that the / HGF = / AEL, and the Z FGK = / LEB, and that, therefore, the two / AEL, LEB, are together equal to the two / HGF, FGK. But if GF do not coincide with EL, it will coincide with some other straight line EC, in the same plane with AEB and EL: and it is manifest that the two / AEC, CEB are together equal to the three ſ. AEL, LEC, CEB, of which the // LEC, CEB are together equal to the L LEB ; wherefore the // AEC, CEB are together equal to the // AEL, LEB; that is, the // HGF, FGK are to- gether equal to the / AEL, LEB. Next, let the / AEC, CEB, which are the sup- plements each of the other, and the / AED, DEB, which are also the supplements each of the other, have their containing straight lines in the same two straight lines AB, CD, so that all these angles have a common summit E. And let the // HGF, FGK Straight Lines and Angles. 17 be any other two angles, which are the supplements _^ TE: II G. each of the other. Then, as hath been shewn in the former case, the two / AEC, CEB, are together equal to the two / HGF, FGK, as are also the two / CEA, AED, and the two / AED, DEB; wherefore (Def. 11. Cor. 1.) the / AEC, CEB, which are the supplements each of the other, are together equal to the / AEC and AED, which are the supplements each of the other; and the same / AEC, CEB, are also together equal to the // AED, DEB, which are supplements each of the other. Therefore, any two angles, &c. a. E. D. CoR. From the demonstration it is manifest that angles which are the supplements of equal angles, or of the same angle, are equal to one another. ScholIUM. . If AED be any given angle, and if the two straight lines AE, DE be produced so as to contain the / CEB, the two / AED, CEB, which are the supplements of the same LAEC, are commonly called opposite or vertical angles, as are, also, the / AEC, C IX. 18 Elementary Theorems ºf Plane Geometry, DEB. In the course of the demonstration of the A —> 18 H G. preceding theorem, it was shewn that the two / AEC, CEB are together equal to the two / AEC, AED; and taking from these equal aggregates the common / AEC, it follows that the L CEB = L AED; it was also shewn that the two / AEC, AED are together equal to the two / AED, DEB; and taking from these equal aggregates the common L AED, it follows that the / AEC = / DEB. If, therefore, two straight lines cut one another, the ver- tical, or opposite, angles shall be equal : this is what is expressed in Th. 1. Cor. 1; and it is more fully explained here, on account of the frequent use which is made of it, in the doctrine of Plane Geometry. For the same reason, it may not be improper to shew how this proposition may be proved in a somewhat different manner. Let, then, the / CEB, AED be the supplements of any the same / CEA ; suppose, also, the L FGH to be equal to the 4 CEA, and let the / FGK be the supplement of the / FGH. And if the L FGH be applied to the equal / AEC, so as that the summit Straight Lines and Angles. 19 G shall coincide with E, and the straight line GH with EA, and that GF shall be on the same side of AEB that EC is, and that the planes of the two angles shall coincide, then (Def. 11. Cor. 3.) GF shall coincide with EC; also (Def. 8. Cor. 1.) GK will then coincide with EB; therefore (Def. 11. Cor. 3.) the / FGK= / CEB. ... º. Again, if the / HGF be applied to the equal L AEC, so that the summit G shall coincide with E, the straight line GH with EC, and the plane of the 1 FGH with the plane of the / CEA, and so as that GF shall be on the same side of CED, that EA is, then (Def. 11. Cor. 3.) GF shall coincide with EA; also, HG coinciding with CE, GK (Def. 8. Cor. 1.) coincides with ED, and therefore (Def. 11. Cor. 3.) the 4 FGK = / AEB; and it has been shewn that the / FGK= / CEB; wherefore (Def. 11. Cor. 1.) the / CEB = 4 AED. Therefore, angles which are the supplements, &c. a. E. D. ScholIUM. The angles made with each of the two containing straight lines of a given angle, by producing the other containing straight line, are both of them (Def. 9.) called the supplements of the given angle; thus the // AEC and DEB are, each of them, the supplements of the L AED; and they have been shewn (Th. 1.) to be equal to one another: if, there- fore, an angle be equal to one of it's two sup- plements, it is equal to the other; so that there is no ambiguity in describing an angle, as being equal 20 Elementary Theorems of Plane Geometry, to it's supplement, without particularizing to which of the two supplements it is equal. Likewise, if an angle be less than one of it's two supplements, it is less than the other, and if it be greater than the one of them, it is greater than the other. Hence, it is evident that there is no absurdity in the next follow- ing definition. • DEF. XIII. Of plane rectilineal angles, a Right Angle is that which is equal to it's supplement; and either of it's containing straight lines is said to be a perpendicular to the other, or to be at right angles to the other: an Acute Angle is that which is less than it's sup- plement; and either of it's containing straight lines is said to be inclined toward the other: an Obtuse Angle is that which is greater than it's supplement. And two straight lines which contain any other angle than a right angle, are said to be oblique to one another. - CoR. 1. It is manifest, therefore, that if the first of two straight lines, which cut one another, be at right angles to the second, the second shall also (Th. 1.) be at right angles to the first. - CoR. 2. The angles which one straight line makes with another, upon the one side of it, are either two right angles, or are (Th. 1.) together equal to two right angles. CoR. 3. All the angles, which any number of straight lines make with a given straight line, at one Straight Lines and Angles. 2] point in it, and on the one side of the line, are together equal to two right angles. CoR. 4. If two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles. CoR. 5. Therefore all the angles made by any number of straight lines, all in the same plane, and all meeting in one point are together equal to four right angles. THEOREM II. All right angles are equal to one another. For, if any two right angles be unequal, one of them must be greater than the other and therefore together with it's supplement (Def. 13.) it must also, be greater than the other together with it's supplement; which (Th. 1.) is impossible; therefore no two right angles can be unequal; wherefore all right angles are equal to one another. CoR. It is manifest, therefore, that from a given point, in a given straight line, there cannot, on the same side of it, be drawn more than one straight line at right angles to it. Theorem III. If two straight lines on opposite sides of another straight line, and in the same plane with it, meet at a point in that line, so as to make the opposite angles equal to one another, these two straight lines shall be in one and the same straight line. 22 Elementary Theorems of Plane Geometry, At the point Ein AEB, let, CE, DE, on opposite {} A.— ~ ID B TMI sides of AEB and in the same plane with it, make the opposite / AEC, DEB, equal to one another: CE and ED are in the same straight line. For if ED be not in the same straight line with CE, let EM be in the same straight line with it; and, because CEM is a straight line, (Th. 1.) the Ł AEC= / MEB; but (hyp.) the / AEC= / DEB; therefore the LMEB = A, DEB; i.e. the less is equal to the greater, which is impossible; therefore EM is not in the same straight line with CE; and, in like manner, it may be shewn that no other can be in the same straight line with it but ED; which is, there- fore, in the same straight line with CE. Wherefore, if two, &c. a. E. D. THEOREM IV. If at a point in a straight line, two other straight lines upon opposite sides of it, and in the same plane with it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. Straight Lines and Angles. 23 At the point B, in AB, let CB, DB, on opposite A. C Tºš D at , sides of AB, and in the same plane with it, make the adjacent / ABC, ABD, together equal to two right angles: BD is in a straight line with CB. For if BD be not in a straight line with CB, let BK be in a straight line with it; therefore the two / CBA, ABK are (Def. 13. Cor. 2.) together equal to two right angles, as are also (hyp.) the two / CBA, ABD; therefore the two / CBA, ABK, are together equal to the two / CBA, ABD; take away the common A CBA, and there remains the / ABK equal to the Z. ABD, the less to the greater; which is impossible: wherefore, BK is not in the same straight line with CB; and in like manner, it may be shewn that no other can be in the same straight line with it but BD; which is, therefore, in the same straight line with CB. Wherefore, if at a point, &c. a. E. D. THEOREM. V. If a straight line, falling upon two other straight lines, makes the alternate angles equal to one another, • these two straight lines, being produced ever so far both ways, shall not meet. 34 Elementary Theorems of Plane Geometry, Let EF, which falls upon AB and CD, make the C /. *— T) alternate / AEF, EFD, equal to one another: AB and CD being produced ever so far both ways, do not meet. For if it be possible, let AB and CD, produced toward B and D, meet in the point G. Then, if the figure GEF be applied to AEFC, so that the point F may fall on E, and FE on EF, the straight line FDG will coincide with EA, and the straight line EBG with FC, because (hyp.) the / AEF= / EFD, and (Th. 1. Cor.) the / BEF= / EFC; wherefore EA and FC, being produced toward A and C, must also meet in some point H, coinciding with the point G.; and thus AB and CD, being produced, will meet both ways, so that there will be two straight lines, in a plane, which do not coincide with another, lying between the two points G and H ; which (Def. 8. Cor. 1.) is impossible; wherefore AB and CD can no where meet toward B and D. In the same manner, it may be shewn that they can no where meet toward A and C. Wherefore, AB and CD, being produced ever so far both ways, shall not meet. Therefore, if a straight line, &c. a. E. D. Straight Lines and Angles. 25 DEF. XIV. Straight lines which are in the same plane, and which, being produced ever so far both ways, do not meet, are called Parallel Straight Lines. CoR. If, therefore, a straight line, falling upon two other straight lines, makes the alternate angles equal to one another, these two straight lines shall (Th. 5.) be parallel. ScholluM. A straight line may always be supposed to be drawn parallel to a given straight line, through a given point without it. For, if the given point and any point in the given line be supposed to be joined by a straight line, it is plainly possible for another straight line to make with the joining straight line, at the given point, an angle equal to that which the joining line makes with the given line, and, therefore, (Def. 14. Cor.) to be parallel to the given line. THEOREM VI. If a straight line, falling upon two other straight lines, makes the exterior angle equal to the interior, and opposite, upon the same side of the line ; or makes the interior angles, upon the same side, together equal to two right angles; the two straight lines shall be parallel to one another. D 26 Elementary Theorems of Plane Geometry, Let EF, which falls upon AB and CD, make the exterior / EGB equal to the interior and opposite / G HD, upon the same side; or make the interior // BGH, GHD, on the same side, together equal to two right angles; AB is parallel to CD. For the LEGB = A GHD, and (Th. 1. Cor.) the / EGB- / AGH; therefore the / AGH = / G HD; and they are alternate angles; therefore (Def. 14. Cor.) AB is parallel to CD. -- Again, because (hyp.) the / BGH, GHD, are equal to two right angles; and that (Def. 13. Cor. 2.) the / AGH, BGH, are also equal to two right angles, the / AGH, BGH are equal to the // BGH, G HD : take away the common / BGH; therefore the L AGH = / GHD; and they are alternate angles; therefore AB is parallel to CD. Wherefore, if a straight, line &c. G. E. D. Cor. 1. There cannot be more than one per- pendicular to a given straight line, passing through a given point without it. For, if there can, then two straight lines may meet, that make with another straight line the interior Straight Lines and Angles. 27 angles, upon the same side, together equal (Def. 13.) to two right angles; which (Th. 6.) is impossible. CoR. 2. If a finite straight line AH, meet two straight lines AF and HB of indefinite length, so as F jº –0 y G B IH C to make with them, on the same side, two // FAH, BHA, of which one, the / BHA, is not less than a right angle, and the other, the L FAH, is greater than a right angle, the two straight lines AF, HB, shall not meet on that side of the finite straight line AH, however far they are produced. For, if A.E be supposed to be drawn from A, at right angles to AH, AE will fall within the Z. FAH, which (hyp.) is greater than a right angle; and if the / AHB be a right angle, then (Th. 6.) AE and HB will not meet however far they are produced; wherefore, AE will always lie between A.F and HB, however far all the three lines are produced towards F, E, and B. AF cannot therefore, meet HB, towards F and B, however far those two lines are roduced. Hence it is plain that AF cannot meet HB, if the / AHB be greater than a right angle. CoR. 3. If from a given point, without a given straight line, two straight lines be drawn, one per- 28 Elementary Theorems of Plane Geometry, pendicular, and the other oblique, to the given straight line, the perpendicular shall lie within the aeute angle contained by the given straight line and the line which is oblique to it. AssumpTION I. Through any given point within an acute angle, or within a right angle, or within an obtuse angle, a straight line may be supposed to pass, which shall cut the two straight lines that contain the angle. SCHOLIUM. If BAC be an acute angle, or a right angle, or an obtuse angle, and D a given point within it, and if it A K. G. -º-º-º- H / - gº ` .1B be granted that amongst the infinite number of points without the L BAC, there are two, on contrary sides of the angle, which are in the same straight line with the point D; or, if it be granted, that, inasmuch aS AB and AC are unlimited in length towards B and C, two points G and H may be taken, one of them in AB, the other in AC, so that every point in GH, Straight Lines and Angles. 29 which joins G and H, shall be farther distant from A than D is ; since then GD, which joins G and D, must, if it be produced far enough, cut AC between A and H, in either of these cases, the assumption, made above, will have been allowed. It is not, how- ever, a new assumption. For it is, in like manner, taken for granted, in the twentieth proposition of the eleventh book of Euclid's Elements. THEOREM VII. If a straight line meets two straight lines, so as to make the two interior angles on the same side of it, one of them a right angle, and the other less than a right angle, these two straight lines, being continually produced, shall meet upon that side, on which the angles are one of them a right angle and the other less than it. Let AB meet AC and GD so as to make the two A CAB, ABD, one of them, namely the / ABD a A. IE right angle, and the other CAB less than a right 30 Elementary Theorems of Plane Geometry, angle: AC and GD, being continually produced, shall meet towards C and D. - At the point A, let EA be supposed to make with BA the / EAB= / CAB, which (hyp.) is less than a right angle; therefore, the whole / CAE is less than two right angles; and (Assump. I.) there is some straight line which may be supposed to pass through the point B and to meet AC and AE; let this line be EBF. If EBF coincide with BD, it is plain that BD and AC, being produced far enough, shall meet: and, if BD lie within the / ABF it will, manifestly, being produced, meet AC between A and F. But, if the LABD be greater than the LABF, let DB be supposed to be produced toward G.; then (hyp.) the LABF is less than a right angle, and there- fore (Def. 13. Cor. 2.) the LABE is greater than the right angle ABG; wherefore, BG shall meet AE, if it be produced far enough, between A and E; let it meet AE in the point G. And, if the / BAG be applied to the LBAC, to which (hyp.) it is equal, so that AG may be on AC, BG will coincide with BD, because (hyp. and Def. 13) the LABG = / ABD; wherefore, AG thus coinciding with AC and BG with BD, it is manifest that AC and BD pro- duced shall meet towards C and D, in some point H, which coincides with the point G. Therefore, if a straight line, &c. a. E. D. CoR. If a straight line meets two other straight lines, so as to make each of the interior angles less than a right angle, or one of the interior angles a right Straight Lines and Angles. 31 angle and the other interior angle, on the same side, not a right angle, these two straight lines, being con- tinually produced both ways, shall meet. The first case of the corollary is manifest from Th. 7. And, in the second case, the interior angle, which is not a right angle, is either less than a right angle, or greater than a right angle ; if it be less, the two straight lines will (Th. 7.) meet, if they be far enough produced, on that side on which these angles are ; and if the interior angle which is not a right angle, he greater than a right angle, it's supplement (Def. 13. Cor. 2.) is less than a right angle; but (Def. 13.) the supplement of the other interior angle is in this case, a right angle; wherefore, the two straight lines, being produced, will in that case, meet (Th. 7.) on the other side of the line which makes with them one of the interior angles a right angle and the other greater than a right angle. SCHOLIUM. A perpendicular may always be supposed to be drawn to a given indefinite straight line (XP.) from a given point (A) without it. For at any point B, in XP, it is evidently possible IP A. 32 Elementary Theorems of Plane Geometry, for a straight line to stand at right angles to XP; and if it pass through 4, then is there a perpendicular to XP passing through A: but if PBQ, supposed to be at right angles to XP, do not pass through A, let A, B be supposed to be joined by AB; it is mani- festly possible for a straight line ACZ to make at A with AB an a B.AZ= 4. ABP, so that (Th. 5.) ACZ can no where meet PBQ, but (Th. 7. Cor.) shall meet AP, because each of the // ABP, BAZ, is less than a right angle; and this straight line AZ shall be per- pendicular to XY; for if the angles at C, where AZ meets XP, be not right angles, AZ (hyp. and Th. 7. Cor.) will meet PQ; but AZ, as hath been shewn, does no where meet PQ; wherefore the angles at C cannot be other than right angles; that is, ACZ passes through the given point A and is perpendicular to XP. A method of drawing a perpendicular to a given straight line, from a given point without it, will be shewn in Book II, of this treatise, from prin- ciples that are not dependent upon the Assumption, which has above been made, for the purpose of de- termining the properties of parallel straight lines. THEOREM VIII. There cannot pass two straight lines through the same point, each of them parallel to the same straight line, which do not, coincide with one another. Let ED, which passes through the point A, be parallel to BC; there cannot pass through 4 any * Straight Lines and Angles. 33 other straight line parallel to BC, which does not F * #T *—º G. H- H r G coincide with ED. For if it be possible, let FAG, which does not coincide with ED, pass through A and be parallel to BC; and from A let AH be supposed to be drawn perpendicular to BC: the two // FAH, EAH cannot (Th. 2. Cor.) both of them be right angles; where- fore, since (hyp.) AHB is a right angle and one of the two / FAH, EAH, not a right angle, one of the two lines EAD, FAG, being produced, will (Th. 7. Cor.) meet BC produced, and therefore (Def. 14.) will not be parallel to BC; ED and FG are not, therefore, both of them parallel to BC. Therefore, there can- not pass, &c. a. E. D. CoR. 1. If a straight line cut one of two parallel straight lines, it shall, being continually produced, cut the other also: else there would be two straight lines passing through the same point and both of them parallel to the same straight line; which (Th. 8.) is impossible. CoR. 2. If two straight lines, which meet in a point, be each of them parallel to any one straight line, they shall be in one and the same straight line. E. 34 Elementary Theorems of Plane Geometry, CoR. 3. If three given straight lines, in a plane, meet in one point, and if no two of them be in the same straight line, any straight line which cuts one of them, shall also cut another of them, the straight lines being produced indefinitely. For if the line, which cuts one of the three that meet in a point, cut neither of the other two, then through the same point there pass two straight lines, which do not coincide with one another, and each of which is parallel to the same straight line; which (Th. 8.) is impossible. Wherefore the line, which cuts one of the three given lines, shall also cut another of them. * THEoREM IX. Two straight lines which are parallel to the same straight line, are parallel to one another. For, if the two straight lines, being continually produced both ways, meet on either side, then through the same point there may pass two straight lines, both of them parallel to the same straight line; which (Th. 8.) is impossible; wherefore, the two straight lines, if produced ever so far, can no where meet; that is, (Def. 14.) they are parallel. a. E. D. THEOREM X. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite on the same side; and also the two Straight Lines and Angles. 35 interior angles on the same side together equal to two right angles. Let EF fall upon the parallels AB and CD; the alternate / AGH, G HD are, equal to one another; the exterior / EGB is equal to the interior and op- posite / GHD, on the same side; and the two interior // BGH, GHD, on the same side, are together equal to two right angles. For if the LAGH be not equal to the 4 GHD, at the point G in HG suppose the LHGK to be made equal to the 4 GHD; wherefore (Def. 14. Cor.) GK is parallel to CD ; and (hyp.) GA is also pa- rallel to CD; which (Th. 8.) is impossible; there- fore, the LAGH is not unequal to the / GHD, i. e. , it is equal to it. Again, (Th. 1. Cor.) the / AGH = / EGB; there- fore the / EGB- / GHD; and if to each of these equals be added the LBGH, the / EGB, BGH are together equal to the / BGH, G HD; but (Def. 13. Cor. 2.) the / EGB, BGH are together equal to two right angles; therefore, the / BGH, G HD are also 36 Elementary Theorems of Plane Geometry, together equal to two right angles. Wherefore, if a straight line, &c. a. E. D. CoR. 1. If a straight line (GH) meet two straight lines (KL, CD) so as to make the two interior angles, on the same side, (KGH, GHC) together less than two right angles, these two straight lines shall meet on that side, on which are the two angles (KGH, GHC) that are less than two right angles. Through G let AGB be supposed to be drawn parallel to CD; therefore (Th. 10.) the two / AGH, GHC are together equal to two right angles; and (hyp.) the two / KGH, GHC being less than two right angles, are less than the two / AGH, GHC ; therefore the / KGH is less than the / AGH ; and (Th. 1. Cor.) the z KGH = ſ. EGL, and the / AGH = / EGB; therefore the / EGL is less than the Z EGB, so that GL falls within the / EGB ; it is plain, therefore, that KL cannot meet CD towards L and D without again cutting GB, which (Def. 8. Cor. 1.) it cannot do; therefore KL cannot cut CD towards L and D; but KL is not parallel to CD, for then (Th. 10) the two / KGH, GHC, would be equal together to two right angles, than which (hyp.) they are less; wherefore, since KL is not parallel to CD and cannot meet it towards L and D, it must, if it be far enough produced, meet CD towards K and C; that is on that side of GH, on which are the angles that are less than two right angles. CoR. 2. A perpendicular to any given straight line is also a perpendicular to every parallel to that line. Straight Lines and Angles. 37 For if the perpendicular be produced far enough, it will (Th. 8. Cor. 1.) cut any parallel to the given line; and (Th. 10.) it will cut it so as to make the two interior angles together equal to two right angles: and since (hyp.) one of these angles is a right angle, the other is also a right angle; that is, the cutting line is perpendicular to the parallel. CoR. 3. If two straight lines (AB, CD) be pa- rallel and from any two points (A, B) of the one, (AB), perpendiculars (AC, BD) be drawn to the other (CD), these two perpendiculars shall be equal to one another. - For suppose CD to be "bisected in F, and FE to be drawn from F perpendicular to CD, and therefore A. Tº B vº, ºr— C F T) (Cor. 2.) also perpendicular to AB. Then if the figure AEFC be supposed to be applied to the figure BEFD so that FC may wholly coincide with it's equal FD, and the plane of the one figure with the plane of the other, FE will still be a common bound- ary of the two figures, because (hyp. and Def. 13.) —z— * To divide any geometrical quantity into two equal parts, is commonly expressed by the single word to bisect; and in that sense the word bisect is to be understood, wherever it occurs in this treatise. 38 Elementary Theorems of Plane Geometry, the / EFC = / EFD ; also, the / FEA = / FEB; and the / FCA = LFDB ; therefore, EA and CA will coincide with EB and DB, each with each, and the point A will fall on the point B; wherefore, the point C coinciding with D, the straight line CA with DB, and the point A with B, it is plain (Def. 11.) that AC= BD, CoR. 4. The straight lines (AC, BD) which join equal and parallel straight lines (AB, CD) towards the same parts, are themselves equal and parallel. For, suppose AB and CD to be bisected in E and F, and E, F to be joined by EF ; then (hyp.) AE, A. IE E C T D EB, CF, FD, are all equal to one another; and (hyp. and Th. 10.) the / AEF= / EFD, and the / CFE = / FEB; if, therefore, the figure BEFD be applied to the figure AEFC, so that the point F may fall on E, and FE on EF, the straight line FD will coincide with EA and the point D with A, because the / EFD = / FEA, and FD= EA ; in like manner it may be shewn that EB will also coin- cide with FC, and the point B with C; so that the point D falling on A, and the point B on C, DB will Straight Lines and Angles. . 39 (Def. 8. Cor. 1.) coincide with AC, and (Def. 11.) be equal to it. CoR. 5. If any number of equal straight lines be each Óf them terminated by a given straight line, and make equal angles with it towards the same parts, their other extremities shall be in a straight line which is parallel to the given straight line. For (hyp. and Th. 6.) the equal straight lines are parallel to one another; and (Cor. 4.) the straight lines joining their extremities are parallel to the given line; therefore (Th. 8. Cor. 2.) they are in one and the same straight line. * CoR. 6. If two straight lines (AB, AC) which contain an angle (BAC) be parallel to other two (DE, DF) each to each, which contain an angle (EDF) the two angles shall be equal to one another. For since ED meets DF, it will, produced if it be necessary, also meet (Th. 8. Cor. 1.) AC, which (hyp.) is parallel to DF; let ED meet AC in G ; wherefore (Th. 10.) the Z. EDF = / DGC, and the / DGC= / BAC; therefore the / BAC = / EDF THI E 35itments of 19tarte (Reemtetry. Book I. --—sº--- CHAPTER II. on THE PRIMARY RELATIONS OF THE sides, ANGLES AND SURFACES, OF PLANE RECTILINEAL FIGURES, IN RESPECT OF EQUALITY OR IN EQUALITY. —O— SECTION I. On the primary relations of the sides, angles and surfaces of triangles in respect of equality or in- equality. *** ****&^^**** DEFINITION XV. A TRIANGLE is a plane figure contained by three straight lines. SCHOLIUM. If in the two straight lines, which contain a given angle, two points be taken that are equidistant from Primary Relations of Sides, &c. of Triangles. 41 the summit of the angle, and if these two points be joined by a straight line, there will be formed a tri- angle which has two of it's sides equal to one another. Again, if in the two straight lines which contain an angle two points be taken, of which one is further from the summit than the other is, and if these points be joined, there will be formed a triangle which has two of it's sides unequal. And since (Th. 10. Cor. 1.) any two straight lines, which make with a third straight line, the two interior angles on the same side, together less than two right angles, will meet if they be continually produced, it is plain that the two angles at the base of a triangle may be equal to one another, or they may be unequal; and that they may be both of them acute, or one of them obtuse and the other acute. THEOREM XI. W If one side of a triangle be produced, the eacterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let it's side BC be produced to D; the exterior / ACD is greater than either of the interior opposite / CBA, BAC. . . For if the LACD be not greater than the / ABC it is either equal to it, or less than it. The LACD is not equal to the AABC, because then (Th. 6.) CA and BA could not meet; and (hyp.) they meet in A. But, if it be possible, let the / ACD be less F 42 Elementary Theorems of Plane Geometry, than the LABC: then from the LABC, the greater, A. E. º T) F there may be supposed to be cut off the LEBC equal to the LACD the less; and BE, since it lies between BA and BC will, manifestly, if it be far enough produced, meet AC between the points A and C; but, because (hyp.) the LACD=/ EBC, BE and CA cannot (Th. 6.) meet one another; wherefore the / ACD is not less than the LABC; and it has been shewn, that the two angles are not equal to one another; therefore the LACD is greater than the 4 ABC. In the same manner, if AC be produced to F, it may be shewn that the / BCF, that is (Th. 1. Cor) the LACD, is greater than the LBAC. There- fore, if one side, &c. a. E. D. CoR. 1. Any two angles, (ABC, BCA) of a triangle (ABC) are together less than two right angles. - - For if BC be produced, the / ABC < LACD; to each of these add the / ACB; and the two / ABC, ACB are less than the two / ACD, ACB, that is (Th. 2. Cor. 3.) than two right angles. Primary Relations of Sides, &c. of Triangles. 43 CoR. 2. If the two angles at the base of a triangle be acute, a straight line drawn from the summit per- pendicular to the line of the base, will fall within the triangle; but if one of the angles at the base be obtuse, the straight line drawn from the summit, per- pendicular to the line of the base, will fall without the triangle. CoR. 3. A triangle cannot have more than one angle that is not less than a right angle. THEOREM XII. The three interior angles of every triangle are together equal to two right angles. Let ABC be a triangle; it's three / ABC, BCA, A. E. IB —& I) CAB, are together equal to two right angles. Suppose the side BC to be produced to D, and since (Th. 11.) the LACD » ABC, from the / ACD suppose the / ECD to be cut off equal to the Z. ABC; wherefore (Th. 6.) CE is parallel to BA; and, since AC meets them, the / ECA = / CAB (Th. 10.); wherefore, the whole / ACD = / CAB + / ABC; to each of these add the / ACB;” and the / ACD + A ACB, that is (Def. 13. Cor. 2.) two right angles, 44 Elementary Theorems of Plane Geometry, will be equal to the three / CAB, ABC, BCA. Therefore, the three interior angles, &c. a. E. D. CoR. 1. From the demonstration it is manifest that, if a side of a triangle be produced, the exterior angle is equal to the two interior and opposite angles. CoR. 2. If one angle of a triangle be a right angle, the two remaining angles are together equal to a right angle: if one angle be an obtuse angle, the two remaining angles are together less than a right angle; and if two angles be acute, but together greater than a right angle, the third angle is also acute. - CoR. 3. If two angles of a triangle be given, the magnitude of the third angle is thereby determined : and if two triangles have two angles of the one equal to two angles of the other, their third angles shall also be equal. - ScholIUM. It is manifest from Th. 12. Cor. 2. and from the last Scholium, that a triangle can have one right angle and only one; that a triangle can have one obtuse angle, and only one; and that a triangle may have it's three angles each of them an acute angle. Hence, the following definitions. DEF. XVI. A Right-angled Triangle is that which has a right angle. T - Primary Relations of Sides, &c. of Triangles. 45 xvii. An Obtuse-angled Triangle is that which has an obtuse angle. XVIII. An Acute-angled Triangle is that which has three acute angles. - CoR. Any two angles of an acute-angled triangle are (Th. 12.) together greater than the third angle. THEOREM XIII. If two sides of a triangle be equal to one another, the angles which they subtend shall be equal to one another. Let the sides AB, AC of the triangle ABC be N 2' Tº ID T G. equal to one another, the angles opposite to them, viz. the / ACB, ABC shall be equal to one another. Let BA and CA be supposed to be produced towards D and E, so that AD and AE may be each 46 Elementary Theorems of Plane Geometry, of them equal to AB or AC, and let EDjoin the points E and D: and since also (Th. 1. Cor.) the LEAD = L BAC, if the / BAC be supposed to be applied to the LEAD, so that AC may wholly coincide with it's equal AD, it is plain that AB will wholly coin- cide with AE; and the points B and C thus coin- ciding with E and D, the straight line BC will (Def. 8. Cor. 1.) coincide with ED, and the / ABC with the LAED; wherefore (Def. 11.) the / ABC = / AED: again, if the / BAC be supposed to be applied to the equal / EAD, so that AC may wholly coincide with it's equal AE, it may in like manner be shewn that the / ACB = AE AED; wherefore (Def. 11. Cor. 1.) the / ABC = LACB. If, there- fore, two sides of a triangle, &c. a. E. D. CoR. 1. If a triangle (ABC) have two equal sides (AB, AC) and if they be produced (to Fand G) the // (FBC, GCB) on the other side of the base shall also be equal. For (Def. 13. Cor. 2.) the two / ABC, CBF are together equal to two right angles, as are also the two / ACB, BCG ; wherefore, the two former angles are together equal to the two latter; and (Th. 13.) the LABC= / ACB; if, therefore, these equal angles be taken away, there will remain the angle CBF= A BCG. CoR. 2. If two sides of a triangle be equal to one another, the two angles at it's base being equal and (Th. 11. Cor. 1.) together less than two right angles, are acute angles. Primary Relations of Sides, &c. of Triangles. 47 CoR. 3. The angles at the base of a right-angled triangle, which has two sides equal to one another, being (Th. 12. Cor. 2.) together equal to a right angle, and being (Th. 13.) equal to one another, are each of them half a right angle. THEOREM XIV. If two sides of a triangle be unequal, the angle opposite to the greater side is greater than the angle opposite to the less side. Let ABC be a triangle, of which the side AC is A. T) IB C greater than the side AB : the LABC is greater than the Z. ACD. From AC the greater of the two sides, suppose AD to be cut off equal to AB the less, and BD to be drawn; the exterior / ADB, of the AS, BDC, is greater (Th. 13.) than the interior and opposite / DCB; but (hyp.) AB = AD, and therefore the / ABD = LADB, wherefore the LABD is greater than the / DCB or ACB; much more, then, is the A ABC greater than the / ACB. Therefore, if two sides, &c. a. E. D. 48 Elementary Theorems of Plane Geometry, CoR. If two sides of a triangle, that are unequal, be produced, the angles on the other side of the base shall (Def. 13. Cor. 2. and Th. 14.) be unequal; and that, which is adjacent to the greater of the two sides, shall be the greater. THEOREM XV. If two angles of a triangle be equal to one another, the sides, also, which subtend the equal angles shall be equal to one another. For, if the sides opposite to the equal angles be unequal, the angles which they subtend, will (Th. 14.) be unequal; which is contrary to the hypothesis; wherefore, those sides are not unequal; that is, they are equal to one another. - THEOREM XVI. If two angles of a triangle be unequal, the side subtending the greater angle is greater than the side subtending the less. Let ABC be a triangle of which the / ABC is A. C greater than the LACB : the side AC is greater than the side AB. Primary Relations of Sides, &c. of Triangles. 49 For if it be not greater, either AC=AB, or AC / DAC, and therefore (hyp.) > a right angle; therefore (Th. 11. Cor. 1.) the a DCE > DEC; therefore (Th. 16.) DE-DC; and in the same manner it may be shewn that DC-BC; much more then is DE-BC. And if BF be any straight line drawn from the extremity B of AB to meet AC, produced, in F, it may in like manner be shewn that BFS BC. G. 50 Elementary Theorems of Plane Geometry, ScholIUM. If two straight lines contain an angle, which is the third part of two right angles, the straight line joining any two points equidistant from the summit, will make (Th. 13.) with the two containing straight lines angles that are equal to one another, and therefore (Th. 12.) each of them also equal to the third part of a right angle. Thus there is a triangle which has all it's angles equal to one another, and which has there- fore also (Th. 15.) all it's sides equal to one another. But if two points be taken in the containing straight lines of any other given angle, equidistant from the summit, the straight line joining them, will make with the containing straight lines equal angles, but which are not equal to the given angle. A triangle may, therefore, (Th. 16.) have only two equal sides. Again, if two points be taken in the straight lines containing a right angle, or an obtuse angle, at un- equal distances from the summit, the straight line joining them will be the base of a triangle which (Th. 11. Cor. 1. and Th. 16.) has no two of it's sides equal to one another. It is plain, also, from Th. 10. Cor. 1. and Th. 12. that there may be a triangle, which has no two of it's angles equal to one another, and which (Th. 16.) has therefore no two sides equal to one another. Hence the following definitions. DEF. XIX. - An Equilateral Triangle is that which has three Primary Relations of Sides, &c. of Triangles. 51 equal sides: an Isosceles Triangle is that which has only two equal sides: and a Scalene Triangle is that which has three unequal sides. CoR. An equilateral triangle is (Th. 13.) equi- angular, and (Th. 12.) each of it's angles is the third part of two right angles. THEoREM XVII. Any two sides of a triangle are together greater than the third side. If a triangle be equilateral it is manifest that any two of it's sides are greater than the third. But let ABC be a triangle, which is not equilateral; let AB A. D B C be it's least side, and from either of the two remain- ing sides as AC, suppose AD to be cut off equal to AB, and BD to be drawn. And, since (Def. 13. Cor. 2.) the two / ADB, BDC, are equal to two right angles, they are greater (Th. 11. Cor. 1.) than the Z ABC, that is, than the two / ABD, DBC; but (hyp. and Th. 13.) the LADB= LABD; take away these equal angles, and there remains the LBDC > / DBC; therefore (Th. 16.) BC > DC; to these unequals add the equals BA and AD, and it is plain 52 Elementary Theorems of Plane Geometry, 2 that CB+ BA > CD + DA or CA. In the same manner it may be shewn that CA + ABS-BC; and since (hyp.) AB is the least side, it is manifest that BC+CA-AB. Therefore, if a triangle, &c. a. E. D. CoR. 1. From the demonstration it is evident that any side of a triangle is greater than the difference between the two remaining sides. CoR. 2. The three sides of a triangle are together greater than the double of any one side, and less than the double of any two sides. THEOREM XVIII. If from the ends of the side of a triangle, there be drawn two straight lines to a point within the tri- angle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Let BD, CD, be drawn from the ends B, C, of A Tº the side BC of the As ABC, to the point Dwithin it; BID+ DC < BA+AC, but the / BDC > / BAC. Suppose BD to be produced to meet AC in E; then (Th. 17.) in the AS BAE, BA+AE-BE; add to each of these EC, and it is plain that BA+AE +EC, that is, BA+AC, >BE+EC; likewise, in the Primary Relations of Sides, &c. of Triangles. 53 As DEC, DE-- EC-DC, and therefore, BD+ DE + EC, or BE + EC, >BD + DC; much more then is BA-HAC- BD + DC. Again, because (Th. 11.) the exterior / BDC-> / DEC, of the /S EDC, and the exterior / BEC > / BAE of the ASABE, much more then is the A BDC-> / BAE or BAC. Therefore, if from the ends, &c. a. E. D. CoR. From the demonstration it is manifest that any two sides of a triangle are together greater than any part of one of them, together with the straight line joining the end of that part and the summit of the opposite angle. THEOREM XIX. If two triangles have a side of the one equal to a side of the other, and two angles also of the one equal to two angles of the other, each to each, and if the equal sides be either adjacent to the equal angles, or opposite to equal angles in each, the two triangles shall be equal. Let the /> ABC, DEF, have the / BAC= / EDF and the / ACB = A DFE; and first let the sides AC, DF, which are adjacent to the equal angles, be equal to one another: the ASABC = AS DEF. For if the AS DEF be applied to the ASABC, so that the point D may be on A, and DF on it's equal AC, DE will coincide with AB, and FE with CB, because (hyp.) the LFDE = / CAB, and the / DFE = / ACB; wherefore the point E will be in AB, and also in CB; it must, therefore, coincide with the 54 Elementary Theorems of Plane Geometry, intersection, B, of AB and CB, and thus DE = AB, *. A * A. H B C E. FE = CB, the / DEF= / ABC, and (Def. 11.) the As DEF=/>ABC, with which it wholly coincides. Next, let the sides which are opposite to equal angles, in each triangle, be equal to one another, viz. AB to DE: likewise, in this case, the ASABC = ZS DEF. For in this case, also, the Z ABC = / DEF; else one of them must be the greater; let the / ABC be the greater of the two; and at the point B, in AB, suppose the / ABH to be made equal to the LDEF. Then it may be shewn as in the former case that the Z, AHB = L DFE; but (hyp.) the / DFE = Z. ACB; wherefore the / AHB = Z. ACB, which (Th. 11.) is impossible. Therefore, the LABC is not un- equal to the / DEF, that is, it is equal to it; since then (hyp.) AB = DE, and the L BAC = LEDF, and that the / ABC= / DEF, therefore, as in the former case, the /S, ABC = As DEF. Therefore, if two triangles, &c. a. E. D. CoR. 1. From this it is manifest that if two tri- angles have a side of the one equal to a side of the other, and two angles also of the one equal to two Primăry Relations of Sides, &c. of Triangles. 55 angles of the other, and if the equal sides be either adjacent to the equal angles, or opposite to equal angles, in each, then, the two remaining sides of the one triangle shall be equal to the two remaining sides of the other, each to each, and the third angle of the one shall be equal to the third angle of the other. CoR. 2. If two /> (ABC, DEF) be equal to one another, and have a side (AB) of the one equal to a side (DE) of the other, and an angle (BAC) adjacent to the former side equal to an angle (EDF) adjacent to the latter, the remaining sides and angles of the two triangles, shall be equal, each to each. If the / ABC be not equal to the / DEF, one of them must be the greater; let the / ABC be the greater, and at the point B in AB, suppose the Z, ABH to be made equal to the LDEF; then (hyp. and Th. 19.) the AS, ABH = AS DEF; but (hyp.) the AS DEF = /> ABC; , wherefore the /> ABH = As ABC ; that is, the less is equal to the greater; which is impossible; therefore the / ABC is not un- equal to the L DEF; that is, it is equal to it; where- fore, also (Cor. 1.) the remaining / DFE = / ACB, the side BC= EF, and the side AC= DF CoR. 3. If any point (C) in a straight line (CE) which is perpendicular to a given finite straight line (AD), be equidistant from the extremities (A, D.) of the finite line, the perpendicular shall divide the given finite line into two equal parts. For if CA and CD be drawn, then (hyp.) CA= CD; therefore (Th. 13.) the / CAE = / CDE; 56 Elementary Thebrems of Plane Geometry, and (hyp. and Th, 2.) the LCEA = 1. CED; where- A | fore (Cor. 1.) EA = ED. • THEoREM XX. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one an- other, the two triangles shall be equal. Let the as ABC, DEF, have the sides AB=DB, A. D B C E. and AC= DF, and the L BAC = / EDF': the AS. ABC = AES DEF Primary Relations of Sides, &c. of Triangles. 57 For if the AS, ABC be supposed to be applied to the AES DEF, so that the point A may be on D, and AB on DE, the point B will be on E, because (hyp.) AB = DE; and AB coinciding wholly with DE, AC shall coincide with DF, and the point C with F. because (hyp.) the Z BAC= / EDF, and AC= DF; and it has been shewn that B coincides with E; wherefore (Def. 8. Cor. 1.) BC coincides with EF, and the ASABC with the AS DEF; therefore (Def. 11.) the AS, ABC=/S DEF. Therefore, if two triangles, &c. a. E. D. CoR. 1. From this it is manifest, that if two tri- angles have two sides of the one equal to two sides of the other, each to each, and have also, the angles contained by those sides equal to one another, their bases, or third sides, shall be equal, and their other angles equal, each to each, viz. those to which the equal sides are opposite. CoR. 2. If two 2s (ABC, FGH) have their ver- tical / (BAC, GFH) equal, and each of them not f E. G. H less than a right angle, and have one side (FH) of the one greater than one side (AC) of the other, and the other side (FG) of the one not less than the other side AB of the other, then shall the As(FGH) which H 58 Elementary Theorems of Plane Geometry, has the aggregate of it's two sides the greater, have also the greater base. - - First, let FG >AB, and FH-AC; let AB and AC be supposed to be produced to D and E, so that AD=FG, and AE=FH, and let DE be supposed to be drawn. Therefore (Th. 20.) DE= GH; but (Th. 16. Cor.) DE > BC; wherefore GH > BC. And if FG= AB, it may in like manner be shewn that GH > BC. CoR. 3. If, therefore, two/> (ABC, FGH) having equal vertical / (BAC, GFH) each of them not less than a right angle, have equal bases (BC, GH) and have a side of the one (AB) less than a side (FG) of the other, the third side (AC) of the former triangle, shall be greater than the third side (FH) of the other. For AC cannot be either equal to FH or less than FH, because then (Cor. 2.) GH would be greater than BC; but (hyp.) GH = BC; wherefore in this case AC > FH. - CoR. 4. Any point in an indefinite straight line, which bisects a finite straight line at right angles, is equidistant from the two extremities of the line so bisected. * --- For the distances of any point, in the bisecting perpendicular, from the ends of the line bisected, are the bases of two right-angled triangles which have one of the sides about the right angles common to both, and the other two sides about the right angles, equal to one another; wherefore (Cor. 1.) these dis- tances are equal to one another. Primary Relations of Sides, &c. of Triangles, 59. CoR. 5. If two given points (B, C) be equi- distant from the extremities (A, D) of a given finite straight line (AD), the straight line (BC), which joins the given points, shall bisect the given line at right angles, and shall also bisect each of the angles (ABD, ACD) which the given line subtends at the two given points". i Suppose BA, BD, CA, CD, to be drawn. Then (hyp.) BA= BD, and CA= CD; wherefore (Th. 13.) the Z. BAD = / BDA, and the / CAD= / CDA, so that the whole / BAC= / BDC; wherefore (Th. 20. Cor. 1.) the / ABC = / DBC, and the / ACB = Z. DCB. Again, since, BA = BD and BE is common to the two /> ABE, DBE, and, as hath been shewn, the LABE = / DBE, therefore (Th. 20. Cor. 1.) AE = ED, and the / AEB = A DEB; * In the figure, the two given points are taken on different sides of the given straight line. But the same method is applicable to the demonstration of the corollary, when the two given points are taken on the same side of the given line. * 60 Elementary Theorems of Plane Geometry, that is, each of the 4AEB, DEB is (Def. 13.) a right angle. * * ScholIUM. It is plain that any given finite straight line has a point of bisection, and that a straight line may stand at right angles to it at that point. If, then, three points being given, there is (Th. 20. Cor. 4.) a line, every point of which is equidistant from the first and second of the given points, and there is, also another line of which every point is equidistant from the second and third of the given points. Now if the three given points be in the same straight line, the two perpendiculars, containing the equidistant points, will no where (Th. 6.) meet one another. But if these three points be not in the same straight line, the two perpendiculars (Th. 10. Cor. 1.) will meet. In this latter case, therefore, there is a point that is equidistant from the three given points, THEOREM XXI. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one of them greater than the angle contained by the two sides equal to them of the other, the base of that which has the greater angle, shall be greater than the base of the other. Let ABC, DEF, be two triangles, which have the two sides AB, AC equal to the two sides DE, DF, Primary Relations of Sides, &c. of Triangles. 61 each to each, viz. AB = DE, and AC= DF; and the ID / BAC; Z. EDF; the base BC- EF. At the point A, in BA, suppose the Z. BAK to be made equal to the / EDF; and since (hyp.) the / BAC > / EDF, the / BACP / BAK; wherefore AKfalls within the / BAC; from AK suppose a part to be cut off equal to D.For to AC, the extremity of which will either be within the ZS ABC, or on BC, or without the AS, ABC. First, let it fall within the /S ABC as at G, and suppose BG to be drawn: then (Th. 18.) AC+ CB > AG + GB; take away the equals AC and AG, and it is plain that CB > GB; but (hyp. and Th. 20.) GB = EF; wherefore CB-EF. Next, let AH = DF; then, manifestly, CB > HB; but (hyp. and Th. 20.) HB= EF; wherefore, CB - EF. Lastly, let AK= DF or AC; then (Th. 17.) AH +HC-AC or AK; that is, AH-H HC-AH-- HK; take away 4H from both these equal aggregates, and it is evident that HC-HK; but (Th. 17.) HK-H FIBS BK; wherefore HC-H. HB-BK, that is, CB > BK; and (hyp. and Th. 20.) BK = EF; there- fore CBP EF. If, therefore, two triangles, &c. a. E. D. 62 Elementary Theorems of Plane Geometry, CoR. Hence, any point, which is not in the straight line that bisects a given finite straight line at right angles, is at unequal distances from the two extremities of that given finite line. THEoREM XXII. If two triangles have two sides of the one equal to : two sides of the other, each to each, but the base of the one greater than the base of the other ; the angle also contained by the sides of that which has the greater base, shall be greater than the angle contain- ed by the sides equal to them of the other. Let ABC, DEF be two triangles, which have the two sides AB, AC equal to the two sides DE, D F B C E DF, each to each, viz. AB = DE, and AC= DF; but the base CBS EF; the L BAC > / EDF. For, if it be not greater, it must either be equal to it, or less; but the L BAC is not equal to the / EDF, because then the base BC would (Th. 20.) be equal to EF; but it is not; therefore the / BAC is not equal to the LEDF; neither is it less; because Primary Relations of Sides, &c. of Triangles. 63 then the base BC would be less than the base EF; but it is not; therefore the L BAC is not less than the LEDF; and it was shewn that it is not equal to it; therefore the Z. BAC is greater than the / EDF Wherefore, if two triangles, &c. a. E. D. THEOREM XXIII. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles equal, each to each, then, if the third angles be each of them an acute angle, or each of them a right angle, or each of them an obtuse angle, the two triangles shall be equal. Let the /> ABC, DEF, have the / BAC = / EDF, and the sides about two other angles, viz. A T) JH. C E F ABC, DEF, equal, each to each, viz. AB = DE and BC = EF, and let the two remaining angles, ACB, DFE, be both of the same species, that is, both of them acute, both of them right, or both of them obtuse angles; the AS, ABC = /> DEF For the side AC= DF; if not, one of them is 64 Elementary Theorems of Plane Geometry, the greater; let AC be the greater, and from it suppose AH to be cut off, equal to DF, and BH to be drawn. Then (hyp. and Th. 20. Cor. 1.) the 4 AHB = A DFE, and BH = EF, but (hyp.) EF = BC; therefore BH-BC, and (Th. 13.) the LBHC = / BCH, which (hyp.) is of the same species as the / DFE, or it's equal, the / AHB; wherefore, the three/ AHB, BHC, HCB are of the same species, which is impossible; for accordingly as the / AHB is acute or obtuse, the / BHC is obtuse or acute, and if the LAHB be a right angle, the / HCB is (Th. 11.) an acute angle. Wherefore AC is not un- equal to DF, that is, AC= DF; and since, also, (hyp.) AB= DE, and the L BAC= / EDF, there- fore (Th. 20.) the AS, ABC=/S DEF. Therefore, if two triangles, &c. a. E. D. CoR. From the demonstration it is manifest that of two such triangles, as are compared in the proposi- tion, the third sides, also, are equal to one another, and the remaining angles of the one to the remaining angles of the other, each to each, namely, those to which the equal sides are opposite. THEOREM XXIV. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the two triangles shall be equal. Let the two /> ABC, DEF, have the three sides of the one equal to the three sides of the other, each Primary Relations of Sides, &c. of Triangles. 65 to each; viz. AB = DE, BC= EF, AC = DF; the - A. - ID F P; C E. For the / BAC = / EDF; if not, one of them is the greater, and therefore (Th. 21.) one of the two bases BC, EF must be greater than the other, which is contrary to the supposition. Wherefore, the L BAC is not unequal to the / EDF, that is, the / BAC= / EDF; and (hyp.) AB = DE, and AC = DF; therefore (Th. 20.) the AS ABC = As DEF. There- fore, if two triangles*, &c. a. E. D. CoR. 1. If two triangles have the three sides of the one equal to the three sides of the other, each to each, it is evident from the demonstration, and from Th. 20, that the three angles of the one shall be equal to the three angles of the other, each to each, namely, * If the greatest side of the one triangle be supposed to be ap- plied to the greatest side of the other, so as wholly to coincide with it, and if the two triangles be supposed to be posited in the same plane, on contrary sides of this common line, it may also be shewn, as in Th. 20. Cor. 5, that the opposite angles are equal to one an- other, and that therefore, (hyp. and Th. 20.) the one triangle is equal to the other. * ... • I 66 Elementary Theorems of Plane Geometry, those to which the equal sides are opposite: and therefore (Th. 19. Cor. 1.) the perpendiculars let fall on two sides, that are equal to one another, from the opposite angles, in two such triangles, shall also be equal to each other. CoR. 2. Upon the same base, and upon the same side of it, there cannot be two triangles, not coin-- ciding with another, that have their sides which are terminated in one extremity of the base equal to one another, and also those which are terminated in the other extremity. For it is manifest from Cor. 1. that two such triangles would wholly coincide with one another. CoR. 3. If a given finite straight line be bisected, the straight lines which join the point of bisection, and any other points that are equidistant from the two extremities of the given line, shall be at right angles to the given line, and shall themselves be in one and the same straight line. Let the point B be equidistant from the extremities 4 N I) A, D, of the given finite straight line AD, which is bisected in E. BE is at right angles to AD. Primary Relations of Sides, &c. of Triangles. 67 For suppose BA and BD to be drawn: then (hyp.) BA= BD and EA = ED, and EB is common to the two Os BEA, BED; therefore (Cor. 1.) the / BEA = / BED; that is (Def. 13.) each of the // BEA, BED, is a right angle. And, if C be any other point that is equidistant from A and B, it may in like manner be shewn that CE is at right angles to AD. Therefore, also, (Th. 2. and Th. 3.) BE and CE are in one and the same straight line, whether the points B and C be on the same side, or on contrary sides, of the given line AD. THE 33 lentent; of #lattt (ſtomtetty. Book I. CHAPTER II. —sº- SECTION II. On the comparison of triangles and parallelograms, upon the same base, or upon equal bases, and be- tween the same parallels. —0— DEFINITION XX. OF quadrilateral plane figures, that is, of plane figures contained by four straight lines, a Parallelogram is that of which the opposite sides are parallel; and the Diameter of a Parallelogram is the straight line joining the summits of any two of it's opposite angles. THEOREM XXV. The two triangles, into which a parallelogram is divided by it’s diameter, are equal to one another. Comparison of Triangles and Parallelograms, &c. 69 Let ACDB be a parallelogram, of which AD and A. E C D BC are the diameters. The AS, ABC = As BCD, and the As ACD = & ABD. Because AB is parallel to CD, and BC meets them, the / ABC = / BCD (Th. 10.); likewise, because AC is parallel to BD, and BC meets them, the / ACB = A CBD; therefore the two /> ABC, BCD have two angles of the one equal to two angles of the other, each to each, and the side BC, which is ad- jacent to the equal angles, is common to both ; there- fore (Th. 19.) the AS, ABC = /> BCD. In the same manner it may be shewn that the /SACD = & ABD. Wherefore, the two angles, &c. a. E. D. CoR. 1. The opposite sides and angles of paral- lelograms are equal to one another. For if ACDB be a parallelogram, and if BC be it's diameter, it is manifest from the demonstration, and from Th. 19. Cor. 1. that AB = CD, AC = BD; and it follows from Def. 20. and Th. 10. Cor. 6, that the opposite angles of parallelograms are equal to one another. CoR. 2. A parallelogram is double of either of the 70 Elementary Theorems of Plane Geometry, two triangles into which it is divided by either of it's diameters. CoR. 3. Either of the two triangles into which a parallelogram is divided by one of it's diameters, is equal to either of the two triangles, into which it is divided by the other. For let G be the point in which the two diameters AD, BC, of the D-1 ACDB cut one another: then since (Cor. 1.) AB = CD, and (Th. 10.) the LABG = Z. GCD, and the / BAG = / GDC, therefore (Th. 19.) the /S AGB= As CGD ; but the AS AGB, together with the AS AGC, makes up the whole AS, ABC, and the /S CGD together with the same AS AGC, makes up the whole & ACD: it is mani- fest, therefore, that the AS, ABC = As ACD; and it has been shewn that the /S, ABC=/S BCD, and that the ZS ACD = /> ABD ; wherefore either of the two /> ABC, BCD is equal to either of the two /> ACD, ABD. CoR. 4. If a straight line, CB be divided into any number of equal parts CD, DE, &c. and from it's points of division any parallels CF, D.H., EK, &c. be drawn to meet another straight line, FG, they shall divide that straight line into equal parts, FH, HK, &c. If FG be parallel to AB, this is manifest from the first corollary; but if FG be not parallel to AB, from C, and D, let CL and DM be supposed to be drawn parallel to FG ; therefore (Th. 10.) the 4. EDM= / DCL, and the / DEM= / CD.L.; and (hyp.) CD = DE, therefore (Th. 19. Cor. 1.) DM = Comparison of Triangles and Parallelograms, &c. 71 CL; but (Th. 25. Cor. 1.) HK= DM and FH=CL; therefore FH = HK; and in the same manner, may the other parts, into which FG is divided by the parallels, be shewn to be equal to one another and to FH or HK; whether AC and AF be, respectively, amongst the equal parts into which AB and AG are divided or not. * CoR. 5. Hence, conversely, if two straight lines AB, AC, which meet one another, be each divided into any the same number of equal parts, AC, CD, &c., and AF, FH &c., or if any two straight lines CB, FG, which do not meet, be each divided into any the same number of equal parts, beginning from C and F, then the straight lines (CF, DH, EK, &c.) which join the corresponding extremities of the correspond- ing parts shall be parallel to one another. For if DH be not parallel to CF, let DP be parallel to CF; therefore (Cor. 4.) AF= FP; but (hyp.) AF= FH; ... FH = FP; i.e. the less is equal to the greater, which is impossible; wherefore DP is 72 Elementary Theorems of Plane Geometry, not parallel to CF; therefore no line but DH is parallel to CF; and in the same manner it may be shewn that EK is parallel to DH.; and therefore (Th. 9. Cor.) CF, DH, EK, &c. are all parallel to one another. ScholIUM. It is manifest, from the fourth corollary of Th. 25, that a given straight line may be supposed to be di- vided into any assigned number of equal parts; it being plainly possible for any number of parts equal to one another to be taken in any assumed indefinite straight line, and for straight lines to be drawn (Schol. Def. 14.) parallel to one another. - Also, since (Th. 25. Cor. 1.) the opposite sides and angles of a parallelogram are equal to one another, if two adjacent sides of a parallelogram be equal, as they evidently may be, all it's sides will be equal; and if one of it's angles be a right angle, it is evident from Th. 10, that all it's angles will be right angles: Now if two given finite straight lines meet one another, so as to make any angle whatever, it is manifestly possible, for two straight lines to be drawn, one from the end of the one given line, the other from the end of the other, parallel, each to each, to the given lines; and since (Th. 8. Cor. 1.) the two straight lines, so drawn, will meet one another, there may, therefore, be four-sided figures, which have all their angles right angles, and some of them may be equilateral, and others not equilateral: there may, also, be equilateral Comparison of Triangles and Parallelograms, &c. 73 four-sided figures, none of the angles of which are right angles. Hence, the following definitions may be laid down, in order further to distinguish the dif- ferent kinds of quadrilateral figures, from one another. DEF. XXI. A Rectangle is a quadrilateral figure which has all it's angles right angles: and it is said to be con- tained by the two sides about any one of it’s angles. CoR. A rectangle is (Th. 10. and Def. 20.) a parallelogram. XXII. An equilateral rectangle is called a Square. XXIII. A rectangle which is not equilateral is called an Oblong. r XXIV. A Rhombus is a quadrilateral figure, which has all it's sides equal, but it's angles are not right angles. XXV. A Rhomboid is a quadrilateral figure, which has it's opposite sides equal to one another, but which has not all it's sides equal, nor all it's angles right angles. R '74 Elementary Theorems of Plane Geometry, XXVI. All other quadrilateral figures are called Tra- peziums. SCHOLIUM. It is easily proved, by the help of Th. 24. Cor. 1. and Th. 5, that a rhombus, and a rhomboid, are parallelograms. So that all the quadrilateral figures, above defined, excepting the trapezium, are different kinds of parallelograms. THEOREM XXVI. If two parallelograms have two adjacent sides of the one equal to two adjacent sides of the other; each to each, and have the angles contained by those sides also equal, the two parallelograms shall be equal to one another. Let ACDB, a cdb be two parallelograms which have the sides AC=ac, CD=cd, and the LACD= / acd: the El ACDB = [] acdb. | { For, (Def. 20.) BD is parallel to AC and bá is parallel to ac; therefore (Th. 10.) the 4. ACD+ Comparison of Triangles and Parallelograms, &c. 75 A CDB is equal to two right angles, as is also the La cd + / cab; therefore the / ACD + / CDB = A acd. + / cab; but (hyp.) the / ACD = / acd; there- fore the / CDB = A edb: In the same manner it may be shewn that the / DBA = Z. dba, and the / BAC= / bac; if therefore the E! acdb be applied to the D ACDB, so that the point e may fall on C, and c a on it's equal CA, it is manifest, that the Dº I a cdb will coincide with the DTACDB; therefore the D a c d b = L ACDB; therefore if two paral- lelograms, &c. a. E. D. CoR. 1. From the demonstration it is evident that if two parallelograms have an angle of the one equal to an angle of the other, the remaining angles of the one shall be equal to the remaining angles of the other, each to each. CoR. 2. Two squares, which have a side of the one equal to a side of the other, are equal to one another. For (Def. 22. Def. 21. Cor.) squares are equilateral right-angled parallelograms. CoR. 3. The straight line drawn from the bisection of any side of a parallelogram parallel to either of the adjacent sides, divides the parallelogram into two equal parts. '% From the bisection E of the Ll ACDB, let EF be drawn parallel to AC or BD: EF bisects the D ACDB. For (Th. 9. Cor. 1.) if EF be parallel to one of the two parallel straight lines it is parallel to the other 76 Elementary Theorems of Plane Geometry, also; therefore the figures AEFC, EBDF are paral- lelograms; therefore (Th. 25. Cer. 1.) EF= BD, also (hyp.) AE = EB, and (hyp. and Th. 10.) the LAEF = / EBD ; therefore the D AEFC = C, EBDF. . In the same manner it may be shewn, if HK be drawn from the bisection H of AC, parallel to AB or CD, that the LT AHKB = [T] HCDK. CoR. 4. The LAHKB is equal to the E. ACFE. For (Cor. 3.) the D HCFG = E AHGE, and likewise the D-J EGKB = D AHGE; therefore A . . ." | F JD (Def. 12. Cor. 1.) the E-1 HCFG = E. EGKB; but the L EGKB together with the CIAHGE makes up the whole LAHKB, and the D HCFG together with the same D A H G E makes up the whole Ll ACFE; therefore the LJ AHKB = [] ACFE. Wherefore, also, either of the IJAHKB, HCDK, is equal to either of the ID ACFE, EFDB. ..º. THEOREM XXVII. Either of the two triangles, into which a paral- lelogram is divided by either of it's diameters, is equal to either of the two parallelograms, into which Comparison of Triangles and Parallelograms, &c. 77 it is divided by a straight line drawn from the bi- section of any of it's sides, parallel to either of the two adjacent sides. Let ACDB be a parallelogram, of which BC is a diameter, and let EF be drawn from the bisection C E. E, of AB, parallel to AC or BD: the AS, ABC = Ll AEFC. For, let G be the point in which BC cuts EF; and since AEFC is a parallelogram, therefore (Th. 25. Cor. 1.) CF= AE; also (hyp.) EB = AE ; there- fore CF= EB; again, because AB is parallel to CD, therefore (Th. 10.) the / BEG = L GFC, and the 4. EBG= / GCF; therefore (Th. 19.) the As EGB = ~ FGC; but the /> EGB together with the figure AEGC makes up the whole /> ABC, and the As FGC together with the same figure AEGC, makes up the whole E, AEFC; therefore the As ABC= EIAEFC: and the /s ABC is equal (Th. 25.) to the AS BCD, and (Th. 25. Cor. 3.) to either of the two triangles into which the EACDB is divided by it's other diameter: also, the El AEFC is equal (Th. 26. Cor. 3. and 4.) to the CJ EBDF, and to either of the parallelograms into which the EACDB 78 Elementary Theorems of Plane Geometry, y is divided by a straight line drawn from the bisection of AC, parallel to AB or CD; therefore either of the two triangles, &c. a. E. D. THEOREM XXVIII. Parallelograms upon the same base and between the same parallels are equal to one another. Let the ID AC, FB, be upon the same base BC B C i8 C and between the same parallels AF, BC; the CIAC = [...] FB. First, if the sides AD, DF of the ID AC, FB, opposite to the base BC, be terminated in the same point D, then, AC and FB are, each of them double (Th. 25. Cor. 2.) of the same AS DBC; therefore (Def. 12. Cor. 1.) the D AC = D FB. Secondly, if the sides AD, EF, opposite to the base BC of the [I] AC, FB, ‘be terminated each within the other, then because ABCD is a paral- lelogram, AD (Th. 25. Cor. 1.) is equal to BC; for the same reason EF = BC; therefore AD = EF; and DE is common; therefore (Def. 12. Cor. 3.) AE = DF; also AB = DC; and (hyp. and Th. 10.) the 4. EAB of the AS AEB, is equal to the LFDC of Comparison of Triangles and Parallelograms, &c. 79 the AS DFC; therefore (Th. 20.) the AS AEB = As DFC; and EBCD is common to the two [E] AC, FB; it is manifest, therefore, that the E ABCD may be applied so that it's parts, taken together, shall coin- cide with the CJ EBCF; therefore the El ABCD= L. EBCF. Lastly, let the sides AD, EF opposite to the base BC, of the ID AC, FB, lie wholly without one an- R other, and let BE cut CD in the point P: let DG be drawn parallel to EB or FC, and GHIK parallel to AF or BC, cutting DC in H, EB in I, and FC in K: again, let HL be drawn parallel to EB or FC, and LW parallel to AF or BC, and so on; also let E, K be joined: then since (hyp.) DGKF is a paral- lelogram, therefore (Th. 25. Cor. 1.) DG = FK; for the same reason, AD = BC, and EF= BC; therefore AD=EF; and (hyp. and Th. 10.) the / ADG, of the As ADG, = / EFK, of the AS EFK; therefore (Th. 20.) the As ADG = /> EFK; but (Th. 25. Cor. 2.) the [] AH is double of the AS, ADG, and the D-1 FI is double of the /> EFK; therefore the TI AH = D FI; and in the same manner it may be 80 Elementary Theorems of Plane Geometry, shewn, that the D-1 GM = [] IN; and so on; until at last there will remain two parallelograms on the same base BC, and having their sides opposite to BC terminated in P, or else terminated each within the other; the which parallelograms, as hath been demonstrated, are, also, equal to one another; there- fore the whole D ABCD is equal to the whole D EBCF; therefore parallelograms upon the same base, &c. a. E. D. z Theorem XXIX. Parallelograms upon equal bases and between the same parallels are equal to one another. , Let ABCD, EFGH be parallelograms upon equal A. L T, E ; H T; Fé–H Gr bases BC, FG, and between the same parallels AH, BG; the El ABCD is equal to the CI EFGH. Suppose BE, CH, to be drawn; then, BC = FG, and (Th. 25. Cor. 1.) FG= EH; therefore BC= EH; and they are parallels, and joined towards the same parts by the straight lines BE, CH: but straight lines which join equal and parallel straight lines towards the same parts, are (Th. 10. Cor. 4.) them- Comparison of Triangles and Parallelograms, &c. 81 selves equal and parallel; therefore EB, CH are both equal and parallel, and EBCH is a parallelogram ; and it is equal (Th. 28.) to ABCD, because it is upon the same base BC, and between the same parallels BC, AD: For the like reason, the [T] EFG H is equal to the same EBCH; therefore also the ABCD is equal to EFG-H. Wherefore paral- lelograms, &c. a. E. D. CoR. Equal parallelograms which are between the same parallels, and not upon the same base, are upon equal bases. * Let the [T] BD, FH be between the same parallels AH, BG, but not upon the same base, and let the D BD = [T] FH: then the base BC is equal to the base F.H. For if not, one of them is the greater; let BC > FG ; from BC let BK be supposed to be cut off equal to FG, and let KL be drawn parallel to BA; therefore (Th. 29.) the CI AK = [] EG ; but (hyp.) the D AC = [-] EG; therefore the El AC is equal to the CI AK, the greater to the less; which is impossible; therefore BC cannot be unequal to FG, that is, BC = FG. THEOREM XXX. Triangles upon the same base, or upon equal bases, and between the same parallels, are equal to one another. •. * First, let the /> ABC, DBC, be upon the same L 82 £lementary Theorems of Plane Geometry, base BC and between the same parallels AD, BC: the AS, ABC = As DBC. - For from B and C let there be drawn BE parallel to BA, and CF parallel to BD, meeting AD pro- duced in the points E and F; therefore each of the figures EBCA, DBCF is a parallelogram; also let G be the bisection of BC, and from G let GH be drawn parallel to CA or BE, and let GK be drawn parallel to BD or CF; then (Th. 27.) the As ABC = E. EBGH, and (Th. 28.) the E. EBGH = E. DBGK; but (Th. 27.) the DDBGK = As DBC; therefore the As ABC= & DBC. And, in the same manner, by the help of the twenty-ninth theorem, it may be shewn that triangles upon equal bases, and between the same parallels are equal to one another. Wherefore, triangles upon the same base, &c. a. E. D. CoR. 1. It is manifest, from the demonstration, that a triangle is equal to any parallelogram upon the half of it's base and between the same parallels. CoR. 2. If two triangles ABC, ADC, be upon unequal bases, and between the same parallels /X and PZ, that which is upon the greater base DC, shall be the greater. Comparison of Triangles and Parallelograms, &c. 83 For if from CD, the greater CE be supposed to be cut off equal to CB the less, and if A, E be Y A. Z. v-f-if-É–3—##–F–A–x joined, then the whole & ADC is greater than its part, the AS AEC; which (Th. 30.) is equal to the As ABC; therefore the AS, ABC > /> ADC. SCHOLIUM. Hence, if any number of straight lines B.H., &c. be taken equal to BC, and if any number D.F, FG, &c. be taken each equal to CD, and if AH and AG be drawn, it follows that if CG=CH, the /S ACG = As ACH; if CG & CH, the As ACG & As ACH; and if CG > CK, the /> ACG > /> ACK. And, in like manner, may the same property be shewn to belong to parallelograms that are upon un- equal bases and between the same parallels. By means of Th. 30, a triangle may be found which shall be equal to any given rectilineal figure ABCDE. For let any one of it's sides as CD be produced both ways; let AC and AD be drawn; through B and E let there be drawn BF parallel to AC, and EG parallel to AD, meeting CD produced in Fand G, and let AFand AG be drawn. 84 Elementary Theºrems Čf Plane Geometry, Then since (Th. 30) the As AFC = As ABC, and the As AGD = /> AED, if the As ACD be added to the aggregates of these equals, it is plain that the whole AS AFG is equal to the rectilineal figure ABCDE. C. TheoreM XXXI. Equal triangles, upon the same base and upon the same side of it, or upon equal bases in the same straight line, and towards the same parts, are between the same parallels. First, let the equal /> ABC, DBC be upon the same base BC, and upon the same side of it; they are between the same parallels. Suppose AD to be drawn; AD is parallel to BC; for A. D — T) C B à-É–F if it is not, suppose AE to be drawn parallel to BC, Comparison of Triangles and Parallelograms, &c. 85 and E, C to be joined. Then (Th. 30.) As ABC = As EBC; but (hyp.) the ZS ABC= /> DBC; there- fore the AS DBC is equal to the /S EBC, the greater to the less; which is impossible; therefore AE is not parallel to BC; and in the same manner it can be shewn that no other line but AD is parallel to BC; AD is, therefore, parallel to it. Secondly, let the equal /> ABC, DEF be upon equal bases BC, EF, in the same straight line BF and towards the same parts; they are between the same parallels. Suppose AD to be drawn; AD is parallel to BC; for if it is not, suppose AG to be drawn parallel to BC, and G, C to be joined. Then, (Th. 30.) the As ABC = /> GEF; but (hyp.) the As ABC = /> DEF: therefore the AS DEF is equal to the AS GEF, the greater to the less; which is impossible; therefore AG is not parallel to BC: and, in the same manner it can be shewn, that no other line but AD is parallel to BC; therefore AD is parallel to BC. Wherefore, equal triangles, &c. a. E. D. - THEOREM XXXII. If a parallelogram and a triangle be on the same base and between the same parallels, the paral- lelogram shall be double of the triangle. For (Th. 30. Cor. 1.) the triangle is equal to any parallelogram upon the half of it's base and between the same parallels with it, and therefore, (Th. 26. Cor. 3.) it is equal to the half of the parallelogram 86 Elementary Theorems of Plane Geometry, which is on the same base, and between the same parallels; that is, the parallelogram is the double of the triangle which is on the same base and between the same parallels. a. E. D. Theorem XXXIII. The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another. Let ABCD be a parallelogram, of which the diameter is AC, and EH, FG, the parallelograms about AC, and BK, KD, the other parallelograms, making up the whole figure, ABCD, and which are called the complements: the complement BK is equal to the complement KD. Suppose HI and EL to be drawn parallel to AC; therefore (Th. 8. Cor. 1.) they will cut DC; let EL cut HG in N, BC in M, and DC, produced, in L: therefore the figures AHKE, AKNE, HDFK, HICK, KCLN, EKCM, and EKGB, are parallelograms; therefore (Th. 25. Cor. 1.) HK = AE, and KN=AB, therefore HK = KN, and (Th. 29.) the EIHDFK= Comparison of Triangles and Parallelograms, &c. 87 El KCLN: again, (Th. 25. Cor. 1.) EK = MC, and KN = CL; also (Th. 10. Cor. 6.) the / EKN = 1MCL; therefore (Th. 20.) the /S EKN = & MCL; but the AS EKN, together with the figure NKCM, makes up the CI EKCM, and the AS MCL together with the same figure NKCM, makes up the CI RCLN; therefore L-J EKCM = […] KCLN; also (Th. 28.) the El EKBG = C, EXCM; and it has been shewn that the D., HDFK = [] KCLN; there- fore the D-1 ERBG = PIDFK; therefore the eomplements, &c. a. E. D. SCHOLIUM. By the help of Th. 33, it is easy to shew the possibility of applying to any side (AD) of a given [...] ABCD, a parallelogram, which shall have it's angles equal to the angles of ABCD, each to each, and which shall be equal to another given [1AEFG. For, let the two given parallelograms be supposed A. E Hº H to be placed so as that their bases AE and AB may be in the same straight line and have a common ex- 88 Elementary Theorems of Plane Geometry, &c. tremity A. Let CD, produced, if it be necessary, meet EF in H, and AG in K, let AH be drawn and let it meet GF, produced, in L; also, through L, let LM be drawn parallel to FE, or to GA, and let it meet AB in M, lastly, through M let MVbe drawn parallel to AD or to BC; therefore (Th. 10.) the El AN is equiangular with the CIAC; and (Th. 33.) the D-I GH = [T] HM; therefore D J G E = Did KM; and (Th. 28.) the CI KM = [T] DM; therefore the E] DM = E. G.E. And since (Th. 30. Cor. 1.) a parallelogram may always be found, which shall be equal to a given triangle, it is further manifest, that to any side of a given parallelogram, there may be applied a parallelogram equiangular with it, and which shall be equal to a given triangle. {} Also, since (Schol. to Th. 30.) a triangle may be found, which shall be equal to any given, rectilineal figure, it is evident that two equiangular parallel- ograms, having a common side, may be found that shall be equal to any two given rectilineal figures, each to each. THE £It muttutº of ºlant Gºtomtettp. # Book I. CHAPTER II. SECTION III. On the primary relations of the sides, angles, and surfaces of plane rectilineal figures that are con- tained by more than three sides, whether the figures be parallelograms or not. --- —O— Definition XXVII. A PolyGON is a plane rectilineal figure contained by more than four sides. - Theorem XXXIV. Any side of a plane rectilineal figure, contained by more than three sides, is less than the remaining sides taken together. - M. 90 Elementary Theorems of Plane Geometry, First, let ABCD be a plane figure contained by A. D T3 C four sides; any of it's sides, as AD, is less than the remaining sides AB, BC, CD, taken together. For suppose BD to be drawn; then (Th. 17.) AB+ BD > AD, and BC+ CD > BD; much more, then, AB + BC + CD > AD. And, in the same manner, it may be shewn that any side of a polygon is less than the remaining sides taken together. Therefore, any side, &c. a. E. D. THEOREM XXXV. All the interior angles of any plane rectilineal jigure, contained by more than three sides, together with four right angles, are equal to twice as many right angles as the figure has sides. Let DEFBC be any plane rectilineal figure, con- tained by more than three sides; all it's interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides. For, take any point A within the figure, and from A suppose straight lines to be drawn to each of the angular points B, C, D, E, F, of the figure: thus Primary relations, &c. of Plane Rectilineal Figures. 91 will the figure DEFBC be divided into as many F IB triangles as it has sides: and (Th. 12.) all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure; but these same angles make up all the angles of the figure, together with all the angles at A, that is (Def. 13. Cor. 5.) together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. CoR. 1. All the interior angles of any quadrila- teral rectilineal figure are together equal to four right angles: and if three of the angles of such a figure be right angles, or be together equal to three right angles, the fourth shall also be a right angle. CoR. 2. If all the angles but one, of any recti- lineal figure, contained by more than three sides, be together equal to all the angles but one of another rectilineal figure, contained by the same number of sides, the two remaining angles shall be equal to one another. CoR. 3. All the exterior angles of any rectilineal figure AC are together equal to four right angles. 92 Elementary Theorems of Plane Geometry, Because every interior LABC, with it's adjacent ex- terior LABD, is equal (Def. 13. Cor. 2.) to two right * D–H t angles, therefore, all the interior together with all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides, that is,. (Th. 35.) to all the interior angles of the figure, together with four right angles; take away, from each of these equal aggregates, all the interior angles of the figure, and there remain all the exterior angles equal to four right angles. CoR. 4. If two polygons be each of them equi- lateral and equiangular and have the same number of sides, any angle of the one shall be equal to any angle of the other. For (hyp. and Th. 35.) the aggregate of all the interior angles, of the one is equal to the aggregate of all the interior angles of the other; and (hyp.) each of these equal aggregates is made up of the same number of equal angles; wherefore any angle of the one is equal to any angle of the other. CoR. 5. If two equilateral and equiangular po- lygons have the same number of sides, and have, also, *. Primary relations, &c. of Plane Rectilineal Figures. 93 any side of the one equal to any side of the other, the two polygons shall be equal to one another. For (Cor. 4.) their angles are equal, each to each, as well as their sides ; it is plain, therefore, that either of the two figures may be applied to the other, so as wholly to coincide with it. SchoLIUM. The unlimited space about a given point, may, without any absurdity, be supposed to be divided into any number of equal angles, having that point for their common summit; in the straight lines contain- ing these angles, points may be supposed to be taken equidistant from the common summit; and then it is plain, from Th. 20, that the straight lines, joining the points so taken, will contain an equilateral and equiangular polygon. There may, therefore, be such figures as those which form the subjects of the two immediately preceding corollaries. In the latter of these corollaries the surfaces of two polygons are shewn to be equal in a particular case. But in general, the surfaces of polygons are to be estimated by di- viding them into triangles. £itments of plane Geometry. Book I. —sº- CHAPTER III. ON THE PRIMAR, Y RELATIONS OF THE STRAIGHT LINES which MEET CIRCLEs, AND of THE CHORDS, ARCHES, ANGLEs, AND surf Aces, of cIRCLES, IN RESPECT OF PoSITION, AND OF EQUALITY OR INEQUALITY. —O— SECTION I. On the place of the centre in a given circle. º DEFINITION XXVIII. If a finite straight line be supposed to revolve, in any the same plane, about one of it's extremities, that is fixt, until it return to the place whence it began to move, the surface, which the revolving line may be Place of the Centre in a given Circle. 95 supposed thus to have passed over, is called a Circle; also the linear space which the moving extremity of the revolving line may be supposed thus to have passed over, is called the Circumference; and the fixt point, about which the finite straight line is supposed to revolve, is called the Centre. CoR. 1. A circle is, therefore, a plane figure, contained by one line, namely the circumference; and all straight lines, drawn from the centre to the circumference, are equal to one another. CoR. 2. The circumferences of equal circles are equal to one another. ScholIUM. What is above deduced, in the first corollary, from the definition of a circle, is laid down by Euclid as the definition itself. But when he has so defined a circle, it becomes questionable whether there can be a plane figure having such a property. This possibility, as it is not self-evident, ought not to be assumed. It seems, therefore, that the definition is better founded upon a simple hypothesis, which serves to shew how such a figure may be understood to be described. And this is the method followed by Euclid himself, where he had the same choice of methods, in his definition of a sphere. DEF. XXIX. A Segment of a circle is the figure contained by a straight line and the part of the circumference 96 Elementary Theorems of Plane Geometry, which it cuts off; a part of the circumference, thus cut off, is called an Arch; and the straight line which cuts off, or subténds, an arch, is called a Chord. CoR. Any chord of a circle falls within the circle. If the chord pass through the centre of the circle, it manifestly fails within the circle; for no point in it can be further from the centre, than each of it's two extremities, which are in the circumference. But let the chord AB, in the circle ACBD, not pass through the centre E ; and any point F of AB having been taken, suppose EA, EF and EB to be drawn. And since (Def. 28. Cor.) EA = EB, there. fore (Th. 13.) the LEAB = A. EBAs but (Th. 11.) the / EFB > / EAB, and therefore the / EFB > EBF; therefore (Th. 16.) EB > EF; but the extremity B, of EB, is not further from the centre than the circumference is; wherefore the extremity F of EF is nearer to the centre than the circum , ference is; therefore the point F is within the circle : and the same may, in the same manner, be shewn of any other point of AB. Wherefore, the whole of AB falls within the circle. -> * Place of the Centre in a given Circle. 97 DEF. XXX. A Diameter of a circle is a chord which passes through the centre. CoR. 1. The centre of a circle divides each of it's diameters into two equal parts. CoR. 2. A circle is divided by it's diameter into two equal parts. For if one of the parts, into which any diameter divides the circle, were applied to the other, so that they might have that diameter for a common boundary, it is plain that they would coincide; else the straight lines drawn from the centre to the cir- cumference could not all be equal. CoR. 3. If the diameters, or the halves of the diameters, of two circles be equal, the circles them- selves shall be equal. For if the one circle were applied to the other, so that their centres and their planes might coincide, it is evident that the circles must wholly coincide with one another; else their semidiameters, and there- fore also their diameters, could not be equal, as they are supposed to be. CoR. 4. The semi-diameters of equal circles are equal to one another; and therefore, also, their dia- meters are equal to one another. DEF. XXXI. A Semi-circle is the figure contained by a dia- N 98 Elementary Theorems of Plane Geometry, meter and the part of the circumference cut off by the diameter. DEF. XXXII. A Sector of a Circle is the figure contained by two straight lines, drawn from the centre, and the arch of the circle that is intercepted between them. THEOREM XXXVI. If a straight line, drawn through the centre of a circle, bisect a chord of the circle, which does not pass through the centre, it shall cut it at right an- gles; and, if it cuts it at right angles, it shall bisect 2.É. - Let ABC be a circle, of which E is the centre; and let CD, which passes through the centre E, \". bisect the chord AB, which does not pass through the centre, in the point F: CD cuts AB at right angles. ' ' . For (Def. 28. Cor. 1.) the point E, in CD, is equi- Place of the Centre in a given Circle. $9 distant from A and B; therefore (hyp. and Th. 24. Cor. 3.) the // EFA, EFB are right angles. Where- fore CD, drawn through the centre E bisecting the chord AB, that does not pass through the centre, cuts the same at right angles. Again, let CD cut AB at right angles in F; then (Def 28. Cor.) the point E, in CD, is equidistant from A and B; therefore (Th. 19. Cor. 3.) FA=FB. Wherefore, if a straight line, &c. a. E. D. THEOREM XXXVII. The straight line which bisects at right angles any chord of a circle passes through the centre of the circle. If the chord be a diameter, it’s bisection is (Def. 28. 30.) itself the centre of the circle. And if the chord be not a diameter, the centre is (Def. 28. Cor. 1.) equidistant from it's extremities; the centre cannot, therefore, be out of the straight line which bisects the chord at right angles; for if it were, it would (Th. 21. Cor.) be at unequal distances from the extremities of the chord. Wherefore, the straight line, &c. a. E. D. THEOREM XXXVIII. If in a circle two straight lines cut one another which do not both pass through the centre, they do not bisect each the other. Let ABCD be a circle, and AC, BD two straight 109 Elementary Theorems of Plane Geometry, lines in it which cut one another in the point E, and do not both pass through the centre: AC, BD do not bisect one another. For, if it is possible, let AE = EC, and BE = ED: if one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre: but if neither of them pass through the centre, let F be the centre of the circle, and join EF: and because FE, a straight line through the centre, bisects another AC, which does not pass through the centre, it shall (Th. 36.) cut it at right angles; wherefore FEA is a right angle: again, because the straight line FE bisects the straight line BD which does not pass through the centre, it shall cut it at right angles; wherefore FEB is a right angle : and FEA was shown to be a right angle; therefore the / FEA is equal to the / FEB, the less to the greater, which is impossible: there- fore AC, BD do not bisect one another. Wherefore, if in a circle, &c. a. E. D. THEOREM XXXIX. * If a point be taken within a circle, from which Place of the Centre in a given Circle. 101 there fall morethan two equal straight lines to the circumference, that point is the centre of the circle. If from the point A, taken within the circle BDC, there fall more than two equal straight lines, namely, B - C AB, AC, AD, to the circumference, the point A is the centre of the circle BDC. For suppose BC and CD to be drawn, and to be bisected in E and F, suppose also AE and AF to be drawn. Then since (Th. 24. Cor. 3. and hyp.) AE is perpendicular to BC, the centre of the circle is (Th. 37.) in E4; in the same manner it may be shewn that the centre is also in FA ; therefore, it is in A, the intersection of EA and FA : for (Def. 8. Cor. 1.) EA and FA cannot meet one another in any other point than A. Wherefore, if a point, &c. Q. E. D. e THE 33 ſententº of ºlant Geometry. Book I. Chapter III. SECTION II. On the primary relations of straight lines, drawn jrom a given point, either within or without a circle, to the circumference, in respect of position and of equality or inequality. —O- THEoREM XL. If any point be taken in the diameter of a circle, which is not the centre, of all the straight lines which can be drawn from it to the circumference, the great- est is that in which the centre is, and the other part of that diameter is the least; and, of any others, that which is nearer to the line which passes through the Straight Lines drawn to the Circumference. 103 centre is always greater than one more remote : and from the same point there can be drawn only two straight lines that are equal to one another, one upon each side of the shortest line. LET ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre : let the centre be E; of all the straight lines FB, FC, FG, &c. that can be drawn from F to the eircumference, FA is the greatest, and FD, the other part of the diameter AD, is the least: and of the others, FB is greater than FC, and FC than FG. Suppose BE, CE, GE, to be drawn; and because two sides of a triangle are greater (Th. 17.) than the third, BE, EF are greater than BF; but AE=EB; therefore AE, EF, that is, AF is greater than BF: again, because BE = CE, and FE is common to the as BEF, CEF, the two sides BE, EF are equal to the two CE, EF; but the / BEF is greater than the a CEF; therefore (Th. 21.) the base BF is greater- than the base FC: for the same reason, CF is greater than GF: again, because GF, FE are (Th. 17.) 104 Elementary Theorems of Plane Geometry, greater than EG, and EG = ED; GF, FE are greater than ED: take away the common part FE, and the remainder GF is greater than the remainder FD: therefore FA is the greatest, and FD the least of all the straight lines from F to the circumference; and BF is greater than CF, and CF than G.F. Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD : at the point E in the straight line EF, let the L FEH be supposed to be made equal to the / GEF, and FH to be drawn; then because GE = EH, and EF is common to the two as GEF, HEF; the two sides GE, EF are equal to the two HE, EF; and the / GEF = / HEF; therefore (Th. 20. Cor.) FG= FH: but, besides FH, no other straight line can be drawn from F to the cir- cumference equal to FG ; for, if there can, let it be FK; and because FK= FG, and FG= FH, FK= FH ; that is, a line nearer to that which passes through the centre, is equal to one which is more remote; which is impossible. Therefore, if any point be taken, &c. a. E. D. Theorem XLI. Of the straight lines, which are drawn from any point without a circle to the concave circumference, the greatest is that which passes through the centre; and of the rest, that, which is nearer to the straight. line that passes through the centre, is greater than the more remote. Straight Lines drawn to the Circumference. 105 Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC be drawn to the concave circumference AEFC, whereof DA passes through the centre. The greatest is AD which passes through the centre; and the nearer to it is always greater than the more remote, viz. DE than DF, and DF than DC. Let M be the centre of the circle ABC, and suppose because AM = ME, add MD to each, therefore AD is equal to EM, MD; but EM, MD are greater (Th. 17.) than ED; i. e. AD is greater than ED: again, because ME=MF, and MD is common to the as EMD, FMD; EM, MD are equal to FM, MD; but the / EMD is greater than the / FMD; therefore (Th. 21.) the base ED is greater than the base FD: O 106 Elementary Theorems of Plane Geometry, in like manner it may be shewn that FD is greater than CD; therefore DA is the greatest; and DE greater than DF, and DF than DC. Therefore, of the straight lines, &c. a. E. D. THEoREM XLII. ... Of the straight lines, which are drawn from any point without a circle to the convex circumference, the least is that which being produced passes through the centre; and of the rest, that, which is nearer to the least, is less than the more remote ; and only two equal straight lines can be drawn from the point to the circumference, one upon each side of the least. *Let ABC be a circle and D any point without it, from which let the straight lines DG, DK, DL, DH be drawn to the convex circumference, whereof DG produced passes through the centre : of these the least is DG ; and the nearer to it is always greater than the more remote, viz. DK than DL, and DL than D.H. Let M be the centre of the circle and suppose MK, ML, MH, to be drawn and DG to be produced to M. * Then because MK, KD (Th. 17.) are greater than MD and MK = MG, the remainder KD is greater than the remainder GD, that is, GD is less than KD : and because MK, DK are drawn to the * See the figure in p. 105. Straight Lines drawn to the Circumference. 107 point K within the triangle MLD from M, D, the extremities of its side MD, MK, KD, are less than ML, LD, whereof MK = ML; therefore the re- mainder DK is less than the remainder DL: in like manner it may be shewn, that DL is less than DH: therefore DG is the least, and DK less than DL, and DL than DH. Also there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least: at the point M, in the straight line MD, suppose the / DMB to be made equal to the / DMK, and DB to be drawn; and because MK = MB, and MD is common to the As KMD, BMD, the two sides KM, MD, are equal to the two BM, MD; and the / KMD = / BMD; therefore (Th. 20. Cor. 1.) DK= DB: but, besides DB, there can be no straight line drawn from D to the circumference equal to DK; for, if there can, let it be DN; and because DK is equal to DN, and also to DB ; therefore DB = DN, that is, the nearer to the least is equal to the more remote, which is im- possible. Therefore, of the straight lines, &c. a. E. D. £iententſ of #lattſ Gºtomittty. Book I. Ch APTER III. SECTION III. On the primary relations of the chords of a circle, in respect of position and of equality or inequality. DEFINITION XXXIII. STRAIGHT lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn . to them from the centre are equal: and the straight line, on which the greater perpendicular falls, is said to be farther from the centre. Primary relations of the Chords of a Circle, &c. 109 THEOREM XLIII. Equal chords in a circle are equally distant from the centre; and chords which are equally distant from the centre are equal to one another. Let the chords AB, CD, in the circle ABCD be B D equal to one another: they are equally distant from the centre. Let E be circle's centre, and from E suppose EF and EG to be drawn at right angles to AB and CD, and E, A and E, C to be joined. Then (hyp. and Th. 36.) AB is bisected in F and CD in G; but (hyp.) AB=CD; therefore, (Def. 11. Cor. 6.) AF= CG; also (Def. 28. Cor. 1.) EA= EC, and the / EFA, EGC being right angles are (Th. 2.) equal to one another; and (Th. 11. Cor. 3.) the other angles of the two /> AFE, CGE are acute angles ; therefore (Th. 23. Cor.) EF = EG ; i. e. (Def. 33.) AB and CD are equally distant from the centre, E, of the circle. Again, if the chords AB, CD be equally distant 110 Elementary Theorems of Plane Geometry, from E the centre, i. e. if the perpendiculars EF, EG, be equal to one another, then AB= CD. For, the two right-angled /> AFE, CGE, having the sides AE, EF, equal to the sides CE, EG, each to each, AF (Th. 23. Cor.) is equal to CG. And it may be shewn, as before, that AB and CD are the doubles of AF and CG. Therefore AB = CD. There- fore, equal chords, &c. a. E. D. THEOREM XLIV. The diameter is the greatest chord in a circle ; and of all others, that, which is nearer to the centre is greater than the more remote; and the greater is mearer to the centre than the less. Let ABCD be a circle, of which E is the centre, and AD a diameter; and let the chord BC be nearer A—B to E than is the chord FG ; AD is greater than any chord, which is not a diameter, and BC is greater than FG. A. From E suppose EB, EC and EFto be drawn, and Primary relations of the Chords of a Circle, &c. l l I also EK and EH at right angles to FG and BC respectively. And since EA = EB and ED = EC, EA+ ED, or AD, - EB+ BC; but (Th. 17.) EB + BCS BC; therefore AD - BC. Again, since (hyp.) BC is nearer than FG to E, EH- EK; and the / BHE, FKE, of the Æs EHB, EKF, are right angles, and (Def. 28. Cor.) EB = EF; therefore, (Th. 20. Cor. 3.) BH > FK; and (hyp. and Th. 36.) BC and FG are the doubles of BH and FK; therefore BC-FG". 4. - Lastly, if BC-FG, its half BH-FK, the half of FG; therefore (Th. 20. Cor. 3.) EH& EK; i.e. BC is (Def. 30.) nearer to the centre than FG. Therefore, the diameter, &c. a. E. D. * This may also be proved, by supposing a part equal to EH to be cut off from EK, and a chord to be drawn through the ex- tremity of that part at right angles to EK; which chord will (Th. 43.) be equal to BC, and (Th. 21. and Def. 28. Cor. 1.) greater than FG, Wherefore, also BC- FG. 3Elentents of ºpiattt Geometry, Book I. CHAPTER III. SECTION IV. On the mutual contact of a straight line and a ~ circle. —O— THEOREM XLV. The straight line, which is at right angles to the diameter of a circle, at the eatremity of the diameter, meets the circle only in that point ; and no other straight line, supposed tº be drawn from the same extremity of the diameter, can fall wholly within the space that lies between the circumference of the circle and the straight line at right angles to the diameter. Mutual Contact of a Straight Line and Circle. 113 Let ABC be a circle, of which the centre is D T. F and the diameter AB; and let EAF be at right angles to AB at it's extremity A: EAF shall meet the circle only in the point A. Take any point E, in EAF, suppose ED to be drawn, and let it meet the circumference in C. And since (hyp.) the LDAE is a right angle, the LAED is (Th. 11. Cor. 1.) less than a right angle; therefore (Th. 16.) DE > DA; but (Def. 28. Cor. 1.) DA = DC; therefore DE > DC; and C is in the circum- ference; therefore the point E is without the circle ABC; and in the same manner it may be shewn that every other point, but A, of EAF is without the circle; wherefore EAF meets the circle in no other point than the point A. Again, no other straight line, drawn from A, can fall wholly within the space, which lies betwen EAF and the circumference. For, let AG be any other straight line drawn through A; if a perpendicular be supposed to be P 114 Elementary Theorems of Plane Geometry, drawn from D to AG, it cannot coincide with DA, Gºl II. H||F because then (Th. 2.) AG would coincide with AE, which is contrary to the supposition; and this perpen- dicular may be shewn to be less than D4, in the same manner as DA was shewn to be less than DE in the former part of the proposition; wherefore the extremity of the perpendicular, which is in AG, lies within the circle; therefore AG does not fall wholly within the space that lies between EAF and the circumference. Therefore, the straight line, &c. Q. E. D. SCHOLIUM. The theorem last demonstrated forms a foundation for the next following definition. DEF. XXXIV. A straight line is said to touch a circle, when it Mutual Contact of a straight Line and Circle. 115 meets the circle, and, being produced, does not cut it: and the straight line itself, which touches a circle, is called a Tangent. CoR. 1. A circle and it’s tangent have only one point that is common to them both. For if any point within the circle, but not in the circumference, were common to them both, it is plain that the tangent would cut the circle: and if any other point in the circumference besides that in which the tangent meets the circle, were common to them both, the part of the tangent lying between those two points, would (Def. 29. Cor.) fall within the circle; and, therefore, in this case also, the tan- gent would cut the circle: which is contrary to the definition. CoR. 2. The straight line, drawn at right angles to the diameter of a circle, from the extremity of the diameter, touches (Th. 45. and Def. 34.) the circle: and there cannot be more than one straight line, which touches a circle in the same point. 'ScholIUM. As it is plainly possible for a straight line to be at right angles to another straight line, at any given point in it, a straight line may always be supposed to be drawn touching a circle at any given point in it's circumference. * THEOREM XLVI. If a straight line touches a circle, the straight line, 116 Elementary Theorems of Plane Geometry, drawn from the centre to the point of contact, shall be perpendicular to the line touching the circle. . Let the straight line EF touch the circle ABC in the point A; let D be the centre, and draw the straight line DA: DA is perpendicular to EF. For, if it be not, from the point Flet ACE be supposed to be drawn perpendicular to EF; and because DEA is a right angle, DAC is (Th. 11. Cor. I.) an acute angle; and to the greater angle (Th. 16.) the greater side is opposite; therefore DA > DE; but DA = DC; therefore DC is greater than DE, the less than the greater, which is impossible; wherefore DE is not perpendicular to EF: in the same manner it may be shewn, that no other is per- pendicular to it besides DA, that is, DA is perpen- dicular to EF. Therefore, if a straight line, &c. Q. E. D. e * THEOREM XLVII. If a straight line touches a circle, and if from the Mutual Contact of a Straight Line and Circle. 117 point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line. Let the straight line DE touch the circle ABC, A. F B I) (. E in C, and from C let CA be drawn at right angles to DE; the centre of the circle is in CA. For, if not, let Fbe the centre, if it be possible, and suppose CF to be drawn; because DE touches the circle ABC, and FC is drawn from the centre to the point of contact, FC is (Th. 46.) perpendicular to DE ; therefore FCE is a right angle : but ACE is also a right angle; therefore (Th. 2.) the / FCE is equal to the LACE, the less to the greater, which is impossible: wherefore F is not the centre of the circle ABC: in the same manner, it may be shewn, that no other point, which is not in CA, is the centre; that is, the centre is in CA. Therefore, if a straight line, &c. a. E. D. TH}. 3BItment; of 13 lant Geometry. Book I. CHAPTER III. SECTION W. On the mutual contacts and intersections of circles. THEOREM XLVIII. If the diameters of two circles be in one and the same straight line, and have a common eatremity, the two circumferences, meeting one another in that point, shall not have any other point common to them both. Let the two circles ABC, ADE, have their dia- meters AB, AD, in the same straight line, and ter- Mutual Contacts and Intersections of Circles. 119 minated by the same point A : the two circumferences shall have no point, but A, common to them both. For, if it be possible, let another point C be common to them both, and, first let both the diameters be on the same side of the point A: let F be the centre of the circle ABC, and G the centre of the other circle ADE, and suppose FC and GC to be drawn. Then (Th. 17.) CG + GF - FC, and FC = FA; therefore CG-H. GF- FA; take away the common part G.F. and CG > GA; but (hyp.) G is the centre of the circle ADE; therefore, CG= GA; which is absurd. Therefore, in this case, the circumferences of the two circles have no point, but A, which is common to them both. And when the diameters of the two circles lie on contrary sides of their common extremity, the same proposition may be proved in the same manner. Wherefore, if the diameters, &c. a. E. D. ScholIUM. It is manifest, from Th. 48, that two circles may 120 Elementary Theorems of Plane Geometry, so meet as not to cut one another; and on this pos- sibility is founded the next following definition. DEF. XXXV. Circles are said to touch one another, which meet but do not cut one another. THEoREM XLIX. If two circles touch each other internally, the straight line which joins their centres being produced shall pass through the point of contact. Let the two circles ABC, ADE, touch each other internally in the point A, and let F be the centre of the circle ABC, and G the centre of the circle ADE: the straight line which joins the centres F, G, being produced, passes through the point 4. For, if not, let it fall otherwise, if possible, as FGDH, and suppose AF and AG to be drawn; and because AG, GF are (Th. 17.) greater than F4, that is, than FH, for FA = FH, both being from the same Mutual Contacts and Intersections of Circles. 121 centre; take away the common part FG ; therefore the remainder AG is greater than the remainder GH: but AG = GD; therefore GD is greater than GH, the less than the greater, which is impossible. Therefore the straight line which joins the points F, G cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, &c. a. E. D. THEOREM L. If two circles touch each other externally, the straight line, which joins their centres, shall pass through the point of contact. Let the two circles ABC, ADE touch each other externally in the point A; and let F be the centre of the circle ABC, and G the centre of ADE: the straight line, which joins the points F, G, shall pass through the point of contact A. * For, if not, let it pass otherwise, if possible, as FCDG, and suppose FA, AG, to be drawn: and because F is the centre of the circle ABC, AF= FC: also because G is the centre of the circle ADE, AG= * Q r 122 JElementary Theorems of Plane Geometry, GD; therefore FA, AG are equal to FC, DG ; wherefore the whole FG is greater than FA, AG; but (Th. 17.) it is also less; which is impossible: therefore the straight line which joins the points F, G shall not pass otherwise than through the point of contact A, that is, it must pass through it. There- fore, if two circles, &c. a. E. D. CoR. It is manifest from this proposition, and from Th. 48, that one circle cannot touch another exter- mally in more points than one. THEOREM LI. If two circles touch one another internally, they shall not have the same centre. Let the two circles ABC, CDE, touch one another internally in the point C: they have not the same centre. * For, if they have, let it be F; suppose FC, and any straight line FEB meeting the circumferences in E and B, to be drawn; and because F is the centre of the circle ABC, CF=FB: also, because F is the Mutual Contents and Intersections of Circles. 123 centre of the circle CDE, CF– FE; and it was shown that CF = FB; therefore FE is equal to FB, the less to the greater, which is impossible; wherefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c. a. E. D. THEoreM LII. If two circles cut one another, they shall not have the same centre. Let the two circles ABC, CDG cut one another in the points B, C; they have not the same centre. For, if it be possible, let E be their centre: suppose EC, and any straight line EFG meeting the cir- cumferences in F and G, to be drawn; and because E is the centre of the circle ABC, CE = EF: again, because E is the centre of the circle CDG, CE = EG; but, it was shown that CE = EF, therefore EF is equal to EG, the less to the greater, which is impossible : therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. 124 Elementary Theorems of Plane Geometry, &c. THEOREM LIII. One circumference of a circle cannot cut another in more than two points. If it be posssible, let the circumference FAB cut the circumference DEF in more than two points, viz. in B, G, F; let K be the centre of the circle ABC, and suppose KB, KG, KF to be drawn; and because within the circle DEF there is taken the point K, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF, the point K is (Th. 39.) the centre of the circle DEF; but K is also the centre of the circle ABC; there- fore the same point is the centre of two circles that cut one another, which (Th. 52.) is impossible. There- fore one circumference of a circle cannot cut another in more than two points. a. E. D. THE 3Element; of #latte Cºtontttty. Book I. Charrºw III. SECTION VI. On the primary relations of angles at the centres, and at the circumferences of circles, and of the arches of circles, and the chords which subtend them, in respect of equality or inequality. —()— Definition XXXVI. ^ As Angle in a Segment of a Circle is the angle contained by two straight lines, drawn from any point in the arch of the segment to the extremities of it's chord: and such an angle is said to stand upon 126 Elementary Theorems of Plane Geometry, the arch intercepted between the straight lines which contain the angle, and to be an angle at the cir cumference. g DeF. XXXVII. A rectilineal figure is said to be inscribed in a circle, when all the angles of that figure are at the circumference of the circle. THEOREM LIV. The opposite angles of any quadrilateral recti- lineal figure, inscribed in a circle, are together equal to two right angles. Let ABCD be a quadrilateral rectilineal figure inscribed in the circle. ABCD: any two of it's opposite angles are together equal to two right angles. Let E be the centre of the circle ; and first let E be within the figure ABCD, and suppose EA, EB, EC and ED to be drawn. Then EA = ED; there- Angles at the Centres and Circumferences, &c. 137 fore (Th. 13.) the LEAD = / EDA ; likewise, the A EDC = / ECD, the / ECB = 1 EBC, and the / EBA = L EAB; therefore, the four / EAB, EAD, ECB, ECD are together equal to the four M. EDA, EDC, EBA, EBC; that is, the two A. DAB, DCB, are together equal to the two / ADC, ABC, and therefore make up the half of all the four angles of ABCD, or (Th. 35. Cor. 1.) the half of four right angles; wherefore, the two // DAB, DCB are together equal to two right angles, as are also the two / ADC, ABC. But if the centre be without the figure ABCD, suppose, as before, EA, EB, EC, and ED, to be drawn; then, it may be shewn, as in the former case, that the ſ, EAB, ECB, ECD, are together equal to the // EBA, EBC, EDC; from these equals, take the // EAD, EDA, which (Def.28. Cor. 1. and Th. 13.) are equal to one another, and there will remain the a DAB+ / DCB = LADC+ / ABC; whence it may be shewn, as in the former case, that the // DAB, 128 Elementary Theorems of Plane Geometry, DCB are together equal to two right angles, as are, also, the two / ADC, ABC. And in like manner may the proposition be demonstrated, if the centre of the circle be in any one of the sides of the inscribed quadrilateral figure. Therefore, the opposite angles, &c. a. E. D. THEoREM LV. The angles in the same segment of a circle are equal to one another. Let ABCD be a circle, and let the / BAD, BED, JE be in the same segment BAED: the / BAD = A BED. . ..., Tº For in BCD take any point C, and suppose BC, DC, to be drawn. Then (Th. 54.) the / BAD, BCD are equal to two right angles, as are also the / BED, BCD; therefore the // BAD, BCD are equal to the // BED, BCD, and if from these be taken the common / BCD, there will remain the Angles at the Centres and Circumferences, &c. 129 a BAD = / BED. Wherefore, the angles, &c. G. E. D. CoR. If a circle be divided into two segments, and if an angle in the one of them be a right angle, any angle in the other is also a right angle. But if any angle in either of the two segments be less than a right angle, any angle in the other is greater than a right angle. THEOREM LVI. The angle at the centre of a circle is double of the angle at the circumference, standing upon the same arch. Let AFBC be a circle, BEC an angle at the centre E, and BAC an angle at the circumference, standing upon the same arch BC: the L BEC is the double of the L BAC. Suppose CE to be produced to meet the cir- cumference in F, and FB to be drawn. And since EF= EB (Th. 13.) the / EFB = 1 EBF; but R 130 Elementary Theorems of Plane Geometry, (Th. 12.) the Z BEC= / EFB+ / EBF; therefore the LBEC is the double of the / BFC, or (Th. 55.) of its equal, the / BAC. Therefore, the angle at the centre, &c. a. E. D. = } ( THEoREM LVII. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and centre E; and let CA divide the circle into the segments ABC, ADC, and suppose BA, AD, DC to be drawn; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than-a semicircle, is greater than a right angle. Suppose AE to be drawn, and BA to be produced Angies at the Centres and Circumferences, &c. 131 to F; and (Th. 13.) because BE = EA, the LEAB = Z. EBA ; also, because AE = EC the L EAC = 4. ECA; wherefore the whole / BAC is equal to the two / ABC, ACB; but FAC, the exterior angle of the AS, ABC, is equal (Th. 11.) to the two / ABC, ACB; therefore the / BAC = / FAC, and each of them (Def. 13.) is therefore a right angle: Wherefore the 1. BAC in a semicircle is a right angle”. And because the two / ABC, BAC, of the AS, ABC, are together less (Th. 12. Cor. 1.) than two right angles, and that BAC is a right angle, ABC must be less than a right angle; and therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle. And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (Th. 54.) to two right angles; therefore the // ABC, ADC are equal to two right angles; and ABC is less than a right angle ; wherefore the other ADC is greater than a right angle. CoR. From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right * If this theorem had been placed after Th. 60, (and Th. 58, Th. 59, Th. 60, are none of them made to depend upon it) then the first case of it might have been proved more concisely, as follows. { Since (Th. 60.) the angles in each of the semicircles, that are on contrary sides of any diameter, are equal to one another, and since (Th. 54.) these angles are together equal to two right angles, it is plain that each of them is a right angle. 132 Elementary Theorems of Plane Geometry, angle, because it's supplement is equal to the same two; and when an angle is equal to its supplement, it is a right angle. t SCHOLIUM. If BCD be a given circle, of which the centre is K, and A any given point without it; and if AK be A. C drawn, and upon it as a diameter there be described a circle ABK, the circumference of which cuts the circumference of the circle BCD in B; then if AB be drawn, it shall touch the circle BCD in B. For if BK be drawn, the LABK is (Th. 57.) a right angle, and therefore (Def. 34. Cor. 2.) AB touches the circle BCD in the point B. | THEOREM LVIII. If a straight line touches a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line Angles at the Centres and Circumferences, &c. 133 touching the circle, shall be equal to the angles which are in the alternate segments of the circle. Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD A. I) E B F be drawn cutting the circle: the angles, which BD makes with the touching line EF, shall be equal to the angles in the alternate segments of the circle: that is, the / FBD is equal to the angle which is in the segment DAB, and the / DBE to the angle in the segment BCD. *- From the point B suppose BA to be drawn at right angles to EF, and take any point C in the arch BD, ‘and suppose AD, DC, CB, to be drawn; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line from the point of contact B, the centre of the circle is (Th. 47.) in BA; therefore the LADB in a semicircle is (Th. 57.) a right angle, therefore the other two / BAD, ABD, are equal (Th. 12.) to a right angle: but ABF is likewise a right angle; therefore 134 Elementary Theorems of Plane Geometry, the LABF is equal to the angles BAD, ABD: take from these equals the common LABD ; therefore the remaining / DBF= Z. BAD, which is in the alternate segment of the circle; and because ABCD is a quadrilateral figure in a circle, the opposite A. BAD, BCD, are equal (Th. 54.) to two right an- gles; therefore the / DBF, DBE, being likewise equal (Def. 13. Cor. 2.) to two right angles, are equal to the / BAD, BCD; and DBF has been proved equal to BAD: therefore the remaining / DBE = / BCD in the alternate segment of the circle. Wherefore, if a straight line, &c. a. E. D. THEoREM LIX. In equal circles, equal angles stand upon equal arches, whether they be at the centres or circum- ferences. Let ABC, DEF, be equal circles, of which the A ~D K S. 27 L centres are G and H, and the equal / BGC, EHF, at their centres. The arch BKC is equal to the arch ELF. Angles at the Centres and Circumferences, &c. 135 For if the circle ABC be applied to the circle DEF, so that the centre G may coincide with H, and GB with it’s equal HE, GC will coincide with it's equal HF, because (hyp.) the LBGC= 1 EHF, and it is manifest that the circle ABC will wholly coincide with the equal circle DEF; and the point B being on E, and the point C on F, the arch BKC will coincide with the arch ELF and be equal to it. Again, let BAC, EDF be equal angles at the circumferences of the equal circles ABC, DEF, of which the centres are G and H. Suppose GB, GC, HE, HF, to be drawn; and if BAC, EDF, be each of them less than a right angle, then (Th. 56.) the // BGC, EHF, being the doubles of the equal angles BAC, EDF, are equal to one another, and therefore, by the first case, they stand upon equal arches; therefore the // BAC, EDF, stand upon equal arches. But if the / BAC, EDF, be not less than right angles; let each of them be supposed to be bisected : then, as hath been shewn, the arches which subtend their halves are equal, each to each; and, therefore the whole arches, on which the whole angles stand, are equal to one another. CoR. In the same manner it may be shewn that, in the same circle, equal angles stand upon equal arches: and it is evident, also, that in equal circles, or in the same circle, the greater angle stands upon the greater arch. 136 Elementary Theorems of Plane Geometry, THEOREM LX. In equal circles, the angles which stand upon equal arches are equal to one another, whether they be at the centres or circumferences. Let the / BGC, EHF at the centres, and BAC, EDF at the circumferences of the equal circles ABC, A D DEF, stand upon the equal arches BC, EF: the / BGC = / EHF, and the / BAC = / EDF. For if the / BGC be not equal to the LEHF, one of them is the greater: let BGC be the greater, and at the point G, in the straight line BG, let the / BGK be supposed to be made equal to the / EHF; but equal angles stand upon equal arches (Th. 59.) when they are at the centre; therefore fºx == EF : but fift =#C ; therefore also fºx is equal to #C, the less to the greater, which is impossible: there- fore the / BGC is not unequal to the / EHF; that is, it is equal to it; and (Th. 56.) the angle at A is half of the LBGC, and the angle at D half of the / EHF: Angles at the Centres and Circumferences, &c. 137 therefore the angle at A is equal to the angle at D. Wherefore, in equal circles, &c. a. E. D. CoR. 1. In the same manner it may be shewn that in the same circle, the angles which stand upon equal arches are equal to one another: and it is evident that in equal circles or in the same circle, the angle which stands upon the greater arch is the greater. CoR. 2. The chord of any arch of a circle is parallel to the straight line that touches the circle in the point in which the arch is bisected. For (Th. 58.) the angle made by the chord of either of the halves of the arch with the touching line is equal to the angle in the alternate segment; and this angle (Th. 60. Cor. 1.) is equal to the angle in the other segment, made by the chord of the arch, with the chord of it's half; therefore (Th. 5.) the chord of the arch is parallel to the line which touches the arch at it's bisection. THEOREM LXI. In equal circles, equal chords cut off equal arches, the greater equal to the greater, and the less to the less. º Let ABC, DEF be equal circles, and BC, EF, equal chords in them, which cut off the two greater arches BAC, EDF, and the two less BGC, EHF: * centres; therefore BGC = EHF. But the whole circumference ABGC is equal to the whole DEHF; the remaining part therefore of the circumference, __T-S .--> viz. BAC, is equal to the remaining part EDF. Therefore, in equal circles, &c. a. E. D. CoR. In the same circle equal chords cut off equal arches. SCHOLIUM. It is evident, from the definitions of a circle and of a diameter, that any two diameters of a circle bisect one another; and it has been shewn (Th. 38.) that no two chords of a circle, which are not diameters, can bisect each the other. Any two equal chords of a circle, which cut one another, have, however, this Angles at the Centres and Circumferences, &c. 139 property, “that the segments of the one, between the point of intersection and the circumference, are equal to the segments of the other, between the same point of intersection and the circumference, each to each.” This property obtains, whether the two equal chords cut one another within the circle, or whether it be necessary to produce them, in order that they may meet one another without the circle: and it is easily demonstrated by means of Th. 61, Th. 59, and Th. 15. .* THEoREM LXII. In equal circles equal arches are subtended by equal chords. Let ABC, DEF be equal circles, and let the ~" A. C *N 7° Gr H arches BGC, EHF, also be equal; and join BC, EF: the chord BC is equal to the chord E.F. Let K, L, be the centres of the circles, and suppose BK, KC, EL, LF to be drawn; and because £Gö. = EHF, the / BKC = / ELF (Th. 60): and because the circles ABC, DEF are equal, the straight 140 Elementary Theorems of Plane Geometry, &c. lines from their centres are equal : therefore BK, KC, are equal to EL, LF, and they contain equal angles: therefore the base BC is equal (Th. 20.) to the base EF. Therefore, in equal circles, &c. a. E. D. CoR. In the same circle equal arches are sub- tended by equal chords. SCHOLIUM. The circumference of a circle evidently admits of being divided into any number of equal parts; and if the points of division be supposed to be joined by straight lines, there will result (Th. 62.) an equilateral rectilineal figure which is inscribed in the circle. There is not, therefore, any absurdity in supposing it possible for such a figure to be inscribed in any given circle. THE Blemtetttg of ºlant (ºtotitettp. Book I. CHAPTER III. SECTION VII. On the comparison of segments, and of sectors of circles, in respect of equality or inequality. —O- DEFINITION XXXVIII. Sr MILAR segments of circles are those in which the angles are equal. And similar sectors are those of which the arches subtend equal angles at the centres of their circles. CoR. The segments into which two equal circles are divided by equal chords, are similar (Th. 61.60.) each to each. 142 Elementary Theorems of Plane Geometry, ScholIUM. Since (Th. 55.) the angles in the same segment of a circle are equal to one another, if an angle in any one segment be equal to an angle in another segment, any other angle in the one shall be equal to any other angle in the other segment. There is therefore no ambiguity in the last definition. THEOREM LXIII. Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another. If it be possible, let the two similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one an- other: then, because the circle ACB. cuts the circle ADB in the two points A, B, they cannot (Th. 53.) cut one another in any other point: one of the seg- ments must therefore fall within the other; let ACB fall within ADB; suppose any straight line BCD, and also AC and AD, to be drawn; and because the segment ACB is similar to the segment ADB, and that similar segments of circles contain (Def, 37.) Comparison of Segments and Sectors of Circles, &c. 143 equal angles; the / ACB = LADB, which (Th. 11. Cor.) is impossible. Therefore, there cannot be two similar segments of a circle upon the same side of the same straight line, which do not coincide. G. E. D. CoR. From this, and from Th. 55, it is manifest, that equal angles, which are subtended by the same straight line and are on the same side of it, cannot be in two segments of circles that do not coincide with one another: if, therefore, one of the equal angles be in any segment of a circle, the other is in the same segment. \ THEOREM LXIV. Similar segments of circles upon equal straight lines, are equal to one another. Let AEB, CFD, be similar segments of circles, E IF | / \ ^ A. B C D upon the equal straight lines AB, CD; the segment AEB is equal to the segment CFD. For, if the segment AEB be supposed to be ap- plied to the segment CFD, so as that the point A may be on C, and the straight line AB upon CD, the point B shall coincide with the point D, because AB is equal to CD: therefore the straight line AB coinciding with CD, the segment AEB must (Th.63.) 144 Elementary Theorems of Plane Geometry, coincide with the segment CFD, and therefore is equal to it. Wherefore similar segments, &c. a. E. D. THEOREM LXV. In equal circles the sectors which are bounded by equal arches are equal to one another. Let the sectors KBGC, LEHF, in the equal circles ABC, DEF, be bounded by the equal arches A. L’s 2"> J3N C E F Gr H BGC, EHF: the sector KBGC is equal to the sector LEHF In £go, fiſh, take any points G, H, and suppose BC, BG, GC, EF, EH, HF, to be drawn. 2-\ 2-\ º Then (hyp.) BGC= EHF, and the circles are equal; therefore (Th. 62.) BC= EF, also BK, KC, are equal to EL, LF, each to each; therefore (Th. 24.) the AS, BKC= /> ELF. Again, since the circles are equal, their whole cir- cumferences (Def. 28. Cor. 2.) are equal; and (hyp.) 2-> BGC = £HF, therefore flat-Édi, and (Th.60.) Comparison of Segments of Circles, &c. 145 the / BGC = / EHF; therefore (Def, 38.) the seg- ments BGC, EHF, are similar ; and as hath been shewn, they are upon equal straight lines BC, EF; therefore (Th. 64.) they are equal to one another; and the /> BKC, ELF, have been proved to be equal to one another: therefore, the sector KBGC is equal to the sector LEHF. Wherefore, in equal circles, &c. a. E. D. CoR. In equal circles, of two sectors bounded by unequal arches, that which is bounded by the greater arch is the greater. £Ittittittº of #lattt (ºtonittty. Book I. -º- CHAPTER III. -º- SECTION VIII. On the comparison of circles, and sectors of circles, with plane rectilineal figures in respect of equality or inequality. —O— DEFINITION XXXIX. A RECTILINEAL figure is said to be described about a circle, when each side of that figure touches the circle. Comparison of Circles with Rectilineal Figures. 147 DEF. XL. The Perimeter of a rectilineal figure is the ag- gregate of the straight lines by which the figure is bounded. DEF. XLI. The Altitude of a rectilineal figure is the straight line drawn perpendicular to the base, from the summit of the angle opposite to the base. THEOREM LXVI. An equilateral rectilineal figure, inscribed in a circle, is also equiangular; and if through the sum- mits of it’s angles, straight lines be drawn touching the circle, the figure thus described about the circle shall also be equilateral and equiangular. Let ABCD be a circle, of which K is the centre; and let AB, BC, CD, &c. be sides of an equilateral 2 148 Elementary Theorems of Plane Geometry, figure inscribed in the circle; the / ABC, BCD, &c. are equal to one another. For (hyp.) AB = BC = CD; therefore (Th. 61.) AB - #6– ČD, and ABC- fici) 3 wherefore, also 2-> 2-> ADC = BAD, and (Th. 60.) the / ABC = 1. BCD. And, in the same manner, may the remaining angles of the inscribed figure be shewn to be equal to one another and to the / ABC, BCD ; wherefore, the inscribed figure is equiangular. Again, through the points A, B, C, D, &c. of the inscribed figure, suppose EAF FBG, GCH, HDI, &c. to be drawn touching the circle; and K, B, and K, C, to be joined. And since (hyp. and Th. 46.) the M. KBG, KCG, are right angles, the / GBC, GCB, are less than right angles, and therefore (Th. 10. Cor. 1.) the tangents drawn through B and C will meet one another. In the same manner it may be shewn that the tangents drawn through A and B, through C and D, &c. will meet one another: they will, therefore, inclose a figure; and this figure shall be equilateral and equiangular. For suppose KA, KF, KG, and KH, to be drawn. The // KAF, KBF. KBG, KCG, KCH are (Th. 46.) right angles; KG is common to the two right-angled DNKBG, KCG ; and KB- KC; therefore (Th. 23. Cor.) GB= GC, and the LBKG = / CKG; i. e. the / BKG is the half of the Z BKC : in the same manner, it may be shewn that the / BKF is the half of the / AKB; but (hyp. Th. 61. Th. 60.) the / AKB = A BKC; therefore, the / BKF- / BKG; Comparison of Circles with Rectilineal Figures. 149 and KB is common to the two right-angled /~ KBF, KBG; therefore, (Th. 19. Cor. 1.) FB = BG. In like manner, it may be shewn that GC = CH; and it has been proved that BG = GC; wherefore, FG, the double of BG, is equal to GH, the double of GC. And in the same manner may the other sides of the circumscribed figure be shewn to be equal to one another and to FG or GH; the figure is there- fore equilateral. And, since the sides of the AS, AFB are equal to the sides of the /S BGC, each to each, (Th. 24. Cor. 1.) the / AFB = LBGC. In the same manner may the remaining angles of the circumscribed figure be shewn to be equal to one another and to the / AFB, BGC: wherefore the circumscribed figure is also equiangular, as well as equilateral. Therefore, an equilateral, &c. a. E. D. CoR. 1. From the demonstration it is manifest, that if two tangents of a circle meet one another, the parts of them, between the point in which they meet, and the points of contact, are equal to one another : and that the triangles, contained by the sides of an equilateral figure inscribed in a circle, and the tangents passing through the extremities of those sides, are (Th. 24.) equal to one another, and have their sides and angles equal, each to each. CoR. 2. It is also evident from the demonstration, that if any rectilineal figure EFGHI be described about a circle ABCD, each of it's / EFG, FGH, &c. is bisected, by the straight line drawn from the centre to the summit of the angle: if, therefore, the cir- 150 Elementary Theorems of Plane Geometry, cumscribed figure be equiangular, it will be equila- teral; for, KG is common to the two /> KFG, KHG and the Z KFG, KGF, KGH, KHG are equal, and therefore (Th. 19. Cor. 1.) FG = GH: and so on; and each of it's sides will (Th. 19. Cor. 1.) be bisected in the point in which it touches the circle; wherefore, also, if the several points, in which a circle is touched by an equilateral and equiangular rectilineal figure described about it, be joined, the figure, thus inscribed in the circle, will (Th. 20. Cor. 1.) be equilateral, and consequently (Th. 66.) equiangular. ſ SCHOLIUM. It has been shewn (Th. 66. and Cor. 2.) that an equilateral rectilineal figure, inscribed in a circle, is also equiangular, and that an equiangular rectilineal figure described about a circle is also equilateral. But the propositions which are the converse of these two are not necessarily true. First, let AB, BC, CD, &c. be the sides of an equiangular figure inscribed in the circle ACD. Then (hyp.) the / ABC = / BCD; therefore (Th. 59.) 3DC- BAD, and consequently, ABC = ÉCD; take from these equals fºc, and AB = CD; and therefore (Th. 62.) AB = CD. Thus it may be shewn that, the first, third, fifth, &c. sides, of the in- *- - VI-w -* * See the fig. in page 147. Comparison of Circles with Rectilineal Figures. 151 scribed equiangular figure, are equal to one another, and that the second, fourth, &c. sides are equal to one another. If, therefore, the inscribed equiangular figure have an odd number of sides, it will be equi- lateral; or having an even number of sides, if any two contiguous sides be equal, all it's sides will be equal to one another. But, the circumference of a circle being supposed to be divided into any number of equal arches, since each of these arches may be supposed to be similarly divided into two unequal parts, the greater part of any one arch equal to the greater part of any other, and the less to the less, it is evident, if the chords of these several parts be drawn, that (Th. 60.) the figure thus inscribed in the circle, will be equi- angular, but (Th. 62. Cor.) will not be equilateral. Secondly, let E.F, FG, GH, &c. be the sides of an equilateral figure described about the circle ACD; and let KF, KG, KH, be supposed to be drawn. Then (Th. 66. Cor. 2.) the LKGF= LKGH, and (hyp.) GF= GH, and KG is common to the two as KFG, KHG; therefore (Th. 20. Cor. 1.) the / KFG = / KHG; i. e. the half of the / EFG is equal to the half of the LGHI; therefore, the LEFG = / GHI. And thus it may be shewn that the first, third, fifth, &c. angles, of an equilateral figure, described about a circle, are equal to one another, and that the second, fourth, &c. angles are also equal to one another. If, therefore, the circumscribed equilateral figure have an odd number of sides, it will be equiangular; or, having an even number of sides, 152 Elementary Theorems of Plane Geometry, if any two contiguous angles be equal, all it's angles will be equal to one another. But, independently of this last condition, if a figure, described about a circle, have an even number of sides, it may be equilateral, without being at the same time equiangular. Lastly, it was shewn (Th. 66. Cor. 2.) that the sides, of an equilateral and equiangular figure described about a circle, are each of them bisected, in the points in which they severally touch the circle : and it may easily be proved, conversely, that if the sides of a recti- lineal figure, described about a circle, be each of them bisected in it's point of contact, the figure shall be equilateral and equiangular. THEOREM LXVII. An equilateral rectilineal figure inscribed in a circle is equal to a triangle which has it's base equal to the perimeter of the inscribed figure, and it's altitude equal to the perpendicular drawn to any of the sides of that figure from the circle's centre : and any rectilineal figure described about a circle is equal to a triangle which has its base equal to the perimeter of that figure, and it's altitude equal to the circle's semi-diameter. Let ABC be a circle, of which K is the centre, and let the inscribed figure ACB be equilateral; also suppose KD to be perpendicular to any side BC, of the figure ACB : ACB is equal to a triangle, which has it's base equal to the perimeter of ACB and its altitude equal to KD. Comparison of Circles with Rectilineal Figures. 153 Suppose KA, KB, KC, to be drawn; the altitudes of the DS KAB, KAC, KBC, are (hyp. and Th. 43.) #–% º Q. S T 7-5-yc F-I & f it. T-W-S P equal to one another: suppose, also, that LM—BC, and that in LM produced, MV= CA, and NP = AB; therefore, LP= BC+CA+AB, the perimeter of the inscribed figure ACB. And if upon LM, MN, NP, the isosceles as QLM, SMN, TNP, having their sides equal to KA, or KB, or KC, be supposed to be described and the perpendicular QR to be drawn, them (Th. 24.) the /S QLM = As KBC, QL = KB, the / QLR = / KBD and (Th. 19. Cor. 1.) QR = KD; in the same manner it may be shewn, that the Os SMN, TNP, are equal to the As KCA, KAB, and have their altitudes each equal to that of the /> KBC, or KCA, or KAB; and since the equal /> QLM, SMN, TNP, have equal alti- tudes, they are (Th. 10. Cor. 4. and 5.) between the same parallels; if, therefore, • QV and QP be drawn, (Th. 30.) the /S, SMN = As QNM, and the AS TNP = ~ QNP; therefore the aggregate of the /> QLM, *—— -z-ºr- * The straight lines 2N, 2P, are wanting in the figure. U 154 Elementary Theorems of Plane Geometry, QMM, QNP, is equal to that of the as QLM, SMN, TNP; therefore the whole As QLP is equal to the whole rectilineal figure ABC, and it has it's base LP equal to the perimeter of ABC, and it's altitude QR equal to KD, the perpendicular drawn from the centre K to any side BC of ABC. Again, let EFGH be any rectilineal figure de- scribed about the circle ABC, touching it in the points B, I, C, 4; and, from the centre K, suppose straight lines, KA, &c., KE &c., to be drawn to the several points of contact, and to the summits of the angles of EFGHI: then will the figure be divided into as many triangles as it has sides, each of them having (Th. 46.) a semi-diameter of the circle for it's altitude; so that these triangles, which make up the whole figure EFGH, have equal altitudes: wherefore, it may be shewn, as in the former case, that they are equal to a triangle having it's base equal to the peri- meter of EFGH, and it's altitude equal to KA, the semi-diameter of the circle. Therefore, an equilateral, &c. G. E. D. CoR. 1. A figure, which is bounded by two semi-diameters, and any number of equal chords, of a circle, is equal to a triangle, having it's base equal to the aggregate of the chords, and it's altitude equal to the perpendicular drawn to any of them from the centre: also, a figure, which is bounded by any number of tangents to a circle and any two straight lines drawn to them from the centre, is equal to a triangle, having it's base equal to the aggregate of the tangents, and it's altitude equal to the circle's semi-diameter. Comparison of Circles with Rectilineal Figures. 155 CoR. 2. An equilateral rectilineal figure inscribed, in a circle, is less than a triangle which has it's base greater than the perimeter of that figure, and an alti- tude equal to the perpendicular let fall, from the centre, on any side of the figure : and a rectilineal figure de- scribed about a circle is greater than a triangle which has it's base less than the perimeter of that figure, and it's altitude equal to the semi-diameter of the circle. Theorem LXVIII. lf from the greater of two unequal magnitudes there be taken it’s half, and from the remainder it’s half, and so on, there shall at length remain a mag- nitude less than the less of the two proposed magni- tudes. Let AB and CD be two unequal magnitudes, of which AB is the greater; if from AB there be taken A. G H T} F C ij É it’s half, and from the remainder it's half, and so on, there shall at length remain a magnitude less than CD. To CD let there be added a magnitude equal to itself, as DE, and to DE another such magnitude, EF, and so on, until the whole CDEF be greater, as it manifestly may be, than AB: also, let AG be 156 Elementary Theorems of Plane Geometry, the half of AB, GH the half of GB, and so on, until there be as many such divisions in the magnitude AB, as there are parts in the whole magnitude CDEF: and if there be only two divisions in AB, the half of AB & CD, because (hyp.) the double of CD > AB. But if there be more than two divisions in AB, and also in CDEF, take AG from AB, and EF from CDEF, and, since CF-AB, there will have been taken from AB it's half, and from CDEF less than it's half; therefore, CE > GB; again, from GB take it's half GH, and from CDE take DE which is not greater than it's half, and there will remain HB- CD. Therefore, if from the greater, &c. a. E. D. CoR. If from the greater of two unequal mag- nitudes there be taken more than it’s half, and from the remainder more than it's half, and so on, there will at length remain a magnitude less than the less of the two proposed magnitudes: or, if there be a series of magnitudes, of which the number is unlimited, and the second of which is not greater than half the first, the third not greater than half the second, and so on, there shall be amongst them a magnitude less than any given magnitude of the same kind with them. Theorem LXIX. A polygon may be inscribed in a given circle, and another polygon may be described about the circle, so that the difference of the two polygons shall be less than any given finite plane surface. Let ABCD be the given circle, and X a given Comparison of Circles with Rectilineal Figures. Hö7 finite plane surface: a polygon may be inscribed in iFT C G. the circle ABCD, and another polygon may be described about it, so that their difference shall be less than the surface X. Suppose, first, the circumference of the eircle to be divided into four equal parts in the points A, B, C, D, and AB, BC, CD, DA to be drawn; also, through A, B, C, D, suppose there to be drawn the tangents HAE, EBF, FCG, and GDH, which may be shewn to meet one another, as in Th. 66; and the difference between the figures ABCD and EFGH, is the aggregate of the /> AEB, BFC, CGD, DHA, which (Th. 66. Cor. 1.) are equal to one another. If, then, this aggregate be not less than X, suppose the arch AB to be bisected in L, and LA, and LB, and QLM touching the circle in L and meeting EA and EB in M and N, to be drawn; wherefore (Th. 60. Cor. 2.) PQ is parallel to AB; and if AP and Bø be drawn perpendicular to AB, PABQ will be a paral- lelogram, and (Th. 32.) the As. ALB is the half of 158 Elementary Theorems of Plane Geometry, the D-1 PABQ but the D PABQ - AMNB; there- fore the AS ALB is greater than the half of AMNB; much more, then, is the AS BNL + /> LMA less than the half of the As EBA: and if straight lines be drawn touching the circle at the bisections of the remaining arches BC, CD, DA, the same consequence will follow ; so that there will thus be a polygon inscribed in the circle ABCD, and another polygon described about it, having their difference less than half of the difference of the two polygons first de- scribed. Likewise, if tangents be drawn through the bisections of BL, £3, &c., the same consequence will again follow; and so on. Wherefore (Th. 68.) there will at length be two polygons one inscribed in the circle ABCD, the other described about it, the dif- ference of which shall be less than the given surface A. Therefore, a polygon may, &c. a. E. D. CoR. 1. Much more, then, may there be a po- lygon inscribed in a given circle, and another polygon described about it, so that the difference between the circle itself and either of the polygons, shall be less than any given finite plane surface. Aſ For the circle is, in all cases, manifestly greater than the inscribed and less than the circumscribed polygon. * CoR. 2. If a plane figure be greater than any polygon that can be inscribed in a given circle and less than any polygon that can be described about the circle, it shall be equal to the circle. For, if the plane figure exeeed the given circle by Comparison of Circles with Rectilineal Figures. 159 any given difference, every polygon described about the circle must (hyp.) exceed the circle by a greater difference, which (Th. 69.) is absurd. And if the plane figure be less than the circle, by any given difference, then every polygon inscribed in the circle must (hyp.) be less than the circle by a greater difference, which also (Th. 69.) is absurd. Wherefore, the plane figure and the given circle cannot be unequal; i.e. they are equal to one another. sº CoR. 3. In the same manner, it may be shewn, in general, that, if each of two given invariable magnitudes be thus a limit between the same two variable magnitudes, i.e. if there always be a difference, between each of the variable and each of the invariable magnitudes, which yet may become less than any assignable, then, the two given invariable magnitudes shall be equal to one another. CoR. 4. It is manifest from the demonstration of Th. 69, that a polygon may be inscribed in a given circle, which shall exceed any rectilineal figure that is less than the circle; and also that a polygon may be described about a given circle, which shall be less than any rectilineal figure that is greater than the circle. Assumption II. The circumference of a circle is greater than the perimeter of any rectilineal figure inscribed in the circle; and it is less than the perimeter of any recti- lineal figure described about the circle. 160 Elementary Theorems of Plane Geometry, { ScHolium. That a circular arch is greater than the chord, and less than the aggregate of the two tangents, between which it lies, has been assumed, both by Archimedes and also by many subsequent writers, as if the truth of that proposition were sufficiently evident of itself. A demonstration of it may, however, be arrived at, by the following steps. (§. 1.) If a given arch of a circle be bisected, if it's half be again bisected, and so on continually, an arch may be found the chord of which, as well as each of the two tangents between which it is contained, shall be less than any given finite straight line. Either the given arch is less than the semi-circum- ference, or it is not. First, let the given arch ACB, of the circle ACBG, A. L- T} Tº ~ E F {X Gº - IH * B be less than the semi-circumference, and let K be the Comparison of Circles with Rectilineal Figures. 161 centre; suppose Ach to be bisected in C, AC in D, and AB, AD, and the diameters DKG, CKE, to be drawn, of which EC cuts AB in F: and since the point C is equidistant (hyp. and Th. 62.) from A and B, as is also (Def. 28. Cor.) the point K, KC bisects (Th. 20. Cor. 5.) AB at right angles in F; suppose also DF to be drawn and produced to meet the cir- cumference in H, and from H. H.L to be drawn at right angles to EC, and therefore (Th. 6.) parallel to BA, so that HL will meet the circumference between the points E and A. And since (Th. 1. Cor.) the 1. EKG= | DKc, ÉG=5C (Th. 39); and (hyp.) po-AD. therefore £&=1D, and GL1 > LAD; therefore (Th.60. Cor. 1.) the / ADG > / LHD; but HL is parallel to BA; therefore (Th. 10.) the / LHD = / AFD : wherefore the / ADG > / AFD; much more then is the / ADFP. Z.AFD; therefore (Th. 16.) AF, which is the half of AB, is greater than AD; and in the same manner a chord may be found, which shall be less than the half of AD, and so on, con- tinually; wherefore (Th. 68. Cor.) a chord may thus at length be found which shall be less than any given finite straight line. Next, let AD, BD, touch the given arch ACB in the points A, B, and let FE touch it in it's bi- section C, and meet DA in F, and DB in E, and suppose AC and BC to be drawn. Then (hyp. and Th.66. Cor. 1.) AF = FC= CE = EB, and D4 = X 162 Elementary Theorems of Plane Geometry, DB; therefore DF=DE ; but (Th. 17.) DF+ DE T) * - > FE; therefore DFX, FA, and DE-EB; i. e. AF and BE are each of them less than the half of DA or DB; and in the same manner a tangent may be found, which shall be less than the half of AF, and so on ; therefore (Th. 68. Cor.) a tangent may thus be found which shall be less than any given straight line. But, if the given arch be greater than the semi- circumference, it's half will be less than the semi- circumference; and this case will thus be reduced to the former case. (§. 2.) A given arch of a circle having been bi- sected, and it's halves having been bisected, and so on continually, and tangents having also been drawn through the extremities of the given arch and through the several points of bisection, so as to meet one an- other, the given arch is the common limit * of the * The mathematical meaning of the word limit has been already explained, in Th. 69. Cor. 3. Comparison of Circles with Rectilineal Figures. 163 aggregate of the tangents, and of the aggregate of the chords of the arches, into which it will thus have been divided. º Either the given arch is less than the semi-circum- * & & - ſº - g ference of the circle, or it is not. First, let ACB be less than the semi-circumference; and suppose both AB to be drawn, and AD, BD, touching the circle, of which ACB is an arch, at the extremities A, B, of AcB. suppose, also, K to be the centre of the circle, and KA, KB, KD to be drawn, of which KD cuts the circumference in C, and AB in E. Then (Th.66. Cor. 1. and Def. 28. Cor.) DA = DB, and KA = KB; therefore (Th. 20. Cor. 5.) the angles at Eare right angles, and the LAKD= LBKD; there- fore (Th. 59) AC= Éd, therefore (Th. 62.) AC= BC, these lines being supposed to be drawn. Let now KG be any straight line drawn from K, cutting AB in F, and ACB in G. and since (hyp. and Th. 57.) the LACB is obtuse, the LABC is (Th. 11. 164 Elementary Theorems of Plane Geometry, Cor. 3.) acute; therefore (Th. 16.) AB >AC; in like manner, the / AEC having been shewn to be a right angle, it may be proved that AC-EC, and KF >KE; but KC=FG, and of KC the part KE- KF the part of KG; therefore the remainder EC > FG: and it has been shewn that AB > AC, and AC-EC; much more then is AB > FG; i. e. the distance between any point in an arch, that is less than the semi-circumference, and it's chord, measured in a straight line drawn from that point to the centre, is less than the chord. But if the arch ACB be bisected, if it's halves be also bisected, and so on continually, it may thus be divided into arches the chords of which shall (§. 1.) be each of them less than any given straight line; and therefore the distance between every point in each of those arches and the cor- responding point of it’s chord, measured in the direction of a straight line drawn from the centre, will be less than any given straight line. It is ma- nifest, therefore, that the arch ACB is the limit of the aggregate of the chords of the parts into which it may be divided by that continual bisection. Again, let KHI be any straight line, drawn from f{, cutting AcB in H, and BD in I. And, since the is KCB is isosceles, the / KCB is (Th. 13. Cor. 2.) acute, and therefore the LDCB is obtuse and (Th. 11. Cor. 3.) therefore greater than the / DBC; where- fore (Th. 16.) CD & BD ; also since (hyp. and Th. 46.) the / KBD is a right angle, the / KID is an obtuse angle, and KD > KI; but KC, a part of º Comparison of Circles with Rectilineal Figures. 165 § - KD, is equal to KH, a part of KI; therefore HI T. < CD, and, as hath been shewn, CD- BD; much more, then is H13 B.D. And if tangents be supposed to be drawn through C, and again through the bi- sections of AC and BC, and so on, tangents may thus be found (§. 1.) which shall each of them be less than any given straight line: much more, then, will the distance, between every point in each of the resulting arches and the corresponding point of it's tangent, measured in the direction of a straight, line drawn from the centre, be less than any given straight line. It is manifest, therefore, that the arch ACB is the limit of the aggregate of the tangents of the parts into which it may be divided by that continual bisection. But if ABQ be the given arch, and it be greater 166 Elementary Theorems of Plane Geometry, than the semi-circumference, suppose it to be bisected © * e e- e- * in B; then will AB, BQ be each of them less than the semi-circumference; and this case will thus be reduced to the former case. (§. 3.) If two variable magnitudes, of which the one continually increases whilst the other continually decreases, have a common limit, and if the decreasing magnitude be always greater than the other variable magnitude, it shall always be greater than their common limit; and the other, viz. the continually in- creasing magnitude, shall always be less than their common limit. - Let A be the common limit of the two variable magnitudes X and K, of which X is always greater than P, and continually decreases, whilst P increases: AC-A, and P-3 A. For, by the definition of a limit, neither X nor P. can ever be equal to A; therefore, either X > A, or X- A: if it be possible let X- A; and since (hyp.) P may be greater than any magnitude that is less than A, F may be greater than X, which is contrary to the supposition: wherefore, since X cannot be either equal to A or less than 4, X > A. And in like manner, it may be shewn that F-3 A. (§. 4.) The arch of a circle, lying between its chord and the two tangents, drawn from it's ea:- tremities and produced so as to meet one another, is greater than the chord, and less than the aggregate of the two tangents. Comparison of Circles with Rectilineal Figures. I67 \ ,-\ * * Let ‘ACB be the arch of a circle, having AB for it's chord, and let AD, BD, which meet in the point D, touch the circle, of which ACB is an arch, in the e a TN – ~TS — — points A, B : ACB > AB; and ACB< AD+BD. Let ECF be supposed totouch the circle at the point of bisection C of ACB, and AC, BC, to be drawn. Suppose, also, tangents to be drawn at the points of bisection of AC and fo, and so on con- tinually. Then (Th. 17.) AD+ DB-AB, AF4. FC>AC, CE4-EB-> BC, also (Th. 18.) AD-I-DB > AC+CB, and (Th. 34.) AF4 FC+CE + EE > AB. And, if G and H be the points of bisection of AC and #C, and the chords AG, GC, BH, HC, be supposed to be drawn, then (Th. 17.) AG + GC + CH+ HB > AC+ CB. Thus it may be shewn that the aggregate of the tangents is always greater than the aggregate of the chords of the arches, and that the aggregate of the tangents continually de- creases, whilst that of the chords continually increases: and (§. 2.) the arch 2CB is the common limit of these two aggregates; wherefore (§. 3.) the aggregate of the tangents is always greater than the arch, and the aggregate of the chords is always less than the arch. But (Th. 17. and Th. 34.) AB is always less * See the fig. in page 162. 168 Elementary Theorems of Plane Geometry, than the aggregate of the chords of the arches, into which ACB is subdivided by the continual bisection; and (Th. 17.) AD+DB is always greater than the aggregate of the tangents drawn at the extremities of 2-TS — 2-> — those arches : therefore, ACB > AB, and ACBq C+ C ; therefore (hyp. and Th. 71.) nA-q C; and q C was taken greater than n B ; therefore nA >q C; and n B : q C. Wherefore, of two unequal magnitudes, &c. a. E. D. Elements of plant Geometry. y Book I. Chartºn IV. SECTION II. On the comparison of ratios. —O— DEFINITION XLV. Of two magnitudes of the same kind, if the less can be multiplied, so as to exceed the greater, the two magnitudes are said to have a Ratio to one another. DEF. XLVI. There are three kinds of ratio: if two finite homo- 180 Elementary Theorems of Plane Geometry, *. geneous magnitudes, be equal, they are said to have to one another a Ratio of Equality; if they be un- equal, the greater is said to have to the less a Ratio of Majority, and the less is said to have to the greater a Ratio of Minority. DEF. XLVII. Any two finite homogeneous magnitudes are called the Terms of the ratio, which those magnitudes have to one another: and the first named, or the first written, magnitude, is called the Antecedent term, and the other the Consequent. DEF. XLVIII. Two Ratios are said to be equivalent to one an- other, when the antecedents, taken any, the same number of times, have toward the consequents taken any the same number of times, the same kind of ratio. Thus, if A, B, C, D, be four magnitudes, and p24, p C, be any equi-multiples of A and C, and if like- wise q B, q D, be any equi-multiples of B and D, then if p C > = < q D accordingly as p A > = < q B, the ratio of A to B is said to be equivalent to the ratio of C to B. SCHOLIUM. This definition of equivalent ratios, which forms, as it were, the basis of our reasoning on the comparison of ratios, has been adopted, principally on account of Comparison of Ratios. 181 the facility with which it may be applied, as a criterion, for determining the proportionality of geometrical magnitudes. The subject of proportionality, however, seems to require more elucidation than can well be given to it, by the mere demonstration of the pro- positions in which it is usually comprised. We shall, therefore, endeavour, more at length, to explain the nature of proportionality, in this preliminary Scho- lium. # f * (§. 1.) The principal object of the science of Geo- metry is the measurement of extension: and this measurement consists entirely in ascertaining the relation, as to quantity, which a proposed magnitude bears to some other magnitude of the same kind with itself, which may be considered as a standard, or, at least, as a known magnitude. When, therefore, two like magnitudes are compared, it may first be considered, whether they are equal; that is, whether the one, or all the parts into which it may have been divided, can be applied to the other, so as wholly to coincide with it. In this case, the two magnitudes are said to have to one another a ratio of equality. But if the two magnitudes under comparison be unequal, the greater is said to have to the less a ratio of majority ; and the less is said to have to the greater a ratio of mi- nority. Thus there are three kinds of ratio: and, if (A) and (B) be two homogeneous magnitudes, and (C) and (D) two other homogeneous magnitudes, then if A = B and C= D, or, if A-B and C-D, or, lastly, if A * B and C* D, in each of these cases, A is said 182 Elementary Theorems of Plane Geometry, to have to B the same kind of ratio, that C has to D. (§. 2.) Now almost all the theorems, which have hitherto been demonstrated, in this book, concerning the comparison of like magnitudes, as to their relative quantity, can serve only to ascertain whether of two such magnitudes, the one has to the other a ratio of equality, or not; that is, whether the one is equal to, or greater than, or less than, the other. But this conclusion is not sufficient for the purpose of measur- ing extension. For that purpose, if two like mag- nitudes be not equal, it is of importance further to ascertain, whether the greater of them contains the less, an exact number of times; or whether, if one of them be divided into a certain number of equal parts, the other is equal to some exact multiple of one of those parts: and thus Geometry connects itself with Arithmetic. (§. 3.) If a known magnitude can be shewn to contain another, or to be contained in it, an exact number of times, that other magnitude, manifestly, may be considered as being thereby determined, in respect of it's quantity. If, for example, 64 = B, whenever the quantity of (A) is known, the quantity of (B) is known also: or if A = 3 C, and if (A) be a standard, or known magnitude, the quantity of C is thereby determined, as being equal to the third part of (A). This, in reality, is nothing more than an expression of the result of actual measurement, as it is practised in the case of lines. If a yard (4) be taken as the standard measure, and if, by an actual Comparison of Ratios. - 183 application of it to the successive parts of another straight line (B), it is found to be contained in that line (B) exactly six times, (B) is said to be mea- sured, and it's quantity is considered as being known. And so of any less straight line (C) which is found to be contained exactly three times in the standard lineal measure (A). (§. 4.) In such instances as these, then, there is a certain multiple of the one of the two magnitudes, under comparison, which is exactly equal to the other magnitude: and if of these equals any equi-multiples be taken, it is evident that they, also, must be equal to one another; and that they, likewise, will exhibit the relative quantity of the two magnitudes, which are to be compared. This relation, therefore, when it has once been determined, by finding a multiple of one of the two magnitudes, that is equal to the other, may afterwards be expressed in a number of ways, to which there is no limit. If m A = B, and if pa, and q B, be any equi-multiples of (mA) and of (B), then pA = qB, and the quantity of (B) compared with that of (A) will thence be as completely determined, as it is by the equation m A = B. Conversely, also, if pA= qB, and if the number (p) be a multiple of (q), it is evident that there is some number (m), such that m A= B. And, likewise, if (p) be a part of (q.), there is some number (n) such that A=n B. (§. 5.) But, further, if (A) and (B) be two like mag- nitudes, if (p) and (q) be two numbers, neither of which is a multiple of the other, and if p A = q B, 184 Elementary Theorems of Plane Geometry, then, also, will the known quantity of (A) absolutely determine the quantity of (B). ... Let, for example, 3 A=5 B, and let A be a known magnitude: therefore, the fifth part of (A) is also known; let, then (C) be the fifth part of (A); therefore, 3 A = 15 C; but (hyp.) 3A = 5 B ; therefore 15 C=5 B, and (B) itself is, therefore, equal to three times the known magnitude (C). So, in general, if p and q be put for any numbers whatever, if p &l= q B, and if (C) be the q" part of the known magnitude (A), then B = p(X: it is evident, also, that, if of these equals, (B) and (pC) any equi-multiples be taken, they will be equal to one another, and will likewise exhibit the relation, as to quantity, which the mag- nitude B has to the known magnitude (C), (C) being a definite part of the known, or standard, magnitude (A). In this case, therefore, the one magnitude is determined, by it's being equal to an exact number of certain equal parts of the other. (§. 6.) If, then, in the comparison of two homo- geneous magnitudes, there could always be found a multiple of the one equal to the other magnitude, or if there could always be found a multiple of the one equal to a multiple of the other, the known quantity of the one of the two magnitudes would always exactly determine the quantity of the other. . It becomes, therefore, very important to know whether a possibility of finding equal multiples of two proposed magnitudes, of the same kind, does in all cases obtain, or not. (§. 7.) Now, if two homogeneous magnitudes, (A) and (B), have a common measure, (M), equal Comparison of Ratios. 185 multiples of them may always be found. For if (4) contain (M) any number of times, exactly, as p times, and if (B) contain (M) any other number of times, exactly, as q times, it is evident that p times (M), taken q times, is equal to q times (M), taken p times; according to a well known arithmetical theorem ; and therefore q times (A) is equal to p times (B). (§. 8.) On the other hand, if two proposed homo- geneous magnitudes AB, and CD, have not a common $º A | - B * T º E. - ~. ...” C Tiš F. measure, no multiple of the one can be found which shall be equal to a multiple of the other. For, if it be possible, let some multiple AE, of AB, be equal to a multiple CF, of CD; let three times AB, for example, be equal to twice CD. Then, it is evident that the half of AB will be found six times in AE; therefore, the half of AB will, also, be found six times in CF, which (hyp.) is equal to AE; therefore, the half of AB will be found three times in CD which (hyp.) is the half of CF; therefore, AB and CD have a common measure, which is contrary to the hypothesis. And the same absurdity may, in like manner, be demonstrated, whatever be the multiples of AB and CD, which are supposed to be equal. Therefore, if two magnitudes have not a A A 186 Elementary Theorems of Plane Geometry, common measure, there are no multiples of them that are equal to one another. sº The question, whether a multiple of the one of two proposed homogeneous magnitudes can always be found that shall be equal to some multiple of the other, is, therefore, reduced to this, namely, whether two such magnitudes have always a common measure. Now, to shew that there are homogeneous magnitudes which have not a common measure, we have only first to demonstrate that if magnitudes have to one another a certain kind of relation, they cannot have a common measure; and then, to prove, that the side of any given square has to it's diameter that particular kind of relation. (§. 9.) First, then, AB, the less of two proposed homogeneous magnitudes, having been taken, as often A Q O B F D C O E. as it can be, from CD the greater, the remainder as often as it can be from AB the less, and so on, if a remainder can be found, that is less than any given magnitude of the same kind with it, the two proposed magnitudes, AB and CD have no common measure. For, if it be possible, let E measure AB and CD: and let AB be taken, as often as it can be, from CD; ket the remainder FD be taken as often as it can be Comparison of Ratios. 187 from AB, and so on, until (hyp.) a remainder GB is found, which is less than E: and since E measures AB, it necessarily measures CF, which is a multiple of AB; it also measures CD; and it, therefore, measures FD, the difference between CD and CF: in like manner, it may be shewn that E measures GB; which, since GB is less than E, is impossible. Therefore AB and CD have not any common ImeaSure, (§. 10.) Let, now, AB be the side of the square ABCD, of which AC is the diameter, and let L be A D ..]...NG H ANE 㺠| F º | B Ö any straight line: let the angle ACB be bisected by CF, and from F let FE be drawn at right angles to AC; wherefore (Th. 19.) the two triangles CEF, CBF are equal, CE is equal to CB, and to AB, and EF is equal to BF; and since the angle AEF is a right angle, and the angle EAF half a right angle, the angle AFE is (Th. 12.) half a right angle, and (Th. 15.) the side AE is, therefore, equal to EF and consequently equal, also, to FB: so that AE is the 188 Elementary Theorems of Plane Geometry, remainder, when AB has been taken from AC; and since (Th. 16.) AF is greater than AE, much more is AB, or CE, greater than AE; therefore AB is greater than the half of AC, and cannot be taken more than once from AC. Again, if FG be drawn bisect- ing the angle AFE, and if GH be drawn at right angles to AF, it may be shewn, as beforé, that HF is equal to EF, and to FB, that AH is equal to HG and to GE, and that AG is greater than AH, and than GE ; and that, therefore, AF, which is greater than AE, is also greater than AH; so that AH is the remainder, when AE has been taken as often as it can be from AB; and, if the same process be repeated by bisecting the angle AGH, since the side AHI of a square will then again have to be taken from it's diameter AG, it is manifest that there will still be a remainder; and so on continually. But since from AC, in the first place, and then from each of the successive remainders, there will thus have been taken more than it's half, a remainder will (Th. 68. Cor.) at length have been found, that is less than L, how- ever little L may have been assumed. Wherefore, (§. 9.) AB and AC have no common measure; and therefore (§. 8.) there is no multiple of the one that is equal to a multiple of the other. It appears, then, that there are homogeneous magnitudes, which have no common measure, and of which, therefore, no multiples can be taken, so as that from the known quantity of one of the magnitudes the quantity of another of them may thereby be absolutely deter- mined. Such magnitudes are called incommensurables; Comparison of Ratios. 189 and it would be easy, if it were necessary to our present purpose, to exhibit other instances of them, besides that of the side and the diameter of a square. (§. 11.) But although when two incommensurable magnitudes, the quantity of one of which is known, are to be compared, no multiples of them can be taken, which, being equal to one another, may serve to determine absolutely and exactly the quantity of the other, yet multiples of the two magnitudes may be so taken, as very nearly to determine the quantity of the other. For if the less of two such magnitudes be taken as often as it can be from the greater, if the remainder be likewise, taken as often as it can be from the less, and so on, continually, a remainder will at last be found, that shall be less than any given magnitude, of the same kind, how little soever it be. But each of these successive remainders is, in reality, the difference between two certain multiples of the two proposed magnitudes; wherefore, two multiples of those magnitudes may, it is evident, be found, the difference between which shall be less than any assigned magnitude of the same kind: and, thus, if the quantity of one of the proposed magnitudes be known, that of the other will, thereby, be very nearly determined. Let A and B, for example, be any two incommen- surable magnitudes, of which A is the greater; when B has been taken as often as it can be from A, let the remainder be R, which is consequently less than B: again, when R has been taken as often as it can be from B, let the remainder be R'; and when R' has been taken as often as it can be from R, let the 190 Elementary Theorems of Plane Geometry, e f remainder be R"; and so on. None of these re- M | | | A B R R’ R” mainders will exactly measure that which immediately precedes it; for if it did, it would also measure the other preceding remainders, and the two magnitudes A and B, which (hyp.) have no common measure. There will, therefore, always be a remainder, however far the operation of taking one remainder, as often as it can be from the next preceding remainder, be carried. And since R is the remainder, when more than it's half has been taken from A, since R’ is the remainder when more than it's half has been taken from R, and so on, a magnitude will (Th. 68. Cor.) at length be found, that shall be less than any given magnitude M, of the same kind as A and B. But, from the operation, it is manifest that R is the dif- ference between A and a multiple of B; that R is the difference between B and a multiple of R, that is, of Comparison of Ratios. 191 the former difference between A and a multiple of B; wherefore R' is the difference between a multiple of A and a multiple of B: in the same manner, it may be shewn that R", and every succeeding remainder, is the difference between a multiple of A and a mul- tiple of B. And it has been shewn that a remainder may be found, that shall be less than any assigned magnitude M, however little it be. There are, there- fore, certain multiples of the two incommensurables A and B, which have a difference that is less than any given magnitude whatever; which are, con- sequently, very nearly equal to one another; and which thus very nearly determine the mutual relation of the two magnitudes in respect of quantity. (§. 12.) What has hitherto been said, in this Scholium, applies only to that mutual relation, singly considered, of two homogeneous magnitudes, which is called Ratio. But, further, the ratio, which one magnitude has to another of the same kind, may be compared with the ratio which a third has to a fourth of the same kind. And the most simple and general comparison, of two ratios, is that which goes no farther than to determine whether they be both of the same kind; that is, whether they be both of them ratios of equality, both of them ratios of majority, or both of them ratios of minority. Ratios admit, however, of a mode of comparison more particular, and more exact, and which, at the same time, leads to results of much greater importance to the science of Geometry. And, first, if A, B, C, be magnitudes of the same kind, and if A contain B as often, exactly, as B contains C; or 192 Elementary Theorems of Plane Geometry, if A be contained in B as often, exactly, as B. is contained in C; or if p times A be equal to q times B, and p times B be also equal to q times C; it is plain that the mutual relation of A and B, in respect of quantity, is the same as that of B and C: likewise, if A and B be homogeneous magnitudes, and if C and D be magnitudes of the same kind with A and B, or, at least, of the same kind with one another, and if A be the same part, or multiple, of B, that C is of D, or if p A = qB, and, at the same time, p C= q D, in any of these cases, the mutual relation of A and B, in respect of quantity, is manifestly the same as that of C and D. - ... (§. 13.) This identity of mutual relation is called Proportionality; the magnitudes between which it obtains, are called Proportionals; and the two ratios, thus compared, are said to constitute a Proportion, and they have by some been called equal, by others similar ; by others the same. Objections may, how- ever, be made to each of these terms, when they are so applied. The term equal ought to be confined, in Geometry, to such magnitudes as can be supposed to be so placed, as wholly to coincide with one an- other. The word similar conveys an indefinite idea; for similarity admits of degrees. And things, which are really separate, cannot with strict propriety be called the same. Let, then, the ratios, which con- stitute a proportion, be called equivalent ratios. The proportionality of four magnitudes, which are denoted by A, B, C, D, is more concisely expressed by saying, that A is to B as C is to D; or by writing Comparison of Ratios. 193 the letters, which denote the magnitudes, with inter- vening signs; viz. A : B : C : D. If, then, of four magnitudes, either all of the same kind, or two and two of the same kind, the first be the same multiple of the second, or the same part of the second, that the third is of the fourth, or if the first contain the same number of certain equal parts of the second, exactly, that the third does of the fourth, the ratio of the first to the second is said to be equivalent to the ratio of the third to the fourth: and this may be considered as the primary motion of proportionality. (§. 14.) But, if four magnitudes A, B, C, D, either all of the same kind, or two and two of the same kind, be incommensurables, no such identity of mutual relation, in respect of quantity, between the first two and the second two, as that which has been spoken of, can be exhibited. For A does not, in that case measure B, nor does B measure A ; neither can any multiples of A and B be found that are equal to one another: the same may be said of C and D. In magnitudes, therefore, that are incommensurable, we must look for some other mark of the existence of this kind of identity of mutual relation in respect of quantity. (15.) Now, it has already been shewn, in the Scholium to Th. 30, that four magnitudes A, B, C, D, may be so related, as that when any equi-multiples of the first and third, as p.1 and pC, have been taken, and any equi-multiples, as q B and q D, of the second and fourth, have likewise been taken, p and q being put for any numbers whatever, if p A3 q B then p C B B 194 Elementary Theorems of Plane Geometry, qD; which is also denoted more concisely thus, if p A > = < q B, then p C >= < q D: and this pro- perty will obtain whether the four magnitudes A, B, C, D, be incommensurables or not. To such a rela- tion of four magnitudes it is expedient to give a name: and if it can be shewn to include in it, the primary notion of proportionality, we may, then, without impropriety, make use of that same appellation, whereby to express it. One of the cases of proportionality, as that term has been already explained, the relation last described, evidently includes; namely, that in which if p A = q B and p C = q D, the magnitudes A, B, C, D, are said to be proportionals. Again, let A, B, C, D, be called proportionals, when, if p AP = < q B, p C P = < q D : then A, B, C, D, being proportionals according to that definition, if A = m B, C= m. D.; or if B = m.A, then D = m C. For, first, let A = m B: let A and C be taken any the same number of times, as twice; and let m B and m D be taken, each of them, that same number of times, namely, twice; so that the multiples of the four magnitudes are 2A, 2 m B, 2 C, 2 m D. It is mani- fest that equi-multiples have thus been taken of the first and third, and again of the second and fourth : and (hyp.) A=m B, and therefore 2A = 2m B; there- fore (def) 2C=2m D, and therefore C=m D. In like manner, if A, B, C, D, be proportionals according to the last definition, and if B = m.A, it Comparison of Ratios. 195 may, by taking for the equi-multiples 2 m 4, 2 B, 2m C, 2 D, be shewn that D=m C. Conversely, also, when four magnitudes A, B, C, D, are so related as that m A = n B and m C-m D, if any multiples whatever be taken of A and B, as p A and q B, then, accordingly as p A > = < q B, p C- = < q D. - For, first, let p A-q B; therefore pm. A - qm B; but (hyp.) m A= n B; therefore pn B-qm B; there- fore pn - qui; therefore pn D > qm D; but (hyp.) n D=m C; therefore pm C - qm D; therefore p C > q D. In the same manner it may be shewn, that p C < q D, when p A ºr q B, and that p C= q D, when p A=q B, if m A = n B, and at the same time m C- n D : the same mode of proof, also, applies to the cases in which either m, or n, denotes unity; that is, to the cases, in which A is either a multiple or a part of B. Thus it appears, that the extended notion of pro- portionality, which was exemplified in the Scholium to Th. 30, does really comprehend that which is more familiar to the minds of the greater number of men, and which was first explained. Whenever, therefore, we can prove four magnitudes to be propertionals, according to our forty-eighth definition, we may thence conclude, if the magnitudes be commensura- bles, that the third is the same multiple, the same part, or the same number of certain equal parts, of the fourth, that the first is of the second. And in this case, the propriety of calling the ratio of the first to 196 Elementary Theorems of Plane Geometry, the second, equivalent to the ratio of the third to the fourth, is very manifest. But when four incommensurable magnitudes are related to one another in the manner specified in the forty-eighth definition, the definition does not of itself directly indicate any such identity of relation, between the two first and the two last of them, as hath been shewn to obtain in the case of commensurable mag- nitudes, that are proportionals. Certain multiples may, however, be taken, as was shewn, of the first and second, the difference of which multiples shall be less than any assigned magnitude, of the same kind with them : and so of the third and fourth. If, then, we can demonstrate, in virtue of our forty-eighth de- finition, that there are equi-multiples of the first and third of four incommensurable magnitudes, and equi- multiples again of the second and fourth of them, such that the multiples of the two first magnitudes are very nearly equal to one another, and the multiples of the two second are, also, very nearly equal to one another, we may consider the two ratios under com- parison, as being very nearly equivalent, even ac- cording to the sense, in which the ratios of commen- surable magnitudes have been considered as equi- valent. Now that there are such equi-multiples of incommensurable proportionals, may easily be shewn, taking for granted certain propositions which will afterwards be demonstrated, and to which we shall refer. For, let A, B, C, D, be four incommensurable magnitudes, that are proportionals according to the Comparison of Ratios. 197 forty-eighth definition: let B be found at most p times in A, and, when p B has been taken from A, let the remainder be R ; so that A = p B+ R; like- wise, let C= q D + S: then shall q be equal to p. For, if it be possible, let p > q; therefore (q+1) is not greater than p ; and (hyp.) A =p B+ R ; there- fore A > (q+ 1) B ; also (hyp.) A : B : C ; D, therefore (Th. 93. Cor.) A : (q+1) B : C : (q+1) D; and it has been shewn that A > (q + 1) B, there- fore (Th. 84.) C - (q + 1) D: but (hyp.) C= q D+ S; therefore C< q D + D, that is (Th. 72.) C-: (q+ 1) D: but it has been shewn that C- (q+1) D: which is absurd; therefore p is not greater than q : and, in like manner it may be shewn that p is not less than q; therefore p=q; therefore if A = p B+ R, C =p D+S. Again, since A : B : C : D, ... (Th.93. Cor.) A : pH : C : p1); but (hyp.) R is the difference between A and p B, and S is the difference between C and p D; there- fore (Th. 88.) R : p B :: S: p D; ... (Th. 96. Cor.) R : B :: S : D; ... (Th. 79.) B : R : D : S. And thence it may be shewn, that R is found as often in B, as S is found in D, in the very same manner, as B was shewn to be contained in A, as often as D is contained in C: and so of the subsequent remainders. It is manifest, therefore, from Ş. 11, that, in the same number of steps, multiples of A and B may be found, which have a very small difference, and multiples of C and D may also be found which 198 Elementary Theorems of Plane Geometry, have a very small difference; and that of these, the first and third are equi-multiples of A and C, and the second and fourth are equi-multiples of B and D. There can, therefore, be taken equi-multiples of the first and third of four incommensurables, and equi-multiples of the second and fourth of them, such that the multiple of the first shall be very nearly equal to the multiple of the second, and that the mul- tiple of the third shall, also, be very nearly equal to that of the fourth. And although these multiples have not exactly equal differences, yet may their differences be made to become so small, as, in both cases, to be less than any finite magnitude of the same kind with themselves. In our forty-eighth definition, then, there is also included that which we have called the primary notion of proportionality, so far as, by approximation, it is applicable to incommensurable magnitudes. DEF. XLIX. Any two equivalent ratios are called a Proportion : and magnitudes, of which the ratios are equivalent to one another, are called Proportionals. ScholIUM. When three magnitudes are proportionals, it is usually expressed by saying, the first is to the second as the second to the third ; in like manner as when four magnitudes are proportionals, the first is said to be to the second, as the third to the fourth : and, conversely, Comparison of Ratios. 199 when the first of three magnitudes is said to be to the second as the second to the third, or when the first of four magnitudes is said to be to the second as the third to the fourth, it is meant, that the magnitudes are proportionals. The proportionality of three mag- nitudes A, B, C, is also expressed, in writing, as follows, viz. A : B :: B : C; in like manner as that of four magnitudes A, B, C, D, is expressed; viz. A : B :: C ; D. y 4. DEF. L. Magnitudes are said to be Continual Proportionals, when the first is to the second as the second to the third; and the second is to the third as the third to the fourth; and so on. DEF. L.I. When three magnitudes are continual propor- tionals, the second of them is said to be a mean Pro- portional between the other two. Dr. LII. The first and the last of any number of propor- tionals are called the Eartremes, and the others are called the Means. DEF. LIII. In proportionals the antecedent terms are said to be homologous to one another: and the consequents are also said to be homologous to one another. 200 Elementary Theorems of Plane Geometry, THEOREM LXXVII. The ratios, which equal magnitudes have to the same magnitude, are equivalent to one another; and the ratios, which the same has to equal magnitudes, are also equivalent to one another. Let A and B be equal magnitudes, and C any other: then A : C : B : C ; also C : A : C : B. Let p A and pH be any equi-multiples of A and B, and let q C be any multiple of C: then since (hyp.) A = B, therefore p A=p B; therefore if p A- = < q C, p B > = < q C; therefore (Def. 48.) A : C :: B : C. Again, since p A = p B, if q CP = < p A, q C-> = < p B; and therefore C : A :: C : B. Wherefore, the ratios, &c. a. E. D. THEoREM LXXVIII. Magnitudes which have equivalent ratios to the same magnitude, are equal to one another : and those to which the same magnitude has equivalent ratios are, also, equal to one another. First, let A : C : B : C; then A= B. If not, one of them is the greater; let A be the greater; then (Th. 76.) there are some equi-multiples, p A and p B, of A and B, and some multiple q C, of C, such that pA-q C and pH~ qC; wherefore A is not to C as B is to C; which is contrary to the hypo- thesis; therefore A is not unequal to B; i. e. A = B. Comparison of Ratios. 201 Secondly, let C : A : C : B; then, if 4 be not equal to B, it may be shewn, in the same manner as before, that multiples q C, p.A, p B, may be taken, such that q C.< p A and >p B; and therefore C is not to A, as G is to B; therefore in this case also, A=B. Wherefore, magnitudes, &c. a. E. D. CoR. It is manifest, from the demonstration, that the ratios which unequal magnitudes have to the same magnitude, are notequivalent to one another: neither are the ratios which the same has to unequal magnitudes, equivalent to one another. THEoREM LXXIX. If the first of four magnitudes be to the second as the third to the fourth, then shall the second of them be to the first as the fourth is to the third. Let A : B : C : D ; then B : A :: D : C. Let there be taken of A and C any equi-multiples pA, p C; and of B and D any equi-multiples q B q D : then (hyp.) A : B :: C : D ; therefore (Def.48.) p C > = < q D accordingly as p & P = < q B; there- fore if q B > = < p A, then q D-= = < r B, q times p BP = < rB, and therefore (Def. 43. Cor. 3.) q times p > = < r ; and therefore q times p1) > = < r D; or, since (hyp.) C-p D, q C - = = < q B, p C> = < q D; also since C : D : E : F, if p C > = < q D, p B >= < q F; therefore if p A >= < q B, p EP = < q F; therefore (Def. 48.) A : B : E : F, wherefore, ratios that are, &c. a. E. D., , - CoR. Ratios, that are equivalent to equivalent ratios, are also equivalent to one another. THEOREM LXXXII. If the first of four proportional magnitudes be a multiple, or a part, of the second, the third is the same multiple, or the same part, of the fourth. If A : B : C ; D, and if A = p B, then C= p D. For (hyp. and Th. 80.) A : B :: p D : D; and (hyp.) A : B :: C : D; ... (Th. 81.) C : D :: p1) : D; - ... (Th. 78.) C-p D. Again, if A : B : C ; D, and if p A= B, then, p C= D. * - - For (hyp.) A : B :: C : D; ... (invertendo) B : A : D : C; and (hyp.) B=pA; therefore, as in the former case, D=p C; that is, C is the same part of D that A is of B. Therefore, if the first, &c. a. E. D. THEoREM LXXXIII. If any number of magnitudes be proportionals, as one of the antecedents is to it's consequent, so shall all the antecedents taken together be to all the con- sequents. -ºf 204 Elementary Theorems of Plane Geometry, Let any number of magnitudes A, B, C, D, E, F. be proportionals; then A : B :: A + C+ E : B+ D+ F. * Take of A, C, E, any equi-multiples p 4, p C, p E, and of B, D, F any equi-multiples q B, q D, q F: then (hyp. and Def. 48.) if p A >= < q B, p CP = < q D; and if p C2 = < q D, p EP = < q F; there- fore if p A >= < q B, p A + p C+p E > = < q B + q D+q F; i. e. (Th. 71.) all the antecedents taken p times > = < all the consequents taken q times; therefore (Def. 48.) A : B :: A+C+E : B+D+ F. Therefore if any, &c. a. E. D. THEOREM LXXXIV. If the first of four proportional magnitudes be greater than the second, the third is also greater than the fourth; and if equal, equal; and if less, less. * r * * Let A : B : C : D and first let A-B; then, C - D. Take the doubles of the four magnitudes; and since (hyp.) A- B, twice AP twice B; therefore (hyp. and Def. 48.) twice C - twice D; therefore C-D. In like manner it can be shewn if A = B, that C-D; and if A & B, that C-3 D. If, therefore, the first, &c. Q. E. D. * THEOREM LXXXV. If four magnitudes, of the same kind, be pro- portionals, then if the first be greater than the third, Comparison of Ratios. 205 the second shall be greater than the fourth ; and if equal, equal ; and if less, less. Let A, B, C, D, be four magnitudes of the same kind, and let A : B :: C : D; also first let A-C; then B- D. Since (hyp.) A-C, there may be taken (Th. 76.) of A and C, some equi-multiples p A, p C, and of B some multiple q B, such that pA-q B, and p C3 q B; and since (hyp.) A : B :: C ; D, and that pA > q B, therefore (Def. 48.) p C-q D; but q B->p C; much more then is q B-q D; therefore (Def. 44. Cor. 4.) B> D. - In like manner, it may be shewn, if A - C, that B & D. . And, if A=C, then (Th. 78.) B- D. If there- fore, four magnitudes, &c. a. E. D. CoR. If A, B, C, D be magnitudes of the same kind, and if A : B :: C : D, then accordingly as B> = < D, AP = < C. For (hyp. and Th. 79.) B : A : D : C; ... (Th. 85.) accordingly as B--- D, AP = < C. THEoREM LXXXVI. The ratio which two magnitudes have to one an- other, is equivalent to the ratio which their equi- multiples have to one another. If p A and p B be any equi-multiples of the two magnitudes A and B, then A : B : p_A : p B. For since A : B :: A : B :: A : B, and so on, therefore (Th. 83.) A : B :: A + 4 + A (p times) : 206 Elementary Theorems of Plane Geometry, B+B+B+&c. (p times); i. e. A : B :: p.A : pK. Wherefore, the ratio, &c. a. E. D. CoR. If four magnitudes be proportionals, the first shall be to the second, as any multiple of the third is to the same multiple of the fourth; and any multiple of the first shall be to the same multiple of the second, as the third is to the fourth; also any multiple of the first shall be to the same multiple of the second, as any multiple of the third is to the same multiple of the fourth. Let A : B :: C : D, then (Th. 86. and 81. Cor.) A : B :: q C : q D; likewise p A : pH :: C : D; ... (hyp. and Th. 81. Cor.) p4 : plb :: q C : q D. THEOREM LXXXVI I. If four magnitudes of the same kind be propor- tionals, the first shall also be to the third as the second is to the fourth. If A, B, C, D, be four magnitudes, of the same kind, and if A : B :: C : D, then, also, A : C :: B : D. Af For (hyp. and Th. 86. Cor.) p4: pK :: qC : q D; therefore (Th. 85.) if p A- = < q C, p B-3– « q D; ... (Def. 48.) A : C : B : D. Wherefore, if four magnitudes, &c. a. E. D. Def. Lv. When it is said of four proportional magnitudes Comparison of Ratios. 207 of the same kind, that they are proportionals, taken alternately, it is meant that the first of them is to the third as the second is to the fourth: and the making such a change in the order of four proportionals is called alternando. THEOREM LXXXVIII. If the first together with the second, of four magnitudes, be to the second, as the third together with the fourth is to the fourth, then shall the first be to the second as the third is to the fourth. If A + B : B : C+D : D, then A : B :: C : D. Take of (A+B) and of (C+ D) any equi-multiples p (A+B) and p (C+D), and of B and D any equi- multiples q B and q D; then (Th. 71.) p4+p B, and p C+p Dare equi-multiples of (A+B) and of (C+D); also q B+p B, and q D+p D, are equi-multiples of B and D; wherefore, (hyp. and Def. 48.) if p A+ p BP = < q B+pH, then p C+p DP = q, it is manifest, that p 4+p B-q B, and p C +p D-q D; and (Th. 71.) p4+pb, and p C+p D, are equi-multiples of A+ B and of C+D, respectively, as q B and q D are of B and D. In the same manner, if p=q, it may be shewn that, p times A+B >q B, and p times C+ DX-q D. Lastly, if p < q, let their difference be r, then (Def. 43. Cor. 1.) plb+r B= q B; also, since (hyp.) A : B :: C : D, therefore (Def. 48.) if p.4- = < r B, p C > = < r D; therefore adding ph to the two former of these multiples, and p D to the two latter, if p A+p BP =

= < p D+ r D; i.e. (Th. 71.) if p times A+B-> = < q B, then p times C+ DX = < q D. Therefore whatever equi-multiples be taken of A+ B and of C+ D, and whatever equi-multiples be taken of B and D, if the Comparison of Ratios. * 209 first multiple> = < the second, the third multiple> = < the fourth; therefore (Def. 48.) A + B : B :: C+ D : D. Wherefore, if the first, &c. a. E. D. ScholIUM. If it be granted, according to the assumption of Euclid (E. ii. 12.), that three finite magnitudes being given, of which the second and third are of the same kind, there is always some magnitude, which shall be to the first as the second is to the third, the pre- ceding proposition may thence be demonstrated in a more succinct manner, as follows. - Let A : B :: C : D, then A+ B : B :: C+D : D. For there is (assump.) some other magnitude, which is to B as C+D is to D, and which (Th. 84.) is greater than B, because C+D-D; let it, then, be E+ B. ... E-H B : B :: C+ D : D ; . ... (dividendo) E : B : C : D; and (hyp.) A : B :: C : D; ... (Th;31.) E : B :: A : B; ... (Th. 78.) E=A, and ... E--B-A-FB; ... A+B : B :: C+D : D. a. E. D. DEF. LVII. When it is said of four proportionals, that they are proportionals taken jointly, or by composition, it is meant, that the first together with the second is to the second, as the third together with the fourth is to ID D 210 Elementary Theorems of Plane Geometry, the fourth ; and the making of this inference is de- noted by the word Componendo. THEoREM XC. If a whole magnitude be to a whole, as a magni- tude taken from the first, is to a magnitude taken from the other; the remainder shall be to the re- mainder, as the whole to the whole. Let a magnitude which is made up of the two parts A and B, be to a magnitude, which is made up of the two parts C and D, as the part B is to the part D; then shall the remainder A be to the re- mainder C, as the whole A+ B is to the whole C+ D. For since (hyp.) A+ B : C+D :: B : D, ... (alternando) A+B : B :: C+D : D; . ... (dividendo) A B :: C : D; ... (alternando) A : C : B : D; -". (hyp. and Th. 81.) A C :: A+B : C+ D. If, therefore, &c. a. E. D. - CoR. If the magnitudes A+ B, C+D, B and D be proportionals, it has been shewn, that A : C : B : D; wherefore, if any number of mag- nitudes, of the same kind, be proportionals, the dif- ferences of the homologous terms shall, also, be like proportionals. THEoREM XCI. If four magnitudes be proportionals, the first is to it’s excess above the second, as the third is to it’s excess above the fourth. Comparison of Ratios. -- 211 If A + B : B :: C+ D : D, then A+ B : A :: C + D : C. For (dividendo) A : B : C : D; ... (invertendo) B : A : D : C; -- ... (componendo) B+A : A : D + C : C. Wherefore, if four, &c. G. E. D. THEOREM XCII. If four magnitudes of the same kind are propor- tionals, the greatest and least of them together are greater than the other two together. Let the four magnitudes A, B, C, D, be of the same kind; also, let A : B : C : D ; and let A be the greatest of them, and consequently (Th. 84, 85.) D the least; then shall A+DP B+ C. For let A= E + C, and B = F4 D; then since (hyp.) E4 C : F+ D : C ; D, ... (Th. 90. Cor.) C : D : E : F. But (hyp.) C-D; therefore (Th. 84.) E-F; there- fore if C+D be added to both E4 C+ D-F-H D + C; but E-H C= A; and F-- D = B; therefore A+ D-B + C. If, therefore, &c. a. E. D. CoR. 1. If three magnitudes be continual pro- portionals, the aggregate of the extremes is greater than the double of the mean term. CoR. 2. It is manifest, from the demonstration, that if three magnitudes be continual proportionals, the excess of the greatest of them above the mean, is greater than the excess of the mean above the least. 212 Elementary Theorems of Plane Geometry, DEF. LVIII. When it is said of four proportional magnitudes, the first of which exceeds the second, that they are proportionals by conversion, it is meant, that the first is to it's excess above the second as the third is to it's excess above the fourth; or that the two consequents may be changed each of them into the remainder left, when it has been taken from it's antecedent: and the making of this inference is denoted by the word Convertendo. THEoREM XCIII. lf four magnitudes be proportionals, and if any equi-multiples be taken of the first and third of them and likewise of the second and fourth, the multiple of the first shall be to the multiple of the second, as the multiple of the third is to the multiple of the fourth. Let A : B :: C : D, and let there be taken of A and C any equi-multiples p A, p C, and of B and D any equi-multiples q B, q D; then p A : q B :: p C : q D. For let there be taken, of p A and p C, any equi- multiples E and F, and of q B and q D any equi- multiples G and H ; then (Th. 73.) E and F are also, equi-multiples of A and C, and G and H are equi-multiples of B and D; and since (hyp.) A : B : C : D, therefore (Def. 48.) if E > = < G, F-- < H; therefore (Def. 48.) p A : q B ::, p C : q D. If, therefore, four magnitudes, &c. a. E. D. Comparison of Ratios. 213 CoR. In the same manner it may be shewn that the first of four proportionals is to any multiple of the second, as the third is to the same multiple of the fourth : and, likewise, that any multiple of the first is to the second, as the same multiple of the third is to the fourth. - Theorem xcIv. *- If there be three magnitudes and other three which, taken two and two in their order, have equi- valent ratios, then if the first be greater than the third, the fourth shall be greater than the sixth; if equal, equal; and if less, less. Let there be three magnitudes A, B, C, and other A, B, C, D, E, F. ; : ; ; ; ; three D, E, F, and let B : C :: E : F Then if A > = < C, D >= < F. First, let A-C; then (Th. 76.) there are some equi-multiples, p A and p C, of A and C, and some multiple q B, of B, such that pA-q B, and q B->p C; since then, p.4-q B, and (hyp.) A : B : D : E, therefore (Def. 48.) p D-q E ; likewise, since q B-> p C, and that B : C : E : F, therefore (Def.48.) q E >p F; and it has been shewn that p D-qE; much more, then, is ply-ph'; therefore (Def. 44. Cor. 4.) DS. F. * 214 Elementary Theorems of Plane Geometry, Secondly, if A3 C, it may, in the same manner be shewn that D* F. - Lastly, let A= C; then (hyp.) B : C : E : F ... (Th. 79.) C : B : F : E ; and, since C= A, ... (hyp.) C : B : D : E; ... (Th. 81.) D : E :: F : E; ... (Th. 78.) D=F. Wherefore, &c. a. E. D. DEF. LIX. When there are any number of magnitudes, and as many others, and the first is to the second of the former set, as the last but one to the last of the others, and the second is to the third of the former set, as the last but two to the last but one of the others, and so on, the magnitudes are said to be proportionals when taken in a perturbate order. THEoREM XCV. If there be three magnitudes, and other three, which, taken two and two in a perturbate order, have equivalent ratios, then if the first magnitude be greater than the third, the fourth shall be greater than the sixth ; and if equal, equal, and if less, less. Let there be three magnitudes A, B, C, and other B A, , C, D, F E. 2 ; : ; ; ; ; ; B : C :: D : E three D, E, F, and let } Comparison of Ratios. 315 Then, if AS = < C, D > = < F. First, let A-C ; then (Th. 76.) there may be taken of A and C, certain equi-multiples p A, p C, and of B some multiple q B, such that p A-qB, and q B->p C; since then, p A-q B, and (hyp.) A : B :: E. : F, therefore (Def. 48.) ple>q F; in like manner, since q B-p C, and (hyp.) B : C : D : E, therefore q D->ple; and it has been shewn that ple>q F; much more, then is q D-q F, therefore (Def. 44. Cor. 4.) D.P. F. - - Secondly, if A-3 C, it may, in the same manner be shewn that D& F. * Lastly, let A=C; then since (hyp.) B : C : D : E, - ... (Th. 79.) C : B : E : D; also since C= A, ... (hyp.) C : B : E : F; ... (Th. 81.) E : D : E : F; • . (Th. 78.) D= F. Wherefore, if there be three magnitudes, &c. a. E. D. THEoREM XCVI. If there be any number of magnitudes and as many others, which, taken two and two, in their order, have equivalent ratios, the first shall be to the last of the former magnitudes, as the first is to the last of the others. First, let there be three magnitudes A, B, C, and A, B, C, , E, F as many others D, E, F, and let 216 Elementary Theorems of Plane Geometry, A : B :: D : E B : C :: E : F: then shall A : C : D : F. For, let there be taken of A and D any equi-multi- ples p.A, p D; of B and E any equi-multiples q B, q.E; and of C and Fany equi-multiples r C, r F, then * ſp.A : q B :: p D : q E . (hyp. and Th. 93.) § : 4. :: q E : #} ... (Th. 94.) if p AP = < r C, p1) > = < r F; ... (Def. 48.) A : C : D : F. Next, let there be four magnitudes A, B, C, D, and other four E, F, G, H, which taken two and two in A, B, C, D, E, F, G, H. their order have equivalent ratios; then shall A : D :: E : H. * For, by the first case, A : C : E : G, and (hyp.) C : D :: G : H, ... (by the first case) A : D :: E : H. And, in the same manner, may the proposition be demonstrated, whatever be the number of magnitudes. If, therefore, there be any number, &c. a. E. D. CoR. If p A : q B :: p C : q D, then A : B :: C : D. .* - For since (hyp.) p4 : q B :: p C : q D, and (Th. 80.) B : B :: q D : D; ... (Th. 96.) p A : B :: p C : D; ... (Th. 79.) B : p_A : D : p C. º Comparison of Ratios. 217 Again, (Th. 80.) p A : A :: p C : C; ... (Th. 96.) B : A : D : C; ... (Th. 79.) A : B C : D And, in the same manner it may be shewn, if A : q B : C : q D, that A : B :: and, also, if 3. pA : B : po D, that A : B :: C: D, C ; D. ScholIUM. If there be any number of magnitudes A, B, C, D, and as many others E, F, G, H, which, taken A, B, C, D, E, F, G, H. A, B, C, D'. two and two in their order, have equivalent ratios, it has been proved that A : D : E : H. And, if the magnitudes A, B, C, D, be changed into A', B, C, D', so that A : B : A' : B'; B : C : B' : C'; C : D :: C': D'; still (Th. 81.96.) A': D’: E : H. However, therefore, the magnitudes A, B, C, D may have been changed, if only the ratios of the first to the second, of the second to the third, and so on, continue to be equivalent to the ratios of the first to the second, of the second to the third, and so on, respectively of the original terms, the ratio of the first to the last of the magnitudes will continue to be equivalent to the same ratio. This sort of permaneney, belonging to the ratio of 2 E . 218 Elementary Theorems of Plane Geometry, the first to the last of any number of homogeneous magnitudes, which is dependent upon the ratios of the first to the second, of the second to the third, and so on, continuing to be equivalent, each to each, has, perhaps, given rise to the notion of Compound Ratio, which is defined as follows. ſ * DEF. LX. When there are any number of magnitudes, of the same kind, the ratio of the first to the last of them, is said to be compounded of the ratio which the first has to the second, and of the ratio which the second has to the third, and so on, unto the ratio which the last but one has to the last, inclusively. CoR. It follows, therefore, from Th. 96, that ratios which are compounded of equivalent ratios, the terms being taken in their order, are equivalent to one another. x DEF. LXI. When three magnitudes are proportionals, the ratio which the first has to the third, being com- pounded of two equivalent ratios, namely, of the ratio which the first has to the second and of that which the second has to the third, is said to be the duplicate ratio of either of those two ratios: and, when four magnitudes are continual proportionals, the ratio which the first has to the fourth, being compounded of three equivalent ratios, is said to be the triplicate ratio of that which it has to the second. In like manner, the first of five continual proportionals is said Comparison of Ratios. 219 to have to the last, a ratio quadruplicate of that which it has to the second; and so on. ScholIUM. A ratio is not, strictly speaking, a species of quan- tity; any more than the kind of connexion, indicated by the word fraternity, is a species of quantity. A ratio is, indeed, a relation founded on quantity; but yet it is merely a relation, and not a quantity. One ratio cannot, therefore, with strict propriety be said to be the double, or the triple, or any other multiple, of another ratio. It is only in virtue of the definition previously laid down, that the terms duplicate, tri- plicate, and so on, can be applied to ratios; and all that they really signify is the number of ratios which precede the last ratio in a set of equivalent ratios, the terms of which are continual proportionals. & Theorem XCVII. If there be any number of magnitudes, and as many others, which taken two and two, in a pertur- bate order, have equivalent ratios, the first shall be to the last of the former magnitudes, as the first is to the last of the others. . . . First, let there be three magnitudes A, B, C, and A, B D, E, 2 * C, F as many others, D, E, F, and let, 220 Elementary Theorems of Plane Geometry, {; ; ; ; ; ; ; then, A c : D : F. B : C :: D : E For let there be taken of A, B, and D, any equi- pa. pR, qC, pID, qB, qF. multiples p.4, plb, ply, and of C, E, and F, any equi-multiples q C, q E, q F; then (hyp. and Th. 86. Cor) p4 = p B : q E : q F. and (hyp. Th. 93.) p B : q C :: p1) : q E ; -". (Th. 95.) if p A- = < q C, p DP = < q F; ... (Def. 48.) A : C : D : F. Next, let there be four magnitudes A, B, C, D, and other four E, F, G, H, which taken two and two, E, F, G, H. A, B, C, D, H in a perturbate order, have equivalent ratios; then, also, shall A : D :: E : H. For (1st Case) A : C :: F : H, and (hyp.) C : D :: E : F; ... (1st Case) A : D : E : H. And, in the same manner may the proposition be demonstrated, whatever be the number of magnitudes. If, therefore, there be, &c. a. E. D. CoR. Ratios which are compounded of equivalent ratios, when their terms are taken in a perturbate order, are equivalent to one another. Comparison of Ratios. 221 DEF. LXII. When there are any number of magnitudes and as many others, which, taken two and two, either in the order wherein they stand, or in a perturbate order, have equivalent ratios, the inference, that the first is to the last of the former, as the first is to the last of the others, is expressed by the term et aequali, or ea aequo. - THEOREM XCVIII. If the first has to the second the same ratio which the third has to the fourth ; and the fifth to the second, the same ratio which the sixth has to the fourth; the first and fifth together shall have to the second, the same ratio which the third and sixth tol gether have to the fourth. * Let A : B : C : D, and E : B :: F: D, then, A+E : B :: C+ F : D. For since (hyp.) E : B ... (convertendo) B : E also (hyp.) A : B :: ... (ea aequo) A : E :: C ... (componendo) A+E : E also (hyp.) E : B ... (ew aequo) A+E : B If, therefore, the first, &c. a. E. CoR. Hence, if A : B : C : D, and A : E :: C: F, then, A : B+E :: C : D + F. 222 Elementary Theorems of Plane Geometry, For (hyp. and invertendo) B : A : D : C, and E. : A :: F : C, ... (Th. 98.) B+E : A :: D+ F : C; ... (invertendo) A : B+E :: C : D+F. SCHOLIUM. Whatever has been proved, in the preceding section, of proportional magnitudes, might have been in like manner demonstrated of proportional numbers, or of any other proportional quantities, of which it is possible for multiples to be taken. Two of the four terms, therefore, or all the four terms of a proportion, may be numbers. In computing the numerical value of a proposed geometrical magnitude, the magnitude is, in reality, compared with some standard magnitude of the same kind, and two numbers are found, the ratio of the former of which to the latter is equivalent to the ratio which the proposed magnitude has to the standard. Thus if the value of B is to be computed by referring it to the standard A, and if it is found that A : B :: 1 : 3, then B-3A ; for (hyp. and Th. 93. Cor.) 3A : B :: 3 : 3; * * w ... (Th. 84.) 3 A=B. In the same manner, if A : B :: 3 : 1, it may be shewn that 3B = A, or B = #4. Also, if A : B :: 2 : 3, then (Th.93.) 3A : 2 B :: 6 : 6 ... 2 B-3A; or B-4A: and if A : B :: 3 : 2, then B =#4. What has thus been proved, in some particular cases, Comparison of Ratios. * 223 may, it is manifest, be demonstrated when any other numbers whatever constitute the two terms of the second ratio; and there results this short rule, which is applicable to all such cases, and which is known by the name of the Single Rule of Three, viz. that the product of the extremes is equal to the product of the II16an.S. * . . . . . . . * ... Again, when a ratio is compounded of other ratios to each of which an equivalent numerical ratio can be found, then is that compound ratio, equivalent to the ratio, which the product of all the numerical ante- cedents has to the product of all the numerical con- sequents. . . . . . . -- - * } For, first, let there be three magnitudes A, B, C, and four numbers m, n, p, q, such that A : B :: m : m B : C :: p : q. Then (Th. 36. Cor) { . # ºf ; ... (ew aequo) A : C :: mp : nq. Secondly, let there be four magnitudes A, B, C, D, and let A : B :: m : n, B : C ::p : q, C : D :: r: s, then shall A : D :: mp r : nqs. A : B :: mpr: npr For (hyp. and Th. 86. Cor.) } : C :: pnr : q n r 47 C : D :: rq n : sqm ... (ea aequo) A : D :: mpr: nqs. And the same proof is applicable, whatever be the number of magnitudes. - Hence, if two proposed ratios be compounded of the same number of ratios which are equivalent, each 224 Elementary Theorems of Plane Geometry, of the one set, to each of the other, taken in any order whatever, and if numerical ratios can be found equi- valent to the compounding ratios, each to each, then shall the two proposed ratios be equivalent to one another. For, it has been shewn, that each of the proposed ratios is equivalent to the ratio which the product of the numerical antecedents has to the pro- duct of the numerical consequents, and therefore (Th. 81.) they are equivalent to one another. Besides the methods, delivered in the preceding section, of investigating whether four quantities be proportional, the following may also, sometimes be used with great advantage. It is founded on an as- sumption which has been already mentioned (Schol. Th. 89.) and which seems to be very admissible; viz. that if there be three quantities of which the two last are of the same kind, then, there is some quantity of the same kind, as the first, which is to the first as the second is to the third. The additional criterion of proportionality is this. • If there be four quantities A, B, C, D, and tw others E, F, such that E : B :: F: D, and if E may be greater or less than A, by a difference less than any assignable, and if also F is greater or less than C, accordingly as E is greater or less than A; then, A : B :: C : D. For if not, let (assump.) G : B : C ; D, and first let G be less than A: then (hyp.) E may be greater than G and less than A, and therefore, also, F less than C; and (hyp.) E : B : F : D. ... (Th. 79.) B : E : D : F: Comparison of Ratios. 225 also, G : B :: C : D; ... (Th. 97.) G : E :: C : F. But (hyp.) G is less than E.; therefore (Th. 85.) C is less than F; but it is also greater than F: which is impossible. There cannot, therefore, be a quantity less than A, which is to B as C is to D. In the same manner it may be shewn that there cannot be a quantity greater than A, which is to B as C is to D. Wherefore (assump.) A : B : C ; D. F F THE 3Elements of plant Grometry, Book I. —sº- CHAPTER v. ON THE PRopoRTIon ALITY OF LINEs, of PLANE FIGURES A N D OF PI, AN E RECTILINEAL A N G. L.E.S. —O— SECTION I. On the proportionality of the sides, and the segments of the sides, and of the surfaces, of triangles. *&^^^^^^42 wººdºº-wº THEoReM XCIX. Ir a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those produced, proportionally ; and if the sides, or Proportionality of Sides, &c. of Triangles. 227 the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle. Let GH be drawn parallel to BC, one of the sides of the As ABC; BG: GA : CH : HA, and AG : GB :: AH : HC. For, let there be taken of AB and AC any equi- multiples AK and AL; and of AG and AH any equi-multiples AM and AN; and suppose KL, and MN, to be drawn; then (Th. 25. Cor. 5.) KL and MN are parallel to BC and GH, and therefore (Th. 9.) they are parallel to one another; therefore if AKP = < AM, ALP = < AN; otherwise KL would cut MN to which it is parallel; therefore (Def. 48.) AB : AG : Ab : AH; therefore (dividendo, or convertendo, if GH and BC 228 Elementary Theorems of Plane Geometry, are on the same side of A, or (componendo) if GH and BC be on contrary sides of the point A) BG : GA : CH : HA; ... (invertendo) AG : GB : AH : HC. Next, let the sides AB, AC, or these produced, of the As ABC, be cut proportionally in the points G, H, and let G, H be joined; GH is parallel to BC. For if not, suppose GB to be parallel to BC; then as hath been shewn, BG : GA :: CP : PA; and (hyp.) BG : GA : CH : PA; ... (Th. 81.) CP : PA :: CH : HA; ... (comp. conv. or divid.) CA : PA :: CA : HA; ... (Th. 78.) AP = AH; which is absurd. Wherefore GP is not parallel to BC; and in the same manner it may be shewn that no other line, drawn through G, but GH, can be parallel to BC. Wherefore, if a straight line, &c. a. E. D. CoR. 1. From the demonstration it is manifest, that the sides of a triangle, and the segments, which a straight line drawn parallel to the base cuts off from those sides, are proportionals. For, let GH be drawn parallel to the base BC, of the As ABC; it has been shewn that AB : AG :: AC : AH; ... (alternando) AB : AC :: AG : AH. CoR. 2. Two given straight lines are cut pro- portionally by any number of parallel straight lines. If the given straight lines be parallel to one another, Proportionality of Sides, &c. of Triangles. 229 the opposite segments, will (Th. 25. Cor. 1.) be equal to one another. But let the straight lines AB, AC, which meet in A, be cut, first, by the three parallels DE, FG, BC; then AD : DF: AE : EG, and DF : FB :: EG : G.C. For, suppose EB to be drawn, and let it cut FG in H. Then (hyp, and Th. 99.) § : FB :: EH : HB (EH : HB :: EG : GC ... (Th. 81.) DF: FB :: EG : G.C. And, in like manner, by the help of this first case, may it be shewn that the straight lines AB, AC are cut proportionally, whatever be the number of paral- lels. AD : D F :: AE : #, ScholIUM. It is manifest from the foregoing proposition, that, if three finite straight lines be given, there is a fourth straight line, which is a fourth proportional to them. 230 Elementary Theorems of Plane Geometry, For if AG, GB' placed in the same straight line, be the two first, and if the third AH be placed so as to make any angle with AB, then, G, H being sup- posed to be joined, and BC to be drawn parallel to GH, meeting AH produced in C, HC is (Th. 99) a fourth proportional to AG, GB and AH; or if AH = GB, then HC is a third proportional to AG and G.B. Further, if a straight line AB+ be divided into any number of parts, another finite straight line AC, placed so as to make an angle with AB, may be similarly divided (Th. 99. Cor. 2.) if B, C be joined, and through the points of division, straight lines be drawn parallel to BC: and, if AB be taken any mul- tiple of AD, so that the parts into which it is divided shall be equal to one another, then if B, C be joined, and DE drawn parallel to BC, as before, a part AE will be cut off from AC, which is the same part of it, that AD is of AB. Thus, any part required may be supposed to be cut off from a given straight line. THEOREM. C. If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base ; the segments of the base shall have the same ratio which the other sides of the triangle have to one another. And if the segments of the base have the same ratio which the other sides of the triangle have *— * See the figure in p. 227. + See the figure in p. 229. Proportionality of Sides, &c. of Triangles. 231 to one another, the straight line drawn from the vertea to the point of section, divides the vertical angle into two equal angles. Let the angle BAC of any triangle ABC be divided into two equal angles by the straight line AD: then B D C BD : DC :: BA : AC. Through the point C suppose CE to be drawn. parallel to DA, and BA produced to meet CE in E. Because AC meets the parallels AD, EC, the LACE is (Th. 10.) equal to the alternate angle CAD: but (hyp.) the / CAD, = L BAD; wherefore the LBAD = / ACE. Again, because the straight line BAE meets the parallels AD, EC, the outward angle BAD is (Th. 10.) equal to the inward and opposite angle AEC: but the angle ACE has been proved equal to the angle BAD; therefore also the / ACE= / AEC, and con- sequently (Th. 15.) the side AE is equal to the side AC: and because AD is parallel to one of the sides of the triangle BCE, viz. to EC, (Th. 99.) BD : DC : BA : AE ; 232 Elementary Theorems of Plane Geometry, but AE = AC; ... (Th. 77.) BD : DC : BA : AC. Let now BD be to DC, as BA to AC, and suppose AD to be drawn; the LBAC is divided into two equal angles by the straight line AD. The same construction being supposed to be made, since, BD : DC :: BA : AC; and (hyp. and Th. 99.) BD : DC :: BA : AE, ... (Th. 81.) BA : AC :: BA : AE; ... (Th. 78.) AC=AE, and (Th. 13.) / the AEC = LACE. But (Th. 10.) the / AEC is equal to the outward and opposite / BAD; and the LACE is equal to the alternate / CAD: wherefore also the L BAD = Z. CAD: therefore the angle BAC is cut into two equal angles by the straight line AD. Therefore, if the angle, &c. a. E. D. TheoreM CI., The sides about the equal angles of equiangular. triangles are proportionals ; and those which are opposite to the equal angles are homologous sides. Let ABC, DEF, be equiangular triangles, having the L ABC = / DEF, and the / ACB = A DFE, and consequently (Th. 12.) the / BAC = / EDF Then shall the sides about the equal angles of the As ABC, DEF, be proportionals, and those shall be the homologous sides, which are opposite to the equal angles. Proportionality of Lines, Plane Surfaces, &c. 233 For the / EDF may be applied to the equal L BAC, A. D G TH / \ E E B C so that DE shall coincide with AB, and DF with AC; let DE and DF when so applied be terminated by the points G and H, in AB, and AC, produced if it be necessary, and let EF coincide with GH; there- fore (hyp. and Th. 6.) GH is parallel to BC; ... (Th.99. Cor. 1.) AB : AC :: AG : AH; i. e. AB : AC :: DE : DF And, in the same manner it may be shewn, by applying first the LDEF to the equal / ABC, and then the / DFE to the equal / ACB that AB : BC :: DE : EF; and that BC : CA :: EF : FD. Wherefore, the sides, &c. a. E. D. CoR. 1. It is manifest, from the demonstration, that a straight line drawn parallel to the base of a given triangle, cuts off from it a triangle, the sides of which and the sides of the given triangle, about equal angles, are proportionals. CoR. 2. The sides of two equiangular /> ABC, DEF, are also proportionals each to each, to which the equal angles are opposite. G G 234 Elementary Theorems of Plane Geometry, For it has been shewn that AB : AC :: DE : DF; ... (alternando) AB : DE :: AC : DF; and so of the remaining sides. - CoR. 3. Two triangles which have the three sides of the one parallel to the three sides of the other, each to each, being (Th. 10. Cor. 6.) equiangular, have the sides about the equal angles, proportionals. CoR. 4. Two parallel straight lines AB, CD, which are cut by any number of straight lines PAC, P yº–ee C F H D PEF, PGH, PBD, that meet in the same point P, are cut by them proportionally. For (hyp. and Cor. 1. and 2.) {# . ; #. : # ... (Th. 81.) AE: CF: EG : FH; ... (alternando) AE: EG :: CF: FH. And in the same manner it may be shewn that EG : GB :: FH : HD. ScholIUM. The last Corollary, of Th. 101, indicates another method, by which a given straight line may be divided similarly to a given divided straight line; viz. by Proportionality of Sides, &c. of Triangles. 235 placing the two given lines parallel to one another, and by drawing, from a point in which the straight lines joining their extremities meet, straight lines through the points of division of the given divided line: or else, if the straight lines joining the ex- tremities of the given lines be parallel, by drawing, through the given points of division, straight lines parallel to them. THEoREM CII. If the sides of two triangles about each of their angles be proportionals, the triangles shall be equi- angular. Let the Ds ABC, DEF have their sides pro- portionals; so that AB : BC :: DE : EF; BC : CA A. D Gr TH / ſ \ F. E. B C :: EF: FD; and therefore (ea aequo) BA : AC :: ED : DF; then the LABC– L DEF; the L BCA = / EFD; and the / BAC= L EDF, For, if AB= DE, then, (hyp. and Th. 85.) BC= EF, and AC= DF, and therefore (Th, 24. Cor. 1.) the DS, ABC, DEF are equiangular. 236 Elementary Theorems of Plane Geometry, But, if AB be not equal to DE, one of them is the greater; let AB > DE, and from AB let there be cut off AG= DE ; also, from G let there be drawn GH parallel to BC: then (Th. 10.) the two /> AGH, ABC are equiangular; ... (Th. 101.) AB : AC :: AG : AH, and (hyp.) AB : AC :: DE or AG : DF; ... (Th. 81.78.) AH = DF: Again (Th. 101.) BC : CA :: GH : HA, and (hyp.) BC : CA :: EF : FD; and it has been shewn that FD = HA; therefore (Th. 81.78.) GH = EF. Wherefore the three sides of the AS DEF are equal to the three sides of the /S AGH, each to each; therefore (Th. 24. Cor. 1.) the angles of the AS DEF are equal to the angles of the AS AGH, each to each, to which the equal sides are opposite; and the angles of the AES AGH have been shewn to be respectively equal to the angles, of the As ABC; therefore the /> ABC, DEF are equi. angular, and have their equal angles opposite to the homologous sides. If, therefore, the sides, &c. a. E. D. THEOREM CIII. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular and shall have those angles equal that are opposite to homologous sides. Let the Ds ABC, DEF’s have the / BAC in the -mºsasº. * See the figure in p. 235. Proportionality of Sides, &c. of Triangles. 237 one equal to the / EDF in the other, and the sides about those angles proportionals; that is, BA : AC :: ED : DF; the Os ABC, DEF, are equiangular and have those angles equal that are opposite to ho- mologous sides. For the / EDF may be applied to the equal A BAC, so as to coincide with it; let then the As DEF, so applied, coincide with the AS AGH ; And since (hyp.) BA : AC :: GA : AH; ... (alternando) BA : GA :: AC : AH; ... (dividendo) BG : GA : CH : HA; therefore (Th. 99.) GH is parallel to BC; therefore (Th. 10.) the AS AGH is equiangular to the AS, ABC; i. e. the AS DEF is equiangular to the AS, ABC, and the / DEF= / ABC, and the / EFD= / BCA. If, therefore, two triangles, &c. a. E. D. SCHOLIUM. Equiangular triangles have been shewn to have their sides about the equal angles proportionals. There may, also, be quadrilateral figures, and polygons, which are both equiangular to one another, and have the sides about their equal angles proportionals. For let ABCDE be any given rectilineal figure having more than three sides, and let FG be any given finite straight line. Let AC and AD be drawn from any one of the angular points of ABCDE to the opposite angular points C, D : at the point F let the / GFH be supposed to be made equal to the / BAC, the / HFK = / CAD, the / KFL = / DAE; and 238 Elementary Theorems of Plane Geometry, so on, if there be more angles; so that the whole C I) TH K / GFL is equal to the whole / BAE : also let FH be a fourth proportional to BA, AC, GF; let FK be a fourth proportional to CA, CD, HF; and FL a fourth proportional to DA, AE, KF; and let GH, HK, and KL be drawn; then (hyp. and Th. 103.) the AS, ABC is equiangular to the AS FGH, the As ACD to the /> FHK, the AS, ADE to the /S FKL ; therefore the figure ABCDE is equi- angular to the figure FGHKL. - - Again, since the triangles into which the tw figures are divided, have been shewn to be equi- angular, each to each, ... (Th. 101.) CB : BA :: HG : GF also AE : ED :: FL : LK. Again, BA : AC :: GF : FH, AC : AD :: FH : FK, AD : AE :: FK : FL ; ... (ew aequo) BA : AE :: GF : FL; and in the same manner it may be shewn, that ED : DC :: LK : KH, and DC : CB :: KH : HG. Proportionality of Sides, &c. of Triangles. 239 Wherefore, the figure FGHKL, which was made equiangular to the given figure ABCDE, has its sides proportional to the sides about the equal angles of the figure ABCDE. It is allowable, therefore, to suppose the existence of rectilineal figures of any the same number of sides which have these properties, and to make the following definition. DEF. LXIII. Similar Rectilineal Figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals. , *. CoR. Rectilineal figures which are similar to the same rectilineal figure are similar to one another. For (Def. 12. Cor. 1.) they are equiangular; and (Th. 81.) their sides about the equal angles are pro- portionals. THEOREM CIV. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals, then, if each of the re- maining angles be either less, or not less, than a right angle; or if one of them be a right angle : the tri- angles shall be equiangular, and have those angles equal about which the sides are proportionals. Let the two /> ABC, DEF’s have the LACB = / DFE, the sides about the two / A, and D, pro- * See the figure in p. 235. 240 Elementary Theorems of Plane Geometry, portionals, viz. BA : AC :: ED : DF, and let the two remaining / ABC, DEF be of the same species: the AS DEF is similar to the AS, ABC. For let the AS EDF be applied to the AS BAC, so that DF may coincide with AH, and let FE coincide, in direction, with HG, which meets AB in G; and since (hyp.) the / DFE, or AHG, - / ACB, therefore (Th. 8.) GH is parallel to BC; therefore (Th. 101. Cor. 1.) BA : AC :: GA : AH or DF; but (hyp.) BA : AC :: ED : DF; ... (Th. 81.) GA : DF: ED : DF; therefore (Th. 78.) AG = DE : and it has been shewn that GH is parallel to BC; therefore (Th. 10.) the / AGH = / ABC; and therefore the Z AGH is of the same species with the / DEF; therefore (Th. 23. Cor.) HG = FE; therefore (Th. 20. Cor.) the angles of the AS AGH are equal to the angles of the AS DEF, each to each; therefore the angles of the AS ABC are equal to the angles of DEF, each to each ; therefore (Th. 101. and Def. 63.) the /> ABC, DEF, are similar to one another. If, there- fore, two triangles, &c. a. E. D. THEOREM CV. Two triangles are similar, which have two sides of the one proportional to two sides of the other, and have the homologous sides parallel, and the two sides of the one either tending toward the same parts Proportionality of Sides, &c. of Triangles. 241 * as the two proportional sides of the other, or towards opposite parts”. º In the Ds ABC, DEF, let BA : AC :: ED: DF; let AB be parallel to DE, and AC parallel to DF, º * **. tº e. C and let AB and AC, DE and DF all tend to the same parts, or else, let AB and AC tend to parts opposite to those, to which DE and DF tend : the AS, ABC is similar to the /> DEF. Since B4 and BC cut one of the parallels AC, they shall (Th. 8. Cor. 1.) also cut the other DF; let them, all the lines having been produced if it be necessary, cut DF in G and H ; in like manner, let CB cut DE in K5 and since GH is parallel to the * This restriction is not made in E. xxxii. 6; without it how- ever the two triangles, which it is the object of the proposition to compare with one another, might not be equiangular ; the two angles, contained by the sides that are proportionals, instead of being equal, might be so related as, together, to make up two right angles. - H H 242 Elementary Theorems of Plane Geometry, base AC of the As ABC, therefore (Th. 101. Cor. 1.) B4 : AC :: BG : GH; likewise, because DK is parallel to GB, the base of the AS BHG, - KD : DH :: BG : GH; ... (Th. 81.) KD : DH :: BA : AC; but (hyp.) ED : D F :: BA : AC; ... (Th. 81.) KD : DH :: ED : DF; and the / KDH, EDF, are either the same angle, or else vertical, and therefore (Th. 1. Cor.) equal angles; therefore (Th. 103.) the DS KDH, EDF are similar; and (hyp. and Th. 10.) the AS KDH is equi- angular with the AS BHG, and the As BHG is equiangular with the AS, ABC; therefore (Th. 101.) the /S, ABC is similar to the ZS DEF. Therefore two triangles, &c. a. E. D. CoR. If two triangles have two sides of the one proportional to two sides of the other, and have the homologous sides parallel, and the two sides of the one either tending toward the same parts as the two sides of the other, or towards opposite parts, the bases, or third sides, of the two triangles, shall either be parallel, or they shall be in the same straight line. For it has been shewn that the sides of the As KDH, EDF, are proportional; therefore (Th. 99.) EF is parallel to HK or BC; or else if EF meets BC, it shall (Th. 8. Cor. 2.) be in one and the same straight line with BC. THEOREM CVI. º * tº º º A right-angled triangle is similar to each of the Proportionality of Sides, &c. of Triangles. 243 triangles into which it is divided by a perpendicular drawn from the right angle to the base. In the As ABC let the 1. BAC be a right angle, and from A let AD be drawn perpendicular to BC: A. * B B- C the Dºs ABC, ABD, ADC, are similar to one another. Because (hyp. and Th. 2.) the / BAC= LADB, and the angle at B is common to the two /> ABC, ABD, therefore (Th. 12.) the remaining / ACB is equal to the remaining / BAD, and the AS, ABC is equiangular to the AS, ABD: likewise, because the angle at C is common to the two right-angled as ABC, ADC, these two triangles are equiangular, and the / CAD equal to the angle at B; therefore, also, since the angles at D are right angles, the AS, ABD is equi- angular to the As ADC; therefore (Th. 101. and Def. 63.) the DS, ABC, ABD, ADC, being equi- angular, are similar to one another. Wherefore, in a right-angled triangle, &c. a. E. D. + CoR. I. The perpendicular drawn from the right angle of a right-angled triangle to the base is a mean proportional between the segments of the base: and each of the sides is a mean proportional between the 244 Elementary Theorems of Plane Geometry, base, and the segment of the base adjacent to that side. For since the Os ABD, ADC are similar, there- fore (Def. 63.) BD : DA :: DA : DC. Also, because the Os ABC, ABD, ADC, are similar, ... BC : BA :: BA : BD; and BC : CA :: CA : CD. CoR. 2. If from any point (D) in a given straight line (BC) a straight line (DA) be drawn at right angles to it, which is a mean proportional between the segments (BD, DC) of the given line, the given line shall subtend a right angle at the extremity A of the mean proportional. For the circumference of a circle described upon BC as a diameter shall cut DA in A; if not let it cut DA in any other point than A; then since (Th. 57.) BC will subtend a right angle at that point; the perpendicular drawn from that point, which is either greater or less than AD, will (Th. 106. Cor. 1.) be a mean proportional between BD and DC; which is contrary to the hypothesis. Where- fore the circumference of a circle described upon BC as a diameter cannot cut the perpendicular AD in any other point than 4; therefore (Th. 57.) BC subtends a right angle at A. CoR. 3. If from any point (A) in the circum- ference of a semicircle (BAC) a perpendicular (AD) be drawn to the diameter (BC) it shall be a mean Proportionality of Sides, &c. of Triangles. 245 º proportional between the segments (BD, DC) into which it divides the diameter. For if A, B and A, C be joined, the / BAC is (Th. 57.) a right angle, and therefore AD is a mean proportional between BD and DC. SCHOLIUM. It is manifest from Th. 106, that if any two finite straight lines, BD, DC, be a given, there is a third straight line which is a mean proportional between them. For, if BD, DC, be supposed to be placed in the same straight line, and if A be the point in which DA, drawn from D perpendicular to BC, cuts the circumference of a circle described upon RC as a diameter, and if AB and AC be drawn, then (Th. 57.) the / BAC is a right angle; and therefore (Th. 106. Cor.) BD : DA :: DA : DC; i. e. DA is a mean proportional between BD and DC. THEOREM CVII. Triangles of the same altitude are to one another as their bases. Let the /> ABC, ACD, have the same altitude; then BC : CD :: /> ABC : As ACD. For if there be taken CH any multiple of CB, and CG any multiple of CD, and if AH and AG be drawn, the As ACH is (Th. 30.) the same mul- tiple of the As ACB, that CH is of CB; and the /S ACG is, likewise, the same multiple of the As ACD, that CG is of CD; and (Th. 30. and 246 Elementary Theorems of Plane Geometry, Th. 30. Cor. 2.) if CH >= < CG, the As ACH > A. H B C D F G = < ACG ; therefore (Def. 48.) BC : CD :: /> ABC : /> ACD. CoR. 1. In the same manner it may be shewn that triangles, which have equal altitudes, are to one another as their bases. CoR. 2. Triangles upon equal bases are to one another as their altitudes. For if, on the equal bases, right-angled triangles be supposed to be constructed equal to the two given tri- angles, each to each, the equal bases may be taken for the altitudes of the two triangles so constructed, which (Cor. 2.) will therefore be to one another as their remaining sides about the right angles, that is, (Th. 31. and Th. 25. Cor. 3.) as the altitudes of the two given triangles; wherefore the given triangles are to one another as their altitudes. ScholIUM. It is manifest, from Th. 107, that, inasmuch as any required part can be cut off from the base of a triangle, so may any required part be cut off from the Proportionality of Sides, &c. of Triangles. 247 triangle; as the base may be divided similarly to a given divided straight line, so may the triangle be divided into triangles, which shall have to one another ratios equivalent to any given ratios; also, two tri- angles of the same or equal altitudes being given, a third proportional to them, or a mean proportional between them, may be found, inasmuch as a third proportional to their bases, or a mean proportional between them, may be found: and, likewise, if three triangles, of the same, or of equal altitudes be given, a fourth proportional to them may be found. Or, if two triangles of the same altitude, and any straight line, be given, another straight line may be found which shall be a fourth proportional to them. And what is here said of triangles, it will afterwards appear, is applicable also to parallelograms. DEF. LXIV. Two magnitudes are said to be reciprocally pro- portional to two others, of the same kind, when one of the former two is to one of the latter two as the remaining one of the latter two is to the remaining one of the former. THEOREM CVIII. Triangles, having one angle of the one equal to one angle of the other, if they have their sides about the equal angles reciprocally proportional, are equal to one another; and if they be equal to one another, 248 Elementary Theorems of Plane Geometry, they have their sides about the equal angles recipro- cally proportional. Let ABC, ADE, be equal triangles, which have the L BAC equal to the / DAE; the sides about the T3 D A. C | º, E F GrºL equal angles of the triangles are reciprocally pro- portional; that is, CA : AD, ... E4 : AB. Let the triangles be placed, so that their sides CA, AD be in one straight line; wherefore also EA and AB are in one straight line; and suppose BD to be drawn. Because the AS, ABC = As ADE, and that ABD is another triangle ... (Th. 77.) - AS CAB : /> BAD :: /> EAD : As DAB; but (Th. 107.) /> CAB : /S, BAD :: CA : AD, and /> EAD : As DAB :: EA : AB; ... (Th. 81.) CA : AD :: EA : AB : wherefore the sides of the ſº ABC, ADE, about the equal angles are reciprocally proportional. 3. But let the sides of the 2s ABC, ADE, about the Proportionality of Sides, &c. of Triangles. 249 equal angles be reciprocally proportional, viz. CA to AD, as EA to AB; the AS, ABC is equal to the AS AD.E. f Having joined BD as before; because, (hyp.) CA : AD : EA : AB, and (Th. 107.) CA : AD :: /> ABC : Zºe BAD, and EA : AB :: As EAD : /S BAD, ... (Th. 81.) ... /> BAC : As BAD :: /> EAD : As BAD; n ... (Th. 78.) the AS, ABC= As ADE. Therefore equal triangles, &c. a. E. D. THEoREM CIX. Triangles which have an angle of the one equal to an angle of the other, have to one another a ratio that is compounded of ratios equivalent to the ratios of the sides about the equal angles. . Let ABC, ADE, be triangles which have the angles at A equal: the /S ABC has to the CŞ ADE T3 . D é || E F Gr}I the ratio that is compounded of the ratios, equivalent I I 250 Elementary Theorems of Plane Geometry, to the ratios which AC has to AD, and which AB has to AE. - Let the triangles be placed, so that their sides CA, AD, may be in one straight line; therefore (Th. 3.) EA, AB, are, also, in one straight line: let BD be drawn, and, any straight line F having been taken, let F: G :: CA : AD; and let G : H :: BA : AE; therefore the ratio of F to H is compounded of ratios which are equivalent to the ratios of the sides; but (Th. 107.) As ABC : As ABD :: CA : AD; and (hyp.) CA : AD :: F : G : ... (Th. 81.) As ABC : As ABD :: F : G. In like manner, /> ABD : As ADE :: G : H ; ... (ew aequo) & ABC : As ADE :: F : H. That is, the As ABC has to the As ADE a ratio which is compounded of ratios equivalent to the ratios of the sides: Wherefore triangles, &c. a. E. D. THEOREM CX. Similar triangles are to one another in the du- plicate ratio of their homologous sides. Let ABC, DEF, be similar triangles, having the / BAC = / EDF, and let BA : AC :: ED : DF, so that the side AB is homologous to the side DE: the As ABC has to the AS DEF the duplicate ratio of that which AB has to D.E. g For, any straight line G having been taken, let AB : AC :: G : H; and DE : DF :: H : K ; Proportionality of Sides, &c. of Triangles. 251 then (Th. 109) as ABC : & DEF: G : K: but (hyp.) AB : Ac :: DE: DF; ... (Th. 81.) G : , H : H : K: therefore G has to K the duplicate ratio of that which ,- * A B - | G has to H, or of that which AB has to DE; and it has been shewn that & ABC : As DEF: G : K; therefore the As ABC has to the AS DEF the duplicate ratio of that which AB has to D.E. Where- fore similar triangles, &c. a. E. D. CoR. I. If a third proportional be found to two homologous sides of two similar triangles, as the former side is to the third proportional, so is the former triangle to the latter. ſ For, the former side has to the third proportional, the duplicate ratio of that which it has to the latter side, which is equivalent to the ratio of the former triangle, to the latter. CoR. 2. If two triangles, which have an angle of the one equal to an angle of the other, be to one 252 Elementary Theorems of Plane Geometry. another in the duplicate ratio of a side of the one, adjacent to the one of the equal angles, to a side of the other, adjacent to the other of the equal angles, the two triangles shall be similar. Let the L BAC of the As ABC, be equal to the a. EDF, of the AS DEF, and let the S. ABC be to A. D T B C T. the AS DEF in a ratio which is the duplicate of that of AB to DE : the AS, ABC shall be similar to the & DEF: - . For if the Z. ABC= / DEF, the two triangles will (Th. 12.) be equiangular and therefore (Th. 101.) they will, also, be similar: but if the / ABC be not equal to the Z DEF, let the / ABH = / DEF; therefore the Dºs ABH, DEF are similar; therefore (Cor. 1.) if G be a third proportional to AB and DE, AS, ABH : /S DEF :: AB : G : and (hyp.) - As ABC : As DEF:: AB : G; ... (Th. 81.) r • As ABH : /> DEF :: AS, ABC : AS DEF; therefore (Th. 78.) As ABH = & ABC; which is impossible; therefore the L ABC is not unequal to the L DEF, and the two /> ABC, DEF are similar. THE 3Bltments of latte Geometry). Book I. CHAPTER V. —e— SECTION II. On the proportionality of the sides, and of the sur- faces, of parallelograms, and of polygons. —O— THEoREM CXI. The parallelograms about the diameter of any parallelogram, are similar to the whole, and to one another ; and similar parallelograms, which have a common angle, and are similarly situated, are about the same diameter. LET ACDB be a parallelogram, of which the dia- C T meter is BC, and EK, HF, the parallelograms about 254 Elementary Theorems of Plane Geometry, the diameter: the IL EK, HF, are similar both to the whole L, ABCD and to one another. Because HG and AB are both parallel to CD, they are (Th. 9.) parallel to one another; therefore (Th. 10.) the LCHG = / CAB ; therefore (Th. 26. Cor. 1.) the El HF is equiangular to the C1 AD. . Again, because HG is parallel to AB, the AS CHG is (Th. 101. Cor. 1.) similar to the As CAB; for the same reason, because FG is parallel to DB, the As CFG is similar to the /> CDB ; ... CH : HG :: CA : AB, HG : GC :: A B : BC, GC : CF :: BC : CD; ... (ew aequo) CH : CF :: CA : CD. Wherefore (Th. 25. Cor. 1.) the sides about the equal angles of the [I] HF AD, are proportionals; and therefore (Def. 63.) the El HF is similar to the D AD. In the same manner it may be shewn that the D-I EK is similar to the C1 AD; therefore (Def. 63. Cor.) the EJ HF is, also, similar to the [] EK. Next, let the [I] ACDB, HCFG, be similar and similarly situated, and have the LACD Common: ACDB and HCFG are about the same diameter. For, let HG and FG, produced, meet BD and AB, in K and E, respectively, and let CG and BC be drawn. Then (hyp. and Th. 25. Cor. 1.) AB = HK, and (hyp.) * CA : AB or HK :: CH : HG, ... (Th. 90. Cor.) AH : KG :: CH ; HG, Proportionality of Sides, &c. of Polygons. 255 Asºº i. e. (Th. 25. Cor. 1.) BK : GK : CH HG; and (Th. 10.) the / BKG = / CHG; therefore (Th. 103.) the / BGK = / H.GC; therefore (Th. 3.) BG and GC are in one and the same straight line; that is, the TJ HF and AD are about the same dia- meter CGB: Wherefore, the parallelograms, &c. Q. E. D. THEOREM CXII. If an equilateral and equiangular polygon have the same number of sides as another equilateral and equiangular polygon, the two figures shall be similar to onother. -- For (Th. 35. Cor 4.) any angle of the one, is equal to any angle of the other figure: and (hyp.) the sides about equal angles, in each, are to one another in the same ratio of equality; therefore (Def.63.) the figures are similar. CoR. If through the angular points of an equi- lateral and equiangular rectilineal figure, inscribed in a circle, straight lines be drawn touching the circle, the figure, thus described about the circle, shall be similar to the inscribed figure. For (Th. 66.) it will be equilateral and equi- angular; and (hyp.) it has the same number of sides as the inscribed figure; therefore (Th. 112.) the two figures are similar. I 256 Elementary Theorems of Plane Geometry, --- Theorem CXIII. The perimeters of similar polygons, inscribed in circles, are to one another as their diameters. Let ABCDE, FGHKL, be two circles, and in these the similar polygons ABCDE, FGHKL; and A. cs—zo HS Tzk let BM and GN be diameters of the circles: as BM is to GN so is the perimeter of the polygon ABCDE to the perimeter of the polygon FGHKL. For, let BE, AM, GL, FN, be drawn; and since (hyp.) BA : AE :: GF: FL and that the LBAE= L GFL, therefore (Th: 103.) the As ABE is similar to the /S FEL, so that the / AEB = L FLG ; but (Th. 55.) the LAMB = LAEB, and the LFNG= A FLG; therefore the / AMB = L FNG; and (Th. 2.) the L BAM = L GFN, each of them (Th. 57.) being a right angle; therefore (Th. 12.) the AS, ABM is equiangular to the AS FGN; therefore (Th. 101. Cor. 2.) AB : FG :: BM : GN. Again (hyp.) AB : BC :: FG : GH; ... (alternando) AB : FG :: BC : GH; Proportionality of Sides, &c. of Polygons. 257 in the same manner it may be shewn that BC: GH :: CD : HK; and so on; therefore (Th. 83.) the aggregate of the antecedents is to the aggregate of the consequents, as AB is to FG, or (Th. 81.) as BM to GN; that is, the perimeter of the polygon ABCDE is to the perimeter of the polygon FGHKL as the dia- meter BM is to the diameter G.N. Wherefore, the perimeters, &c. a. E. D. CoR. 1. If in each of two circles an equilateral and equiangular polygon, be described, of the same number of sides, and if through the angular points of these polygons straight lines be drawn touching the circles, the polygons so described about the circles will. (Th. 112. Cor. and Def. 63. Cor.) be similar to the inscribed polygons, and to one another: and it may be shewn, as in the proposition, that their peri- meters are to one another as their homologous sides, and therefore, also, as the diameters of the circlés. - CoR. 2. If a straight line be divided into any two parts, and if upon the whole line and each of the parts there be described similar polygons, the perimeter of the polygon which is described upon the whole line, shall be equal to the perimeters taken together of the polygons which are described upon the two parts. For it was shewn, in the demonstration, that the perimeters of similar polygons are to one another as their homologous sides; therefore (Th. 98.) the ag- gregate of the perimeters of the polygons described on the two parts into which the line is divided, is to the perimeter of the polygon on the whole line, as the K K 258 -Elementary Theorems of Plane Geometry, A aggregate of the two parts is to the whole, that is (Th. 84.) in a ratio of equality. > CoR. 3. Hence, also, if upon each of two given unequal straight lines and upon their difference, there be described similar polygons, the perimeter of that - which is described upon the difference shall be equal to the excess of the perimeter of the greater of the other two above the perimeter of the less. ScholIUM. It may be easily shewn, also, that the perimeters of any similar polygons described about circles, are to one another as the diameters of the circles. Theorem CXIV. Parallelograms of the same altitude, or of equal altitudes, are to one another as their bases. For (Th. 25. Cor. 2.) they are the doubles of the triangles into which they are divided by their dia- meters; and (Th. 107.) these triangles are to one another as the bases of the parallelograms; therefore (Th. 86. Cor.) the parallelograms are to one another as their bases. CoR. Parallelograms upon equal bases, are (Th. 107. Cor. 2. and Th. 86. Cor.) to one another as their altitudes. ScholIUM. Since (Schol. to Th. 33.) two parallelograms having Proportionality of Sides, &c. of Polygons. 259 a common side, and therefore having also equal alti- tudes may be found that shall be equal to two given rectilineal figures, each to each, it is manifest, from Th. 112, that two straight lines may always be found that shall be to one another as any two given recti- lineal figures are. º Theorem CXV. Parallelograms, having one angle of the one equal to one angle of the other, if they have their sides about the equal angles reciprocally proportional, are equal; and if they be equal, they have their sides about the equal angles reciprocally proportional. For, in the first case, the triangle contained by the diameter and the two sides of the one, is equal (Th. 108.) to the triangle contained by the diameter and the two reciprocally proportional sides of the other; and therefore (Th. 25. Cor. 2.) the parallel- ograms are equal to one another. And, in the second case, the triangles, which are the halves of the equal parallelograms, are equal, and have an angle of the one equal to an angle of the other; and therefore (Th. 108.) the sides about those equal angles, which are also the sides of the parallelograms, will be reciprocally proportional. '', CoR. 1. If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means : and if the rect- angle contained by the extremes be equal to the 260 Elementary Theorems of Plane Geometry, rectangle contained by the means, the four straight lines are proportionals. * 3. For, if four straight lines be proportionals, the rectangle contained by the extremes and the rectangle contained by the means, will be parallelograms, which have their sides about equal angles (Th. 2.) recipro- cally proportional; therefore (Th. 115.) they are equal to one another : and if the two rectangles be equal to one another, they will (Th. 115.) have their sides about the equal angles reciprocally proportional ; so that the two sides of the one rectangle are the extremes, and the two sides of the other the means, of the four proportionals. * * CoR. 2. Hence, if three straight lines be pro- portionals, the rectangle contained by the extremes is equal to the square of the mean: and if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals. For, if the mean of three proportionals be taken twice, there will then be four proportionals, and (Cor. 1.) the rectangle contained by the extremes is equal to that contained by the means, that is, to the square of the mean proportional. Also, if a square be equal to a rectangle, the square may be considered as a rectangle contained (Cor. 1.) by two equal means, that is, by one mean proportional between the two sides of the other rectangle. CoR. 3. Hence if any three straight lines A, B, C, be continual proportionals, the first shall be to the third, as the square of the first is to the square of the second. . . Proportionality of Sides, &c. of Polygons. 261 For (Th. 114.) A is to C, as the square of A is to the rectangle contained by A and C, or (Cor. 2.) as the square of A is to the square of B. THEOREM CXVI. Equiangular parallelograms have to one another the ratio, which is compounded of ratios equivalent to the ratios of their sides. For (Th. 25. Cor. 2.) they are the doubles of the triangles, into which they are divided, by their dia- meters; and (Th. 109.) these triangles are to one another in a ratio that is compounded of ratios equi- valent to the ratios of the sides about the equal angles; which sides are also the sides of the parallel- ograms; therefore (Th. 86. Cor.) the equiangular parallelograms are to one another in a ratio that is compounded of ratios equivalent to the ratios of their sides. CoR. i. Equiangular parallelograms are to one another as the rectangles contained by their sides. For the ratio of the parallelograms, and the ratio of the rectangles will (Th. 116) be each equivalent to the same compounded ratio, and therefore (Th. 81.) they will be equivalent to one another. . . . CoR. 2. Hence also triangles, which have an angle of the one equal to an angle of the other, are to one another, as the rectangles contained by the sides about the equal angles. gº For (Cor. 1.) the parallelograms, which are their. doubles, will be to one another in that ratio; there- f 262 Elementary Theorems of Plane Geometry, fore (Th. 86. Cor.) the triangles themselves will be to one another in that ratio. - CoR. 3. If the bases of four rectangles be pro- portionals and their altitudes be also proportionals, the rectangles themselves shall, likewise, be proportionals. For the ratio which the first rectangle has to the second, is (Th. 116.) compounded of ratios equi- valent to those of which the ratio of the third rect- angle to the fourth is compounded; therefore (Def. 60. Cor.) the first rectangle is to the second as the third is to the fourth. ScholIUM. It was shewn, in the Scholium to Th. 98, that a ratio, compounded of other ratios, to each of which an equivalent numerical ratio can be found, is equivalent to the ratio which the product of all the numerical antecedents has to the product of all the numerical consequents: it follows, therefore, from Th. 116, that equiangular parallelograms are to one another as the products of the numbers which are pºportional to their sides: and thus, two squares are to one another as the squares of any two numbers which are pro- portional to their sides. & *g THEOREM CXVII. Similar polygons have to one another the duplicate ratio of that which their homologous sides have. Proportionality of Sides, &c. of Polygons. 263 Let ABCDE, FGHKL, be similar polygons, of which AB and FG, BC and GH, &c. are homologous A. IF - E. * º Kºy C D . º sides: ABCDE has to FGHKL the duplicate ratio of that which any side AB, of the former, has to the homologous side FG, of the latter. From A and F, the summits of the equal angles BAE, GFL, let there be drawn to the opposite equal angles AC, AD, FH, FK, dividing the polygons, since they have the same number of sides, into the same number of tri- angles: and since (hyp.) AB : BC: FG : GH, and that the LABC = / FGH, therefore (Th. 103.) the & triangles ABC, FGH, are similar; therefore the / BCA = / GHF; in the same manner it may be shewn that the Os AED, FLK, are similar; and since (hyp.) the / BCD = / GHK, the parts of which, viz. the A BCA, and / GHF, have been shewn to be equal, therefore the remainders are equal; i. e. the / ACD = L FHK; but, since the Os ABC, FGH, are similar, ... AC : CB :: FH : HG; and (hyp.) BC : CD :: GH : HK; ... (ea aequo) AC : CD :: FH : HK; and it has been proved that the / ACD = / FHK; 264 Elementary Theorems of Plane Geometry, } therefore (Th. 103.) the /> ACD, FHK, are similar. In the same manner, the remaining triangles, into which the polygons are divided, if there be more of them, may be shewn to be similar: and (Th. 110.) the AS, ABC has to the similar /S FGH the duplicate ratio of that which AC has to FH, and the As ACD has also the same ratio to the similar /> FHK; therefore (Th. 81.) AS, ABC : As FGH :: As ACD : As FHK: in the same manner it may be shewn, that & 4CD : As FHK :: & ADE : & FKL; therefore (Th. 83.) the AS, ABC is to the ZS FGH as the aggregate of the antecedents is to the aggregate of the consequents, i.e. as the polygon ABCDE is to the polygon FGHKL: but (Th. 110.) the As ABC has to the AS FGH the duplicate ratio of that which AB has to FG ; therefore (Th. 81.) ABCDE has to FGHKL the duplicate ratio of that which AB has to FG, . \ - CoR. 1. It appears from the demonstration, that similar polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have. CoR. 2. In like manner it may be proved, that similar four-sided figures are to one another in the duplicate ratio of their homologous sides, as are also (Th. 110.) similar triangles: Wherefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. CoR. 3. Similar rectilineal figures are to one an- other as the squares of their homologous sides. Proportionality of Sides, &c. of Polygons. 365 For (Th. 117.) the rectilineal figures, and the squares on their homologous sides, have to one an- other the same duplicate ratio; therefore (Th. 81.) the rectilineal figures are to one another as the squares of their homologous sides. t CoR. 4. Similar rectilineal figures are equal to one another, if two of their homologous sides be equal: and, conversely, rectilineal figures which are similar and equal, have their homologous sides equal, each to each. CoR. 5. If three straight lines be proportionals, as the first is to the third, so is any rectilineal figure on the first, to a similar and similarly described figure on the second. *A For (Def. 61.) the first of these lines has to the third the duplicate ratio of that which it has to the second. * SCHOLIUM. It is manifest, from Th. 117. Cor. 3, and from the latter part of the Scholium to Th. 116, that similar rectilineal figures are to one another as the squares of any two numbers that are proportional to two homo- logous sides. - - THEoREM CXVIII. If four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals ; and if the similar rectilineal jigures similarly described upon four straight lines L L 266 Elementary Theorems of Plane Geometry, be proportionals, those straight lines shall be pro- portionals. * . Let the four straight lines AB, CD, EF, GH, be proportionals, viz. AB to CD, as EF to GH; and K 2^ 2^ M N S \ \_o W \ E F. G. H. P R upon AB, CD, let the similar rectilineal figures KAB, LCD, be similarly described; and upon EF, GH, the similar rectilineal figures MF, NH, in like manner. The rectilineal figure KAB. is to LCD, as MI' to NH. To AB, CD, suppose X to be a third proportional, and O to be a third proportional to EF, GH. Then (hyp.) AB : CD :: EF : GH; and {Th. 81. and hyp.) CD : X :: GH : O : ... (Th. 96.) AB : X :: EF : , O. But (Th. 117. Cor. 5.) AB : X :: KAB : LCD, and EF : O :: MF : NH; ... (Th. 81.) KAB : LCD :: MF : NH. And if the rectilineal figure KAB be to LCD as MF to NH, then AB : CD :: EF: GH. Proportionality of Sides, &c. of Polygons. 267 For let PR be a fourth proportional to AB, CD and EF; and upon PR let there be supposed to be described the rectilineal figure SR, similar and simi- larly situated, to either of the figures MF, NH. Then (hyp.) AB : CD :: EF : PR; ... (first case and hyp.) - KAB : LCD :: MF: SR; also (hyp.) KAB : LCD :: MF : NH; ... (Th. 78.) NH = SR: And NH and SR are, also, similar rectilineal figures and similarly situated; ... (Th. 117. Cor. 4.) GH=PR. But (hyp.) AB : CD :: EF : PR; ... (Th. 77.) AB : CD :: EF : GH. If, therefore, four straight lines, &c. a. E. D. CoR. 1. . If a rectilineal figure be similar to each of two other rectilineal figures, and be also a mean pro- portional between them, its side shall be a mean portional between the sides, that are homologous to it of the other figures: and if the side of a rectilineal figure be a mean proportional between the sides that are homologous to it, of two other similar figures, that figure shall also be a mean proportional between the two others. CoR. 2. If four straight lines be proportionals, the squares described upon them shall, also, be pro- portionals: and if four squares be proportionals, their sides shall, also, be proportionals. THEOREM CXIX. Similar polygons inscribed in circles are to one another as the squares of their diameters. 268 Elementary Theorems of Plane Geometry, Let ABCDE, FGHKL, be two circles, and in them the similar polygons ABCDE, FGHKL ; and let BM, GN be diameters of the circles: as the square of BM is to the square of GN, so is the polygon ABCDE to the polygon FGHKL. For let BE, AM, GL, FN, be drawn. Then, it may be shewn, as in Th. 113, * that BM : GN :: AB : FG; therefore (Th. 118. Cor. 2.) the square of BM is to the square of GN as the square of AB is to the square of FG, that is (hyp. and Th. 117. Cor. 3.) as the polygon ABCDE is to the polygon FGHKL; there- fore (Th. 81.) the square of BM is to the square of GN as the polygon ABCDE is to the polygon FGHKL. CoR. If in each of two circles an equilateral and equiangular polygon be inscribed of the same number of sides, and if through the angular points of these polygons straight lines be drawn touching the circles, the polygons so described about the circles, will be (Th. 112. Cor. and Def. 63. Cor.) similar to the inscribed polygons, and to one another. They will, Proportionality of Sides, &c. of Polygons. 269 therefore, also be to one another as the squares of the diameters of the circles. ScholIUM. The homologous sides of any similar polygons, described about circles, are to one another as their diameters; and, therefore, (Th. 117. Cor. 3. Th. 118. ' and Th. 81.) similar polygons described about circles are to one another as the squares of their diameters. THII. HElements of 12lant Geometry. Book I. CHAPTER V. —-sº- SECTION III. On the proportionality of angles and circular arches. —O— THEor EM CXX. In equal circles, angles, whether at the centres or circumferences, are to one another as the arches which subtend them. LET ABC, DEF, be equal circles; and at their centres the A. BGC, EHF, and the / BAC, EDF at their circumferences; as BC to EF, so is the / BGC to the / EHF, and the L BAC to the Z EDF, 2-N 2-\ Suppose CK, KL, to be any number of arches, 2- 2-N 2-N each equal to BC, and FM, MN, to be any number Proportionality of Angles and Circular Arches. 271 2-> A whatever, each equal to EF: suppose, also, GK, GL, H.M. HN, to be drawn. /*S 2-N 2-N - Because BC, CK, KL, are all equal, the / BGC, CGK, KGL are (Th. 60.) also all equal: Therefore 2-N 2-N what multiple soever BL is of BC, the same multiple is the / BGL of the / BGC: For the same reason, 2-N 2-N whatever multiple EN is of EF, the same multiple + 2-N is the / EHN of the / EHF: And (Th.60.) if BL:- 2-\ = < EN, the / BGL - = < the / EHIV. There se 2-N 2-\ are, then, four magnitudes, the two arches BC, EF, 2-N and the two / BGC, EHF; of the arch BC and of the LBGC, there have been taken any equi-multiples 2-N whatever, namely, BL and the / BGL; also, of the 2-N arch EF, and of the LEHF, there have been taken 2-S any equi-multiples whatever, namely, EN, and the 272 Elementary Theorems of Plane Geometry, - 2-N A. E.HN; and it has been proved that if BL - = < £N, then the BGL = < the EHN. 2- 2-S ... (Def. 48.) BC : EF :: / BGC : A EHF. But (Th. 56. and Th. 86.) / BGC : / EHF :: / BAC : A EDF; 2-S 2-N ... (Th. 81.) BC : EF :: / BAC : , EDF, Wherefore, in equal circles, &c. a. E. D. CoR. 1. Since, as an angle at the centre, or at the circumference of a circle, is to a right angle, so is the arch subtending the angle to the fourth part of the circumference, therefore as an angle at the centre, or at the circumference, is to four right angles, so is (Th. 93. Cor.) the arch subtending the angle to the whole circumference. CoR. 2. Hence, in unequal circles, the arches (A), (a), subtending equal angles (B), are to one an- other as the whole circumferences (C), (c), of the circles. - For (Cor. 1.) A : C : B : four right angles :: a c; ... (Th. 81.) A : C :: a c; ... (alternando) A : a ... C : c. ScholIUM. By producing the base of a triangle indefinitely, by cutting off, from the part produced, straight lines Proportionality of Angles and Circular Arches. 273 equal to the base, and by joining the vertex of the triangle and the extremities of those lines it is easy (Th. 30.) to exhibit any assigned multiples of a given triangle. But with angles, and sectors of circles, the case is not the same. For a multiple of a given angle may exceed the whole angular space about a given point; and as a multiple of a given arch of a circle may exceed the whole circumference, so may a mul- tiple of a given sector exceed the whole circle. Such kind of multiples, although they may be conceived by the mind, cannot be geometrically exhibited to the eye. It may be proper, therefore, to give another demonstration of Th. 120, founded on the criterion of proportionality established, in the latter part of the scholium which follows Th. 98. Let, then, ABC, DEF, be equal circles, and let 2-N 2-N BGC, EHF, be angles at their centres; BC : EF: T) IS LBGC: L EHF By continually bisecting the arch EF and the arches into which it is subdivided, let EF be supposed to be divided into any number ºf equal parts, of which EK is one; suppose BL to be M M 2.74 Elementary Theorems of Plane Geometry, : * : * * 2-\ ... t • ? - f equal to EK; and since (Th.68.) there is no limit to the diminution of these parts, it is evident that in BC there may be taken a multiple, BM, of EK, or of BL, which is equal to EK, such that MC, the tº : . 2-N /*k * s - difference between BM and BC, shall be less than any assignable arch: let BM be supposed to be so taken, and HK, GL, GM, to be drawn. Then it is manifest, from Th.60, that the / EHFis the same multiple of the 4. EHK that EF is of EK; likewise that the LBGM 2-N is the same multiple of the / BGL that BM is of BL, and that the / EHK = / BGL, because (hyp.) EK=BL; therefore (Th. 80.) 2-N 2-N BM : BL :: / BGM. : / BGL, fºr or fill £º :: / EHK or BGL : / EHF; ... (Th. 94.) fin : ſº :: / BGM : / EHF. 2-\ 2-N Also (Th.60. and Cor. 1.) if BMP = < BC, the / BGM- = < the LB GC ; there are therefore, four magnitudes, viz. BC, EF, the 1. BGC, and the 1. 2-> EHF, and two others, viz. BM and the L BGM, Proportionality of Angles and Circular Arches. 275 such that BM: EF:: / BGM : / EHF, and that 2-> * * >\ • * if BM, the difference of which from BC may be made less than any assignable arch, be-a=< BC, then the 4. BGM- = ~~ BGC : EHF :: sector KBGC: sector LEHF For it is manifest from Th. 65, and it's corollary, --~ that if any multiple be taken of BGC, it will be the arch of a sector, that is the same multiple of the sector KBGC; and if any multiple be taken of £HPit will, 278 Elementary Theorems of Plane Geometry, likewise, be the arch of a sector that is the same mul. tiple of the sector LEHF; and, also, that accordingly A as the multiple of BëC>= = < the multiple of the latter; therefore (Def. 48.) HGö EHF :: sector KBGC : sector LEHF: Wherefore in equal circles”, &c. a. E. D. CoR. I. Sectors of the same circle are to one an- other as their arches. CoR. 2. A sector of a circle is to the whole circle as the arch of the sector is to the whole circum- ference. t For (Cor. 1.) any sector is to a quadrant of the circle as the arch of that circle is to a fourth part of the circumference; therefore (Th. 93. Cor.) any sector is * This proposition might also easily be deduced from Th. 70. Cor. 1. For the sectors of equal circles are, according to that corollary, equal to the triangles of the same altitude, which have the arches of the sectors for 'their bases: Therefore the sectors are r- to one another as their arches. Proportionality of Circles, &c. and of Segments. 279 to the whole circle as the arch of the sector is to the whole circumference. - CoR. 3. In equal circles, or in the same circle, sectors are to one another (Th. 122. 120.) as the angles. subtended by their arches, at the centres, or at the circumferences. CoR. 4. In unequal circles (K), (k), similar sectors, (S), (s), the arches of which (A), (a), subtend (Def, 38.) equal angles (B) at the centres, are to one another as the circles. r For (Th. 122. Cor. 2. and Th. 120. Cor. 1.) S : K :: / B : four right angles :: s : k ; ... (Th. 81.) S’: K :: s : k ; ... (alternando) S : s :: K : k. - CoR. 5. If a straight line be divided into any two parts, the circumference of the circle described upon the whole line as a diameter, shall be equal to the circumferences of the circles described upon each of the parts as a diameter. t ScholIUM. The preceding proposition may, also, be proved in the same manner, as it was shewn, in the Scholium which follows Th. 120, that, in equal circles, angles at the centres are to one another as the arches which subtend them. 1. * , - 280 Elementary Theorems of Plane Geometry, Theorem CXXIII. Circles are to one another as the squares of their diameters. - Let ABC, DEF, be two circles, and BC, EF N |M A. T) T- tº TH - . . —- G. L their diameters: as the square of BC is to the square of EF, so is the circle ABC to the circle DEF For if GH and LM be supposed to be equal to the circumferences of ABC and DEF, and if HK be sup- posed to be drawn perpendicular to GH and made equal to the half of BC, and MN to be drawn perpendicular to LM and made equal to the half of EF; and if K, G and N, L be joined, then (Th. 70.) the circle ABC is equal to the /S KHG, and the circle DEF to the AS NML ; also (hyp. and Th. 121. Cor.) GH: HK :: LM : MN, and (Th. 2.) the / KHG = 4. NML, each of them being a right angle; therefore Proportionality of Circles, &c. and of Segments. 281 ... (Th. 103.) the As KHG is similar to the AS NML; therefore (Th. 117. Cor. 3.) the AS KHG is to the As NML as the square of KH to the square of NM; but (hyp. and Th. 86.) KH : NM :: BC : EF; therefore (Th. 118.) the square of KH is to the square of NM as the square of BC is to the square of EF; therefore (Th. 81.) the AS KHG is to the AS NML as the square of BC is to the square of EF; that is, (hyp.) the circle ABC is to the circle DEF, as the square of BC to the square of EF. Wherefore circles, &c. a. E. D. e CoR. 1. From the demonstration it is manifest, that circles are also to one another as the squares of their semi-diameters. CoR. 2. It is manifest, from Th. 123, and Th. 122. Cor. 4, that in unequal circles, similar sectors are to one another as the squares of the dia- meters, or (Th. 123. Cor. 1.) as the squares of the semi-diameters. CoR. 3. As the one of two circles is to the other, so is any polygon inscribed in the one (Th. 119.) to a similar polygon inscribed in the other. CoR. 4. If three straight lines be proportionals, as the first is to the third, so is the circle which has the first for it's diameter, or for it's semi-diameter, to the circle which has the second for it's diameter, or its semi-diameter; so also is any sector of a circle which has the first for it's semi-diameter, to a similar sector of a circle which has the second for it's semi-diameter. N N 282 Elementary Theorems of Plane Geometry, a tº Theorem CXXIV. Similar segments of circles are to one another as the squares of their chords. Let ABC, DEF, be any two circles of which BCG, EFH, are similar segments, that is, such that K. - L “sº “spºº y H 1. any two points G and H being taken in BGC, EHF, the // BGC, EHF, are equal to one another; the segment BCG is to the segment EHF as the square of BC is to the square of EF. - For let K and L be the centres of the circles ABGC, DEHF and let. KB, KC, LE, LF, be drawn. And since (hyp.) the Z. BGC = / EHF therefore (Th. 54.) the angles subtended by BC and EF at any points A, D, in BAC, EDF, are equal to one another, therefore the / BKC, ELF, which (Th. 56.) are their doubles, are also equal to one another; ºtherefore (Th. 123. Cor. 2.) the sector KBGC is to the sector LEHF, as the square of KB is to the square of LE: but since the / BKC= Proportionality of Circles, &c. and of Segments. 283 * ELF, and that the as KBC, LEF, are isosceles, they are (Th. 12. and Th. 13.) equiangular; therefore (Th. 101. Cor. 2.) KB : LE:: BC : EF; therefore (Th. 118. Cor. 2.) as the square of KB is to the square of LE so is the square of BC to the square of EF; therefore (Th. 81.) the sector KBGC is to the sector LEHF as the square of BC is to the square of EF: and (Th. 117. Cor. 3.) the As KBC is to the AS LEF as the square of BC is to the square of EF; therefore (Th. 81.) the sector KBGC is to the sector LEHF as the AS KBC is to the AS LEH ; there- fore (Th.90.) the segment BCG is to the segment EFH as the sector KBGC is to the sector LEHF, that is (Th. 81.) as the square of BC to the square of EF: Wherefore similar segments, &c. a. E. D. CoR. I. It is manifest, from the demonstration, that similar segments of two circles are to one another as the squares of the semi-diameters, or (Th. 123. Cor. 1.) as the circles themselves, or as (Th. 123.) the squares of the diameters. CoR. 2. If three straight lines be proportionals, as the first is to the third, so is any circular segment, which has the first for it's chord, to a similar circular segment, which has the second for it's chord. TIII, 33ittittittg of #lattt (ºrontºttp. CHAPTER WI. ON . THE COM PARISON OF R.ECTANGL ES CONTAIN ED BY sTRAIGHT LINEs AND THEIR seGMENTs, AND of sIMILAR FIGURES, DESCRIBED UPON THE SIDEs of TRIANGLES AND PRoportion AL LINEs. –O— SECTION I. On the comparison of rectangles contained by straight lines and their segments. THEOREM CXXV. If there be two straight lines, one of which is divided into any number of parts; the rectangle con- tained by the two straight lines, is equal to the Comparison of Rectangles, &c. 285 rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines ; and let BC be divided into any parts in the points D, E: the A B T) E. C G- -º- F K. L. H. rectangle contained by the straight lines A, BC, is equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. From the point B suppose BF to be drawn at right angles to BC, and BG to be made equal to A; and through G suppose GH to be drawn parallel to BC: and through D, E, C, suppose DK, EL, CH to be drawn parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH ; and BH is contained by A, BC, for it is contained by GB, BC, and GB=A; and BK is contained by A, BD, for it is contained by GB, BD, of which GB = A; and DL is contained by A, DE, because DK, being (Th. 25.) equal to BG, is equal to A ; and in like manner the rectangle EH is contained by A, EC: therefore the rectangle contained by A, BC is equal 286 Elementary Theorems of Plane Geometry, to the several rectangles contained by A, BD, and by A, DE; and also by A, EC. Wherefore, if there be two straight lines, &c. a. E. D. * * CoR. 1. Hence, if a straight line be divided into any two parts, the rectangle contained by the whole line and a straight line equal to it, that is, the square of the line, is equal to the rectangles contained by the whole, and each of the parts. For, if A = BE, which is divided into two parts in D, then (Th. 125.) the rectangle contained by A and BE, i. e. the square of BE, is equal to the rectangle contained by A and BD together with the rectangle contained by A and DE; or (Th. 26.) to the rectangle contained by BE and BD, together with the rectangle contained by BE and D.E. CoR. 2. Hence, also, if a straight line be divided into any two parts, the rectangle contained by the whole, and a straight line equal to one of the parts, is equal to the square of that part, together with the rectangle contained by the two parts. For if BC be divided into the parts BE and EC, and if A = BE, then (Th. 125.) the rectangle contain- ed by A and BC, is equal to the rectangle contained by A and BC, that is, the square of BE, together with the rectangle contained by A, or BE, and E.C. Theorem CXXVI. If two straight lines, which cut one another, be produced so as to meet the circumference of a circle, the rectangle contained by the segments of one of Comparison of Rectangles, &c. 287 them is equal to the rectangle contained by the segments of the other. - Let the two straight lines EA, ED, cut one another in E, and either each meet the circumference of the circle ADF in two points A, B, and D, C, as in fig. 1, 2, or let one of them EA meet the circum- ference in two points B, A, and the other ED in one point, D, as in fig. 3, or else let each of them meet the circumference in one point only, as in fig. 4, the rectangle AEx EB is equal to the rectangle DEx EC. For, let AC and DB be drawn : them, since the A. EDB = A. EAC, because they are in the same segment, or the / EDB, contained by the tangent ED and DB, is equal (Th. 58.) to the L EAC in the alternate segment, or else the / EDB, EAC, con- tained by the tangents, EA, ED, being each of them equal to the angle in the alternate segment BFD, are equal to one another; and that the angles AEC, 288 Elementary Theorems of Plane Geometry, DEB, are either vertical and (Th. 1. Cor.) equal angles, F F or they are the same angles, therefore (Th. 101.) the As AEC, DEB, are similar, and AE : EC :: DE : EB; •". (Th. 115. Cor. 1.) AEx EB = EC X DE. - Wherefore, if two straight lines, &c. a. E. D. CoR. 1. Since (Th. 36.) a chord which is per- pendicular to the diameter of a circle is bisected by it, the rectangle contained by the segments into which the diameter is divided by a chord that is per- pendicular to it, is equal to the square of half the chord : and hence, and from Th. 44, it follows that the square of the half of a given straight line is greater than the rectangle contained by any two unequal parts into which the given line can be divided. CoR. 2. If from any point, E, without a circle AFCB, (fig.3) two straightlines, EBA, ED, be drawn, one of which EBA cuts the circle, and the other, ED, touches it, the rectangle contained by the whole line Comparison of Rectangles, &c. 289 E4, which cuts the circle, and the part of it, EB, without the circle, is equal to the square of the tangent, E.D. - CoR. 3. If the rectangles contained by the seg- ments of two straight lines, AB, CD, (fig. 1.) which cut one another in E, be equal, and the circum- ference of a circle pass through three of the ex- tremities, A, B, C, of the two lines, it shall also pass through the fourth extremity, D. For if not, let it pass through H; therefore (Th. 126.) HEx EC= AEx EB; but (hyp.) DEx EC= AE x EB; therefore HEx EC is equal to DEx EC, that is, a part is equal to the whole, which is im- possible: Wherefore the circumference which passes through A, C, B, cannot pass otherwise than through D. CoR. 4. If from any point, E, (fig. 3.) without a circle, AFDB, two straight lines be drawn, EBA, ED, one of which, EBA, cuts the circle, and the other, ED, meets it, if the rectangle contained by the whole line, E4, which cuts the circle, and the part of it, EB, without the circle, be equal to the square of the line, ED, which meets it, the line, ED, which meets, shall touch the circle. For, if ED meet the circumference again, it may be shewn as in the preceding corollary, that two unequal rectangles are each of them equal to the rectangle AEx EB, and therefore equal to one another, which is absurd: wherefore ED, when pro- duced, does not cut the circle, that is, (Def. 34.) it touches the circle. - º O O 290 Elementary Theorems of Plane Geometry, Der. LXV. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure, about which it is described. - THEOREM CXXVII. If an angle of a triangle be bisected by a straight line, which likewise cuts the base; the rectangle con- tained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bi- secting the angle. | Let ABC be a triangle, and let the LBAC be bisected by the straight line AD; the rectangle BA, / A. sº …ſº E AC is equal to the rectangle BD, DC, together with the square of AD. - • Comparison of Rectangles, &c. 291. Suppose the circle ACB to be described about the triangle, and AD to be produced so as to meet the circumference in E, and EC to be drawn: then, because (hyp.) the / BAD= L CAD, and (Th. 55.) the Z. ABD = / AEC, for they are in the same segment, the ſº ABD, AEC, are equiangular to one another: therefore (Th. 101.) as BA to AD, so is EA to AC, and consequently (Th. 115. Cor. 1.) the rectangle BA, AC is equal to the rectangle EA, AD, that is, (Th. 125. Cor. 2.) to the rectangle ED, DA, together with the square of AD: but (Th. 126.) the rectangle ED, DA is equal to the rectangle BD, DC. Therefore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD. Wherefore, if an angle, &c. a. E. D. THEOREM CXXVIII. If from the summit of any angle of a triangle a straight line be drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle. Let ABC be a triangle, and let AD be the per- 7– * It is manifest, from the Scholium which follows Th. 20, that there is a point which is equidistant from the three angular points of any given triangle; and that therefore, it is possible to describe a circle, the circumference of which shall pass through the three angular points of a triangle. - 292 Elementary Theorems of Plane Geometry, pendicular from the summit of the LA to the base E BC; the rectangle BA, AC is equal to the rectangle contained by AD and the diameter of the circle de- scribed about the triangle. r Suppose the circle ACB to be described about the triangle, and its diameter AE to be drawn, and E, C, to be joined: because (Th. 57. and Th. 2.) the right angle BDA is equal to the LECA in a semi- circle, and the LABD to the / AEC in the same segment; the triangles ABD, AEC, are equi- angular: therefore (Th. 101.) as BA to AD, so is EA to AC; and consequently the rectangle BA, AC is equal (Th. 115. Cor. 1.) to the rectangle EA, AD. If therefore, from an angle, &c. a. E. D. THEoREM CXXIX. The rectangle contained by the diagonals of a quadrilateral rectilineal figure inscribed in a circle, is equal to both the rectangles contained by it's opposite sides. * Let ABCD be any quadrilateral rectilineal figure comparison of Rectangles, &c. 293 inscribed in a circle, and AC, BD its diagonals; the ? - rectangle AC, BD is equal to the two rectangles con- tained by AB, CD, and by AD, BC. Suppose the LABE to be made equal to, the * DBC; add to each of these the common / EBD, then the Z. ABD = / EBC: and (Th. 55.) the a BDA = LBCE, because...they are in the same segment; therefore the AS, ABD is equiangular to the AS BCE: wherefore (Th. 101.) as BC is to CE, so is BD to DA; and consequently the rectangle BC, AD is equal (Th. 115. Cor. 1.) to the rectangle BD, CE: again, because the LABE= L DBC, and (Th. 55.) the / BAE = / BDC, the AS, ABE is equi- angular to the As BCD: as therefore B4 to AE, so is BD to DC; wherefore the rectangle BA, DC is equal to the rectangle BD, AE; but the rectangle BC, AD has been shewn equal to the rectangle BD, CE; therefore the whole rectangle AC, BD(Th. 125.) is equal to the rectangle AB, DC, together with the rectangle AD, BC. Therefore the rectangle, &c. Q. E. D. 294 Elementary Theorems of Plane Geometry. Con. If D be in the bisection of ADö, then, wherever the point B is taken, in ABC, if BC, BD, BA, AC, AD, DC be drawn, BD : AB + BC :: AD : A.C. For, in this case, (hyp. and Th.62.) AD=DC; but (Th. 123.) BDx AC=ABx CD+BC x AD; therefore (Th. 125) BDx ACis equal to the rectangle . contained by AB+ BC and AD; therefore (Th. 115. Cor. 1.) - BD : AB+ BC :: AD : AC. '''H ſº £itments of plant cºrontettp. CHAPTER VI. SECTION II. On the comparison of similar figures, described upon the sides of triangles and proportional lines. —O— Theorem CXXX. n In right-angled triangles, any rectilineal, or cir- cular figure", described upon the side subtending the right angle, is equal to the two similar and similarly posited figures, which are described upon the sides containing the right angle. LET the AS, ABC be right-angled at 4; upon BC, CA, AB, let there be described three similar, and * By circular figures, in this and the subsequent propositions, are to be understood either whole circles, or similar segments, or similar sectors of circles. 296. Elementary Theorems of Plane Geometry, similarly placed, rectilineal or circular figures, H, hº K. f * * * K, L ; H on BC, K on CA, and L on AB: then H= K+ L. From A let AD be supposed to be drawn per- pendicular to BC; then, since (Th. 106. Cor. 1.) BC, CA, CD, are proportionals, as are, also, CB, BA, DB, therefore (Th. 117. Cor. 5. Th. 123. Cor. 4. Th. 124. Cor. 2.) - H : K :: CB : CD, and H : L :: CB : DB ; ... (Th. 98. Cor.) H . K+ L :: CB : CD + DB. But CB = CD + DB ; therefore (Th. 84.) H = K+ L. Therefore, in right-angled triangles, &c. a. E. D. THEoREM CXXXI. If upon each of two straight lines, upon their aggregate, and upon a mean proportional between them, there be described similar and similarly posited rectilineal or circular figures, that which is described upon their aggregate shall be equal to the two figures described upon the lines, together with the double of Comparison of similar Figures upon Triangles, &c. 297 that described upon the mean proportional between them. ~ \ Between two straight lines "BD, DC, placed in the same straight line BC, let DA placed perpendicular to BC, be a mean proportional ; and let there be described the similar and similarly posited rectilineal or circular figures, H on BD, K on DC, M on DA, S on BC: then S= H+ K+ twice M. Suppose AB and AC to be drawn; then (hyp. and Th. 106. Cor. 2.) the / BAC is a right angle; upon AB and AC let there be described figures P and Q, similar and similarly posited, to those described on the given lines; therefore (Th. 130.) S = P + Q; also, P= H-H M, and Q = K+ M, therefore S= H + K + twice M. If, therefore, upon each, &c. Q. E. D. & CoR. 1. If the two given straight lines be equal, the mean proportional between them being then equal to either of them, the figure described upon their aggregate will be quadruple of the similar and similarly described figure upon either of them. CoR. 2. If a straight line be divided into any two. parts, the square of the whole line is equal to the squares of the two parts together with twice the rectangle contained by the parts. - For (Th. 131.) the square of the whole line is equal to the squares of the two parts, together with twice the square of the mean proportional between * See the figure in p. 296. P P 298 Elementary Theorems of Plane Geometry, them; i. e. (Th. 115. Cor. 2.) together with twice the rectangle contained by the parts. - THEOREM CXXXII. If upon each of two given straight lines, upon their difference, and upon a mean proportional between them, there be described similar and similarly posited rectilineal or circular figures, that which is described on the difference, shall be equal to the eacess of the two figures described on the two given lines above the double of the figure described on the mean pro- portional. | * Let BD be the difference of the two given straight lines BC, CD: if, upon BC, as a diameter, a semi- A. B I) E. & circle BAC be described, the circumference of which is met in A by D4 drawn from D at right angles to BC, and if AB, AC be drawn, then (Th. 57. Th. 106. Cor. 1.) AC is a mean proportional between BC and CD : and, if there be described the similar and similarly posited rectilineal or circular figures H on BD, K on BC, L on CD, M on AC, H, is equal to ^ Comparison of similar Figures upon Triangles, &c. 299 the excess of K+ L above twice M ; that is, H + twice M-K-- L. g - For upon AD and AB let there be described ) figures P and Q, similar and similarly posited to those described on the given lines: then (Th. 130.) Q=P +H; to these equals add M-H Li 4. ... Q+ M-H L = M-H-L-I-P-H H; but (Th. 130.) Q+ M= K, and L + P=M; ... K+ L = twice M + H. If, therefore, upon each, &c. a. E. D. ** CoR. If a straight line (BC) be divided into any , two parts in (D) the squares of the whole line (BC) and of one of the parts (CD) are equal to twice the rectangle contained by (BC and CD) the whole and that part, together with the square of (BD) the other part. - For (Th. 115. Cor. 2.) the square of AC is equal to the rectangle contained by BC and CD; therefore (Th. 132.) the squares of BC, CD are equal to twice the rectangle contained by BC and CD, together with the square of B.D. THEOREM CXXXIII. If upon the aggregate and the difference of two unequal given straight lines and upon a mean pro- portional between the given lines, there be described similar and similarly posited rectilineal or circular jigures, that which is described on the aggregate shall be equal to the figure described on the difference together with the quadruple of the figure described on the mean proportional. 300 Elementary Theorems of Plane Geometry, Let BC* be the aggregate of the two straight lines BD, DC, of which DC is the greater, and between which DA is a mean proportional; from DC let there be cut off DE = DB, so that EC is the difference between BD and DC; and let there be de- scribed the similar and similarly posited rectilineal or circular figures, K on BC, L on EC, M on AD; then K= L + four times M. For upon BD and DC let there be described figures P and Q similar and similarly posited to those described on the given lines; therefore (Th. 131.) K= P + Q+twice M.; and (Th. 132.) P--Q= L-H. twice M ; therefore K = L + four times M. If, there- fore, upon, &c. a. E. D. CoR. If a straight line (CD) be divided into any two parts (in E) four times the rectangle contained by the whole line (CD) and one of the parts (DE), together with the square of the other part (EC) is equal to the square of the straight line (CB) which is made up of the whole (CD) and that part (DE). For (Th. 115. Cor. 2.) the square of AD is equal to the rectangle contained by CD and DB or DE; and therefore the quadruple of the square of AD is equal to the quadruple of the rectangle contained by CD and DE; therefore (Th. 133.) the square of CD is equal to the square of EC together with four times the rectangle contained by CD and DE. THEOREM CXXXIV. 3. If, upon each of two unequal given straight lines, * See the figure in p. 298. Comparison of similar Figures upon Triangles, &c. 301 and upon a mean proportional between their aggregate and their difference, there be described similar and similarly posited rectilineal or circular figures, the difference of the figures described on the two given lines, shall be equal to the figure described on the mean proportional. A- Let AB be the greater of the two given straight lines, and AC a part of it, the less; let BA be pro- I) A C E B duced to D so that AD=AB, and therefore DC is the aggregate, and CB the difference of AB and AC; let CE, drawn perpendicular to DB, be a mean pro- portional between DC and CB; and let there be described the similar and similarly posited rectilineal or circular figures, K on AB, L on AC, and M on CE; then shall the difference of K and L be equal to M. f Suppose ED and EB to be drawn, CB to be bisected in F, FG to be drawn at right angles to AB, meeting EB in G, and G, A to be joined; and let there be described on GF a figure P, similar and similarly posited to the figures described on the given lines. Then (hyp. and Th. 106. Cor. 2.) the / DEB is a right angle; also, since the angles at C and F are 302 Elementary Theorems of Plane Geometry, right angles, therefore (Th. 6.) EC is parallel to GF; but (hyp.) CF= FB; therefore (Th. 99.) EG=GB; also (hyp.) DA = AB; therefore (Th. 99.) GA, is parallel to ED; and therefore (Th. 10.) the LAGB is a right angle ; therefore (hyp. and Th. 106. Cor.) GF is a mean proportional between AF and FB: also, because the triangles ECB, GFB, are (Th. 101. Cor. I.) similar, • ... (Th. 101. Cor. 2.) CB : FB :: EC : GF; but (hyp.) CB is the double of FB; therefore (Th. 82.) EC is the double of GF; therefore (Th. 131. Cor. 1.) M is quadruple of P; but (Th. 133.) the quadruple of P is the excess of K above L (for hyp. BF=FC); wherefore, M is equal to the excess of K above L, that is, to the difference between K and L. If, therefore, upon each, &c. a. E. D. CoR. The difference of two squares is equal to the rectangle contained by the aggregate. (DC) and the difference (CB) of their sides AB and AC. - For (Th. 115. Cor. 2.) the square of EC is equal to the rectangle contained by DC and CB; and it is also (Th. 134.) equal to the difference of the squares of AB and AC. The difference of these squares is, therefore, equal to the rectangle contained by DC and CB. y THEOREM CXXXV. If upon each of two given unequal straight lines, upon their aggregate, and upon their difference, there be described similar and similarly posited rectilineal Comparison of similar Figures upon Triangles, &c. 303 or circular figures, those which are described on the aggregate and on the difference are, together, the double of the figure described on the two given wnequal lines. Upon the two given unequal straight lines AB, BC, upon their aggregate AC, and (BD being cut off equal to BC) upon their difference AD, let there be described similar and similarly posited rectilineal or circular figures, K on AB, L on BC, M on AC, and tº tº e L e is wº * * * *****".... assee'sses **s, ea" *e **s, • *** *se • * • *** **e * ** *e º º e” *... ºf * • **. º º A: **** #C e * -> º *::s, ...” * •A we eº & *A*s, .** sº **, **see • ** * *e. ** a s a e s s a “* *- - sº ^e - s? o •e *s *e • se ***, º º **s, ... • ** & º *sse te • * * •ow **s--------------" t- M N on AD : then M and N together are the double of K and L taken together. - For on a mean proportional between AB and BC let there be described a figure P, similar and similarly posited to the figures described on the given lines: then (Th. 133.) M-N-H the quadruple of P; to each of these equals add N; therefore M+ N= the double of NH-the quadruple of P; but (Th. 132.) K+ L=N +the double of P; therefore the double of K and L taken together is equal to the double of N together with the quadruple of P; therefore M and N, together, are the double of K and L taken together. If, therefore, upon each, &c. G. E. D. 304 Elementary Theorems of Plane Geometry, THEOREM CXXXVI. If upon each of the three sides of an obtuse-angled triangle, and upon a mean proportional between a side about the obtuse angle and the part of it produced between that angle and a perpendicular drawn to it from the opposite angle, there be described similar and similarly posited rectilineal or circular figures, that, which is described on the side subtending the acute angle, shall be greater than the figures on the sides containing the obtuse angle, by twice the figure on the mean proportional. In the /> ABC let the / ABC be an obtuse angle, and let CD drawn from C, perpendicular to AB C * •. e ". e e Q e * º © º _r © º • produced, meet AB produced in D; also let BE be a mean proportional between AB and BD; and let there be described the similar and similarly posited rectilineal or circular figures, H on AC, K on CB, L on BA, M on BE : then H = L + K+ twice M. Comparison of similar Figures upon Triangles, &c. 305 For let there be described figures similar and similarly posited to those on the given lines, namely, N on BD, P on AD, Q on CD: then since (hyp.) the LADC is a right angle, therefore (Th. 130.) H = P + Q; but (Th. 131.) P = L-H N + twice M ; therefore H= L-H N+ Q+ twice M ; also (Th. 130.) K= N + Q ; therefore H = L +/K+ twice M: If, therefore, upon each, &c. a. E. D. CoR. In obtuse-angled triangles, the square of the side (AC) subtending the obtuse angle, is equal to the squares of the sides (AB, BC) containing that angle, together with twice the rectangle con- tained by a side (AB) about the obtuse angle, and the part of it (BD), produced, between that angle and a perpendicular drawn to it from the opposite angle. For (Th. 115. Cor. 2.) the square of EB is equal to the rectangle contained by AB and BD; therefore (Th. 136.) the square of AC is equal to the squares of AB, BC, together with twice that rectangle. THEOREM CXXXVII. If, upon each of the three sides of any triangle, and upon a mean proportional between a side about an acute angle, and its segment terminated by that angle and a perpendicular drawn to the side from the opposite angle, there be described similar and similarly posited rectilineal or circular figures, the figure upon the side subtending the acute angle shall be less than the two figures on the sides containing the acute angle, by twice the figure on the mean proportional. Q Q. 306 Elementary Theorems of Plane Geometry, In the As ABC let the / B be an acute angle, and from the / C opposite to AB, a side about the acute a B, let there fall on AB the perpendicular CD ; let BE be a mean proportional between AB and BD ; and let there be described the similar and similarly posited rectilineal or circular figures, H on AC, K on CB, L on BA, and M on BE; then H+ twice M- K -- L. For let there be described figures similar and similarly posited to those on the given lines, namely, E M. N on DB, P on AD, and Q on CD. Then since (hyp.) the / ADC, BDC, are right angles, therefore (Th. 130.) H=Q+ P; add, to each of these, twice M ; therefore H+ twice M-Q+P.+ twice M, but (Th. 132.) P + twice M = L + N: therefore H+ twice M=Q+ L--N; also (Th. 130.) K = Q - N3 wherefore H + twice M = K+ L; so Comparison of similar Figures upon Triangles, &c. 307 that H is less than K+ L by twice M. If, there- fore, upon each, &c. a. E. D. CoR. 1. Hence, and from Th. 115. Cor. 2, it is manifest that in every triangle, the square of the side subtending any of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either of those sides and the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle. CoR. 2. If upon the three sides of a triangle there be described any similar and similarly posited recti- lineal or circular figures, of which one is equal to the remaining two, the triangle is right-angled. For if not, it is either acute-angled, or else obtuse- angled. But (hyp. and Th. 136, 137.) it cannot be either acute-angled or obtuse-angled; therefore it is right-angled. THI E 3: It murn tº of 13 latte (ſºtomuttty. —O— Book II. ELEMENTARY PROBLEMS. O F P L A N E G E o M ET R Y. —sº- CHAP. I. PROB LEMS RELATING TO STRA IGHT LIN ES AND p LAN. F. RECTILIN EAL A N G L ES. —O— POSTULATES. I. LET it be granted, that a straight line may be drawn from any one point to any other point. II. Also, that a terminated straight line may be pro- duced to any length, in a straight line, on either side. Problems relating to Straight Lines, &c. 309 III. Also, that a circle may be described from any centre, at any distance from that centre. PROBLEM I. To find a point the distances of which from each of two given points shall be equal to one another, and shall each of them be equal to the distance between the given points. d Let A, B be the two given points, and let (Pos- tulate 1.) AB be drawn : it is required to find a point, the distances of which from A and B shall be equal to one another, and shall each of them be equal to AB. , From the centre A, at the distance AB, describe (Postulate 3.) the circle BCD ; also from the centre B, at the distance BA, describe the circle ACE ; and let the circumferences of the circles cut one an- other in C and F: either of the two points, C and F, is equidistant from A and B; and the distance of either of them from A, or from B, is equal to AB 310 Elementary Problems of Plane Geometry, For let CA and CB (Postulate 1.) be drawn; and, because the point A is the centre of the circle BCD, therefore (Def. 28.) AC= AB; likewise, because B is the centre of the circle ACE, BC = B.A.; but it has been shewn that AC = AB; therefore (Def. 11. Cor. 1.) AC = BC. Wherefore CA, AB, BC, are equal to one another. And if FA and FB be drawn, it may in like manner, be proved that FA, AB, BF, are equal to one another. Therefore either of the two points C and F, &c. a. E. F. ScholIUM. Since (Th. 53.) one circumference of a circle cannot cut another in more than two points, it is manifest that there are only two points, which answer to what is required in Prob. 1. PROBLEM II. From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line: it is required to draw from the point A a straight line equal to BC. From A to B draw (Postulate 1.) AB; find (Prob. 1.) a point D, the distances of which from A and B shall be equal to one another, and each of them equal to AB. Draw DA and DB, and produce them to E and F. From the centre B at the distance BC describe the circle CGH, cutting DB produced in Problems relating to Straight Lines, &c. 3]. I G; and from the centre D, at the distance DG, describe the circle KGL, cutting DA produced in L. AL shall be equal to BC. For (construction) DA = DB : likewise (con- struction and Def. 28.) DL = DG ; therefore (Def. 11. Cor. 4.) AL = BG: But (construction and Def. 28.) BC = BG; therefore (Def. 1 1. Cor. 1.) AL = BC. Wherefore from the given point A, a straight line has been drawn equal to the given straight line BC. Q. E. F. SCHOLIUM. If, in the preceding problem, the given point be in the given line, that part of the construction which directs the given point and the extremity of the given line to be joined, becomes superfluous : and if the given point be at either extremity of the given line, it will only be necessary to describe a circle from that point, as a centre, at the distance of the given line; any semidiameter of which circle will be the line 312 Elementary Problems of Plane Geometry, required to be drawn. Again, the given point may be such as that the first part of the construction shall shew it's distances from the two extremities of the given line to be equal to one another and to the given line; in which case, it is plain, that the straight line joining the given point and either extremity of the given line, is that which was required to be drawn. PROBLEM III. From the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less. From the point A draw (Prob. 2.) AD=C; and from the centre A, at the distance AD, describe (Postulate 3.) the circle DEF, and let it's circum- ference cut AB in E, AE = C. For (construction and Def. 28.) AE=AD; also (construction) AD= C; therefore (Def. 11. Cor. 1.) AE= C. Wherefore, from AB the greater, &c. a. E. F. Problems relating to straight Lines, &c. 313 ScholIUM. If, in the preceding problem, the two given straight lines have a common extremity, the only construction required is, from that point, as a centre, at the distance of the less of the two lines, to de- scribe a circle. PROBLEM IV. To bisect a given finite straight line. Let AB be the given finite straight line: it is XK sk -- j}~ required to bisect it. * Find (Prob. 1.) two points, C, D, on contrary sides of AB, each of them equidistant from the points A and B, and from C to D draw (Postulate 1.) CD. Then (Th. 20. Cor. 5.) shall AB be bisected in E, the point in which it is cut by CD. a. E. F. R. R. 314 Elementary Problems of Plane Geometry, \ - PROBLEM V. To bisect a given rectilineal angle. Let BAC be the given rectilineal angle: it is required to bisect it. * Take any point D in AB, and from the centre A at the distance AD describe a circle DFE cutting AC in E; find (Prob. 1.) a point G, which shall be equidistant from D and E; and join A, E, by the straight line AG: AG bisects the / BAC. For join D, E; and since (construction and Def.28.) AD=AE, the point A is equidistant from D and E, as is also (construction) the point G; therefore (Th. 20. Cor. 5.) AG bisects the / BAC. a. E. F. PROBLEM VI. From a given point to draw a straight line which shall be at right angles to a given indefinite straight line. First, Let the given point be in the given straight line. - Problems relating to Straight Lines, &c. 315 Let AB be the given straight line, and C the given G / C C IH point; it is required to draw from C a straight line at right angles to AB. Take any point D in CA, and from CB cut off (Prob. 3.) CE = CD; find (Prob. 1.) a point G equi- distant from D and E, and from C to G draw CG. CG is at right angles to AB. For draw GD and GE; and since (construction) GD = GE, therefore, (Th. 13.) the 4 GDC = Z. GEC; also (construction) DC = EC; therefore (Th. 20. Cor. 1.) the / DCG = / ECG ; and they are supplements each of the other, therefore each of them is a right angle. Secondly, Let the given point (C) be not in the given straight line (AB). - Take any point D upon the other side of AB; from the centre C, at the distance CD, describe a circle EDF, the circumference of which cuts AB in E and 316 Elementary Problems of Plane Geometry, F; find (Prob. 1) a point G, on the same side of AB that D is, equidistant from E and F, and draw CG cutting AB in H; wherefore (construction and Th. 20. Cor. 5.) CH cuts AB at right angles in the point H. Wherefore, from the given point, C &c. a. E. F. ScholIUM. In the preceding problem, the straight line AB is supposed to be indefinite both ways. But if it be required to draw from the eatremity of any given straight line, a straight line which shall be at right angles to it, the given straight line must first be produced. PROBLEM VII. From a given point to draw a straight line, which shall make, with a given indefinite straight line, a rectilineal angle equal to a given rectilineal angle. If the given rectilineal angle be a right angle, the problem is the same (Th. 2.) as the next preceding problem. But, first, let the given point (C) be in the given straight line (AB) and let the given angle (DEF) be not a right angle. Take any point D, in DE, and from D draw (Prob. 6.) DF perpendicular to EF; also from CB cut off (Prob. 3.) CG = EF; from G draw (Prob. 6) GH perpendicular to CG: from GH cut off GK= FD, and join K, C, by KC; then (construction and Th. 20. Cor. 1.) the / KCG = / DEF: Problems relating to Straight Lines, &c. 317 Secondly, Let the given point (C) be not in the given straight line (AB), and let the given angle (DEF) be not a right angle. KW I) A. C. G. B p-F— I) ATI-. CE. F. From C draw (Prob. 6.) CG perpendicular to AB; and from any point D, in ED, draw DF per- pendicular to EF; at the point C in GC, make (first, case) the LGCH= LFDE; wherefore (hyp. and Th. 7.) CH will meet AB; let it meet AB in H; and since (construction) the two / H.GC, GCH, of the AS CHG, are equal to the two / EFD, FDE, of the AS DEF, therefore (Th. 12. Cor. 3.) the LCHG = LDEF. Therefore, from the given point C, &c. Q. E. F. CoR. Hence, from the greater of two given angles a part may be cut off, which shall be equal to the less. ſ For at the summit of the given angle, which is the greater of the two, make (Prob. 7.) with either of it's 318 Elementary Problems of Plane Geometry, containing straight lines an angle, lying within it, which shall be equal to the less; and the angle, so made, shall be a part of the greater of the two given angles, and shall be equal to the less of them. PROBLEM VIII. To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given E A. T E D C straight line; it is required to draw a straight line through the point A, parallel to the straight line B.C. In BC take any point D, and join AD; and at the point A, in the straight line AD make (Prob. 7.) the LDAE= / ADC; and produce the straight line EA to F. * Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate // EAD, ADC, equal to one another, EF is (Th.6.) parallel to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given straight line B.C. Which was to be done. PROBLEM IX. From a given straight line to cut off any part required. - Problems relating to Straight Lines, &c. 319 Let AB be the given straight line; it is required to cut off any part from it. - From the point A draw a straight line AC making C B any angle with AB; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it; join B, C, and draw (Prob. 8.) DE parallel to it: then AE is the part required to be cut off. Because ED is parallel to one of the sides of the As ABC, viz. to BC, as CD is to DA, so is (Th. 99.) BE to EA; and, by composition (Th. 89) CA is to AD, as BA to AE: but CA is a multiple of AD; therefore (Th. 82.) BA is the same multiple of AE: whatever part therefore AD is of AC, AE is the same * If from the centre D, at the distance DA, a circle be de- scribed cutting DC in F, and if, again, from the centre F, at the distance FD another circle be described, and so on, it is evident that any assigned multipſe of AD may thus be found. 320 Elementary Problems of Plane Geometry, - part of AB : Wherefore, from the straight line AB the part required is cut off. a. E. F. % PROBLEM X. To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another, which the parts of the divided given straight line have. Let AB be the straight line given to be divided, B and CD the divided line, divided in the points E, F: it is required to divide AB similarly to CD. Join A, C and in AC produced take any point G; through A draw (Prob. 8.) AH parallel to CD; through G, D, and G, F and G, E. draw GDH, GFL, GEK, cutting AH in the points H, L, K; join H, B, and through K, and L, draw KM, and LN, each of them parallel to HB. Then (Th. 101. Cor. 4.) CE. : EF :: AK : KL ; Problems relating to Straight Lines, &c. 321- and (Th. 99. Cor. 2.) AK : KL :: AM : MN; ... (Th. 81.) CE : EF :: AM : MN. And in the same manner it may be shewn that, EF : FD :: MN : NB. Therefore the given straight line AB is divided simi- larly to CD. a. E. F. CoR. It is manifest that, by the help of the above problem, a given finite straight line may be divided into two parts, which shall have to one another a ratio equivalent to that which the one of any two given straight lines has to the other. PROBLEM XII. To find a third proportional to two given straight : lines. - - Let AB and C be the two given straight lines: it |D is required to find a third proportional to AB and C. From the point A draw a straight line AD, making any angle with AB, and from AD cut off (Prob. 3.) a part AE = C: produce AB to F and make BF-C; S S 322 Elementary Problems of Plane Geometry, from B to E draw BE, and through Fdraw (Prob.8.) FD parallel to BE, and let it meet AD in E. Then (construction and Th. 99.) AB : BF :: AE : ED; i. e. AB : C :: C : E.D. Wherefore, to the two straight lines AB and C a third proportional ED has been found. a. E. F. PROBLEM XII. | To find a fourth proportional to three given straight lines. Let A, B, C, be the three given straight lines; it is required to find a fourth proportional to A, B, C. Take two straight lines DE, DF, containing any angle EDF; and upon these make DG equal to 4, T} ET F GE equal to B, and DH equal to C; and having joined G, H, draw EF(Prob. 8.) parallel to it through the point E.: and because GH is parallel to EF, one of the sides of the triangle DEF: DG is to GE, as DH to HF (Th. 99.); but DG = A, GE= B, and Problems relating to Straight Lines, &c. 323 DH = C; ... A : B : C : HF. Wherefore to the three given straight lines, A, B, C a fourth pro- portional HF is found. a. E. F. PROBLEM XIII. To find a mean proportional between two given straight lines. Let AB and C be the two given straight lines: T A #-É G it is required to find a mean proportional between them. Produce AB to D and make (Prob. 3.) BD=c, bisect (Prob. 4.) AD in E.; from the centre E, at the distance EA, or ED, describe the semicircle AFD; from B draw (Prob. 6.) BF perpendicular to AD, and let it meet the arch of the semicircle in F. Then is BF (Th. 106. Cor. 3.) a mean proportional between AB and BD, i. e. (construction) between A and C. Therefore between the two given straight lines, &c. Q. E. F. $ CoR. Hence may be found the side of a square which shall be equal to a given rectangle: for 324 Elementary Problems of Plane Geometry, (Th. 115. Cor. 2.) the side of such a square is a mean proportional between the sides which contain the rectangle. PROBLEM XIV. To divide a given straight line into two parts, that shall be reciprocally proportional to two given straight lines. - Let A, B, be the two given straight lines, and CD the straight line given to be divided: it is required G- A. Hi–AK B– T. C F L D to divide CD into two parts, that shall be reciprocally proportional to A, B. Find (Prob. 13.) a mean proportional E between A and B; bisect CD (Prob. 4.) in F; from the centre F, at the distance FC, or FD, describe the semicircle CGD; from F draw (Prob. 6.) the semi-diameter FG perpendicular to CD; from FG" cut off (Prob. 3.) * If E be greater than FG the construction fails; neither is the problem in that case possible : for then the square of E- FG3 or CF"; i. e. (Th. 115. Cor. 2.) the rectangle contained by 4 and B. is greater than CF, which square (Th. 126. Cor. 1.) and Problems relating to Straight Lines, &c. 325 FH= E; through H draw (Prob. 8.) HK parallel to CD and let it meet the arch of the semicircle in K; also from K draw KL parallel to GF, and let it meet the diameter in L. Then CL : A :: B : LD. For (construction) the figure HLisa parallelogram; therefore (Th. 25. Cor. 1.) KL = HF; but (con- struction) HF = E.; therefore KL = E; also (con- struction and Th. 10.) KL is perpendicular to CD; ... (Th. 106. Cor. 3.) CL : LK :: LK : LD; but LK = E; ... CL : E :: E : LD; also (construction and Th. 79.) - E. : A :: B : E. ... (Th. 97.) CL : A : B : LD. Therefore (Def. 64.) CD is divided in the point L into two parts that are reciprocally proportional to A and B. a. E. F. CoR. Hence, and from Th. 115. Cor. 1. and 2, it is manifest that a given straight line may be divided into two parts that shall contain a rectangle equal to any given rectangle not greater than the square of the half of the given line. PROBLEM XV. To produce a given finite straight line, so that is greater than the rectangle contained by any two unequal parts into which CD can be divided; it is impossible, therefore, that any such rectangle can be equal to that contained by 4 and B; and therefore it is impossible (Th. I 15. Cor 1.) that CD can be divided into two parts that shall be reciprocally proportional to A and B. The problem is also impossible if E=FG, unless, at the same time A = B. 326 Elementary Problems of Plane Geometry, another given straight line shall be a mean pro- portional between the whole line thus produced and the part of it produced. - Let AB and C be the two given straight lines: it is required to produce AB, so that C may be a Gr N A. E. B H D R mean proportional between the whole line thus pro- duced and the part of it produced. Produce AB to D; bisect (Prob. 4.) AB in E; from the centre E, at the distance EA, or EB, describe the circle AFB; from B draw (Prob. 6.) BG perpendicular to AB, and make (Prob. 3.) BG=C; join E, G; from ED cut off EH=EG : then AH : C : C : BH. For draw HF, from Hto the point Fin which EG cuts the circumference; also join A, F, and F, B. And since (construction and Def. 28.) the two sides GE, EB, of the AS EBG, are equal to the two sides HE, EF, of the AS EFH, each to each, and that the A GEH is common to the two triangles, therefore (Th. 20. Cor. 1.) FH = BG, and the / EFH = /. EBG; but (construction) the / EBG is a right angle; ºtherefore the Z. EFH is also a right angle; Problems relating to Straight Lines, &c. 327 therefore (Def. 34. Cor. 2.) FH touches the circle at H; and FB cuts it; therefore (Th. 58.) the L HFB = / FAH; and the / FHA is common to the two DS FBH, AFH: therefore (Th. 12. Cor. 3.) the two /> FBII, AFH are equiangular; ... (Th. 101.) AH : HF :: HF: HB. But it has been shewn that HF = BG; and (con- struction) BG=C; ... (Th.87.) AH : C : C : HB. Therefore, the straight line AB has been produced, &c. a. E. D. º CoR. 1. From this and from Th. 115. Cor. 2, it is manifest, how a given straight line may be produced so that the rectangle contained by the whole line so produced and the part of it produced, shall be equal to the square of the given line, or to any other given square. CoR. 2. If the straight line C be equal to AB, it is manifest that the straight line AH is divided in extreme and mean ratio in the point B. PROBLEM XVI. To cut a given straight line in eartreme and mean ratio. Let AB be the given straight line: it is required to cut it in extreme and mean ratio; that is, to Á f B C divide it into two parts the one of which shall be a 328 Elementary Problems of Plane Geometry, * mean proportional between the whole and the re- maining part. . . . . Produce (Prob. 15.) AB to C, so that AC : AB :: AB = BC, and from AB cut off BD = BC; then AB : BD :: BD : AD. For (construction) AC : AB :: AB : BC; ... (construction and Th. 88.) * BD : AB :: AD : BD; ... (Th. 79.) AB : BD :: BD : AD. Therefore the straight line AB is cut in extreme and mean ratio in D. a. E. F. CoR. 1. It is manifest from the demonstration, that if a straight line be cut in extreme and mean ratio, and if from the greater of its two segments a part be cut off equal to the less, the greater segment will thus be cut in extreme and mean ratio. . For, by the construction, AC is cut in extreme and mean ratio in B; and BD, which is equal to BC the less, having been cut off from AB the greater segment, it was shewn that AB is thus divided in extreme and mean ratio. * i 4. CoR. 2. When a straight line is cut in extreme and mean ratio, it is divided into two parts, so that (Th. 115. Cor. 2.) the rectangle contained by the whole line and one of the parts is equal to the square of the other part. SCHOLIUM. From the last corollary and from the principles established in the scholium to Def. 48, it is evident, Problems relating to Straight Lines, &c. 329 that if a given straight line be cut in extreme and mean ratio, the line itself, and the parts into which it is thus divided, compared with one another, are incommensurable magnitudes. & t T T THE 3Elements of plant &tometry, Book II, —O— CHAP. II. Problems relating to Triangles, and other Plane Rectilineal Figures. —sº- PROBLEM XVII. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third. Let A, B, C, be the three given straight lines, of IK J) F G- H. E. A. T. . tº B— C; which any two whatever are greater than the third, Problems relating to Triangles, &c 331 viz. A and B greater than C; A and C greater than B; and B and C than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each. Take a straight line D.E terminated at the point D, but unlimited towards E, and make (Prob. 3.) DF= A, FG = B, and GH = C; and from the centre F, at the distance FD, describe (Post. 3.) the circle DKL ; and from the centre G, at the distance GH, describe another circle HLK; and join KF, KG; the AS KFG has its sides equal to the three straight lines, A, B, C. Because the point F is the centre of the circle DKL, FD = FK; but (construction) FD = A; therefore FK = A; again, because G is the centre of the circle LKH, G H = GK; but GH = C; therefore also GK = C; and FG = B; therefore the three straight lines KF, FG, GK, are equal to the three A, B, C: and therefore the /S KFG has its three sides KF, FG, GK equal to the three given straight lines, A, B, C. a. E. F. *' CoR. From this and from Th. 24, it is manifest how to make a triangle which shall be equal to any given triangle; viz. by making a triangle which shall have its three sides equal to the three sides of the given triangle each to each. ^ ScholIUM. If, in the preceding problem, the three given straight lines be equal to one another, it will only be 332 Elementary Problems of Plane Geometry, necessary, as in Prob. 1, to describe a circle from each of the extremities of any of them, at the distance of that straight line, and then to join those two ex- tremities, and, either of the points in which the two circumferences cut one another. PROBLEM XVIII. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallel- A. If G. / D B T. C ogram that shall be equal to the given & ABC, and have one of its angles equal to D. Bisect (Prob. 4.) BC in E, join AE, and at the point E in the straight line EC make (Prob. 7.) the angle CEF equal to D; and through A draw (Prob. 8.) AG parallel to EC, and through C draw CG parallel to EF: therefore FECG is a parallel- ogram : and because BE = EC, the As ABE = As AEC, since they are upon equal bases BE, EC, and between the same parallels. BC, AG; therefore Problems relating to Triangles, &c. 333 the AS, ABC is double of the AS AEC. And the D FECG is likewise double (Th. 32.) of the AS AEC, because it is upon the same base, and between the same parallels: therefore the EI FECG = /> ABC, and it has one of its / CEF, equal to the given A D: wherefore there has been described a D-I FECG equal to a given /> ABC, having one of its / CEF, equal to the given / D. G. E. F. PROBLEM XIX. ! To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It TN P th F|\—* K. | C * M Gr º º FI A. L D L-1 is required to apply to the straight line AB a parallel- ogram equal to the AS C, and having an angle equal to D. 334 Elementary Problems of Plane Geometry, On AB produced describe (Prob. 17. Cor.) the As PBN equal to the As C; make (Prob. 18.) the D BEFG equal to the AS PBN, and, therefore, equal, also, to the As C, and having the L EBG = the A. D.; produce FG to H; and through 4 draw (Prob. 8.) AH parallel to BG or EF; and join H, B. * Then because the straight line HF falls upon th parallels AH, EF, the / AHF, HFE, are together equal (Prob. 10.) to two right angles; wherefore the // BHF, HFE, are less than two right angles: but straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet (Th. 10. Cor. 1.) if produced far enough; therefore HB, FE, shall meet if produced; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB, to the points L., M.: then HLKF is a parallelogram of which the diameter is HK, and AG, ME, are the parallelograms about HK; and LB, BF, are the complements; therefore LB is equal (Th. 33.) to BF: but BF = /> C; therefore LB- /> C; and because (Th. 1. Cor.) the / GBE = / ABM, and is likewise equal to the / D; the / ABM is equal to the LD: therefore the El LB is applied to the straight line AB, is equal to the AS C, and has the / ABM equal to the / D: Q. E. F. PROBLEM XX. To describe a parallelogram equal to a given Problems rélating to Triangles, &c. 335 rectilineal figure, and having an angle equal to a given rectilineal angle. º Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe A. T) T G. L. \\ E. | E C I. H. M. a parallelogram equal to ABCD, and having an angle equal to E. Draw DB, and describe (Prob. 18.) the D-1 FH = ~ ADB, and having the L HKF= / E; produce KH to M, and to GH apply (Prob. 19.) the El GM = As DBC, having the angle at H equal to the / E, which, since (construction and Th. 10.) the 4 GHM is equal to the / FKH and also to the / E, will coincide with the / GHM; also (con- struction and Th. 8. Cor. 2.) GL is in the same straight line with FL, and (Th. 9.) LM is parallel to FK; therefore KFLM is a parallelogram; and because the As ABD = D FH, and the AS, DBC = D.HL, the whole figure ABCD = E, FKML; therefore the El FKML has been described equal to the rectilineal figure ABCD, and having the / FKM = /. E. a. E. F. a CoR. From this it is manifest how to a given straight line to apply a parallelogram, which shall 336 Elementary Problems of Plane Geometry, have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by first applying (Prob. 19.) to the given straight line apa- rallelogram equal to the first As ABD, and having an angle equal to the given angle. ScholIUM. If two unequal rectilineal figures be given, it is evident, that their difference may be exhibited by the help of the preceding proposition and its corollary. For, first, let a parallelogram be described equal to the one of the given figures, and then let there be applied to one of its sides a parallelogram, equal to the other given figure, and having an angle equal to and coinciding with an angle of the first described parallelogram; it is manifest that the one of these parallelograms will be a part of the other; and that the remaining part will be equal to the difference of the two given rectilineal figures. PROBLEM XXI. To describe a square upon a given straight line. Let AB be the given straight line. It is required to describe a square upon A.B. t From the point A draw (Prob. 6.) AC at right angles to AB; and make (Prob. 3.) AD = AB, and through the point D draw DE parallel (Prob. 8.) to AB, and through B draw BE parallel to AD; there- fore ADEB is a parallelogram: whence (Th, 25. Cor. 1.) Problems relating to Triangles, &c. 337 \ AB = DE, and AD= BE; but BA = AD; therefore C p TE A B the four straight lines BA, AD, DE, EB, are equal to one another, and the D A DEB is equilateral; likewise all its angles are right angles; because, the straight line AD meeting the parallels AB, DE, the // BAD, ADE, are equal (Th. 10.) to two right angles; but BAD is a right angle; therefore also ADE is a right angle; but the opposite angles of parallelograms are (Th. 25. Cor. 1.) equal; therefore each of the opposite/ ABE, BED, is a right angle; wherefore the figure ADEB is rectangular; and it has been demonstrated that it is equilateral; it is there- fore a square, and it is described upon the given straight line AB. a. E. F. PROBLEM XXII. bpon a given straight line to describe a rec- tilineal figure similar to, and situated similarly to, a given rectilineal figure. Let ABCDE be the given rectilineal figure, and U U. 338 Elementary Problems of Plane Geometry, FG the given straight line: it is required upon the A ; * G I) H straight line FG to describe a rectilineal figure similar to, and situated similarly to, ABCDE. From any one of the angles, as A, of ABCDE, let AC, and AD be drawn to the opposite angles of the figure; at the point F, in FG, make (Prob. 7.) the / GFH = / BAC, the / HFK= / CAD, the / KFL = / DAE; and so on, if there be more angles; so that the whole / GFL may be equal to the whole / BAE; also, make (Prob. 12. and 3.) FH equal to a fourth proportional to BA, AC, GF, and FK to a fourth proportional to CA, AD, HF, and FL a fourth proportional DA, AE, KF; and let GH, HK and KL be drawn. Then (construction and Th. 103.) the AS, ABC is equiangular to the As FGH, the /> ACD to the /> FHK, the /> ADE to the AS FKL; therefore the whole figure FGHKL is equiangular to the whole figure ABCDE. And since the triangles, into which the two figures ABCDE, FGHKL, are divided, have been shewn to be equiangular each to each, ... (Th. 101.) CB : BA :: HG : GF, \ Problems relating to Triangles, &c. 339 and AE : ED :: FL. : LK. Again, BA : AC :: GF : FH, AC : AD :: FH : FK, and, AD : AE :: FK : FL ... (ew aequo) BA : AE :: GF : FL. And, in the same manner, it may be shewn that *. ED : DC :: LK : KH, - and that DC : . CB :: KH : HG. Wherefore, because the figures ABCDE, FGHKL are equiangular and have their sides about the equal angles proportionals, they are similar; and upon the given straight line FG a figure has been described &c. G. E. F. - PROBLEM XXIII. To describe a rectilineal figure which shall be similar to one, and equal to another given rectilineal jigure. Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, 340 Elementary Problems of Plane Geometry, and D that to which it must be equal. It is required to describe a rectilineal figure similar to ABC, and equal to D. * * Upon the straight line, BC describe (Prob. 20. Cor.) the E BE equal to the figure ABC; also upon CE describe the DCM equal to D, and having the Z FCE = the / CBL ; therefore BC and CF are in a straight line (Th. 10. and Th. 4.), as also LE and EM: between BC and CF find (Prob. 13.) a mean proportional GH, and upon GH describe (Prob. 22.) the rectilineal figure KGH similar and similarly situated to the figure ABC: and because . BC : GH :: GH : CF, and if three straight lines be proportionals, as the first is to the third, so is (Th. 117. Cor. 5.) the figure upon the first to the similar and similarly described figure upon the second; therefore as BC to CF, so is the rectilineal figure ABC to KGH: but (Th. 114.) BC : CF :: L BE : DEF: ... (Th. 81.) fig, ABC : fig. KGH :: E BE : DEF; and the figure ABC = E BE; ... (Th. 85.) the figure KGH = E. E.F. But El EF is equal to the figure D; therefore KGH = D; and it is similar to ABC. Therefore the rectilineal figure KGH has been described similar to the figure ABC and equal to D. a. E. F. CoR. From this it is manifest how to describe a square that shall be equal to a given rectilineal figure; which may, also, be more easily done by the following construction. Let ABC be any rectilineal figure, and let it be Problems relating to Triangles, &c. 341 required to describe a square that shall be equal to ABC. - To any side BC, of the figure ABC, apply (Prob. 20. Cor.) the rectangle BE equal to the figure ABC; produce LE to N, and make EN= EC; upon LN as a diameter describe the circle LPN, and produce CE until it meets the circumference in P: the square described (Prob. 21.) upon EP shall be equal to the figure ABC. For (Th. 106. Cor. 3.) EP is a mean proportional between LE and EN or EC; therefore (Th. 115. Cor. 2.) the rectangle BE is equal to the square of EP; but (construction) the rectangle BE = ABC; therefore the square of EP is equal to the figure ABC. DEF. LXVI. If two parallelograms be described, having a common angle and lying between the same parallels; the one on a given straight line, the other on that line produced, that which stands upon the produced line is said to be applied to the given line exceeding by the parallelogram which is the difference of the two so described, and which is called the eaccess: and if two parallelograms be described having a common angle, and lying between the same parallels, the one on a given straight line, the other on a part of it, that which stands upon the part, is said to be ap- plied to the given line deficient by the parallelogram, which is the difference of the two so described, and which is called the defect. 342 Elementary Problems of Plane Geometry, SCHOLIUM. If to a given straight line AB it were required to apply a parallelogram having its defect similar and DN, I Tºy L S G Histºff— | ſ T) IE .A. Tº H B similarly situated to the given DJ CE; upon AB describe (Prob. 22.) the El ABFG similar and similarly situated to CE, and draw its diameter GB: then, if through any point P, in GB, there be drawn (Prob. 8.) QPS parallel to AB, and PR parallel to GA, the D QR will be applied to AB deficient by the CI RS, which (Th. 111.) is similar to CE : and it follows from Th. 111, that no parallelograms can be applied to AB deficient by parallelograms similar to CE, but such as have one of their angular points in the diameter of ABFG. If, then, DE, the base of CE, be equal to half of the given line AB, the EI CE shall be greater than any El QR that can be applied to AB deficient by a parallelogram similar to CE. For bisect (Prob. 4.) AB in H; through H draw (Prob. 8.) HKL parallel to AG, meeting GB in K and QPS in L; and through K draw MKN parallel to AB. And since (construction) AH= HB, therefore G. Q Problems relating to Triangles, &c. 343 (Th. 25. Cor. 1.) QL = LS; therefore (Th. 29.) the El ML = E KS; but (Th. 33.) the E. KS = […] KR ; therefore the D-1 KR = [...] ML; and the DT, ML - C, MP; therefore the D-J KR - DT MP; to each of these add the common El MR, and the [] AK - D QR : but (construction and Th. 29.) the CI AK = E HN; and (Th. 117. Cor. 4.) the DJ HN = [] CE; therefore the D-1 CE > D H QR. And in the same manner may the proposition be proved, if the point P be taken between K and B. Wherefore, of all parellelograms applied to the same straight line deficient by parallelograms, similar and similarly situated to a parallelogram described upon the half of the line, that which is applied to the half, and which is similar to its defect, is the greatest. Thus, an infinite number of parallelograms may be applied to a given line each of them deficient by a parallelogram similar and similarly situated to a given parallelogram; but the greatest of them is that which stands on the half of the given line: and, henée, there is a manifest limitation to the next following problem. PROBLEM XXIV. To a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by, or eaceeding by, a parallelogram similar to a given parallelogram : but in the former case the given rectilineal figure must not be greater than a parallel- ogram described on the half of the given line, similar to the given parallelogram. 344 Elementary Problems of Plane Geometry, Let AB be the given straight line, C a given rectilineal figure, and D a given parallelogram : it w" E IP N\O G F T. B B *L*-*. IC L P O N is required to apply to AB a parallelogram equal to C, and deficient, in the first case, or exceeding, in the second case, by a parallelogram similar to D; but in the first case, the figure C must not be greater than a parallelogram described on the half of AB similar to D. Bisect (Prob. 4.) AB in E, and upon EB describe (Prob. 22.) the CJ EF similar and similarly situated to D, and draw its diameter GB; describe also (Prob. 23.) the E. KM similar to D, or to EF, and equal to C; and let KL be the side of it which is homo- logous to EB. And, in the first case, since (hyp.) C is not - EF, neither is KM > EF; therefore (Th. 117. Cor. 4.) KL is not - EB, the half of AB : divide, therefore, in the first case, AB in H (Prob. 14.) Problems relating to Triangles, &c. 345 so that AH : KL :: KL : HB; and, in the second case, produce (Prob. 15.) AB to H, so that AH : KL :: KL : HB; through H draw (Prob. 8.) HN parallel to BF, meeting (Th. 8. Cor. 1.) GB, or in the second case GB produced, in N; and through N draw NP parallel to AB, meeting AP, drawn through A parallel to BF, in P, and FB in O. The D HP = C, and it is applied to AB, de- ficient in the first case, and exceeding in the second case, by the D HO which is similar to D. For (Th. 114.) t - HB : AH x HB :: HB : HA, & and El HO : [-] HP :: HB : HA; * ... (Th. 81.) - Li HO : El HP :: HB : AH x HB. But (Th. 115. Cor. 2. and construction,) AH X HB - KL”; ... C. HO : Li HP :: HB : KL*. Again, because (Th. 111.) the E1 EF is similar to the D-I HO, ... (Th. 117. Cor. 3.) E. EF: D HO : EB : HB3; and it was shewn that El HO : Li HP :: HB3 : KL*, ... (Th.97.) E, EP: El HP :: EB : KL’; But (construction and Th. 117. Cor. 3.) * * L EF: D KM: EB : KL’; therefore (Th. 81. and 78.) D HP = D KM, and (construction) El KM = C; therefore D HP = C; and it is applied to AB, deficient, in the first case, X X * 346 Elementary Problems of Plane Geometry, and exceeding, in the second case, by the E HO, which (Th. 111.) is similar to the E! EF, and there- fore (construction and Def. 63. Cor.) similar also to the D. D. a. E. F. PROBLEM XXV. Upon a given finite straight line to describe an equilateral and equiangular pentagon. Let AB be the given finite straight line: it is D required to describe an equilateral and equiangular pentagon on AB. Produce (Prob. 15.) AB to C, so that AC : AB :: AB : BC, and in BA produced make also AD = BC, and therefore BD = AC; from the centres A and B, at the equal distances AC, BD, describe the circles EF, EG, cutting one another in E; join E, A and E, B; and from EA and EB cut off (Prob. 3.) EH, and EK each equal to AB; also join A, K and A, H, and let AK and AH, produced, meet the circumferences of the circles EF and EG in the points F and G.; join E, F and F, B, and A, Problems relating to Triangles, &c. 347 G, and G, E: the figure AGEFB is an equilateral and equiangular pentagon described on AB. For since (construction) AC : AB :: AB : BC; and that AC = EB, and EK = AB, therefore KB = BC, and EB : BA :: BA : BK; therefore since the /> EBA, KBA, have their sides about the common L EBA proportionals, they are (Th. 103.) equiangular; therefore the / AKB = L EBA, and Z. BAK = L BEA; therefore (Th. 15.) AK– AB; but (construction) EK = AB; therefore EK = KA and therefore (Th. 13.) the / KEA = L KAE; but it was shewn that the Z BAK = / BEA or KEA ; therefore each of the // BAF, FAE is equal to the A BEA, and they are equal to one another: in the same manner it may be shewn that each of the // ABG, GBE is equal to the / BEA and that they are equal to one another; therefore (construction and Th. 20. and Cor. 1.) the three As AEB, E.A.F. EBG are equal to one another, and have their bases AB, EF and EG equal, and their other angles equal each to each ; and since the /S EGB = As AEB, therefore (Th. 31.) GA is parallel to EB and therefore (Th. 10.) the / AGB = A GBE; and it has been shewn that the Z GBE = / GBA; therefore the L AGB = / GBA, and (Th. 15.) AG = AB: in the same manner it may be shewn that BF = AB ; and it was shewn that AB = EF = EG ; wherefore the figure AGEFB is equilateral : it is, also, equiangular : for the sides BF, FA of the /s ABF, are equal to the sides AG, GB, of the /S BAG, each to each, and AB is common to the two triangles, therefore \ 348 Elementary Problems of Plane Geometry, (Th. 24. Cor. 1.) the LABF= L BAG: in the same manner, it may be shewn that the LABF= / BFE, and the A BAG = z_AGE; also, since FE = FB, the / FEB= / FBE, and it was shewn the / BEG, of the AS E.BG, is equal to the / EBA, of the /S EBA ; therefore the whole / FEG = / FBA; wherefore, the five / ABF, BFE, FEG, EGA, GAB, are equal to one another, and upon AB there has been described the equilateral and equiangular pentagon ABFEG, a. E. F. CoR. From this it is manifest how to describe an isosceles triangle (EBA) having each of the angles at the base double of the third angle. PROBLEM XXVI. Upon a given finite straight line to describe an equilateral and equiangular hewagon. Let AB be the given straight line: it is required £—D / N/\ A. R to describe an equilateral and equiangular hexagon on AB, Problems relating to Triangles, &c. 349 Upon AB describe (Prob. 17.) the equilateral triangle ACB; produce AC and BC to D and E, and make CD and CE (Prob. 3.) each equal to CA or CB; also, bisect (Prob. 5.) the / ACE by CF, and produce FC to G, so that (Th. 1. Cor.) CG also bisects the / DCB; make CF and CG each equal to CA or CB; and draw AF, FE, ED, DG, GB: then is ABGDEF an equilateral and equiangular hexagon. For (Th. 12.) the exterior / ECA, of the AS ACB, is equal to the two interior / CBA, BAC, which angles (construction and Def. 19. Cor.) are equal to one another and to the / ACB; therefore the A ACB is equal to the half of the / ECA, that is, (construction) to each of the / ECF, FCA; therefore (Th. 1. Cor.) the six angles at C are equal to one another; and the sides of the Os ACB, BCG, GCD, DCE, ECF, FCA, about the equal angles at C, are (construction) equal to one another; therefore (Th. 20. Cor. I.) their bases, AB, BG, GD, DE, E.F, FA, are equal to one another, and the angles at the bases are also equal; wherefore the figure ABGDEF, is both equilateral and equiangular; and it is described on A. Wherefore on AB there has been described, &c. G. E. F. THE, £itments of plane &rometry. * Book II. —O— CHAP. III. Problems relating to circles, their arches, chords, p tangents, and segments. —sº- PROBLEM XXVII. An arch, or the whole circumference, of a circle being given, to find the centre. LET BCD be an arch, or the whole circumference, of a circle: it is required to find the centre of the circle, of which BCD is an arch or the circumference. Problems relating to Circles, Arches, &c. 351 2-N In BCD take any three points B, C, D; join B, C and C, D; bisect (Prob. 4.) BC in E, and CD in F; from E and Fdraw (Prob. 6) EA perpendicular to BC and FA perpendicular to CD, and let them "meet in A: A is the centre of the circle, which was to be found. For (Th. 37, and construction), the centre of the circle is in EA, and it is, also, in FA; therefore it is in a point common to EA and FA ; but (Def. 8. Cor. 1.) the two straight lines EA and FA cannot have more than one point common to them both, namely the point A, in which they meet; therefore the point A is the centre of the circle of which ABC is an arch or the circumference, and a point A has been found which, &c. a. E. F. CoR. 1. From this it is manifest, if a segment of a circle be given, how to describe the circle of which it is the segment; viz. by finding the centre, and at the distance of any point in the arch, that bounds the segment, describing the circle. CoR. 2. It is, also, manifest, that by the help of this problem a circle may be described, the cir- cumference of which shall pass through any three given points, that are not in the same straight line. * If E, F be supposed to be joined by a straight line, it is manifest that AE, AF will make with that straight line two angles, on the same side of it, each of which is less than a right angle; and therefore (Th. 10. Cor. 1.) E4 and FA will meet if they be produced far enough. 352 Elementary Problems of Plane Geometry, Af SCHOLIUM. If the whole circumference of a circle be given, its centre may more readily be found, by bisecting the chord which joins any two assumed 'points in the circumference and drawing through the point of bisection another chord at right angles to it. For (Th. 37.) this latter chord will be a diameter, and therefore (Def. 30. Cor. 1.) its bisection will be the centre of the circle. PROBLEM XXVIII. To bisect a given arch of a circle. Let ADB be the given arch; it is required to º, bisect it. -- Join A, B, and bisect (Prob. 4.) AB in C; from the point C draw (Prob.6.) CD at right angles to AB: the arch ADB is bisected in the point D. Join A, D and D, B, and because (construction) CD bisects AB at right angles, therefore (Th. 20. Cor. 5.) DA=DB; but (Th. 61.) equal chords cut off equal arches, the greater equal to the greater, 2-N 2-\ . and the less to the less; and AD, DB are each of Problems relating to Circles, Arches, &c. 363 them less than a semicircle, because DC passes (Prob. 27.) through the centre. Wherefore the arch AD is equal to the arch DB : therefore the given arch is bisected in D. a. E. F. ~ *. DEF. LXVII. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. * PROBLEM XXIX. In a given circle to place a straight line, equal to a given straight line not greater than the dia- meter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. It is required to place in ABC a straight line equal to D. * Draw (Prob. 27.) BC the diameter of the circle ABC; then, if BC is equal to D, the thing required Y Y 354 Elementary Problems of Plane Geometry, is done; for in the circle ABC a straight line BC is placed equal to D; but, if it is not, BC is greater than D; make (Prob. 3.) CE = D, and from the centre C, at the distance CE, describe the circle AEF, and join C, A : therefore, because C is the centre of the circle AEF, CA = CE; but D = CE; therefore D = CA: wherefore, in the circle ABC, a straight line is placed equal to the given straight line D, which is not greater than the diameter of the circle. a. E. F. ScholIUM. The last problem might be otherwise enunciated, as follows. “From a given point in the circumference of a given circle to draw a chord, which shall be equal to a given straight line not greater than the diameter of the circle.” The necessity for introducing the limitation, which appears in the enunciation of the problem, is evident from Th. 44. PROBLEM XXX. To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle. First, let A be a given point without the given circle BCD; it is required to draw a straight line from A which shall touch the circle. Find (Prob. 27.) the centre E of the circle, and Join A, E; and from the centre E, at the distance Problems relating to Circles, Arches, &c. 355 EA, describe the circle AFG ; from the point D draw Xº, (Prob.6.) DFat right angles to EA, and draw EBF, AB. AB touches the circle BCD. Because E is the centre of the circles BCD, AFG, EA = EF, and ED=EB; therefore the two sides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two /> AEB, FED; therefore (Th. 20. Cor. 1.) DF= AB, and the / EBA = / EDF: but EDF is a right angle; wherefore EBA is a right angle: and EB is drawn from the centre : but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle (Def. 34. Cor. 2.): therefore AB touches the circle; and it is drawn from the given point A. But, if the given point be in the circumference of the circle, as the point D, draw DE to the centre E and DF at right angles to DE; DF touches the circle. a. E. F. 356 Elementary Problems of Plane Geometry, PROBLEM XXXI. To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment 2% E- B R from the circle ABC that shall contain an angle equal to the angle D. Draw (Prob. 30.) the straight line EF touching the circle ABC in the point B, and at the point B, in the straight line BF make (Prob. 7.) the angle FBC= LD; therefore, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the L FBC is equal (Th. 58.) to the angle in the alternate segment BAC of the circle: but the / FBC= / D; therefore the angle in the segment BAC is equal to the / D: wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given a D. a. E. F. Problems relating to Circles, Arches, &c. 357 PROBLEM XXXII. Upon a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle. , 2. Let AB be the given straight line and CDE the given rectilineal angle: it is required to describe, on Gr H AY IF B AB, a segment of a circle containing an angle equal to the / CD.E. In DC take any point C, make DE = DC, and join C, E; bisect (Prob. 4.) AB in F; from F draw (Prob. 6.) FG perpendicular to AB; at the point A, in AB, make (Prob. 7.) the / BAH equal to the / DCE, or the / DEC; and since (construction and Th. 13. Cor. 2.) the / DCE, DEC, are each of them less than a right angle; therefore the / BAH is less than a right angle, and (Th. 10. Cor. 1.) AH will meet FG; let it meet FG in H; and describe (Prob. 27. Cor. 2.) the circle ABH, the circumference of which 358 Elementary Problems of Plane Geometry, &c. passes through the three points A, B, H: the segment AHB shall contain an angle equal to the / CD.E. For join H, B; then (construction and Th. 20. Cor. 4.) HB = HA; therefore (Th. 13.) the / HBA = / HAB; and the / HAB was made equal to one of the equal / DCE, DEC, of the As CDE ; there- fore (Th. 12. Cor. 3.) the third / AHB, of the As AHB, is equal to the / CDE: wherefore, upon the given straight line AB, the segment AHB of a circle has been described, which contains an angle equal to the given / CD.E. a. E. F. THE 33ſtment; of ºlaut Geometry. Book II. CHAPTER IV. Problems relating to the method of inscribing a rectilineal figure in a circle, and a circle in a rectilineal figure, and of describing a rectilineal Jigure about a circle, and a circle about a rectilineal Jigure. —O— PROBLEM XXXIII. In a given circle to inscribe a triangle equiangular to a given triangle. LET ABC be the given circle, and DEF the given triangle; it is required to inscribe in the circle ABC a triangle equiangular to the AS DEF. Draw (Prob. 30.) the straight line GAH touching the circle in the point A, and at the point A, in the straight line AH, make (Prob. 7.) the L HAC = 360 Elementary Problems of Plane Geometry, / DEF; and at the point A, in the straight line AG, make the / GAB = A DFE, and join B, C: therefore, because HAG touches the circle ABC, and AC is drawn from the point of contact, the A. HAC is equal (Th. 58.) to the LABC in the alternate segment of the circle: but the LHAC= A DEF; therefore also the / ABC= / DEF; for the same reason, the / ACB= / DFE ; therefore the remaining / BAC is equal (Th. 12. Cor. 3.) to the remaining / EDF: wherefore the AS, ABC is equi- angular to the AS DEF, and it is inscribed in the circle ABC. a. E. F. PROBLEM XXXIV. About a given circle to describe a triangle equi- angular to a given triangle. Let ABC be the given circle, and DEF the given triangle: it is required to describe a triangle about the circle ABC, equiangular to the AS DEF: In the given circle inscribe (Prob. 33.) the Inscribing a Rectilineal Figure in a Circle, &c. 361 * 2 & ABC equiangular to the AS DEF; find the l, - “iº, centre G of the circle, and from G draw (Prob. 6.) the semi-diameters GH, GI, GK, perpendicular to AB, BC, CA, respectively; also through H, I, K draw (Prob. 30.) the straight lines LM, MN, NL, touching the circle ABC, in the points H, I, K, and since GH is (Th. 46. and construction) perpendicular to each of the two straight lines AB and LM, they are (Th. 6.) parallel to one another; in the same manner it may be shewn that MN is parallel to BC, and NL to CA; therefore (Th. 10. Cor. 6.) the / LMN = / ABC, the / MTVL = / BCA and the 1. NLM = / CAB ; therefore (construction) the AS LMW is equiangular to the AS DEF; and it is described about the circle ABC. a. E. F. DEF. LXVIII. A circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure. Z Z { ** Ç 362 Elementary Problems of Plane Geometry, PROBLEM XXXV. To inscribe a circle in a given triangle. Let the given triangle be ABC: it is required to inscribe a circle in ABC. - Bisect (Prob. 5.) the / ABC, BCA, by the straight lines BD, CD, meeting one another in the point D, from which draw (Prob. 6.) DE, DF, DG per- pendiculars to AB, BC, CA; and because the / EBD = / FBD, for the LABC is bisected by BD, and that the right angle BED is equal to the right angle BFD, the two /> EBD, FBD, have two angles of the one equal to two angles of the other, and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore (Th. 19. Cor. 1.) their other sides shall be equal; therefore DE = DF: for the same reason, DG = DF; there- fore the three straight lines DE, DF, DG are equal to one another, and the circle described from the centre D, at the distance of any of them, shall pass through the extremities of the other two, and touch Inscribing a Rectilineal Figure in a Circle, &c. 363 the straight lines AB, BC, CA, because the angles at the points E, F, G are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches (Def. 34. Cor. 2.) the circle: therefore the straight lines AB, BC, CA do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC. a. E. F. PROBLEM XXXVI. To describe a circle about a given triangle. Describe (Prob. 27. Cor. 2.) a circle, the circum- ference of which shall pass through the three an- gular points of the given triangle, and (Def 65.) it will be described about the triangle. a. E. F. PROBLEM XXXVII. e To inscribe a square in a given circle. Let ABCD be the given circle; it is required to A. GBC, GCD, and that (construction) the / GCB = A GCD, therefore (Th. 20. Cor. 1.) the / GBC= / GDC, and GB = GD ; but it has been shewn that GD = GC; therefore GB = GC: and, since the / CDE is double of the A CDG, and that the / CD.E= / CBA, and, as hath been shewn, the / GDC = / GBC, therefore the A CBA is double of the / GBC; that is GB bisects the / CBA. It may, therefore, in the same manner be shewn, that GA = GB, and that GA bisects the A BAE; and so on. Wherefore, the straight lines, drawn from G to the angular points of the figure, are all equal to one another; and therefore a circle described from the centre G, at the distance GC, or GD, will be described about the figure. a. E. F. PROBLEM XL. In any given equilateral and equiangular recti- lineal figure, having more than three sides, to inscribe a circle. - Inscribing a Rectilineal Figure in a Circle, &c. 367 Let ABCDE be an equilateral and equiangular *> rectilineal figure, of more than three sides: it is required to inscribe a circle in ABCDE. º Bisect (Prob. 5.) the / BAE, AED, by the straight lines AF, EF; and from the point F, in which they meet, draw (Prob. 6.) FG perpendicular to A.E. Describe a circle from the centre F, at the distance FG, and it will be inscribed in the figure ABCDE. For the sides of the figure are (Prob. 39) the chords of a circle described from the centre F at the distance FA ; and (hyp.) they are equal to one another; therefore (Th. 43.) they are equally distant from the centre; that is, the perpendiculars drawn to them from the centre are all equal; therefore a circle described from the centre F, at the distance FG will (Def. 34. Cor. 2.) touch each of the sides of the figure ABCDE, and will (Def. 68.) be inscribed in it. CoR. It is manifest from Prob. 39. and Prob. 40, that the centre of the circle, which is described about an equilateral and equiangular rectilineal figure is also the centre of the circle to be inscribed in it. 368 Elementary. Problems of Plane Geometry, PROBLEM XLI. To inscribe an equilateral and equiangular pen- tagon in a given circle. Let ABCDE be the given circle; it is required to T inscribe an equilateral and equiangular pentagon in the circle ABCDE. * - Describe (Prob. 25. Cor.) an isosceles & FGH, having each of the angles at G, H, double of the angle at F; and in the circle ABCDE inscribe (Prob. 33.) the /S ACD equiangular to the AS FGH, so that the LCAD be equal to the angle at F, and each of the / ACD, CDA, equal to the angle at G or H; wherefore each of the angles ACD, CDA, is double of the / CAD. Bisect (Prob. 5.) the / ACD, CDA, by the straight lines CE, DB; and join AB, BC, DE, EA, ABCDE is the pentagon required. , Because each of the / ACD, CDA, is double of CAD, and is bisected by the straight lines CE, DB, the five / DAC, ACE, ECD, CDB, BDA, are equal to one another; but equal angles stand upon equal (Th. 59.) arches; therefore the five arches AB, BC, Inscribing a Rectilineal Figure in a Circle, &c. 369 CD, DE, EA, are equal to one another: and equal arches are (Th. 62.) subtended by equal straight lines; therefore the five straight lines AB, BC, CD, DE, EA, are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also (Th. 66.) equiangular: wherefore, in the given circle, an equi- lateral and equiangular pentagon has been inscribed. Q., E. F. CoR. From this, and from Th. 66, it is manifest how to describe an equilateral and equiangular pentagon about a given circle; namely, by inscribing in the circle such a pentagon and then, through its angular points, drawing (Prob. 30.) straight lines which shall touch the circle. PROBLEM XLII. To inscribe an equilateral and equiangular her- agon in a given circle. Let ABCDEF be the given circle: it is required |D to inscribe an equilateral and equiangular hexagon in it. - 3 A 370 Elementary Problems of Plane Geometry, Find (Prob. 27.) the centre G of the circle ABCDE, and in it place (Prob. 29.) a straight line AB equal to its semi-diameter, draw AG, BG, and produce them to meet the circumference in D and E; also bisect (Prob. 5.) the / BGD, by the dia- meter CF, and draw BC, CD, DE, EF, F.A. Then (Prob. 26.) ABCDEF is an equilateral and equi- angular hexagon; and it is inscribed in the circle ABCDE. G. E. F. CoR. 1. From this, and from Th. 66, it is mani- fest how to describe an equilateral and equiangular hexagon about a given circle. CoR. 2. The side of a hexagon inscribed in a circle is equal to the circle's semi-diameter. PROBLEM XLIII. To inscribe an equilateral and equiangular quin- decagon in a given circle. Let ABCD be the given circle: it is required to I. A. * inscribe an equilateral and equiangular quindecagon in the circle ABCD. Inscribing a Rectilineal Figure in a Circle, &c. 371 Let AC be the side of an equilateral triangle ACD, inscribed (Prob. 33.) in the circle, and AB the side of an equilateral and equiangular pentagon ABEFG, in- scribed (Prob.41.) in the same: then (hyp.) AB, BE, AG, and GFare equal; therefore (Th. 61.) ABB = AGF. takeaway from these equals, the arches ABCandſ 4GD, which (construction and Th. 61.) are equal to 2-S 2-\ one another, and there remains CE = FD. Again (con- c 2-S 2-> - struction and Th. 61.) ABC = CFD; take from these equals the equal arches EF and AB, and there , 2-N 2-S 2-N * remains BC = CE + FD; but it has been shewn 2-N 2-S 2-N 2-> that CE = FD; therefore BC is the double of CE; 2-S * * and if BC be bisected (Prob. 28.) in H, and BH, *mº-sº ºmºm-as 2-N 2-N 2-S HC, CE, be drawn, BH = HC = CE, and therefore (Th. 62.) BH= HC = CE. In like manner, if a straight line BK, equal to EC, be placed (Prob. 29.) 2-N 2-N in BA, and if KA be bisected in L, the arch BKA will be divided into three equal arches, which will have their chords (Th. 62.) equal to one another, and to BH, HC, CE; and so of the three remaining arches AG, GF FE. In this manner, therefore, will an equilateral quindecagon be inscribed in the circle, which (Th. 66.) is also equiangular. a. E. F. CoR. From this, and from Th. 66, it is manifest how an equilateral and equiangular quindecagon may be described about a given circle. - 372 Elementary Problems of Plane Geometry, ScholIUM. The circumference of a circle evidently admits of being divided into any number of equal parts, and, therefore (Th. 62.) it may be considered as possible to inscribe in a circle an equilateral polygon, of any number of sides, which (Th. 66.) will also be equi- angular. But no geometrical solution has yet been given of the general problem—“To divide the cir- cumference of a circle into any given number of parts.” By the help of Prob. 28, Prob. 37, Prob. 41, Prob. 42, and Prob. 43, there may be inscribed in a circle equi- lateral and equiangular polygons of 4, 8, 16 &c. sides; of 5, 10, 20 &c. sides; of 6, 12, 24 &c. sides; and of 15, 30, 60 &c. sides. It is evident, also, that if an isosceles triangle could be described, having each of the equal angles at the base any required multiple of the angle at the vertex, then might an equilateral and equiangular polygon of any odd number of sides be inscribed in a circle: and such a polygon, of an even number of sides, might be inscribed in a circle, if after dividing that number by the greatest power of 2, which will divide it, the quotient be either unity, or an odd number, such that there is a known method of inscribing an equilateral and equiangular polygon of that odd number of sides in a circle. The quotient will always be, either unity, or an odd number. If it be unity, the polygon may be inscribed by the help of Prob. 28; and if it be an odd number, such as has been mentioned, a polygon of that odd number of Inscribing a Rectilineal Figure in a Circle, &c. 373 sides must first be inscribed in the circle; and then each of the arches subtended by its sides having been bisected, and those parts again bisected, and so on, the polygon proposed to be inscribed will have for its sides the chords of the several arches, after the last bisections. In general, if a number be the product of several factors, an equilateral and equiangular polygon, of that number of sides, may be inscribed in the circle, if the circumference can first be divided into as many . equal parts as are indicated by one of the factors, and if then each of those equal parts can be divided into as many equal parts as are indicated by another of the factors, and so on. Thus, in the case of the quindecagon (Prob. 43.) the number 15 is the product of the numbers 3 and 5; the circumference is first divided into five equal arches by the sides of the inscribed pentagon; then, each of those arches is divided into three equal parts: and by this method the whole circumference is divided into fifteen equal arches, the chords of which are the sides of the quin- decagon proposed to be inscribed. Polygons, which are equilateral and equiangular, are called Regular Polygons. Now it is evident, from what was said in the beginning of this scholium, that a regular polygon may be inscribed in a given circle, having the number of its sides greater than any given number; and therefore, also, (Th. 66.) a regular polygon may be described about the circle having the number of its sides greater than any given number. But (Assump. II. Schol.) the circumference of a circle is greater than the perimeter of any * * 374 Elementary Problems of Plane Geometry, regular polygon inscribed in it, and less than that of any regular polygon described about it; and the ... greater the number of the sides of the inscribed and circumscribed polygons, the more nearly do their perimeters and the circumference of the circle ap- proximate to the relation of equality. If, therefore, a circle be given, we are in possession of a method of finding, geometrically, a polygon, such that the difference between its perimeter and the circumference . of the circle shall be indefinitely small. But it is of more importance, to be enabled to approximate to the numerical value of the circumference of a circle, when the numerical value of its diameter is given, or has been ascertained. To shew how this may be done, we shall have recourse, as the case requires it, to the usual language of calculation. We shall first lay down the general forms, which are requisite for the solution of the problem, and then point out the manner in which they may be applied; leaving it to the reader himself to make the application, by means of the common rules of arithmetic. First, then, if 2 a denote the side of any regular polygon inscribed in a given circle, p the perpen- dicular drawn to it from the centre, and 2 the dia- meter of the circle, d the side of a regular polygon, of the same number of sides described about the circle, and h the side of a polygon of double the number of sides inscribed in the circle, & (I.) d = ** = —“–1 * , * p (1 – a”)# (II) h = 2%. (1-p)} = 2* {1-(1-a')}}}. Inscribing a Rectilineal Figure in a Circle, &c. 375 The above equations are easily deduced from the common property of similar triangles, and from Th. 36. and Th. 130. Now (Prob. 42. Cor. 2.) the side of a regular hexagon, inscribed in the circle, which has 2 for its diameter, is unity. Hence, from Equation II, The side of an inscribed polygon of 12 sides = 2-ya. of 24 sides = ~/2-x/2+,73. of 48 sides g = J.E.7sº of 96 sides = —F C— IC -D not in the same plane with it; AB shall be parallel to CD. In EF take any point G, from which let there he supposed to be drawn passing through EF and AB, the straight line GH at right angles to EF; and in the plane passing through EF, CD, let GR be Mutual Intersections of Straight Lines &c. 395 supposed to be drawn at right angles to the same EF And because EF is perpendicular both to GH and GK, EF is perpendicular (Th. 139.) to the plane HGR passing through them : and EF is pa- rallel to AB; therefore AB is at right angles (Th. 141.) to the plane HGK. For the same reason, CD is likewise at right angles to the plane HGK. There- fore AB, CD are each of them at right angles to the plane HGK. But if two straight lines are at right angles to the same plane, they shall be parallel . (Th. 142.) to one another. Therefore AB is parallel to CD. Wherefore two straight lines, &c. a. E. D. Theorem CXLV. If two straight lines meeting one another be pa- rallel to two others that meet one another, and are not in the same plane with the first two; the first two and the other two shall contain equal angles. Let the two straight lines AB, BC, which meet one another be parallel to the two straight lines DE, EF, that meet one another, and are not in the same plane with AB, BC. The / ABC = / DEF Suppose BA, BC, ED, EF, to be taken all equal to one another, and AD, CF, BE, AC, DF, to be drawn : because BA is equal and parallel to ED, therefore AD is (Th. 10. Cor. 4.) both equal and parallel to BE. For the same reason, CF is equal and parallel to BE. Therefore AD and CF are each of them equal and parallel to B.E. Therefore (Th. 144.) AD is parallel to CF; and it is equal to 396 Elementary Theorems of Solid Geometry, &c. it, and AC, DF, join them towards the same parts; B •e ^. ñ F and therefore (Th. 10. Cor. 4.) AC is equal and parallel to D.F. And (Th. 24.) because AB, BC, are equal to DE, FE, each to each, and the base AC to the base DF; the Z ABC = / DEF. Therefore, if two straight lines, &c. a. E. D. THE Elements of $olity grometry. CHAPTER II. On the mutual intersections and inclinations of planes. —O— THEOREM CXLVI. Planes to which the same straight line is perpen- dicular do not meet one another, though produced. LET the straight line AB be perpendicular to each of the planes CD, EF; these planes do not meet one another though produced. If it be possible, let them meet one another when produced ; their common section (Def. 8. Cor. 2.) shall be a straight line GH, in which let any point K be taken, and AK, BK drawn: then, because AB is perpendicular to the plane EF, it is perpendicular (Def. 65.) to BK which is in that plane. Therefore ABK is a right angle. For the same reason, BAK is 398 Elementary Theorems of Solid Geometry, A a right angle; wherefore the two / ABK, BAK, of s i, N the ZS ABK, are equal to two right angles, which (Th. 12. Cor. 1.) is impossible: therefore the planes CD, E,F, though produced, do not meet one another. Therefore planes, &c. a. E. D. DEF. LXXI. Parallel Planes are such as do not meet one another, though produced. * CoR. Planes to which the same straight line is perpendicular, are (Th. 146.) parallel to one another. THEOREM CXLVII. If two straight lines meeting one another, be Mutual Intersections and Inclinations of Planes. 399 parallel to two straight lines which meet one another, but are not in the same plane with the first two ; the plane which passes through these is parallel to the plane passing through the others. Let AB, BC, two straight lines meeting one another, be parallel to DE, EF, that meet one an- D other, but are not in the same plane with AB, BC: the planes through AB, BC, and DE, EF, shall not meet, though produced. From the point B suppose BG to be drawn per- pendicular to the plane which passes through DE, EF, and let it meet that plane in G ; and through G, suppose GH to be drawn parallel to ED, and GR parallel to EF; and because BG is perpen- dicular to the plane through DE, EF, it shall (Def. 65.) make right angles with every straight line dº 400 Elementary Theorems of Solid Geometry, meeting it in that plane. But the straight lines GH, GK, in that plane meet it: therefore each of the // BGH, BGK, is a right angle: and because BA is parallel (Th. 144.) to GH (for each of them is parallel to DE, and they are not both in the same plane with it) the 4 GBA, BGH, are together (Th. 10.) equal to two right angles: and BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA: for the same reason, GB is perpendicular to BC: since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B, GB is (Th. 139.) perpendicular to the plane through BA, BC: and it is perpendicular to the plane through - DE, EF; therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: but planes to which the same straight line is perpen- dicular, are parallel (Th. 146. and Def. 71.) to one another: therefore the plane through AB, BC, is parallel to the plane through DE, E.F. Wherefore, if two straight lines, &c. a. E. D. CoR. If three equal perpendiculars be drawn to a given plane, and if their extremities do not lie in the same straight line, the plane which passes through them shall be parallel to the given plane. For (Th. 142. Th. 144.) the three perpendiculars will be parallel to one another; and (hyp.) they are equal; wherefore (Th. 141. Cor. and Th. 10. Cor. 4.) the straight lines, which join the upper extremities of Mutual Intersections and Inclinations of Planes. 401 any two of them and the third, shall be parallel to the straight lines which join the lower extremities of, the same two and of the third; therefore (Th. 147.) the plane that passes (Th. 138.) through the upper ex- tremities of the three perpendiculars, shall be pa- rallel to the given plane. ScholIUM. A plane may always be supposed to be drawn, which shall be parallel to a given plane * (DF) and which shall pass through a given point (B) without the given plane. For (Th. 141. Schol.) a perpendicular BG may be drawn from B to the plane DF; and at the point B, in any two planes BK, BH, both passing through BG, which are not in the same plane, there may be drawn BA, and BC, perpendicular to BG; and then the plane CA, that is drawn through BA and BC, will be parallel to the plane DF. For, since (hyp.) BG is perpendicular to BC and to BA, it is (Th. 139. and Def. 69.) also perpen- dicular to the plane CA: wherefore (Th. 146. and Def. 71.) the plane CA, which passes through the point B, is parallel to the given plane DF. Or, if BC and BA be supposed to be drawn parallel to any two straight lines EF and ED, which meet one another, in the plane DF, then (Th. 147.) *- : *- - ~f * See the figure in p. 399, 3 E 402 Elementary Theorems of Solid Geometry, the plane which passes through BC and BA will be parallel to the given plane DF. THEOREM CXLVIII. If two parallel planes be cut by another plane, their common sections with it are parallels. For (Def. 8. Cor. 2.) the two common sections of the planes are straight lines, and they are both in the same cutting plane; and since the parallel planes, in which they are, do not (Def. 71.) meet when pro- duced, neither do these straight lines any where meet one another; wherefore (Def. 14.) they are parallels. CoR. 1. Two parallel straight lines, which are terminated by two parallel planes, are equal to one another. Let BG be parallel to CK, and let BG and CK lie between the parallel planes CA and FD; BG= CK. For through the parallels BG, CK, let there be supposed to pass the plane BCKG, cutting the planes CA and FD, in BC and GK; therefore (Th. 148.) BC is parallel to GK; and (hyp.) BG is parallel to CK; therefore BCKG is a parallelogram, and (Th. 25. Cor. 1.) BG = CK. CoR. 2. Hence, the straight lines drawn from any number of points in either of two parallel planes, perpendicular to the other, being (Th. 142.) parallel, are also (Cor. 1.) equal to one another. Mutual Intersections and Inclinations of Planes. 403 CoR. 3. . If a straight line (BG) be perpendicular E * * * * * * * * * * * * * * * * * * a tº e a tº gº º º IC & º A. * * * * * * * * * * * * * * * * * * *- a s gº º º º a a e s tº se is to (AC) one of the two parallel planes (AC), (DF), it shall be perpendicular to the other (DF) also. For let any plane BK, passing through BG, cut the planes AC, DF, in BC, GK, and let any other plane BH, also passing through FG, cut the planes AC, DF, in BA, GH: then (hyp. and Def. 69.) the 4 GBC, GBA, are right angles, and (Th. 148.) GK is parallel to BC, and GH to BA; therefore (Th. 10. Cor. 2.) the / BGK, BGH, are right angles: and therefore (Th. 139. and Def.69.) BG is perpendicular to the plane DF. e CoR. 4. Hence, two planes, which cut one: an- other, cannot each of them be parallel to the same plane. { For if from any point in the common section of the two planes, which cut one another, a perpendicular 404 Elementary Theorems of solid Geometry, be supposed to be drawn to the third plane, it will (hyp. and Cor. 3.) be perpendicular to each of the two intersecting planes; which (Def. 69. Cor. 1.) is impossible. CoR. 5. If two straight lines be cut by parallel planes, they shall be cut in the same ratio. Let the straight lines AB, CD, be cut by the parallel planes GH, KL, MN, in the points A, B, B; / … / H /LV 7° C, F, D: as AE is to EB, so is CFto FD. Suppose AC, BD, AD, to be drawn, and let AD meet the plane KL in the point X; suppose also, EX, XF to be drawn. Because the two parallel planes KL, MN, are cut by the plane EBDX, the common sections EX, BD, are (Th. 148.) pa- rallel. For the same reason, because the two parallel planes GH, KL, are cut by the plane AXFC, the Mutual Intersections and Inclinations of Planes. 405 common sections AC, XF, are parallel; and because EX is parallel to BD, a side of the As ABD; ... (Th. 99.) AE : EB :: AX : XD; Likewise, AX : XD :: CF : FD, ... (Th. 81.) AE : EB :: CF : FD. THEOREM CXLIX. If two planes cut one another, and if a straight line in either of them, drawn perpendicular to their common section, be at right angles to the other plane, then shall all straight lines drawn in the former plane perpendicular to the common section, be at right angles to the latter plane. Let CE be the common section of the two planes CK, DE, and let AB, drawn in DE, perpendicular G- A \ ID |H * C F H- *E to CE, be also perpendicular to the plane CK; then shall any other straight line in DE, drawn per- pendicular to CE, as GF, be also perpendicular to the plane CK. -- For, since (hyp.) the 4ABF and GFB are right angles, therefore (Th, 8.) GF is parallel to AB; 406 Elementary Theorems of Solid Geometry, but (hyp.) AB is perpendicular to the plane CK; therefore (Th. 141.) GF is also perpendicular to the plane CK. In like manner it may be proved that any other straight line in DE, drawn perpendicular to CE, shall also be perpendicular to the plane CK. Wherefore, if two planes, &c. a. E. D. DEF. LXXII. A Plane is perpendicular to a plane, when the straight lines drawn in one of the planes, perpen- dicularly to the common section of the two planes, are perpendicular to the other plane. THEoREM CL. If two planes, which cut one another, pass through two parallel straight lines, their common section shall be parallel to each of those lines. - Let AB be the common section of the two planes A. C g-f l ; : M . . R *, - N GD and AF, which pass through the two parallel * Mutual Intersections and Inclinations of Planes. 407 straight lines CD and EF; AB is parallel to CD and to EF. ** For if it be not in the plane GD, let AK be supposed to be drawn parallel to CD, and, in the plane AF, AL parallel to EF: and, since AK and EF are parallel to CD, therefore (Th. 144.) AK is parallel to EF; also (hyp.) AL is parallel to EF; therefore (Th. 144.) AK is parallel to AL, which is absurd; therefore no other straight line, passing through A, but AB, is parallel to CD and to EF. THEOREM CLI. If two planes cutting one another be each of them perpendicular to a third plane; their common section shall be perpendicular to the same plane. Let the two planes GD, AF, of which AB is the common section, be each of them perpendicular to the plane MN; AB is also perpendicular to the plane MN. From any two points F, D, in the common sections BF, BD, of the planes AF, MN, and GD, MN, let there be drawn FE and DC perpendicular to BFand BD respectively; therefore (hyp. and Def. 72.) EF and CD are each of them perpendicular to the plane MN; therefore (Th. 142.) EF is parallel to CD; therefore (Th. 150.) AB is parallel to EFor to CD; * See the figure in preceding page. 408 Elementary Theorems of Solid Geometry, and EF and CD have been shewn to be each of them perpendicular to the plane MN; therefore (Th. 141.) AB is also perpendicular to the plane MN. Therefore if, &c. a. E. D. CoR. If a plane be at right angles to the common section of two planes, it shall be at right angles to each of the planes. For, otherwise, it is manifest, from the proposition, that there might be two straight lines both terminated in the same point of the plane, and both at right angles to it; which (Def. 69. Cor. 2.) is impossible. THEoREM CLII. If a plane, which cuts two parallel planes, be perpendicular to the one of them, it shall also be per- pendicular to the other. Let the plane EF cut the two parallel planes AB, CD, in EG and HK, and let EF be per- pendicular to AB; it is also perpendicular to CD. For since (hyp.) the plane CD is parallel to AB and EF cuts them, therefore (Th. 148.) HK is parallel to EG ; from any point L, in EG, let there be drawn LM perpendicular to EG; therefore (Th. 8. Cor. 1.) LM will cut HK; let it cut HK in N, and since (hyp.) ELN is a right angle, and EL is pa- rallel to HN, therefore (Th. 10.) HNL is a right angle; again, from L let there be drawn, in the plane AB, LP perpendicular to EG; through LP and LM, let there pass the plane PM, and let it cut Mutual Intersections and Inclinations of Planes. 409 CD in NQ; therefore (hyp. and Th. 148.) Na is p E. ^ H L N * GS IX F parallel to LP; and (hyp.) the / NLP is a right angle; therefore (Th. 10.) the / LNQ is a right angle; but the / LNH has been shewn to be a right angle; therefore (Th. 139. and Def. 69.) LN is per- pendicular to the plane CD, in which are NH and NQ; and therefore (Th. 149, and Def. 72.) the plane EF, in which is LV is also perpendicular to the plane CD. Wherefore, if a plane, &c. a. E. D. CoR. Hence, it may easily be shewn, ea absurdo, that different planes, which are at right angles to the same plane, are parallel to one another. SCHOLIUM. If two planes cut one another, and from any points in their common section there be drawn perpen- diculars to that line in each of the two planes, the 3 F 410 Elementary Theorems of Solid Geometry, perpendiculars that are in the one of the planes will (Th. 8.) be parallels, as will also the perpendiculars that are in the other; and therefore (Th. 145.) the angle contained by the two perpendiculars drawn. from any one point of the common section, shall be equal to the angle contained by the two perpendiculars drawn from any other point. Hence the following definitions. DEF. LXXIII. The * Inclination of a plane to a Plane is the angle contained by two straight lines drawn from any the same point of their common section at right angles to it, one upon one plane, and the other upon the other plane: and two planes are said to have a like inclination to another, when the said angles of inclination are equal to one another. * THEOREM CLIII. Two planes are parallel, that have like inclina- tions to a third plane and have also their common sections with it, parallel to one another. * Properly speaking one plane can be said to be inclined to another, only when the angle described in the above definition is an acute angle; when it is obtuse, there is a re-clination of the one plane from the other; and when it is a right angle there is neither inclination nor re-clination. It has been found convenient, however, to extend the sense of the word inclination. Mutual Intersections and Inclinations of Planes. 41 l Let the two planes AB, CDs have like inclinations to the plane EF, and cut it in EG, HK; and let EG be parallel to HK: then is the plane AB parallel to the plane CD. For from any point L in EG let there be drawn, in the plane EF, LM perpendicular to EG and cutting HK in N; therefore (hyp. and Th. 10.) the / HNM = / ELM, and therefore (hyp. and Th. 2.) the LHNM is a right angle; also, from L in the plane AB, let there be drawn LP per- pendicular to EG ; let a plane PM be drawn through LM and LP, and let it cut the plane CD in NQ ; then, since EL is perpendicular to LM and LP it is * (Th. 139) perpendicular to the plane PM; and (hyp.) HIV is parallel to EL; therefore (Th. 141.) HV is perpendicular to the plane PM, and therefore (Def. 69.) the / HNQ is a right angle; there- fore (Def. 73.) the / MNQ is the inclination of the plane CD to EF, as the / MLP is, also, to the in- clination of the plane AB to EF; therefore (hyp. and Def. 73.) the / MINQ = / MLP; therefore (Th.6.) WQ is parallel to LP, and (hyp.) HN is parallel to EL; therefore (Th. 147.) the plane AB, which passes through EL and LP, is parallel to the plane CD, which passes through HN and NQ: Wherefore, two planes are parallel, that have, &c. Q. E. D. * See the figure in p. 409. 412 Elementary Theorems of Solid Geometry, &c. Theorem CLIV. If two parallel planes be cut by a third plane, they shall have a like inclination to that plane. Let the two parallel planes "AB, CD, be cut by a third plane EF, in EG and HK: the planes AB, CD have a like inclination to EF. * From any point L in EG, let there be drawn in the plane EF, LM perpendicular to EG, and let LM meet HK in N, also, let there be drawn from L, in the plane AB, LP perpendicular to EG, and through LP and LM let there pass the plane PM, and let it cut the plane CD in NQ; and because the plane EF cuts the two parallel planes AB, CD, in EG and HK, therefore (Th. 148.) EF is parallel to HK; but (hyp. and Th. 139.) EG is perpen- dicular to the plane PM; therefore (Th. 141.) HK is also perpendicular to the plane PM, and therefore (Def. 69.) the / HNM, HNQ, are right angles; therefore (Def. 73.) the / MVQ is the inclination of the plane CD to the plane EF, as (hyp. and Def. 73.) the / MLP is, also, the inclination of the plane AB to the same plane EF: and since (Th. 148.) LP and NQ, being the sections which the plane MP makes with the parallel planes AB, CD, are parallel to one another, therefore (Th. 10.) the / MVQ- / MLP; therefore (Def. 73.) the planes AB, CD, have a like inclination to EF. If, therefore, two planes, &c. Gl. E. D. * See the figure in p. 409. Elements of $olio (ºtomtetry). -Q- CHAPTER III. On Solid Angles. . —O— DEFINITION LXXIV. A Solrd Angle is that which is made by the meeting of more than two plane angles, which are not in the same plane, in one point. º THEOREM CLV. If a solid angle be contained by three plane angles, any two of them is greater than the third. Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB : any two of them are greater than the third. 414 Elementary Theorems of Solid Geometry, If the 4 BAC, CAD, DAB, be all equal, it is l) º º ** *e evident that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A, in AB, let there be made, in the plane which passes through BA, AC, the / BAE = / DAB; take any two points F, G, in AB and AC; let F, G be joined, and let FG cut AE in H; in AD let AK be taken equal to AH, and let FK, GK, be drawn. Then since AF is common to the two /> FAH, FAK, and AK = AH, and that the / FAH = / FAK; therefore (Th. 20. Cor. 1.) FK = FH; but, (Th. 17.) in the As KFG, FK-- KG > FG or FH-i-HG; therefore, since FK = FH, KG > GH: again, since AG is common to the two /> HAG, KAG, and that AH = AK, and KG > HG, therefore (Th. 22.) the On Solid Angles. 415 \ 4. KAG > the / HAG ; that is, the / DAC > the 4. EAC; and (hyp.) the / DAB = L BAE; there- fore the / DAC + / DAB > the / BAE + / EAC, that is, > the / BAC; but (hyp.) the L BAC is not less than either of the // DAB, DAC; therefore the a BAC, together with either of them, is greater than the other. Wherefore, if a solid angle, &c. G. E. D. Theorem CLVI. Every “solid angle that is not divided by any of the planes, produced, which contain it, is contained hy plane angles, which together are less than four right angles. f Let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, D F I} FAB; these together are less than four right angles. Suppose the planes, in which the angles are, to be cut by a plane, and let the common section of it *- *The proposition applies to those solid angles only, in which no two of the containing planes, project toward the interior of the figure. 416 Elementary Theorems of Solid Geometry, $ with those planes be BC, CD, DE, EF, FB: and because the solid angle at B is contained by three plane / CBA, ABF, FBC, of which any two are greater (Th. 155.) than the third, the // CBA, ABF, are . greater than the / FBC: for the same reason, the two plane angles at each of the points C, D, E, F, viz. the angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the angles of the polygon BCDEF: therefore, all the angles at the bases of the triangles are together greater than all the angles of the polygon: and because all the angles of the triangles are together equal (Th. 12.) to twice as many right angles as there are triangles ; that is, as there are sides in the polygon BCDEF: and that (Th. 35.) all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon; there- fore all the angles of the triangles are equal to all the angles of the polygon together with four right angles. But all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved. Wherefore the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, are less than four right angles. Therefore every solid angle, &c. a. E. D. THEOREM CLVII. If each of two solid angles be contained by three plane angles equal to one another, each to each, the On Solid Angles. 417 planes, in which the equal angles are, have like in- clinations to one another. Let the solid angle at A be contained by the three plane / CAD, CAE, EAD; and the solid angle at E. a TH B by the three plane / FBG, FBII, GBH; of which the / CAD = / FBG, the / CAE = L FBH, and the L EAD = / HBG; the planes, in which the equal angles are, have like inclinations to one another. t Let AC, AE, AD, B.F. B.H. BG, be taken equal to one another, and CD, CE, ED, FG, FH, HG be drawn. In AC take any point K, and from K, lettherebe drawn perpendicular to AC, KLin the plane CAD; and KM in the plane CAE; therefore, (Def 73.) the / LKM is the inclination of the plane CAE to the plane CAD; and since the /> CAD, CAE, are isosceles, the // ACD, ACE are acute, and therefore, (Th. 10. Cor. 1.) the perpendiculars KL, KM, will meet CD and CE; let them meet CD and CE in L and M, and in FB take FN = CK; and from N let there be drawn, perpendicular to BF, WP in the plane FBG, and Na r 3 G #18 Elementary Theorems of Solid Geometry, in the plane FBH, meeting FG and FH in Pand Q, respectively; therefore the / PNQ is the inclination of the plane FBIH to the plane FBG; let L, M, and P, Q be joined. . º Then (hyp. and Th. 20. Cor. 1.) the As CAD = As FBG, the / ACD = / BFG, and CD = FG ; likewise, the AS CAE = /> FBPI, the / ACE = / BFH, and CE = FH; and the As EAD = As HBG, and ED = HG; since, then, the three sides of the As CED, are equal to three sides of the As FHG, each to each; therefore (Th. 24. Cor. 1.) the / DCE = / GFH; again, in the right-angled /~ CKL, FNP, the acute / KCL, NFP, have been shewn to be equal, and (hyp.) KC = NF, therefore (Th. 19. Cor. 1.) UL = FP and KL = NP; likewise, in the right-angled /s CKM, FNQ, it may be shewn that CM = FQ, and KW = WQ; since L.C, CM are equal to PF, FQ, each to each, and the / LCM has been proved to be equal to the / PFQ; therefore (Th. 20. Cor. 1.) LIM = P0; wherefore, the two /> KLM, NPQ, have their three sides equal each to each ; therefore (Th. 24. Cor. 1.) the / LKM = / PNQ; that is, (Def. 73.) the plane CAE has to CAD a like incli- nation to that which the plane FBH has to FBG. If, therefore, each, &c. a. E. D. THEOREM CLVIII. If two triangular solid angles have two plane angles of the one, equal to two plane angles of the other, each to each, and if, also, the planes, in which On Solid Angles. 419 the equal angles are, have like inclinations to one another, their remaining plane angles shall be equal; and the planes, in which they are, shall have like in- clinations to the other planes of the solid angles. Let CAD, CAE, two of the three plane angles . which contain the solid angle at A, be equal to FBG, FBH, two of the three plane angles which contain the solid angle at B, namely, CAD to FBG, and CAE to FBH; and let the planes CAE, FBH have like in- clinations to the planes CAD, FBG: the third plane // EAD, HBG are equal; and the planes EAD, HBG have like inclinations to the planes CAE, FBIH, and also like inclinations to the planes CAD, FBG. For the same construction being supposed to be made, as in Th. 157, it may be shewn, as it was in that theorem, that DC = GF, CE = FH, KL = NP, LC= PF, KM = NQ, and CIM = FQ : and be- cause KL = NP, and KM = NQ; and (hyp. and Def. 73.) the / LKM = / PNQ; therefore (Th. 20. Cor. 1.) LM = P0; and because LC– PF, and CIM = FQ, and ML = QP; therefore, (Th. 24. Cor. 1.) the / LCM = L PFC); since, then, DC= GF, and CE = FH, and the / DCE = Z GFH; there- fore (Th. 20. Cor. 1.) DE = GH: lastly, because, . in the As EAD, HBG, AE = BH, and AD = BG, and ED = HG; therefore (Th. 24. Cor. 1.) the / EAD = / HBG; and therefore (Th. 157.) the * See the figure in p. 417. 420 Elementary Theorems of Solid Geometry, &c. planes of the two solid / A and B, in which the equal plane angles are, have like inclinations to one another. If, therefore, &c. a. E. D. SCHOLIUM. In the figures made use of in the two preceding theorems, the plane angles, which contain the solid angles, that are compared with one another, are sup- posed to be placed in the same order in both. But the mode of demonstration there given is equally ap- plicable whether the plane angles be disposed in the same order, in both the solid angles, or not. From Th. 157. it is easy to shew the possibility of making at a given point, in a given straight line, a solid angle, that shall be contained by three plane angles, equal, each to each, to the three plane angles which contain a given solid angle. But this possibility is of itself sufficiently manifest. THEOREM CLIX. If a solid angle be contained by three plane angles, each of which is a right angle, the planes which con- tain it shall be at right angles to one another. This is manifest from Th. 139. and Th. 149. THE \ 32ſtituetttº of $olio Gºtomuttty, * —-º- CHAPTER IV. ON THE SECTIONS OF SOLIDS, —O— SECTION I. On the sections of solids, of which the boundaries are all of them plane surfaces. sº ſºººººººººººººººº- DEFINITION LXXV. A PyRAM ID is a solid figure, contained by planes that are constituted betwixt one plane, called the Base, and one point above it, in which they meet, called the Pertea of the pyramid : and the perpen- dicular drawn from the vertex to the plane of the base, is called the Altitude of the pyramid. 422 Elementary Theorems of Solid Geometry, DEF. LXXVI. A Prism is a solid figure, contained by plane figures, of which two, that are opposite, are equal, similar, and parallel, to one another, and the others are parallelograms; the two opposite figures that are equal, similar, and parallel, are called the Bases, and the other containing figures are called the Sides, of the prism : also, the straight line drawn from any point in one of the bases perpendicular to the plane of the other base, is called the Altitude of the prism. DEF. LXXVII. A Parallelepiped is a solid figure, contained by six quadrilateral plane figures, whereof every op- posite two are parallel: and if two of the containing quadrilateral figures be squares, and perpendicular to one another, the parallelepiped is called a Cube. CoR. It follows, therefore, from Th. 152. Def. 72, Def. 69. and Th. 149, that all the containing planes of a cube are perpendicular to one another; and therefore (Th. 148. Th. 25. Cor. 1. Th. 150. and Def. 69.) they are all of them squares; and (Th. 26. Cor. 2.) they are equal to one another. THEOREM CLX. If a pyramid be cut by a plane, that is parallel to its base, the section shall be similar to the base. Let EFGH be a section of the pyramid ABCD On the Sections of Solids, &c. 423 - S, when it is cut by a plane parallel to its base ABCD: the figure EFGH is similar to ABCD. For since the cutting plane is (hyp.) parallel to the plane of ABCD, therefore (Th. 148.) EF is pa- rallel to AB, FG to BC, GH to CD, and HE to DA; therefore (Th. 145.) the two figures ABCD, EFGH, have their angles equal, each to each : again, since, as hath been shewn, EF is parallel to AB, and FG and BC, therefore (Th. 101. Cor. 1.) the 2s ABS, EFS are similar, as are, also, the /> BCS, FGS; ... AB : BS :: EF: FS, and BS : BC :: FS : FG ; ... ea (aequo) AB : BC :: EF: FG. And, in like manner, it may be shewn, that the sides about the other equal angles of the figures ABCD, EFGH, are proportionals; therefore (Def. 63.) the 424 Elementary Theorems of Solid Geometry, figure EFGH is similar to ABCD. If, therefore, a pyramid, &c. a. E. D. w CoR. 1. If a given pyramid (ABCD–S) be cut by a plane parallel to its base, the base of the given pyramid shall be to the base of the pyramid (EFGH — S) so cut off from it, as the square of the altitude of the given pyramid to the square of the altitude of the pyramid so cut off. For let SK be a perpendicular from S to the base of the pyramid ABCD–S; let it meet ABCD in K, and EFGH in L; and since (hyp.) EFGH is parallel, to ABCD, therefore (Th. 148. Cor. 3.) SL is perpendicular to EFGH; so that (Def 75.) SK is the altitude of the pyramid ABCD-S, and SL the altitude of the pyramid EFGH-S. Let AK and EL be drawn; therefore (Def. 69.) the // SKA, SLE are right angles, and the Z. ASK is common to the two /> ASK, ESL ; therefore (Th. 12.) they are equiangular and (Th. 101. Cor. 2.) SK : SL :: SA : SE; and it has been shewn that the 2s ASB, ESF are similar; therefore (Th. 101. Cor. 2.) SA : SE :: AB : EF; therefore (Th. 81.) SK : SL : AB : EF; but since the two figures. ABCD, EFGH, are (Th. 160.) similar, and that AB and EF are homo- logous sides, they are to one another (Th. 117. Cor. 3.) as the square of AB to the square of EF, or, since SK : SL : AB : EF, as (Th. 118.) the square of SK to the square of SL. CoR. 2. Hence, if two pyramids which have equal bases and equal altitudes, be cut by planes On the Sections of Solids. 425 parallel to the bases, and at equal perpendicular distances from them, the sections shall be equal to one another. THEoREM CLXI. If a prism be cut by a plane, that is parallel to its base, the section shall be equal and similar to the base. Let EFGH be the section of the prism ABCD ... KL, when it is cut by a plane parallel to its base ABCD: the figure EFGH is equal and similar to ABCD. For (hyp.) the cutting plane is parallel to the plane of ABCD; therefore (Th. 148.) EF is pa- rallel to AB, and (Def. 76.) EA is parallel to FB; 3 H 426 Elementary Theorems of solid Geometry, &c. . therefore EABF is a parallelogram, and (Th, 25. Cor. 1.) EF=ABs in like manner, it may be shewn that the other sides of EFGH are equal and parallel to the other sides of ABCD, each to each; where- fore, the two figures EFGH, ABCD, have also (Th. 145.) their angles equal, each to each; and it is evident that the one figure may be applied to the other, so as to coincide with it; they are, therefore, both similar and equal to one another. Wherefore, if a prism, &c. a. E. D. …” - CoR. 1. It is manifest from the demonstration, that, if a solid be contained by six planes, two and two of which are parallel, the opposite planes are similar and equal parallelograms. * CoR. 2. It follows, therefore, from Def. 76. and Def. 77, that every parallelepiped is a prism. THE 33ſentent; of $olio Cºrontettp. CHAPTER IV. SECTION II. On the sections of solids, which are supposed to be described by the revolution of plane figures, about a side that remains fiat. —O— DEFINITION LXXVIII. If a Solid be supposed to be described by the re- volution of a plane figure about one of its sides, which is a fixt straight line, that fixt side is called the Aaris of the Solid. CoR. If a solid, so described, be cut by a plane which passes through its axis, the section of it shall be divided, by the axis, into two equal plane figures, each of them equal to the plane figure, by the re- 428 Elementary Theorems of Solid Geometry, volution of which the solid is supposed to have been described. For it is manifest that the revolving plane figure must, in the course of its revolution, wholly coincide with each of the two parts, into which the section is divided by the axis. Def. LXXIX. A Cone is a solid, supposed to be described by the revolution of a right-angled triangle about one of the two sides containing the right angle, as an axis, which is also called the Altitude of the cone : the circle, described by the other of the two sides of the triangle, is called the Base of the come; and the extremity of the axis which is not in the base, is called the cone's Perter. Also if the axis be equal to the semi-diameter of the base of a cone, it is called a Right Cone. CoR. If a cone be cut by a plane, which passes through its axis, the section (Def 78. Cor.) shall be an isosceles triangle. DEF. LXXX. A Cylinder is a solid, supposed to be described by the revolution of a right-angled parallelogram about one of its sides, as an axis; the figures described by the two sides that are adjacent to the axis, are called the Bases of the cylinder, and the surface described by the other revolving side is called the Curved Sur- face of the cylinder: and the altitude of the cylinder On the Sections of Solids, &c. 429 is the straight line drawn from any point in either of the two bases perpendicular to the other base. CoR. 1. If a cylinder be cut by a plane which passes through its axis, the section (Def. 78. Cor.) shall be a right-angled parallelogram. CoR. 2. If from any point in the perimeter of the base of a cylinder, a straight line be drawn pa- rallel to the axis, and towards the same parts as the axis, it will, manifestly, lie in the curved surface of the cylinder; and all such straight lines, intercepted between the two bases of the cylinder, being each of them (Def 78. Cor. and Th. 25. Cor. 1.) equal to the axis, are equal to one another. DEF. LXXXI. A Sphere is a solid, supposed to be described by the revolution of a semi-circle about its diameter, as an axis; and the point which divides the axis into two equal parts is called the Centre of the sphere. CoR. 1. All the straight lines, which are drawn from the centre to the surface of the sphere are (hyp. and Def. 28. Cor. 1.) equal to one another. CoR. 2. If, therefore, a sphere be cut by a plane, which passes through its centre, the section will be a plane figure, such that all straight lines drawn from the centre to its perimeter, are equal to one another; i. e. it will be a circle, having its centre in the centre of the sphere; and it will, manifestly, divide the sphere into two equal parts. ciº 430 Elementary Theorems of Solid Geometry, THEOREM CLXII. If a finite straight line be at right angles to another fiat straight line, about which the plane containing both the lines is supposed to revolve, the jigure, described by the finite revolving straight line, shall be a circle, having the fict straight line per- pendicular to its plane. _Let the finite straight line AC be at right angles to CZ in the point C, and let ADBE be the figure Z A C 13 y described by CA, when the plane, containing AC and ZC, is supposed to revolve about ZC, which remains fixt: ADBE is a circle, having CZ perpendicular to its plane. * - Let, D, E, &c. be any points whatever in the perimeter of the figure ADBE, and suppose CD, CE, &c., to be drawn: then (hyp. and Def. 8. Cor. 3.) CD, CE, &c. will be in the figure described by On the Sections of Solids, &c. 431 4C, they will be equal to one another, and ZC will be at right angles to each of them; therefore (Th. 140.) they will be all in one plane; wherefore (Def. 28.) the figure ADBE is a circle; and since all the straight lines in it, drawn from C, are at right angles to ZC, ZC (Def. 69.) is perpendicular to its plane. If, therefore, a finite straight line, &c. a. E. D. CoR. 1. The base of a cone (Def. 79.) is a circle, having its centre in the axis; and the axis is per- pendicular to the base. CoR. 2. The bases of a cylinder (Def. 80.) are circles, and the axis is perpendicular to each of them; wherefore (Def 71. Cor.) they are parallel to one an- other: also, since (Def. 80. and Th. 25. Cor. 1.) they have equal semi-diameters, they are (Def. 30. Cor. 3.) equal to one another. *. THEoREM CLXIII. ſf a solid figure described by the revolution of a plane figure, about an aris, be cut by a plane at right angles to the aris, the section of it shall be a circle, having its centre in the aris. Suppose the solid ASB to have been described by the revolution of a plane figure about the axis SC, and let FGH be the section of this solid when it is cut by a plane at right angles to SC: the figure FGH is a circle, having its centre in SC. For let AFSGB be the section of the solid when it is cut by any plane passing through the axis SC, 432 Elementary Theorems of Solid Geometry, - and let AFSGB cut FGH in FG; and since (hyp.) - S … e. IHL IE g- IB -e-Z TX SK is perpendicular to the plane of FGH; therefore (Def. 69.) the / SKF is a right angle; therefore, if the plane figure ASC, which (Def 78. Cor.) is equal to that which describes the solid, revolve about SC, the figure described by KF will (Th. 162.) be a circle at right angles to SK, but (hyp.) the plane of FGH is also at right angles to SK; therefore the plane of the section FGH coincides (Def. 69. Cor. 1.) with the circle described by KF, therefore the section FGH - is a circle, of which K, a point in the axis of the solid, is the centre. If, therefore, &c. a. E. D. A THEOREM CLXIV. If a come be cut by a plane that is parallel to its base, the section of it shall be a circle, having its centre in the aris, and which is to the base of the cone On the Sections of Solids, &c. 433 as the square of the segment of the aris between the vertea and the cutting plane, is to the square of the aris. - • , Let FGH be the section of the cone ABD-S, of which the axis is SC, when it is cut by a plane parallel to the base ABD; which (Th. 162. Cor. 1.) is a circle; and let FGH cut the axis in K : the section FGH is a circle, and it is to the circle ABD as the square of KS is to the square of C.S. From K let there be drawn to the perimeter of the section, any straight lines KH, KF; and let a plane, passing through SC and KH, cut the convex surface of the cone in SHD, and its base in CD; likewise let a plane passing through SC and KF, cut the convex surface of the cone in SFA, and its base in CA; and since (hyp.) the plane FGHisparallel to ABD, therefore (Th. 148.) FK is parallel to AC; 3 I 434 Elementary Theorems of Solid Geometry, therefore (Th. 101. Cor. 1.) KF: CA :: KS : CŞ. In like manner it may be shewn that KHH. : CD: KS: CŞ; therefore (Th. 81.) KF: CA :: KH : CD; but (Th. 162. Cor. 1.) CA=CD; therefore (Th. 85. Cor.) KF = KH; and in the same manner may all other straight lines, drawn from K to the perimeter of the section FGH, be shewn to be equal to one another; therefore the section FGH is a circle, of which K is the centre: and (Th. 123. Cor. 1.) the circle FGH is to the circle ABD as the square of KF to the square of CA; and it has been shewn that KF : CA :: KS: CS; therefore (Th. 115.) the circle FGH is to the circle ABD as the square of SK to the square of SC: wherefore if a cone be cut, &c. a. E. D. CoR. If a cone and a pyramid, which have equal bases and equal altitudes, be cut by planes parallel to their bases and at equal perpendicular distances from them, the sections shall be equal to one another. Theorem CLXV. If a cylinder be cut by a plane, that is parallel to its base, the section of it shall be a circle, equal to the base and having its centre in the axis of the cylinder. | Let FGH be the section of the prism ABD-ST. when it is cut by a plane parallel to the base ABD: the figure FGH is a circle. Let the axis CZ of the cylinder cut the section * On the Sections of Solids, &c. 435 FGH in K, and from Klet there be drawn, to the C D ... • * * * * * * * * : * “..... • * * * * * : * * * * * s • * s a * * * e *s º * W**, g perimeter of the section, any straight lines, KF, KH; and let a plane passing through ZKC and KF cut the curved surface of the cylinder in FA and its base in CA; likewise, let a plane passing through ZKC and KH, cut the curved surface of the cylinder in HD and its base in CD: then (Def. 80. Cor. 1.) FM is parallel to KC; also since (hyp.) the plane of FGHis parallel to the base ABD, therefore (Th. 148.) KF is parallel to CA; therefore the figure AFKC is a parallelogram, and therefore (Th. 25. Cor. 1.) KF= CA. in like manner it may be shewn that KH = 436 Elementary Theorems of Solid Geometry, CD; and so on; but (Th. 162. Cor. 2.) the straight lines CA, CD, &c. are all equal to one another; therefore the straight lines KF, KH, &c. are all equat to one another; therefore the section FGH is a circle having its centre in K. Wherefore, if a cylinder, &c. a. E. D. *. ScholIUM. The following more comprehensive definitions might be given of a Cone and of a Cylinder. “A Cone is a solid figure described by the re- volution of a straight line, one extremity of which remains fixt in a given point above the plane of a given circle, whilst the other extremity moves through the whole circumference of the circle: the given circle is called the base of the cone, and the straight line joining its centre, and the given point above the circle, is called the axis of the cone.” - “If one extremity of a straight line, which line is situated above the plane of a given circle, move through the whole circumference of the circle, the line continuing always parallel to itself, the solid contained by the circle, by the surface described by the revolving line and by a plane cutting that surface and parallel to the given circle, is called a cylinder: the given circle is called the base, and a straight line drawn from the centre of the base parallel to the re- volving line is called the axis of the cylinder.” These more general definitions having been laid down, the two theorems, immediately preceding this On the Sections of Solids, &c. 437 scholium, and the methods of proof which have been applied to them, will be found to apply equally to such cones and cylinders as have been so defined. THEOREM CLXVI. If a sphere be cut ly a plane the section shall be a circle. .* First, if the cutting plane pass through the sphere's centre, the common section is (Def. 81. Cor. 2.) a circle. * But, secondly, let ApHE be a sphere, of which C is the centre; let it be cut by a plane which does not pass through its centre; and let EFH be the common section of the plane, and the sphere: EFH is a circle. For, from C, let CG be supposed to be drawn at 438 Elementary Theorems of Solid Geometry, - right angles to the plane EFH ; and, in that plane, let there be drawn, from G, any two straight lines GE and GF, to the sphere's surface. Then (Def. 69.) CG is at right angles to GE and GF; let, also, the points C, E, and C, F be joined. Then, in the two right-angled triangles CGE, CGF, because the hypotenuse CE is (Def. 81. Cor. 1.) equal to the hypotenuse CF, and the side CG is common to both the triangles, therefore (Th. 23. Cor.) the third side GE, in the one, is equal to the third side GF in the other. In the same manner may any other straight line as GH, drawn in the plane EFH from G to the sphere's surface, be shewn to be equal to GE, or G.F. Therefore (Def 78. Cor. 1.) the figure EFH is a circle. Wherefore the common section, &c. a. E. D. CoR. 1. It has appeared, from the demonstration that the centre G of the circular section, EFH, of a sphere, made by a plane which does not pass through the sphere's centre, is in the perpendicular CG, drawn to that section from the sphere's centre C: and, conversely, (Def. 69. Cor. 2.) the centre C of the sphere is in the straight line GC drawn at right angles to the plane of the section EFH, from its centre G. Therefore CG, the straight line joining the centres of the sphere and of any such section, is perpendicular to the plane EFH of that section : and, a plane which passes through the centre of such a section and cuts it perpendicularly, passes also (Def. 72. and Def. 69. Cor. 2.) through the centre of the sphere. *es, On the Sections of Solids, &c. 439. CoR.2. If a hemisphere (ABC) and a right-angled *COrle (ADC), having the same base (AEC) be cut i by planes parallel to the common base and passing . through points in their axes that are equidistant from the sphere's centre and from the cone's vertex, the two circular sections shall be together equal to the common base of the two solids. - Let K be the centre of the circle AEC: let DK, the axis of the come, be produced to meet the surface of the hemisphere in B; also let ABCD be the section, made by a plane passing through BKD and cutting both the solids, so that AKC is a diameter of the sphere, and (Th. 162. Cor. 1. and Def. 69.) DKB is perpendicular to AC; wherefore KB may be considered as the axis of the hemisphere; and KB (Def, 81. Cor. 1.) is equal to AC, which (hyp. and Def. 79.) is equal to KD; therefore KB = KD. Let DP be 440 Elementary Theorems of Solid Geometry, &c. any part of DK, and let KG be taken, in KB, equal to DP; also let HFL, QRS, be the sections of the solids, when they are cut by planes parallel to the base AEC: the circle HFL together with the circle QRS is equal to the circle AEC. - For, let K, H be joined. Then (hyp. and Th. 166. Cor. 1. and Th. 164.) G and P are the centres of the circles HFL, QRS; and, if HL and QS be sections made by the plane ABCD passing through BKD, with the planes of HFL and QRS, GH and PQ are semi-diameters of the circles HFL, QRS; also (hyp. and Th. 148.) PQ is parallel to KA; therefore (Th. 19.) the / PQD = / KAD; and since (hyp. and Def. 79.) KA = KD, therefore (Th. 13.) the A KAD = / KDA or PDQ ; therefore (Th. 15.) PD = Pa; and (hyp.) PD= GK; therefore GK= PQ, and GK* = PQ”; but (hyp. and Th. 148. Cor. 3.) the / H.GK is a right angle; therefore (Th. 130.) KHP = KG'+GH"; therefore KAP=PQ" + GH"; therefore (Th. 123. Cor. 1. and Th.98. Cor.) the circle AEC is equal to the circle QRS together with the circle HFL. THE £itments of solin &tometry, —— CHAPTER V. 30N THE SURFACES OF SOLIDS. —O— n SECTION I. On the surfaces of solids, the boundaries of which are ł all of them planes. & ***** ************* DEFINITION LXXXII. A RIGHT Prism is that in which the parallelograms, that form its sides, are each of them at right angles to its base. THEoREM CLXVII. The convew surface of a right prism is equal to a rectangle, which has its base equal to the perimeter 3 K 442 Elementary Theorems of Solid Geometry, of the base of the prism, and its altitude equal to the altitude of the prism. For the convex surface of any prism is equal to the aggregate of the parallelograms which compose it; and since (Def. 82.) each of the parallelograms that form the sides of a right prism is perpendicular to the base, therefore (Th. 151.) their common intersections are also perpendicular to the base of the prism, and therefore (Def. 69.) they are perpendicular to the sides of the base; therefore each of the parallelograms which form the sides of a right prism is a rectangle, having one of the sides of the prism's base for its base, and the prism's altitude for its altitude: it is manifest, therefore, that the aggregate of these rectangles, that is, the convex surface of the prism, is (Th. 125.) equal to a rectangle which has its base equal to the perimeter of the base of the prism, and its altitude, equal to the altitude of the prism. a. E. D. CoR. 1. Right prisms, which have equal bases and equal altitudes, have their convex surfaces equal: it is manifest, also, (Th. 114) that, if right prisms have equal altitudes, their common surfaces are to one another as the perimeters of their bases; and if the perimeters of their bases be equal, their convex sur- faces are to one another as their altitudes. CoR. 2. If the altitudes (A) and (a) of two right prisms be reciprocally proportional to the perimeters (B) and (b) of their bases, the convex surfaces (S) and (s) of the prisms shall be equal: and if the convex surfaces (S) and (s) of two right prisms be equal, the On the Surfaces of Solids, &c. 443 altitudes (A) and (a) of the prisms shall be recipro- cally proportional to their bases (B) and (b). First, if A : a ... b : B, the rectangle contained by A and B is (Th. 115. Cor. 1.) equal to the rectangle contained by a and b, and, therefore, (Th. 167.) the convex surfaces of the two prisms are equal to one another. + Again, if, conversely, the convex surfaces of the two prisms be equal to one another, then (Th. 167.) the rectangle contained by A and B is equal to the rectangle contained by a and b, and therefore (Th. 115. Cor. 1.) A : a 3: B : b. SCHOLIUM. The comparison of plane surfaces having been fully treated of in the first part of this book, it is unnecessary further to investigate the subject of our present discussion. The preceding proposition has been demonstrated chiefly on account of the use which is to be made of it in the next following section. * 3Blententſ of $olio (ſºtomtettp. CHAPTER V. SECTION II. -Q- On the surfaces of solids the boundaries of which are not all of them planes. —O— THEOREM CLXVIII. IF from any point in the circumference of the base of a cylinder, two straight lines be drawn, the one touching the circle which is the base, the other parallel to the aris, the curved surface of the cy- linder, and the plane that passes through the straight lines so drawn, shall have no common points but those of the straight line, drawn parallel to the aris, through which the plane passes. ! On the Surfaces of Solids, &c. 445 Let the circle ADB be the base of the cylinder ABD – EF, of which CL is the axis; from any TP RLT F--> G LT& º --- IB point A, in ADB, let there be drawn GAH touching the circle ADB, and AE parallel to the axis CL; therefore, (Def. 80. Cor. 2.) AE is in the curved surface; and let the plane GI pass through GAHand AE: the plane GI and the curved sur- face of the cylinder have no common points, but those of A.E. For, let T be any point in the curved surface which is not in AE, through T, let a plane pass parallel to the plane of ADB, let it cut the plane GI in RPS, AE in P, and the axis CL in K, and let PQ be the section which it makes with the cylinder; therefore . (Th. 165.) PTO is a circle having its centre in K. 446 Elementary Theorems of Solid Geometry let KP and CA be drawn. And since (hyp.) the plane of PTQ is parallel to the plane of ADB, and GI cuts them, therefore (Th. 148.) RPS is parallel to GAH; also since (hyp.) PTO is parallel to ADB, and PA parallel to KC; therefore (Th. 148. Cor. 1.) PA = KC; therefore, (Th. 10. Cor. 4.) TR is pa- rallel to AC; since, therefore, RP is parallel to GA, and PK parallel to AC, therefore, (Th. 147.) the / KPR = / CAG; but (hyp. and Th. 46.) the / CAG is a right angle; therefore, the angle KPR is a right angle; and therefore (Th. 45.) there is no point but P common to RS and the circumference of PTO. It is manifest, therefore, that there is no point but P common to the circumference of PTO and the plane GI; therefore, the point T is out of the plane G1: and, in the same manner it may be shewn that any other point of the curved surface, which is not in AE, is out of the plane GI; where- fore, if from any point, &c. a. E. D. DEF. LXXXIII. If from any point in the circumference of the base of a cylinder two straight lines be drawn, one of them touching the base, the other parallel to the axis, the plane which passes through them is said to touch the cylinder, in the straight line parallel to the axis. CoR. If two planes which cut one *nother both of them touch a cylinder, their common section shall be On the Surfaces of Solids, &c. 447 parallel to the straight line in which either of them touches the cylinder. For (hyp.) each of the planes passes through a straight line which is parallel to the axis, and which (Th. 162. Cor. 2. and Th. 141.) is perpendicular to the plane of the base; therefore, (Th. 149.) each of these planes is perpendicular to the base of the cy- linder; therefore (Th. 151.) their common section is perpendicular to the plane of the base; and, therefore, (Th. 142.) it is parallel to the axis, and (Th. 144.) also to each of the straight lines in which the plane touches the cylinder. a DEF. LXXXIV. * A prism is said to be inscribed in a cylinder, when its bases are rectilineal figures inscribed in the bases of the cylinder: And a prism is said to be described about a cy- inder, when each of the parallelograms, that con- stitute its sides, touches the cylinder. Assumption III. The curved surface of a cylinder is greater than the aggregate of the sides of any prism inscribed in the cylinder, and less than the aggregate of the sides of any prism described about it. DEF. LXXXV. A pyramid is said to be inscribed in a cone, when 448 Elementary Theorems of Solid Geometry, its base is a rectilineal figure inscribed in the base of the come, and its vertex is the same as the vertex of the cone: and a pyramid is said to be described about a cone, when it has the same vertex as the cone, and when the bases of its triangles each of them touch the circle which is the base of the cone. THEOREM CLXIX. The curved surface of a cylinder is equal to a rec- tangle, which has for its base a straight line equal to the circumference of the cylinder's base, and for its altitude a straight line equal to the cylinder's altitude. Let ABC — DEF be a cylinder, of which ABC, DEF, are the bases: the curved surface I. s g º ſº : \; ; - - & IC i S- - # TFT TM - * : : - * - tº - : - : - - º - - * -> º º tº - - : ... • : e - - - - - o : º - - - - - ; H * * * : 23. : - .* *. - .* : : .** * - • * ** ; : ...” *. - * * º : : -. a sº * * * * . e 2. " " ; ,<----------------|---------º. *...* **. *...* * *A*. C .' * **, •. * ...' “.. ".. .." .." “.. .* *::---- •.” tº- ... • *" ** of ABC – DEF is equal to a rectangle, having * On the Surfaces of Solids, &c. 449 for its base a straight line (N) equal to the circum- ference of ABC, or of DEF, and for its altitude a straight line equal to the cylinder's altitude. For let the circumference of ABC be supposed to be divided into three equal parts in the points A, B, C, let AB, BC, CA, be drawn; therefore, (Th. 62.) the /> ABC is equilateral. Through A, B, C let there be drawn HAG, GBI, ICH, touching the circle ABC; then (Th. 66.) the As GHI is also equilateral. From A, B, C and G, H, I, let there be drawn pa- rallel to the axis of the cylinder, AD, B.E, CF, meet- ing the plane of the base DEE in the points D, E, F, therefore, (Def. 80. Cor. 2.) AD, B.E. CF, lie in the curved surface of the cylinder, and the points D, E, F, are in the circumference of the base DEF; also (hyp. and Th. 148. Cor. 1.) the straight lines AD, BE, CF, are equal as well as parallel to one another : if there- fore, DE,TEF, FD, be drawn, the figures ABED, ACFD, BCFE, will (Th. 10.Cor.4.) be parallelograms which (hyp. and Th. 149.) are perpendicular to the plane ABC: and (Def. 82. and 84.) ABC — DEF is a right prism inscribed in the cylinder. Again, let KI, a plane passing through EB and GBI, cut KHa plane passing through DA and GAH in KG, and also cut LI, a plane passing through FC and HCI in MI, and let KM, KL, and LM be the intersections of these three planes with the plane of the base DEF; then, since (hyp. and Th. 141.) EB, DA, and FC, are perpendicular to the plane of ABC; the three planes KI, KH, LI, are (Th. 149.) 3 L 450 Elementary Theorems of Solid Geometry, perpendicular to it; and, therefore, (Th. 151.) their intersections, KG, LH, MI, are also perpendicular to it; and (Th. 142.) parallel to one another; therefore (hyp. and Th. 148. Cor. 1.) KG, LH, MI, are equal as well as 'parallel to one another; therefore, (Th. 10. Cor. 4.) KGIM, KGHL, LHIM, are parallelograms each of them perpendicular to the base of the cylinder: wherefore the solid GHI–KLM is (Def. 82. and 84.) a right prism described about the cylinder. And by continually bisecting the arches AB, BC, 2-\ CA, and drawing tangents through the points of bi- section, right prisms of double, &c. the number of sides may be inscribed in and circumscribed about the cylinder, all of the same altitude with the cylinder: but (Th. 166. Cor. 1.) the convex surfaces of these prisms are to one another as the perimeters of their bases; and (Schol. to Assumption II.) two polygons may be found, one inscribed in the circle ABC, the other described about it, the difference between the primeters of each of which and a straight line (N) equal to the circumference of ABC, shall be less than any given line; therefore, two right prisms may be found, one inscribed in the cylinder, the other described about it, the difference between the convex surfaces of each of which, and a rectangle having (N) for its base and AD for its altitude, shall be less than any given surface. If it be possible, let this rectangle be greater than the curved surface of the cylinder; then, there is someinscribed right prism the convex surface of which differs less from that rectangle than the curved ^. On the Surfaces of Solids, &c. 45l surface of the cylinder does, and which is therefore greater than that curved surface; but (Assumption III.) it is also less; which is absurd. Lastly, if it be possible, let the rectangle contained by (N) and AD be less than the curved surface of the cylinder; then, there is some circumscribed right prism, the convex surface of which differs less from that rectangle than the cylinder's curved surface does, and which is therefore less than that curved surface; but (Assump- tion III.) it is also greater; which is absurd. Where- fore, the curved surface of the cylinder cannot be either greater or less than the rectangle contained by (N) and AD; that is, it is equal to it. Therefore, &c. a. E. D. CoR. 1. Cylinders which have equal bases and equal altitudes have their curved surfaces equal : it is manifest, also, from Th. 114, that if cylinders have equal altitudes, their curved surfaces are to one an- other as the circumferences of their bases; and if they have equal bases, their curved surfaces are to one an- other as their altitudes. CoR. 2. If the altitudes (4) and (a) of two cy- linders be reciprocally proportional to their bases (B) and (b), the curved surfaces (S) and (s) of the cy- linders shall be equal: and conversely. The Corollary may be demonstrated exactly in the same manner as Cor. 2. Th. 167. is demonstrated. ScholIUM. According to the same method, by which a rect- 452 Elementary Theorems of Solid Geometry, &c. angle has been found, that is equal to the curved surface of a given cylinder, it may be shewn, that the curved surface of a cone is equal to a triangle, which has for its base a straight line equal to the circum- ference of the cone's base, and for its altitude any straight line drawn from the vertex of the cone to any point in the circumference of its base: and that the surface of a sphere is the quadruple of the circle by the revolution of the half of which about its diameter, the sphere has been described. “T H E. #lements of ºolin Geometry. — Q- / CHAPTER XVI. ON THE MAGNITUDEs of son, IDs, com PARED witH on E A NOTHER.. Q SECTION I. - On the relative magnitudes of solids the boundaries of which are all of them planes. • sºº ºvººdºº ºp-º-º-º- ºw" Theorem CLXX. Solid figures contained by the same number of equal and similar planes similarly situated, and * By the phrase similarly situated, it is meant, that the similar planes containing each of the solid angles of the similar figures, are in the same order, and that their inclinations have the same kind of direction, in both the solid figures. 454 Elementary Theorems of Solid Geometry, having like inclinations to one another, are equal to one another. # LET AG and KQ be two solid figures contained by the same number of equal and similar planes, simi- TR Q. IH g "-, A. *B IK f [...d larly situated, and having like inclinations to one another; viz, let the plane AC be similar and equal to the plane KM, the plane AF to KP, BG to LQ, GD to QN, DE to NO, and FH to PR: the solid AG is equal to the solid KQ. --- For since (Th. 117. Cor. 4.) similar and equal rectilineal figures have their homologous sides equal, each to each, if the plane AC be applied to KM, so that AB may coincide with the homologous side KL, and A with K, then (hyp.) the plane AC will coincide with KM, so that AD coincides with KN, DC with WM, CB with ML, the point C with M, and D with N; but (hyp.) the planes AF, KP are on the same side of the planes AC, KM, and have like inclinations to them ; therefore the plane AF coincides with the plane KP, the point E with O, and the point F with P. because (hyp.) the figures AF, KP are similar and equal. In the same manner On the Magnitudes of Solids, &c. 455 it may be proved, that the other plane figures of the solid AG coincide with the plane figures, equal and similar to them, of the solid KQ: wherefore, since, all the planes and sides of the solid AG, coincide with the planes and sides of the solid KQ, it is manifest that the one solid must coincide with the other; that is (Def. 12.) they are equal to one an- other. Wherefore, solids, &c. a. E. D. - * CoR. It is manifest from the demonstration and from Th. 157, that two solid angles contained, each of them, by three plane angles, which are equal to one another, each to each, and similarly situated, are equal to one another. Def. LXXXVI. The Insisting Straight Lines of a Prism are the sides of the parallelograms betwixt the two bases of the Prism. CoR. The insisting straight lines of a prism are (Def.76. and Th. 25. Cor. 1.) equal and also (Th.144.) parallel, to one another. DEF. LXXXVII. A Triangular Prism is that of which the bases Q * w e © de are triangles: and also, a Triangular Pyramid is that which has a triangle for its base. THEOREM CLXXI. Two triangular prisms, having two of their three insisting straight lines common to both, and having 456 Elementary Theorems of Solid Geometry, the remaining insisting straight lines parts of one and the same straight line, are equal to one another. Let the two triangular prisms ABC—ab c, ABD — abd. have the two insisting straight lines Aa, Bºb C T) - (? cl .* º → A. common to both, and the remaining insisting straight lines CC, Dd, in one and the same straight line Cd: the prism ABC — a b c is equal to the prism ABD — abd. First, let the two insisting straight lines Co, Dil, have a part common to both: and since (hyp. and Def. 76.) Ac and Ad are parallelograms, therefore (Th. 25. Cor. 1.) Aa- Ce and Aa-Dā; therefore CC–Dd; take from both the common part DC, and CD=cd; therefore (Th. 30.) the AS CAD=/S cad,” and the /> CBD = /> c bal; also (Def. 76.) the As CBA = As cha, and the AS DBA = As dba ; and it is manifest from (Def. 76. and Th. 154.) that the planes which contain the solid figures ABCD, abcd, have like inclinations to one another; and they are similarly situated, and similar as well as equal to one sº On the Magnitudes of Solids, &c. 457 *- another; therefore (Th. 170.) the solid ABCD is equal to the solid abed; add to each the solid ABD-abe, and the prism ABC—abc is manifestly equal to the prism ABD – abd. # Secondly", let the two insisting straight lines CC, Dd of the two triangular prisms ABC— abo, ABD — abd, have no part common to both. Since CA, ca, DA, da, are in the same plane with the two parallels Aa, Cd, they are in the same plane with one another; as are, also, CB, cºb, DB, d5; and since ca meets da, it will (Th. 8. Cor. 1.) also meet DA, which (hyp. and Def. 76) is parallel to da; let * The method that is here followed, in order to shew the equality of the two prisms in this second case, has been borrowed from the second volume of the Elements of Geometry, published at Cambridge in the year 1815, , 3 M 458 Eſementary Theorems of Solid Geometry, ca cut DA in E, which point (hyp.) is manifestly between a and c; from ca cut off any part c P, such that c P is not greater than Ea. Through P let there be drawn a plane parallel to the El ABba, cutting the planes ACda, BCdb, in FH, f, the planes ACB, acb, in Ff. Pºp, and the planes ADB, a db, in Gg, Hh; also, through c let there be drawn a plane parallel to ABD, cutting the planes ACca, BC.cb, in CK and cº, and the plane f PH.h in Kk; and through D let there be drawn a plane parallel to ABC cutting the planes ADda, BDdb, in DL and d7, and the plane ff'Hh in Ll. And since (hyp. and Th. 148.) FH is parallel to Aa, and ſh to Bb and that Aa, Bb, and Cd, are (Def. 86. Cor.) parallel to one another, therefore (Th. 144.) FH, fm, and Cd, are parallel to one another: in the same manner it may be shewn that Fſ, KR, Pp, Gg, Ll, Hh are parallel to one another: and (hyp. and Th. 148, and Th. 25. Cor. 1.) cK is parallel and equal to DG, and to dB, ch to Dg and to dh, cHº to DL, and cp to D7; and (Th. 145.) the z Ke P= A GDL ; therefore (Th. 20.) the AS Ke P is equal and similar to the AS GDL: in like manner it may be shewn that the AS kcp is equal and similar to the As gl)l, the AS cKk to the AS DGg, and the AS chºp to the As DLl; also, since KK, Pp, Gg, Ll, are all parallel to one another, and that KP= GL, therefore (Th. 29.) the El KPok is equal and similar to the El GLlg; and the planes which contain the two - On the Magnitudes of Solids, &c. 459 solids KCP—kep, GDL–gdl, are similarly situated and (Th. 154.) have like inclinations to one another; therefore (Th. 170.) they are equal to one another. In the same manner it may be shewn that the solid CFf—c Kk is equal to the solid DLl—d Hh ; therefore the whole solid CFf— cFp is equal to the whole solid D.Gg—d Hh. Again, if P do not coincide with a, let Pal be taken not greater than Ea, and let there be drawn through Q a plane MNnm parallel to the El Aab B, through P a plane Ppr R parallel to ABD, and through G, a plane Ggt T parallel to ABC: and, in the same manner, as before, it may be shewn that the solid mR'—pC) is equal to the solid sG — nPI; and so on : wherefore it is manifest that the two triangular prisms ABC—abe, ABD — abd, may be separated into the same number of parts, which taken two and two are equal to one another; therefore the whole prism ABC— abo is equal to the whole prism ABD — abd. Wherefore triangular prisms, &c. a. E. D. THEOREM CLXXII. Two triangular prisms, having one of their in- sisting straight lines common to both, and having their other insisting straight lines parts of the same two parallel straight lines, are equal to one another. Let the two triangular prisms ABC-abe, DEC – dec, have the insisting straight line Co common to both, and the insisting straight lines Aa, Dal, Bb, Ee N. 460 Elementary Theorems of Solid Geometry, in the same two parallel straight lines Ad, Bc: the C two prisms are equal to one another. For, let BD, bd, be supposed to be drawn: then (Th. 171.) the prism ABC—abc is equal to the prism DBC—dbc, and also the prism DBC– d.bc is equal to the prism DEC–dec; therefore the prism ABC – abo is equal to the prism DEC-dec. Wherefore, two triangular prisms, &c. a. E. D. THEOREM CLXXIII. Triangular prisms having equal insisting straight lines, that are parts of the same three parallelstraight lines, are equal to one another. Let the two triangular prisms ABC— abo, DEF – def, have their insisting straight lines Aa, Dal, &c. equal to one another, and parts of the same three parallel straight lines Ad, Be, Cf. the two prisms are equal to one another. On the Magnitudes of Solids, &c. 461 For let AF, BF, af, bf, be supposed to be Í) à drawn; then (Th. 171.) the prism ABC—abc is equal to the prism ABF-alf; and the prism ABF–abf is equal to the prism DEF—def; therefore the prism ABC—abc is equal to the prism DEF—def. Wherefore, two prisms, &c. a. E. D. THEOREM CLXXIV. If a parallelepiped be cut by a plane, which passes through the point of bisection of any of its insisting straight lines and is parallel to its base, it shall be divided into two equal parts ; and either of the parts into which a parallelepiped is divided, by a plane so passing through the bisection of any one of its in- sisting straight lines, is equal to either of the parts into which it is divided, by a plane so passing through the bisection of any other of its insisting straight lines. Let K be the point of bisection of any of the 462. Elementary Theorems of Solid Geometry, insisting straight lines, as EA, of the paralleepiped T N G- I. jS. ...----9M. IQ : R!º y : ; L - ...a... ("N B ...' V P|A A . T) * ABCD.... EFGH; and through K let there be supposed to pass a plane, which is parallel to the base ABCD, and which therefore divides (hyp. and Def. 77.) the parallelepiped ABCD... EFGH into two parallelepipeds EM, KC: the solid EM is equal to the solid K.C. ' º For (hyp. and Th. 148.) KL is parallel to AD, and KI to AB, and IM to BC; and (Th. 161.) KM is a parallelogram equal and similar to the CI ABCD; since then (hyp.) EK=KA, therefore (Th. 29.) the also (hyp. and Th. 148. Cor. 1.) FI = EK, and IB= RA; therefore FI = IB, and the E FIMG is equal and similar to the EI IBCM: in like / On the Magnitudes of Solids, &c. 463 manner it may be shewn, that the C. EKIF is equal and similar to the El KABI, and that the Ea HLMG is equal and similar to the D LDCM. Also the planes which contain the two solids EM, KC, are similarly situated, and (Th. 154.) have like inclina- tions to one another; therefore, (Th. 170.) the solid EM is equal to the solid K.C. Again, if FG be considered as one of the insisting straight lines of the parallelepiped AG, and if a plane NP pass through the biection N of FG, parallel to the plane AF or to DG, cutting KM in RS, it may be shewn as before that AN and PG are equal pa- rallelepipeds. And it is manifest, from the demon- stration, that the solid FR is equal to the solid WL, and also to the solid KQ; therefore, the solid NL is equal to the solid KQ ; add to each the solid FR: and it is evident that the solid EM is equal to the solid EQ. Wherefore, if a parallelepiped, &c. a E. D. THEOREM CLXXV. If a parallelepiped be cut by a plane passing through the diagonals of two of the opposite parallelo- grams which contain the solid, it shall be divided into two equal parts. Let AB be a parallelepiped, and DE and CF the diagonals of the ID GB, AH, which are drawn between equal angles in each; and because CD, FE, are each of them parallel to GA, therefore, (Th. 144.) they are parallel, and (Th. 148. Cor. 1.) equal 464 Elementary Theorems of Solid Geometry, to one another; therefore, (Th. 14. Cor. 4.) CF and DE are parallel to one another and in the same plane with CD and FE; and the plane CDEF shall di- vide the solid AB into two equal parts. At the point G, in AG, and in the plane GADC, let the Z. AGK be made equal to the Z HBF, and in the plane GAEF let the / AGL be made equal to the / HBC also, let KM, LN, be made each equal to AG, and let KL, AM, and ANbe drawn. And because the / AGK = / HBF, and the / AGL = Z HBC, and that (hyp. and Th. 157.) the planes in which these angles are, have like inclinations, therefore, the third / KGL = the third / CBF. and since the Z. AGL = Z HBC, and (Th. 10.) the A. GLF = / AGL, and that (Th. 147.) the / GFL = / CBH; there- fore, the / GLF = / GFL, and therefore (Th. 15.) GL = GFor CB. In the same manner it may be shewn, that GK = BF; and the / KGL = Z. CBF; On the Magnitudes of Solids, &c. 465 therefore (Th. 20.) the As KGL is equal and similar to the /> FBC, and the /s. AMN to the AS HED; and the [I] GM, GN, NK, are equal and similar to BE, BD, DF, each to each; and they are similarly situated, and have like inclinations to one another; therefore, (Th. 170,) the prism KGL - MAN is equal to the prism FBC – EHD; but (Th. 172.) the prism KGL - MAN is, also, equal to the prism CGF – DAE; therefore, the prism CGF – DAE is equal to the prism FBC – EHD, Wherefore, if a parallelepiped, &c. a. E. D. - CoR. If both the diagonals AH, DE, of the base AEHD, of the parallelepiped AB, be drawn, and C ^- –B P - º & ſº tº & *: - . . . . . . . . . . . . . 2 : " - - - - - - - - - - - - - - - - - - - - - - - ::: * * ... • * * & ... • * s º ... e. “ gº “. . . . . . . * * * * = . . . . * * g a s “ .*** * * * º also both the diagonals GB, CF, of the opposite tº 6 FBC, and if the points P, Q, in which these diagonals cut one another be joined, then since, as hath been shewn, CF is parallel to DE and CG is parallel to DA, therefore (Th. 145.) the /, GCP = 3 N 466 Elementary Theorems of Solid Geometry, / ADQ; for the same reason the A CGP = / DAQ, and (Th. 25. Cor. 1.) GC= AD, therefore, (Th. 19. Cor.) GP is equal, as well as parallel, to AQ: therefore PQ is (Th. 10. Cor. 1.) equal and pa- rallel to GA, and therefore, (Th. 144.) PQ is also equal and parallel to CD, BH, and FE; wherefore GCP – ADQ, BFP – HEQ are prisms, and they may be shewn to be equal to one another, in the same manner as the two prisms CGF – DAE, FBC – EHD, were shewn to be equal; to each of them add the prism CPB – DQ H, and it is evident that the prism GCB – ADH is equal to the prism CFB – DEH. Wherefore, the prisms into which a paral- lelepiped is divided by a plane passing through two of the diagonals which join equal angles in its two bases, and the prisms into which it is divided by a plane passing through the other two diagonals are all equal to one another. * DEF. LXXXVIII. A prism, which has a triangle or a parallelogram for its base, is said to be contained by any three straight lines, which make angles with one another, at the same point, so as to constitute one of the solid angles of the prism. THEOREM CLXXVI. If a parallelepiped be divided into two prisms, by a plane passing through the diagonals of two of its opposite parallelograms, and also into two parallelepi- peds by a plane passing through the bisection of any e; On the Magnitudes of Solids, &c. 467 of its insisting straight lines, and parallel to its base, either of the prisms shall be equal to either of the parallelepipeds. Let the parallelepiped AG be divided into two prisms by the plane FBDH, passing through the E G. F. *T*. ; H I *—M “..... ; A' ar K%,........]…:*: #L F3 G A. T D diagonals FH, BD of the opposite [I] EG, AC; also let AG be divided into two parallelepipeds by the plane KLMI, which passes through the bisection K of any of the insisting straight lines EA, and which is parallel to the base: either of the prisms EFH – ABD, GHF– CDB, is equal to either of the pa- rallelepipeds KC, KG. w \, For let the plane drawn through K parallel to AC, cut HD in L, GC in M, and FB in I, and let IL be supposed to be drawn. Then it is manifest (from the hyp. and Th. 161, and Th. 154) that the two prisms § A68 Elementary Theorems of Solid Geometry, EFH- KIL, and KIL – ABD, are contained by the same number of equal and similar figures, similarly situated, and having like inclinations to one another ; therefore (Th. 170.) the prism KIL– ABD is equal to the prism EFH- KIL; also (Th. 174) the prism EFH – KIL is equal to the prism FHG—ILM; therefore the two prisms EFH- KIL, KIL – ABD are together equal to the two prisms EFH- KIL, FHG– ILM; that is, the whole prism EFH–ABD is equal to the parallelepiped KG; therefore, (Th. 174, and Th. 175. Cor.) any of the prisms into which a pa- rallelepiped can be divided by planes passing through the diagonals of any two of its opposite parallelograms is equal to any of the parallelepipeds into which it can be divided by a plane passing through the bisection of any of its insisting straight lines, and parallel to its base. G. E. D. CoR. Hence a triangular prism is equal to the parallelepiped contained under any two and the half of the third of any three containing straight lines of the triangular prism. THEoREM CLXXVII. Two parallelepipeds having equal insisting straight lines that are parts of the same four parallel straight dines, or having two of their insisting straight lines, common to both, and the others parts of the same two parallel straight lines, are equal to one another. Let the parallelepipeds BH, LS, have cqual in- # On the Magnitudes of Solids, &c. 469 sisting straight lines, that are parts of the same four H. . . P § parallel straight lines AN, BM, ES, FR: the solid BH is equal to the solid L.S. For let FA, GD, QK, and RN, be drawn. Then (hyp. and Th. 173.) the prism FAB- GDC is equal to the prism QKL – RNM, and the prism EFA — - HG D is equal to the prism. PQR — SRN; whence, it is manifest that the whole solid BH is equal to the whole solid L.S. And in the same manner, by the help of Th. 171. it may be shewn that two parallelepipeds, which have two of their insisting straight lines common to both, and the others parts of the same two parallel straight lines, are equal to one another. Wherefore, two pa- rallelepipeds, &c. a. E. D. THEoREM CLXXVIII. Parallelepipeds having equal bases, and having also their insisting straight lines lying between the same two parallel planes are equal to one another. Let the base of ABCD of the parallelepiped Ac be 470 Elementary Theorems of Solid Geometry, equal to the base EFGH of the parallelepiped Eg; of fº D and let the insisting straight lines Aa, Bb, Cº, Dá, Ee, Ff, G3, Hh, lie between the same two parallel planes, in the one of which are the two bases ABCD, EFGH, and in the other of which are the opposite [I] abcd, efgh: the solid Ac is equal to the solid Eg. For let the E1 EFGH be supposed to be so placed as that its side HE is not parallel to the side DA of the E ABCD; let DA and HE when produced, On the Magnitudes of Solids, &c. 471 meet in the point K, in like manner, let CB and GF, produced, meet in L, and chandgf, in M, and let the plane, in which are KL and LM, meet the plane of Ad in KN, and the plane of ac in NM; therefore (hyp. and Th. 148.) KLMN is a parallelogram: let KP be taken equal to AD; and through P let there be drawn on the plane of AC and EG, Pa parallel to KL, and meeting CL in R, HK in S, and GL in Q; also, through Plet there be drawn, in the plane of Ad, Ppparallel to KNand meeting dM in p: therefore (Th. 147) the plane passing through PQ and Pp is parallel to KLMN; let this plane meet the plane of ac in pr. the plane of Bc in Rr, that of Eh in Ss, that of eg in sq. and that of Fg in Qq; therefore (hyp. and Def. 77.) the solids Kr and Kq are paral- lelepipeds; and, if KL, NM, PR, pr; SQ, sº be considered as their insisting straight lines, they are (Th. 177.) equal to one another; also, since (hyp.) KP = AD, therefore (Th. 29.) the D-1 AC = L. KR; and (Th. 28.) the El KR = [-] KQ; also (hyp.) the LIAC = [T] EG; therefore the D-1 KQ = D EG; and therefore, (Th. 29. Cor.) LQ = FG; therefore (Th. 177.) the solid Kq is equal to the solid Eg, and the solid Kris equal to the solid Ac; and it has been shewn that Kr = Kq; therefore the solid Ac is equal to the solid E.g. Wherefore, parallelepipeds, &c. a. E. D. CoR. 1. Parallelepipeds which have the same base or equal bases, and which have equal altitudes, are equal to one another. 472 Elementary Theorems of Solid Geometry, For their bases having been placed in the same plane, the parallelograms opposite to them, will (Th. 147. Cor.) since their altitudes are equal, be in a plane which is parallel to the plane of the bases; and, there- fore, (Th. 178.) the solids will be equal to one another. CoR. 2. Of two parallelepipeds, which have equal altitudes but unequal bases, that which has the greater base is the greater. For if from the greater base, a similar parallelo- gram be cut off, (Schol. to Prop. 33.) that is equal to the less base, it will, manifestly, be the base of a parallelepiped, which hies between the same two pa- rallel planes, as that which has the greater base, and which is less than that parallelepiped, and equal to the other. - CoR. 3. It is manifest, from Th. 176. and Th. 178. Cor. 1, that a triangular prism, is equal to a parallelepiped, which has an equal base and an equal altitude. º CoR. 4. Hence all prisms that have the same base, or equal bases, and equal altitudes, are equal to one another. For any prism, which is not a triangular prism, may be divided into triangular prisms, by drawing straight lines from any two corresponding angular points in its two bases, to the remaining angular points. Also (Schol, to Th. 32.) a parallelogram may be found On the Magnitudes of Solids, &c. 473 which shall be equal to the base of the prism, and the parts of which shall be equal to the triangles into which the base is divided each to each. If, therefore, upon that parallelogram a parallelepiped be supposed to be constituted of the same altitude as the prism, and if it be divided into parallelepipeds by planes parallel to its sides, and passing through the straight lines which divide its base into parallelograms that are equal to the triangles, into which the prism's base is divided, then (Cor. 3.) the several parts of the two solids will be equal each to each; and, therefore, the whole prism will be equal to the whole parallelepiped. In the same manner it may be shewn that any other prism, which has an equal base, and an equal altitude, is equal to that same parallelepiped; therefore, all prisms which have equal bases and equal altitudes are equal to one another. THEOREM CLXXIX. Parallelepipeds that have equal altitudes are to one another as their bases. Let the parallelepipeds AD, EH, of which the P—º- H. V X. .** * tº- ºf 20 ...” .# .*T †—# †-I. f T ū-É = Š bases are AC and EG, have equal altitudes: the 3 O 474 Elementary Theorems of Solid Geometry, solid AD is to the solid EH as the base AC to the base EG. - For in AB and EF, produced, let any number of parts BK, KL, &c.betaken each equal to AB, and FQ, QR, RS, &c. each equal to EF: let the TICK, ML, &c. and the TI GQ, TR, WS, &c. be supposed to be completed; therefore (Th. 29.) the TI AC, BM, KO, &c. are equal, as are, also, the [T] EG, FT, QM’, RF, &c.; if, therefore, the solids BN, KP, &c., and the solids FP, QX, RZ, &c. be completed, the solids AD, BN, KP, &c. will (Th. 178.) be equal, as will, like- wise, the solids EH, FP, QX, RZ, &c.; therefore whatever multiple the base AO is of the base AC, the same is the solid AP of the solid AD; and whatever multiple EP is of EG the same is the solid EZ of the solid E.H ; but (hyp. and Th. 178. Cor. 1. 2.) if AO > = < EY, AP × = < EZ ; therefore (Def. 48.) AC : EG :: AD : EH. Wherefore pa- rallelepipeds, &c. a. E. D. CoR. It is manifest, therefore, that all prisms which have equal altitudes are to one another as their bases. For parallelepipeds, which, having their altitudes and their bases equal to those of the prisms, are (Th. 178. Cor. 4.) equal to the prisms, are to one another (Th. 179.) as their bases. THEoREM CLXXX. Parallelepipeds, which have equal bases, are to one another as their altitudes. On the Magnitudes of Solids, &c. 475 Let the parallelepipeds DC, HF, be upon equal º ** ; : .* º: A. bases, AC, and EG; and let DI, HK, which are perpendicular to AC and EF, respectively, be the altitudes of the parallelepipeds : the solid HG is to the solid DC as HK is to DI. In the greater altitude HK, let KL be taken equal to DH the less; and through L suppose the plane MN to be drawn perpendicular to HK, and therefore (Th. 146.) parallel to the plane EG ; suppose, also, ML and EK to be drawn, which (hyp. Def. 69. and Th. 5.) are parallel to one another; wherefore MG is a parallelepiped, having its base and altitude re- spectively equal to the base and altitude of the parallelepiped DC; therefore (Th. 178. Cor. 1.) MG is equal to DC: and if the T1HF, MF, be considered as the bases of the solids HG, MG, these solids have the same altitude; *. ... (Th. 179.) HG : MG : [-] HF : [-]MF; but (Th. 114.) El HF: D MF :: HE : ME; 476 Elementary Theorems of Solid Geometry, and (Th. 99) HE : ME :: HK : LK. But the solid DC = MG, and DI= LK ; ... (Th. 77.) HG : DC :: HK : D.I. Therefore, parallelepipeds, &c. a. E. D. CoR. 1. Wherefore, all prisms, which have equal bases, are to one another as their altitudes. For, parallelepipeds may be 'supposed to stand upon bases equal to those of the prisms, and also to have their altitudes equal to the altitudes of the prisms; since, then, these parallelepipeds are to one another as their altitudes, the prisms, which are equal to them, are to one another (Th. 81.) as their altitudes. CoR. 2. It is evident from the demonstration, that, a prism being given of any definite altitude, another similar prism may be found which shall stand upon the same base with the given prism, and which shall be of any other definite altitude. THEOREM CLXXXI. Any two prisms are to one another in the ratio which is compounded of the ratios of their altitudes and of their bases. Let B be the base, and 4 the altitude of any prism X, and let b be the base, and a the altitude of any other prism Y: then is X to Y in the ratio which is compounded of the ratios of A to a and of B to b. - For if Z be a prism which (Th. 180. Cor. 2.) has B for its base, and a for its altitude, then On the Magnitudes of Solids, &c. 477. (Th. 180. Cor. 1.) X : Z :: A : a ; and (Th. 179. Cor.) Z : Y :: B : b ; therefore (Def. 60. Cor.) X is to Y in a ratio compounded of the ratios of A to a and of B to b. Wherefore, any two prisms, &c. a. E. D. CoR. 1. Since (Schol. to Th. 114.) two straight E lines may be found, which shall be to one another as any two given plane rectilineal figures, if the base B, of the prism X, be to the base b of the prism P as D is to E; and if (Schol. to Th. 99.) the altitude A, of the prism X, be to the altitude a, of the prism Y, as E is to F; it is manifest, from the proposition, that the prism X is to the prism P as D is to F. that is, two straight lines can always be found which shall have to one another a ratio equivalent to the ratio of any two given prisms. 478 Elementary. Theorems of Solid Geometry, CoR. 2. The bases and altitudes of equal prisms are reciprocally proportional; and prisms of which the bases and altitudes are reciprocally proportional are equal to one another. Let B be the base, and A the altitude of the prism A, and let b be the base, and a the altitude of the prism F: and, first, let X = P : then B : b : a A. . For if B : b : D : E, and A : a 3: E. : F, then (Th. 181.) X: P : D : F, Af But (hyp.) X= P ; therefore (Th. 84.) D = F; - ... A : a ... E : D ; ... (Th. 79.) a . A :: and (hyp.) B : b :: ... (Th. 81.) B : b :: Secondly, let B : b : a . A ; t A be equal to the prism P. r For, if B : b : D : E, and A : a ... E : F, then (Th. 181) X: P : D : F; - and since (hyp.) A : a .. ... (Th. 79.) a . A :: also (hyp.) B : b :: 2 ... (Th. 81.) D : E :: 2 ... (Th. 78.) D= F, and ... (Th. 84.) X=F. 9 2 en shall the prism D : D : (l, A : h E F (! F SCHOLIUM. If (Schol. to Th. 33. and to Th. 166.) a numerical ratio can be found equivalent to the ratio of the bases of two given prisms, and if, also, the ratio of the altitudes of the two prisms be equivalent to another given numerical ratio, then (Th. 181, and Schol. to On the Magnitudes of Solids, &c. 479 Th. 98.) the two prisms will be to one another as the product of the antecedents of these numerical ratios is to the product of the consequents : and thus two numbers may be found, which shall be to one another, as the two given prisms are to one another. THEOREM CLXXXII. If a triangular pyramid be cut by two planes parallel to its base, the difference between the part of the pyramid between two such planes, and a triangular prism between the same two planes, which has for its base the section nearest to the base of the pyramid, shall be less than a prism having the same altitude as the prism between the two sections, and having for its base the difference of the two sections. Let DEF, GHK, be two sections of the triangular pyramid, S- ABC, when it is cut by planes parallel 480 Elementary Theorems of Solid Geometry, to its base ABC. In HG, the common section of the two planes SAB, GHK, let Hal be taken equal to ED; and, likewise, in HK the common section of the two planes SBC, GHK, let Hf be taken equal to EF, let D, F and d, f be supposed to be joined; and in ED and EF let there be taken Eg = HG, and Ek = HK, and let G, g, and K., k, be joined: then since (hyp. and Th. 148.) HGd is parallel to Eg|D. and HKf to ERF, therefore, (hyp. and Th. 10. Cor. 4.) Ddgg, Dāh B, g(HE, EHKk, EHf F, kkff, Ddf|F, g(; Kh, are parallelograms; and the as DEF, dHſ, having their sides (Th. 25. Cor. 1.) equal, have (Th. 24. Cor. 1.) their angles equal, and are similar and equal to one another, as are, also, for the same reason the />g|Ek, GHK; it is manifest, there- fore, that the figure DgkP is equal and similar to the figure d6 Kf; therefore, (hyp. and Def. 76.) DEF— dHf, GHK –g Bk, are prisms between the two parallel sections DEF, GHK, having those sections for their bases; and Dgh F- d6 Kf is a prism between the same two planes, having DgkH', the difference of the two sections for its base, and being itself the dif- ference of the two prisms DEF—d Hf, GHK - gEk: but the part of the pyramid between the two sections is manifestly greater than the prism GHK– gEk, which lies within it; therefore, the difference between that portion of the pyramid and the prism DEF – dBlf is less than the prism DgkE – d6Kf, which is the difference between the two prisms DEF —d Hf, and GHK–g Ek. Wherefore, if a triangular pyramid, &c. a. E. D. On the Magnitudes of Solids, &c. 481 CoR. 1. In the same manner it may be shewn that the difference between the part of a pyramid, between its base and any plane parallel to its base; and a triangular prism of the same altitude, having the base of the pyramid for its base, is less than a prism of that same altitude, having for its base the difference between the base of the pyramid and its section made by the parallel plane. * CoR. 2. The altitude of a triangular pyramid having been divided into any number of equal parts, if planes parallel to the base be drawn through the points of division, and if the base and the several sec- tions be made the bases of triangular prisms described about and inscribed in the pyramid, as in the demon- stration of Th. 182, it is manifest, from that de- monstration, that the aggregate of the inscribed prisms is equal to the aggregate of all the circumscribed prisms, but that which is on the base of the pyramid, and which is therefore equal to the difference between the aggregates of the circumscribed and of the in- scribed prisms, and is greater than the difference between the aggregate of the circumscribed prisms and the pyramid itself. THEOREM CLXXXIII. Triangular pyramids which have equal bases and equal altitudes, are equal to one another. Let the two triangular pyramids S – ABC, T– DEF have equal bases ABC, DEF, and equal 3 P 482 Elementary Theorems of Solid Geometry, altitudes SG, TH: the two pyramids S- ABC, T- *ru T S; º DEF are equal to one another. For if the equal altitudes SG, TH, be supposed to be divided into the same number of equal parts; and if through the points of division, planes be supposed to be drawn parallel to the bases of the solids, the corresponding sections will (Th. 160. Cor. 2.) be equal; and if upon these several sections as bases, triangular prisms be supposed to be constructed between the parallel planes of the sections, as in Th. 182, the cor- responding prisms will (Th. 178. Cor. 4.) be equal to one another; and, therefore, the aggregate of the one set of these prisms will be equal to the aggregate of the other set: but (Th. 182. Cor. 2.) the difference between the pyramid S – ABC and the aggregate of the prisms about the pyramid S – ABC, is less than the prism having for its base the base ABC, and having On the Magnitudes of Solids, &c. •º 483 one of the equal parts of SG for its altitude; which prism, since the magnitude of the equal parts of SG may (Th. 68.) be diminished indefinitely, may (Th. 180. Cor. 1.) be made less than any given finite prism; wherefore, the pyramid S – ABC is the limit of the aggregate of the prisms so described about it; for the same reason, the prism T – DEF is, also, the limit of the equal aggregate of the prisms similarly described about it; therefore", the pyramid S.– ABC is equal to the pyramid T-DEF. Wherefore, tri- angular pyramids, &c. a. E. D. THEOREM CLXXXIV. Every triangular pyramid is the third part of a prism of the same altitude, and having the same base. Let D — ABC be a triangular pyramid: it is the #3' º third part of a prism of the same altitude, which has ABC for its base. . s *If A and B be each a limit of the same greater variable magnitude X, A= B. If not, one of them is the greater. Let A > B; wherefore A and B have a finite difference; and X > A ; much more, then, have X and B a finite difference; which is contrary to the hypothesis. 484 Elementary T.heorems of Solid Geometry, ‘For, if from B, and A, BE and AF be drawn parallel to CD, and if these three parallels be cut by a plane DEFparallel to the base ABC, it may be shewn, as in Th. 182, that ABC–DEF is a prism, and it has the same altitude as the pyramid D – ABC. From the prism ABC — DEF, let there be taken the py- ramid B — EFD, and there remain the two pyramids B – ADC, B – ADF, which, since (Th. 25.) they F —9 • * 2 * * g * * sº * a * • * s ** • * a” • * .** & e” à s & e s tº e is ºdºº wes we a w w = * * *p, * * * * * * * * * * * * * * * C º have equal bases ADC, ADF, and a common altitude are (Th. 183.) equal to one another; also (Th. 183.) the pyramid B — EFD is equal to the pyramid D– ABC, which is the same as the pyramid B – ADC. Wherefore, the pyramid D – ABC is the third part of the prism ABC – DEF, which (Th. 178. Cor. 4.) is equal to any other prism on the same base ABC, and of the same altitude. Wherefore, every pyramid, &c. a. E. D. * CoR. Every triangular prism may be divided into three equal triangular pyramids. m On the Magnitudes of Solids, &c. 485 THEOREM CLXXXV. \ Every pyramid is the third part of a prism of the same altitude, and having the same base. Let F – ABCDE be any pyramid: it is the third part of a prism, of the same altitude, which has | il K. & A /~i? “N/s. TA B ABCDE for its base. For parallel to any one of the sides of the triangles which terminate in F, as AF, let there be drawn from the other angles B, C, D, E of the base, BG, CH, DI, EK, parallel to AF, and let these parallels be cut by the plane FGHIK parallel to the base ABCDE; also let AC, AD, FH, FI be drawn: then it may be shewn, as in Th. 182, that ABCDE — FGHIK is a prism, and also, if planes be sup- posed to pass through AC and FH, and through AD 486 Elementary Theorems of Solid Geometry, &c. and FI, that these planes will divide the prism ABCDE –FGHIK, into triangular prisms having for their bases the corresponding triangles, into which the bases ABCDE – FGHIK, have been divided by the straight lines drawn from A and F: but these same planes divide the pyramid F – ABCDE into the same number of triangular pyramids, having the same bases as the several triangular prisms; and the prisms and pyramids have all the same altitudes, namely, that of the pyramid F-ABCDE. Wherefore the prism ABCDE – FGHIK, and the pyramid F- ABCDE, are thus divided into the same number of parts, and each part of the pyramid is (Th. 184.) a third part of the prism; therefore, (Th. 83.) the whole pyramid F – ABCDE is a third part of the whole prism ABCDE – FGHIK, which (Th. 178. Cor. 4.) is equal to any other prism on the same base and of the same altitude. Wherefore, every prism, &c. Q. E. D. CoR. It is manifest, from Th. 179. Cor. Th. 180. Cor. 1. Th. 181. Th. 86. and Th. 81. that pyramids of equal altitudes are to one another as their bases; that pyramids of equal bases are to one another as their altitudes; and that any two pyramids are to one another in the ratio which is compounded of the ratios of their altitudes and of their bases; and it is further manifest, from Th. 181. Cor. 2. that the bases and altitudes of equal pyramids are reciprocally propor- tional; and that pyramids which have their bases and altitudes reciprocally proportional are equal to onc another. TfLE 3Elements of $olio (%tomtettp, CHAPTER VII. * SECTION II. . On the relative magnitudes of solids, the boundaries of which are not all of them planes. •ºwº arezºv******** Theorem CLXXXVI. A cylinder and a parallelepiped, having equal bases and equal altitudes, are equal to one another. LET the cylinder Ac and the parallelepiped EH be of equal altitudes, and have, also, their bases AC, EG, equal to one another: the cylinder Ac is equal to the parallelepiped EH. 488 Elementary Theorems of Solid Geometry, For if not, one of them is the greater; let E H be greater than the cylinder Ac; and from EH, the - i. J& F greater, let the parallelepiped EM be supposed to be cut off, by a plane that is parallel to FH, such that EM is equal to the cylinder Ac; and since EL is less than EG, it is, also, (hyp.) less than the circle ADC; therefore, (Th. 69. Cor. 4.) a polygon ABCD may be inscribed in the circle ADC, which shall be greater than EL; and if from the angular points A, B, C, D, of the inscribed polygon there be drawn Aa, Bb, Co, Dä, parallel to the axis of the cylinder and meeting its base ac, in a, b, c, d, and if planes be supposed to pass through these parallels, the solid ABCD – abcd will (Def. 80. Cor. 2. and Def. 76.) alí On the Relative Magnitudes of Solids, &c. 489 be a prism, inscribed in the cylinder; and (hyp.) it is of the same altitude as the solid E.M.; and it has a greater base; therefore (Th. 179. Cor. and Th. 84) the prism ABCD —abcd is greater than the solid E.M.; much more then is cylinder Ac, of which the prism is a part, greater than the solid E.M.; but the cylinder is also equal to EM; which is impossible; therefore EH cannot be greater than the cylinder Ac: and in the same manner, it may be shewn, that a prism may be described about the cylinder which shall be less than any parallelepiped of the same altitude, the base of which exceeds Eg, and that, therefore, the solid EH cannot be less than the cylinder Ac; therefore, the cylinder Ac is equal to the parallelepiped E.H. Therefore, a cylinder, &c. a. E. D. CoR. 1. Wherefore (Th. 178. Cor. 4.) a cylinder and any prism, which have equal bases and equal al- titudes, are equal to one another; and cylinders, which have equal bases and equal altitudes, are also equal to one another. CoR. 2. It is manifest (Th. 179. Cor. Th. 180. Cor. 1. Th. 181. Th. 86. Th. 81.) that cylinders, of equal altitudes, are to one another as their bases; that cylinders, which have equal bases, are to one another as their altitudes; that any two cylinders are to one another in the ratio, which is compounded of the ratios of their altitudes and of their bases: and (Th. 181. Cor. 2.) it is also evident, that the bases and altitudes of equal cylinders are reciprocally propor- tional; and that, conversely, cylinders, which have their bases and altitudes reciprocally proportional, are equal to one another. 3 Q. 490 Elementary Theorems of Solid Geometry, CoR. 3. If the axes of two cylinders be to one another as the diameters of their bases, the solids shall be to one another in the triplicate ratio of their axes, or of the diameters of their bases. For the compound ratio which (Cor. 2.) is equi- valent to the ratio of the solids, will then be com- pounded of three ratios, each of which is the ratio of an axis to an axis, or of a diameter to a diameter; because (Th. 123.) the bases are to one another as the squares of their diameters, that is (Th. 117.) they are to one another in the duplicate ratio of their diameters, or of (hyp.) the axes; therefore (Def. 61.) the cylinders are to one another in the triplicate ratio of their axes, or of the diameters of their bases, CoR. 4. Hence, if the axes of two cylinders be to one another as their bases, the cylinders shall be to one another as two cubes, the sides of the squares containing which, are respectively equal to the axes of the cylinders, or respectively equal to the diameters of the bases of the cylinders. For (Th. 181.) these cubes will be to one another in the same triplicate ratio as that in which the cylinders (Cor. 3.) are to one another; therefore (Th. 81.) the cylinders are to one another as these cubes. - CoR. 5. It follows from Th. 186. Cor. 1. and Th. 185, that, if a pyramid and a cylinder have equal bases and equal altitudes, the pyramid is a third part of the cylinder. On the Relative Magnitudes of Solids, &c. 491 \ THEOREM CLXXXVII. A cone and a triangular pyramid having equal bases and equal altitudes, are equal to one another. Let the cone S- ABCD, and the triangular py- ramid T- EFG, be of equal altitudes, and have their T bases ADB, EFG, equal. The cone S– ADB is equal to the pyramid T-EFG. . * For, if it be possible, let the pyramid T– EFG be greater than the cone, and let a part of it T- KFGH be cut off by a plane TKH, so that the remainder T-EKH is equal to the cone; therefore the circle ADB is greater than the AS EKH, and in the circle there may (Th. 69. Cor. 4) be inscribed a polygon ABCD which is greater than EKH; and if SA, SB, 492 Elementary Theorem: of Solid Geometry, SC, SD be drawn, the pyramid S.– ABCD, which is contained in the cone, will (Th. 185. Cor. and Th. 84.) be greater than the pyramid T-EKH; much more, then, is the cone S-ADB greater than the pyramid T-EKH; but it is also equal to it, which is impossible; therefore the pyramid T-EFG cannot be greater than the come S–ADB : and, in the same manner it may be shewn that it cannot be less than S— ADB ; therefore the cone S – ADB is equal to the pyramid T-EFG. Wherefore, a cone, &c. G. E. D. CoR. 1. A cone and any pyramid having equal bases and equal altitudes, are equal to one another: and cones, also, which have equal bases and equal altitudes are equal to one another. For (Th. 185.) all pyramids, which have equal bases and equal altitudes, are equal to one another. CoR. 2. Whatever was proved of prisms in Th. 179. Cor. Th. 180. Cor. 1, Th. 181. Cor. 2, is true of cones; because it is true (Th. 185.) of py- ramids that have equal bases, and equal altitudes, and are equal to the cones. CoR. 3. Every cone is the third part of a cylinder of the same altitude, and having the same base. For a pyramid, which (Th. 187.) is equal to the cone, is also (Th. 186. Cor. 5.) a third part of a cylinder of the same base and altitude. On the Relative Magnitudes of Solids, &c. 493 Theorem CLXXXVIII. Two sets of cylinders may be described, the one of which shall be greater, and the other less, than a given hemisphere, and the difference between which shall be less than any given cylinder. Let ABD be a given hemisphere, of which CA is the axis, C being the sphere's centre: a set of cy- lº linders may be described, which shall exceed ABD, and another set may be described which shall be less than ABD, such that the difference between the two sets shall be less than any given cylinder W. Let the axis CA be supposed to be divided into any number of equal parts, in E, F, G &c. and let the solid be cut by planes, perpendicular to its axis, passing through E, F, G, &c.; therefore (Th. 166.) the sections will be circles; again, let HEI, KFL, MGN, be the diameters of these circular sections, determined by a plane BAD which cuts the solid, by passing through 494 Elementary Theorems of Solid Geometry, its axis AC. also let two cylinders be supposed to be described, on contrary sides of the circle of which HI is the diameter, having that circle for their common base, and the two equal parts EA, EF, of AC, for their equal axes; and let hI and Ii be the sections of these cylinders, cut by the plane BAD; in like manner, let Ll, and Lºm, Nn and Np, &c. be the sections of the cylinders described on contrary sides of the circles, of which KL, MW, &c. are the diameters, and which have for their axes the several equal parts into which AC is divided: and let Dg be a cylinder having the same base as the given he- misphere, and having for its axis CG the last of the equal parts. Then since (Th. 44. and hyp.) KLPHI, MW-> KL, &c., and BD > MN, it is manifest that the cylinders, of which ply, m L, i I, are the sections, being contained in the hemisphere, are together less than it, and that the cylinders, of which q D, nN, ll, hl, are the sections, each of them containing a portion of the hemisphere, are together greater than the cylinder. Again (hyp. and Th. 186. Cor. 1.) the cylinders Ih, Ji are equal, as are likewise the cylinders Ll, Lºm, and the cylinders Nn, Wo: it is manifest, there- fore, that the difference between the set of inscribed cylinders and the other set is equal to the last cylinder gl). If, then, the cylinderg D be not less than W, let the axis be supposed to be divided into twice as many parts, and let the same construction be made, as before; thus the last cylinder, which will On the Relative Magnitudes of Solids, &c. 495 still be the difference between the set of inscribed cylinders and the other set, will be diminished by its half; and if this diminution be continually repeated, there will (Th. 68.) at last remain a cylinder which is less than the given cylinder W. CoR. It is manifest, from the demonstration, that two sets of cylinders may, in like manner be described, the one of which shall be greater and the other less, than a given cone, and the difference between which shall be less than any given cylinder. THEOREM CLXXXIX. If a hemisphere and a right-angled cone be wpon the same base, the hemisphere shall be double of the CO%26. Let the hemisphere ABC and the right-angled cone ADC, have a common base, of which K is the centre: the hemisphere ABC is double of the come ADC. Let DK be the cone's axis, and let it be produced. to meet the surface of the hemisphere in B; also let EC be a cylinder having the same base AC, and the same axis DK, as the cone. Let EAC1, DAC, ABC, be the sections of these three solids when they are cut by a plane passing through the straight line DKB, in which are their axes; therefore (Def 79. and hyp.) DK= KA, and (Def.81. Cor. 1.) KA=KB; therefore DK= KB : let the equal straight lines KB and KD be divided each into the same number 496 Elementary Theorems of Solid Geometry, of equal parts in F, G, H, and f, g, h; through these points of division let there be drawn planes parallel to the base AC, and let the planes so drawn through f, g, h, meet the cylinder EC in e, c, a ; therefore (Th. 165. and Th. 186. Cor. 1.) the cylinders Eh, ed, cb, a C, into which they divide the cylinder EC, are all equal to one another. And, if cylinders be described in and about the cone and the hemisphere, having the base, and the sections of these solids made by the parallel planes, for their bases, and having the equal parts, into which DKB is divided, for their axes, then, since (Th. 166. Cor. 2.) the first circular section of the cone together with the last of the hemisphere is equal to the common base AC of the cone, and the hemisphere, therefore (Thr186. Cor. 1. On the Relative Magnitudes of Solids, &c. 497 and Th. 98. Cor.) the first cylinder inscribed in the cone together with the last cylinder inscribed in the hemisphere is equal to the cylinder a C; the second in the come together with the last but one in the hemisphere is equal to the cylinder cb, and so on : thus all the cylinders inscribed in the cone and in the hemisphere are, together, equal to the cylinder e C: and it was shewn in the demonstration of Th. 188, that all the cylinders inscribed in the he- misphere together with the cylinder a C, are equal to all the cylinders described about it; and, in like manner, it may be shewn that all the cylinders in- scribed in the cone together with the cylinder a C, are equal to all the cylinders described about it; wherefore if to eC be added the two cylinders a C, a C, or if to Ec be added a cylinder eI, having the same base EI as the cylinder EC, and having its axis D3 equal to one of the equal parts of DK, it is evident that the cylinder eC will be equal to the aggregate of the cylinders described about the cone and the hemisphere: again it may be shewn, as in the proof of Th. 188, that the cylinders e I and e I, may be made to become less than any given cylinder ; that is, the difference between the cylinders EC and each of the cylinders e C and e C may be made less than any given cylinder; but (Th. 188. and Cor.) the differences between the compound solid DABC, and each of the cylinders e C and e C, may also be made less than any given cylinder ; therefore (Th. 69. Cor. 3.) the compound solid DABC is equal to the cylinder EC; but (Th. 187. Cor. 3.) the cylinder 3 R. 498 Elementary Theorems of Solid Geometry, EC is triple the cone DAC; therefore (Th. 77.) the compound solid DABC is to the come DAC as three to one; therefore dividendo (Th. 88.) the hemisphere is to the come as two to one; that is, the hemisphere ABC is double of the cone ADC. Wherefore, if a hemisphere, &c. a. E. D. CoR. 1. From the demonstration, and from Th.86, it is manifest, that a sphere is to a cylinder, having its axis, and the diameter of its base, each equal to a diameter of the sphere, as the number two is to three. CoR. 2. Wherefore (Th. 186. Cor. 3. and Cor. 4. and Th. 81.) spheres are to one another in the tri- plicate ratio of their diameters; or in the ratio of cubes, such that the squares containing them have their sides respectively equal to the diameters of the spheres. * ScholIUM. The relative magnitude of a hemisphere, compared with that of a right-angled cone upon the same base, has been ascertained chiefly by means of the property belonging to those two solids which is demonstrated in Th. 166. Cor. 2. Now a similar property also obtains, in the solid generated by the revolution of the figure, called a parabola, about its axis. For if any two sections of the paraboloid be taken, the one at the same distance from the vertex of the solid, that the other is from the base, they will be together equal to the base. It is easy, therefore, to apply the reason- ing used in Th. 189, so as to shew that a paraboloid On the Relative Magnitudes of Solids, &c. 499 is the half of a cylinder upon the same base and of the same altitude. This conclusion, as well as that of Th. 189, might, indeed, be arrived at, by a less cir- cuitous way; namely, by the Method of Indivisibles. But then the principles of that method have not so manifestly firm a foundation, as that upon which are built the longer demonstrations of the ancient geo- metricians. ADDENDA, —º- The two following theorems may serve to com- plete what has been said in Schol. Def. 48. §. 11. 1. Any two unequal magnitudes (A B) and (AC) being given, an intermediate magnitude may be found which shall be commensurable with a third given magnitude (DE) of the same kind. Bisect DE in F, DF in G, and so on, until 9 B A H D G F E a part DG shall have been found, less (Th. 68.) than CB, the difference between AB and AC; and let (H) be the least multiple of DG which exceeds AC; then H → AB; for if not, a part equal to DG might be taken from (H) and there would remain a multiple of DG greater than AC, because DG & CB; which is contrary to the supposition: wherefore, H & AB and > AC: And DG measures H and D.E. 2. Any two incommensurable magnitudes (AB) and (DE) being given, a third magnitude (H) may be found commensurable with one of them (DE) and the difference between which and the other (AB) shall be less than any assigned magnitude. For let AC and AB have any assigned difference CB, however small; then (1) a magnitude (H) may be found commensurable with DE, such that H > AC; and H & AB; the difference, therefore, between AB and H, which is commensurable with it, is less than the assigned magnitude CB. J A N # I N D E X Shewing the order in which the PRoPositions of EUCLID are arranged in this Treatise. -º- EUCLID. Book I. Proposition Page Proposition * | *śī..." º' ... #::::::::::Pº H. tº e º O e º º : XXXVII. ....—XXX..............81 mi. in............sº XXXVIII. ...—XXX..............81 iv..........th XX • e º 'º e s e º ºs º is e º º 56 XXXIX......—XXXI.............84 v. Tºxiii.......... ... XI. . . . . . . . . —XXXI.............84 vi........... ºv.............. * | XII: . . . . . . . . —XXXII. ...... . . . . .85 vii gº º tº dº nº º sº is º º –xxiv. 'Cor. ºtº e º ſº º ºs 66 XLII... . . . . . . Prob. XVIII......... .332 viii....... ºxiv. o. i., | XIIIſ. ...... Th. XXXIII...........86 ix...........prob. v.............. 3. XLIV........ Prob. XIX. . . . . . . . . . .333 x. Liv.............sis XLV. . . . . . . . XX. . . . . . . . . . . .334 xi.......... VI. . . . . . . . . . . . . .315 XLVI. . . . . ..—XXI........... .336 xii. tº e º º ºs º º is Twº tº º ºs º º º ſº e º & G tº 315 XLVII.----- Th. CXXX. .......... 295. xiii.........Dei.xiii. Corº....... 2, XLVIII......—CXXXVII. Cor. 2.307 XIV. ........Th. IV. . . . . . . . . . . . . . . . 22 XV..........-I. Cor. . . . . . . . . . . . • 17 EUCLID. Book II. XVI. . . . . . . . . —XI. . . . . . . . . . . . . . . . 41 XVII........ —XI. Cor. 1. . . . . . . . .4% He e º 'º º º ºs º ºs º º & Th. gº &...: XVIII — XIV tº 47 ...” ‘’’ ‘’’ tº º se * . COr. 1. . . .286 ix........ xvi........... tº tº III. . . . . . . . . . . —CXXV. Cor. 2. . . .286 XIX. . . . . . . . . sºmsºmºsº “43 IV........... -CXXXI. Cor. 2...297 § • p * s e s a e e Tºi tº º & E º tº gº tº e º ſº tº * V. . . . . . . . . . . . —CXXXIV. Cor... .302 .*.*.*.*.*.*.*.*, º ſº º ſº º te º ſº e T -4° W lllli e s e s e e s e s , e. e. * | VI........... —CXXXIV. Cor... .302 XXII. . . . . . . . Prob. XVII. ........ • .330 | VII. ......... —CXXXII, Cor.....299 XXIII. . . . . . . VII. ........... 316 VIII......... —CXXXIII. Cor. . .300 XXIV. . . . . . .Th. XXI.............. •60 IX........... -CXXXV. ........ 302 XXV. . . . . . . .-XXII. ............ 62 X....... . . . . .-CXXXV. ........ 302. XXVI. . . . . . . -XIX. . . . . .........53 | x1,.......... Prob. XVI. Cor. 2..... .328 XXVII: .....—V........... tº e º ºs e tº 23 | XII..... .....Th. CXXXVI. Cor... .305 XXVIII. ....-VI... "........... .25 XIII. . . . . . . . ..—CXXXVII. Cor. 1.307 sº • * * * * Ex tº dº e º ſº e º e º e º '• • * : XIV. sº e º ſº tº tº ... Prob. XXIII. Cor. • . . .340 xxxi....... Pºviii............. - §...º. º: EUCLID. Book III. XXXIII......—X. Cor. 4. . . . . . . . . . .38 XXXIV.,....—XXV.,...,...,...,68 I. . . . . . . . . . . . Prob. XXVII. . . . . . . . . .350 II. . . . . . . . . . . Def. XXIX, Cor. . . . . . . . .96 3 S 2 . I N D E X. Proposition tº tº " * iii........Th. xxxvi..........: EUCLID. Book V. TV........... — XXXVIII. . . . . . . ...99 || Proposition Page V. . . . . . ...... —LII........ tº $ tº º tº te & 193 * “.........Th. LXXI............173 VI. . . . . . . . . . . - LI. . . . . . . . . . . . . . . 122 | "........... • – LXXII........ ...175 VII.......... -XL............... 102 | III........... – LXXIII. ......... 175 VIII. ........—XLI. and XLII.104–6 IV. .......... —XCIII............212 IX. .......... —XXXIX..........100 V. . . . ....... •- LXXIV...........176 X............ — LIII. ............ 124 VI. .......... - LXXV. ..........177 XI........... —XLIX............ 120 | VII. ......... —LXXVII.........200 XII. .......• ‘T-L. . . . . . . . . . . . . . . . . * | VIII. ........ {º}. and LXXVIII. XIII. . . . . . . . . } Cor. to Th. XLIX, Cor........177 and 201 & and L.... . . . . . 120. 122. IX. . . . . tº 9 º' tº e g -LXXVIII. ... . . . . .200 XIV. . . . . . . . . — XLIII. ........... 109 | XI. .......... – LXXXI. ... . . . . . .202 XV. . . . . . . . . . —XLIV. ........... 110 | XII. ..........—LXXXIII. .......203 XVI. . . . . . . . . --XLV. . . . . . . . . . . . . 112 | XIV. ........ '.- LXXXV......... .204 XVII. . . . . . . . Prob. XXX. . . . . . . . . . . 354 | XV. .......... - LXXXVI. . . . . . . .205 XVIII. . . . . . . Th. XLVI. . . . . . . . . . . . 115 XVI.......... – LXXXVII... . . . . .206 XIX. . . . . . . . . — XLVII. . . . . . . . . . . 116 | XVII......... – LXXXVIII. . . . . .207 XX. . . . . . . . . . –LVI. . . . . . . . . . . . . . 129 | XVIII. ....... — LXXXIX....... .208 XXI. . . . . . . . . —LV. . . . . . . . tº E e º is as 128 XIX........ •-XC...............210 XXII. . . . . ...— LiV... . . . . . . . . . . . 126 XX.......... — XCIV............213 XXIII. . . . . . .-LXIII. . . . . . . . . . . .142 | XXI..........—XCV.............214 XXIV.......—LXIV............ 143 | XXII. ........—XCVI............215 XXV. . . . . . . . Prob. XXVII. Cor. 1. . .351 | XXIII. . . . . . . – XCVII. ..........219 XXVI. ......Th. LIX. ......... . . . . 134 XXIV. ....... -XCVIII. ... . . . . . .221 XXVII. . . . . . —LX.. . . . . . . . . . . . . . 136 | XXV......... —XCTI.............211 XXVIII. ....— LXI. ... . . . . . . . . . . 137 - XXIX. ......— LXII. . . . . . . . . . . . 139 EUCLID. Book VI. XXX.’....... Prob. XXVIII. . . . . . . • 352 *in ºr ºr -r- XXXI. ......Th. LVII. . . . . . . . . . . . . 130 ! I. ... . . . . . . . . . }Tºyº, XXXII. .....—LVIII. . . . . . . . . . . . 132 II ºxcix..........º. XXXIII. .... Prob. XXXII.........'.357 | #. C. ......go XXXIV...... XXXI. . . . . . . . . . 356 ſº. ...—ci...............” XXXV.......Th. CXXVI. . . . . . . . . . . 286 v. —cii.............as YXXVI...... -*- º tº s is tº a tº a tº º ve 386 ºf .......... Tomi............º. xxxvii...—CXXVI. Cor. 4.89 vii...— …........... * . . W. T. VIII. . . . . . . . .-CVI. . . . . . . . . . . . . .242 EUCLID. Book IV. IX. . . . . . . . . . . Prob. TX. . . . . . . . . . . . . .318 I............ Prob. XXIX. . . . . . . . . . . 353 X... . . . . . . . . . . - X.... . . . . . . . . . . . . .320 II. . . . . . * * * * * * — XXXIII.......... 359 | XI. . . . . . . . . . . — XI. . . . . . . . • . . . . . .321 IIT...........— XXXIV. . . . . . . . . . 360 | XII.... . . . . . . . — XII. . . . . . ........ ,322 IV. . . . . . . . . . . — XXXV... . . . . . . . .362 | XIII. . . . . . ... - XIII. . . . . . . . . . . . .323 V. . . . . . . . . . . . — XXXVI. . . . . . . . . . 363 | XIV. . . . . . . . .Th. CXV. . . . . . . . . . . . . .259 VI. . . . . . . . . — XXXVII. . . . . . . . .363 | XV. . . . . . . . . . —CVIII............. 247 VII. ... . . . . . . . — XXXVIII. ....... 364 | XVI. ........—CXV. Cor. 1...... 259 VIII. . . . . . . . . — XL. . . . . . . . . . . . . .366 | XVII. ........-CXV. Cor. 2...... 260 IX. . . . . . . . . . . — XXXIX.... . 365 | XVIII. ...... Prob. XXII. . . . . . . . . . .337 X............— XXV. Cor. ...... 348 XIX.........Th. CX. ............. .250 XI. ..........— XLI. ............ 368 XX. ... . . . . . . .- CXVII. . . . . . . . . . . 262 XII. ... . . . . ...— XLI. Cor......... 369 | XXI. . . . . . . . . Def. LXIII. Cor... . . . .239 XIII. . . . . . . . . —XL. . . . . . . . . . . . . . . 366 | XXII.........Th. CXVIII...........265 XIV. . . . . . . . . . XXXIX......... .365 XXIII. ......- CXVI.'........ . . .261 XV. ... . . . . . . .-XLII.............369 | XXIV........— CXI............. .253 XVI, . . . . . . . . .- XLIII, , , , , , , , , , , ,370 | XXV, , , , , , , , , Prob, XXIII, , , , . . . . . .339 I N D É X. 3 Propositiou • Page Proposition - IPage XXVI........Th. CXI. ... • • • • • • • • • • • 253 | XVIII........TIi. CXLIX...........405 XXVII....... Def. LXXVI..Schol... .342 | XIX. ........ — CXI. ............ 407 XXVIII...... Prob XXIV. . . . . . . . . . .343 || XX........... — CLV.......... ...413 XXIX. ......— XXIV. ... • • • • • • • • 343 XXÍ. s • «• • ® • e e — CLVI &* * • • s * ® • • • * de .415 XXX.. ......—XVI. s • • s • e • • a • • s • 327 XXIV........— CLXI. Cor. i. . . . . 426 XXXI. ......Th. CXXX. . . . . . . • • • • 295 | XXV......... — CLXIX...........473 XXXII.......— CV. Cor..... • • • • • • 242 | XXVIII...... — CLXXV...... . . . .463 XXXIII...... — CXXII........... 277 XXIX. - XXX. $ . . . . — CLXXVIII.Cor. 1.471 EUCLID. Book X. XXXI. ?< r< XXXII...... — CLXXIX.........473 I. • • • • • • • • • • • .Th. LXVIII. . . . ......155 XXXIII. ... .— CLXXXI.........476 XXXIV......— CLXXXI. Cor. 2. 478 EUCLID. Book XI. XL...........— CLXXIX. Cor....474 I.............Def. VIII. Cor. 3........4 ę & II............Th. CLXXXV.........383 EUCLID. Book XII. III...... ..... Def. VIII. Cor. 2. .......4 • p- rww • IV. .... ......Th. CXXXIX....... - , 384 ] I.............Tii. CXIX.......... ... 267 V............. — CXL...... . . ... . . .387 II. •• • • • • • • • • • - CXXIII..........280 VI. .. Iάζ Vi ; ........- CLXXXV. Cor....486 s • • • • e e s ® ® .• •-*-*-*-*-*-* π ę 0I*. s ę 3 i * \ii ę ę • ® • «• * * * - CXLI............ 388 || VII.......... — CLXXXIV. Cor..484 IX. e • e • • — CXLIV o • ω ... 394 VII. Cor. 1....— CLXXXV........485 X. » '. «-* CXLV e • e * ®* ' • «• ę ę 395 Xgt; • © © © • • — CLXXXV. Cor. o .486 XI. • • • • • • • • • • Š Th. 141. ........ : X. ...... • • • • • —-CLXXXVII.Cor.3.492 XII... Agg. :;:;;::::-:-:---- XI. ..........-— CLXXXVII. C. 2. 492 XIII. ..... • • • Def. LXIX. Cor. 2..... 389 XII... • • • • • • • — CLXXXVI.Cor. 3. 490 XIV. . . • • • • • • Th. CXLVI........... 397 XÍÍi` ce • *…*v , e vu/ • XV...........—CXLVII. :........ 398 ] XIV. } ¢ © © ■ ô* * —CLXXXVI.Cor. 2. 489 XVI. . . . . • • • • — CXLVIII........ 402 XV... ) XVII......... — CXLVHII. Οor. 5. 404 XVIII........ —CLXXXIX.Cor.3, 493 ~p*<»-~ * . . ; sº . . ." 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