BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF ftenrs W. Sage 1891 ^..zj.s^g.^. WATHEMATICS g/.-^y^../,?.?..^,.. 7673-2 Cornell University Library QA 453.H19 1907 Rational geometry; a text-book for the sc 3 1924 001 509 557 The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924001509557 WORKS OF PROF. G. B. HALSTED PUBLISHED BY JOHN WILEY & SONS. Elements of Geometry. 8vo, cloth, $1.75. Synthetic Geometry. For Schools and High Schools. 8vo, cloth, $1.50. Rational Geometry. A Text-book for the Science of Space. Second Edition, Revised. i2mo, viii + 273 pages, 238 figures. Cloth, $1.75. Synthetic Projective Geometry. Being No. 2 of the Mathematical Monograph Series. Cloth, $1.00, RATIONAL GEOMETRY A TEXT-BOOK FOR THE SCIENCE OF SPACE BY GEORGE BRUCE HALSTED A.B.and A.M., Princtton; Pk.D., John's Hapkim; F.R.A.S. SECOND EDITION, THOROUGHLY REVISED FIRST THOUSAND NEW YORK JOHN WILEY & SONS London: CHAPMAN & HALL, Limited 1907 'iM- , T Copyright, 1904, 1907, BY GEORGE BRUCE HALSTED. ROBERT DRUMMOMD, PRINTER, NEW YORK. PREFACE TO THE SECOND EDITION. This is the first text-book of geometry to treat jiroportion without any continuity assumption. Likewise it treats 'content' witho\it assuming it to be a magnitude. It discards the circle or compasses as an instrument for problem solving; any un- marked visiting-card is an all-sufficient instrument. The marvelous two-term prismatoid formula dominates mensuration. The beginnings cling close to Hilbert. Geometry at last made rigorous is also thereby made more simple. George Bruce Halsted. Greeley, Colorado. CONTENTS. CHAPTER I. PAGE Association i CHAPTER II. Betweenness 5 CHAPTER III. Congruence 15 CHAPTER IV. Parallels 35 CHAPTER V. Problems of Construction 50 CHAPTER VI Sides and Angles 61 CHAPTER VII. A Sect Calculus 66 CHAPTER VIII. Proportion and the Theorems of Similitude 72 CHAPTER IX. Equivalence 81 V Vt CONTENTS. CHAPTER X. PAGE The Circle lou CHAPTER XI. Length and Content of the Circle 128 CHAPTER XII. Geometry OF Planes 141 CHAPTER Xni. Polyhedrons and Volumes > 165 CHAPTER XIV. Tridimensional Spherics 184 CHAPTER XV. Cone and Cylinder 194 CHAPTER XVI. Pure Spherics 200 CHAPTER XVII. Angloids or Polyhedral Angles 236 APPENDIX I. Proofs of Two Betweenness Theorems 241 APPENDIX II. The Compasses 247 APPENDIX HI. The Solution of Problems 250 Index 267 TABLE OF SYMBOLS. We denote triangle b3'' A ; the vertices by .4, B, C; the angles at A, B, C by a, j3, y; the opposite sides by a. b, c; the altitudes from A, B, C by ha, hb, lu; the bisectors of a, /?, y by ta, h, 1c\ the medians to a, h, c by lUa, nib, ntc', the feet of ha, hb, he hy D, E, F; the centroid by G; the orthocenter by H ; the in-center by /; the in-radius by r; the ex-centers beyond a, h, c by /,, /j, 7,; their ex-radii by r,, r,, r^; the circiimcenter by O; the circumradius by R; angle by ^; angles by ^s; angle made by the rays BA and BC by y^ABC; angle made by the rays a and h both from the point O by ^ (a, h) or :^ ci?i ; bisector by bi'; circle by O; circles by Os; circle with center C" and radius r by OC"(r); congruent by = ; equal or equivalent by completion by = ; for example [exempli gratia] by e.g.; greater than by > ; less than by < ; minus by — ; parallel by ||; parallels by ||s; vil viii TABLE OF SYMBOLS. parallelogram by llg'm ; perimeter [sum of sides] by p ; perpendicular by j.; perpendiculars by xs; plus by + ; quadrilateral by quad' ; right by r't; spherical angle by 'v ; spherical triangle by S; similar by -r ; symmetrical by •!• ; therefore by .". RATIONAL GEOMETRY. CHAPTER I. ASSOCIATION. The Geometric Elements. 1. Geometry is the science created to give under- standing and mastery of the external relations of things; to make easy the explanation and descrip- tion of such relations and the transmission of this mastery. 2. Convention. ^A^e think three different sorts of things. The things of the first kind we call poiiits, and designate them by .4, S, C, ; the things of the second system we call straiiilits, and designate them hy a, b, c, . . . ; the things of the third set we call planes, and designate them by «, /?, ?- 3. We think the points, straights, and planes in certain mutual relations, and we designate these relations by words such as "lie," "between," "par- allel," "conginient." The exact and complete description of these rela- 2 RATIONAL GEOMETRY. tions is accomplished by means of the assumptions of geometry. 4. The assumptions of geometry separate into groups. Each of these groups expresses certain con- nected fundamental postulates of otu" inttiition. I. The first group of assumptions: assumptions of association. 5. The assumptions of this group set up an asso- ciation between the concepts above mentioned, points, straights, and planes. They are as follows: I I. Two distinct points, A, B, always determine a straight, a. Of such points besides "determine" we also em- ploy other turns of phrase; for example, A "lies on" a, A "is a point of" a, a "goes through" A "and through" B, a "joins" A "and" or "with" B, etc. When we say two things determine some other thing, we simply mean that if the two be given, then this third is explicitly and uniquely given. If A lies on a and besides on another straight b weusealsothe expression : "the straights "a" and ' ' b " have the point A in common." I 2. Any two distinct points of a straight determine THIS straight. That is, if AB determine a and AC determine o, and B is not C, then also B and C determine o. 1 ^. On every straight there are at least two points. In every plane there are always at least three non-co- straight points. ASSOCIATION. I 3 I 4. Three points, A, B, C, not costraight, always determine a plane a. We use also the expressions : A, B, C "lie in" a, A, B, C "are points of" a, etc. I 5. Any three non-costraight points A, B, C of a plane a determine this plane a. I 6. If two points A, B of a straight a lie in a plane a, then every point of a lies in a. In this case we say : The straight a lies in a. I 7. // two planes a, /3 have a point A in common, then they have besides at least another point B in common. I 8. There are at least four non-costraight non- coplanar points. 6. Theorem. Two distinct straights cannot have two points in common. Proof. The two points being on the first straight determine (by I 2) that particular straight. If by hypothesis they are also on a second straight, therefore (by I 2) they determine this second straight. Therefore the first straight is identical with the second. 7. Theorem. Two straights have one or no point in common. Proof. By 6 they cannot have two. 8. Theorem. Two planes have no point or a straight in common. Proof. If they have one point in common, then (by I 7) they have a second point in common, and therefore (by I 6) each has in it the straight which (by I i) is determined by these two points. 9. Corollary to 8. A point common to two planes 4 RATIONAL GEOMETRY. lies in a straight common to the two, which may be called, their straight of intersection or their meet. 10. Theorem. A plane and a straight not lying in it have no point or one point in common. Proof. If they had two points in common the straight would be (by I 6) situated completely in the plane. 11. Theorem. Through a straight and a point not on it there is always one and only one plane. Proof. On the straight there are (by I 2) two points. These two with the point not on the straight determine (by I 4) a plane, in which (by I 6) they and the given straight lie. Any plane on this point and straight would be on the three points already used, hence (by I 5) identical with the plane determined. 12. Theorem. Through two different straights with a common point there is always one and only one plane. Proof. Each straight has on it (by I 2) one point besides the common point, and (by 6) these two points are not the same point, and (by I 2) the three points are not costraight. These three points determine (by I 4) a plane in which (by I 6) each of the two straights lies. Any plane on these straights would be on the three points already used, hence (by I 5) identical with the plane determined. CHAPTER II. BETWEENNESS. II. The second group of assumptions: assumptions of betweenness. 13. The assumptions of this group make precise the idea "between," and make possible on the basis of this idea the arrangement of points. 14. Convention. The points of a straight stand in certain relations to one another, to describe which especially the word "between" serves us. II I. If A, B, C are points of a straight, and B lies between A and C, tlioi B also lies between C and A, and is neither C nor A. Fig. I. II 2. If A and C are two points of a straight, then there is always at least one point B, ivliich lies between Fig. 2. A and C, and at least one point D, such tltat C lies between. A and D. S R/tTlONAL GEOMETRY. II 3. Of any three points of a straight there is always one and only one which lies between the other two. 15. Definition. Two points A and B, upon a straight a, we call a segment or sect, and designate it with AB ov BA. The points between A and B are said to be points of the sect AB or also situated within the sect AB. All remaining points of the straight a are said to be situated without the sect AB. The points A, B axQ called end-points of the sect AB. II 4. (Pasch's assumption.) Let A, B, C be three points not costraight and a a straight in the plane ABC going through none of the points A, B, C; if Fig. 3- then the straight a goes through a point within the sect AB, it must always go either through a point of the sect BC or through a point of the sect AC. Deductions from the assumptions of association and betweenness. 16. Theorem. Between any two points of a straight there are always indefinitely many points. [Here taken for granted, and its proof removed to Appendix I.] 17. Theorem. If any finite number of points of BETIVEENNESS. 7 a straight are given, then they can always be ar- ranged in a succession A, B, C, D, E K, such that B lies between A on the one hand and C, D, £,..., K on the other, further C between A, B on the one hand and D, E, . . . , K on the other, then D between A, B, C on the one hand and E, . . . , K on the other, and so on. Besides this distribution there is only one other, the reversed arrangement, which is of the same character. [This theorem is here taken for granted, and its proof removed to Appendix I.] 1 8. Theorem. If A, B, C be not costraight, any straight in the plane ABC which has a point within the sect AB and a point within AC cannot have a point witJiiii BC. Proof. Suppose F, G, H three such costraight points. One, say G, on AB, must (by II 3) lie between the others. Then the straight AB must (by II 4) have a point within the sect EC or the sect CH, which (by 7 and II 3) is impossible. 19. Theorem. Every straight a, which lies in a plane «, separates the other points of this plane a into two regions, of the following character: every 8 RATIONAL GEOMETRY. point A of the one region determines with eiwry point B of the other region a sect AB, within which lies a point of the straight a; on the contrary, any two Fig. s. points A, A' of one and the same region always deter- mine a sect A A' which contains no point of a. Proof. Let A he a. point of a which does not lie on a. Then reckon to one region all points P of the property, that between A and P, therefore Fig. 6. within AP, lies no point of a; to the other region all points Q such that within AQ lies a point of a. Now is to be shown : (i) On PP' lies no point of a. (2) On QQ' lies no point of o. (3) On PQ lies always a point of a. BETIVEENNESS 9 (i) From hypothesis neither within AP nor AP' lies a point of a. This would contradict II 4, if within PP' were a point of a. (2) By hypothesis there lies within AQ a point of a, likewise within AQ' ; therefore (by 18) none within OQ'. (3) By hypothesis AP contains no point of a; AQ on the other hand contains one such. There- fore (by II 4) a meets PQ. 20. Convention. If A, A', 0, B are four costraight points such that is between A and B but not between A and A'; then we say: the points A, A' Fig. 7. lie in the straight a on one and the same side of the point 0, and the points A, B lie in the straight a on different sides of the point 0. 21. Definition. The assemblage, aggregate, or to- tality of all points of the straight a^situated on one and the same side of is called a ray starting from O. Consequently every point of a straight is the ori- gin of two rays. 22. Convention. Using the notation of 19, we say: the points P, P' lie i)t the plane a on one and the same side of the straight a and the points P, Q lie ill the plane a on different sides of the straight a. 23. Theorem. Every two intersecting straights a, b separate the points of their plane a not on either into four regions such that if the end-points of a sect are both in one of these regions, the sect contains no point of either straight. lo RATIONAL GEOMETRY. Proof. Let be their common point and A another point on h, and B another point on a. Then two points both on the A side of a and the B side of b make a sect which (by 19) can contain no point either of a or of 6. So also if both were Fig. 8. on the A side of a and the non-i? side of fe ; or both on the non-A side of a and the B side of b ; or both on the non-A side of a and the non-B side of b. 24. Definition. A system of sects AB, BC, CD, . . . , KL is called a sect-train, which joins the points A and L with one another. This sect-train will also be designated for brevity by A BCD . . . KL. The points within the sects AB, BC, CD, . . . , KL, together with the points A, B, C, D, . . . , K, L are all together called the points of the sect-train. In particular if the point L is identical with the point A, then the sect-train is called a polygon and is designated as polygon A BCD . . . K. 25. The sects AB, BC, CD, ... , KA are called the sides of the polygon. The points A, B, C, D, . . . , K are called the vertices of the polygon. 26. A sect not a side but whose end-points are vertices is called a diagonal of the polygon; BETlVEENhlESS. n 27. Polygons with 3, 4, 5, . , n vertices are called respectively triangles, quadrilaterals, penta- gons ii-goiis. 28. If the vertices of a polygon are all distinct from one another and no vertex of the polygon falls within a side and finally no two sides of the polygon have a point within in common, then the polygon is called simple. By quadrilateral is meant simple quadrilateral. A plane polygon is one all of whose sides are co- planar. A convex polygon is one no points of which a-e on different sides of the straight of any of its sides. 29. Theorem. Every simple polygon, whose ver- tices all lie in a plane «, separates the points of this plane a, which do not pertain to the sect-train of the polygon, into two regions, an inner and an outer, of the following character; if .4 is a point of the inner (interior point) and B a point of the outer (exterior point), then e^'ery sect-train which joins A with B has at least one point in common with the polygon; on the contrary if .4, A' are two points of the inner and B, B' two points of the outer, then there are always sect-trains, which join .4 with A' and B with B' and have no point in common with the polygon. There are straights in n: which lie wholly outside the polygon; on the contrary no such straights which lie wholly within the polygon. Proof. For a triangle ABC there is (by 23) a region with points on A12 A side of BC, the B side of CA, and the C side of AB, i.e., an inner region. 12 RATIONAL GEOMETRY. Moreover, the straight determined by a point on h and a point on c both in non-A lies wholly without the region ABC, since it cannot again meet 6 or c and so cannot (by II 4) have a point in common with BC. Moreover, if any straight has a point within ABC, it has a point on a side. For the Fig. 9. straight determined by the point within and any point on a side has (by II 4) a point on another side, thus making another triangle, in common with one side of which the given straight has a point, and therefore (by II 4) with another side, that is with a side of the original triangle.* 30. Corollary to 29. A straight through a ver- tex and a point within a triangle has a point within the ' opposite ' side. * It can be shown, that any other simple polygon has a diagonal which does not cut the sides, then gives a number of triangles, then satisfies 29. BETIVEENNESS. 13 31. Theorem. Every plane a separates all points not on it into two regions of the following character: every point A of the one region determines with every point B of the other region a sect AB, within which lies a point of a; on the contrary any two points .4 and .4' of one and the same region always determine a sect AA', which contains no point of a. Proof. Let .4 be a pomt which does not he on a. Then reckon to the one region all points P of the property, that between A and P, therefore with- in AP, lies no point of a ; to the other region all points Q such that within AQ Ues a point of a. Now is to be shown: (i) On PP' Hes no point of a. (2) On 0.0' Hes no point of a. (3) On PQ lies always a point of «. (i) From hypothesis neither within AP nor AP' lies a point of a. Suppose now a point of « lay on PP'. Then the plane a and the plane APP' would have in common this point and consequently (by 9) a straight a. This straight goes through none of the points A, P, P' ; it cuts PP' ; it must there- fore (by II 4) cut either AP or AP', which is con- trary to hypothesis. (2) By hypothesis there lies within AQ a. point of a, likewise \\'ithin AO' The intersection straight of the planes a and AQQ' therefore meets two sides of the triangle .4 (_\/; consequently (by 21) it can- not also meet the other side QQ' . (3) AP contains by hypothesis no point of a; AQ on the other hand contains one such. The inter- section straight of the planes a and APQ therefore 14 R/ITIONAL GEOMETRY. meets the side AQ and does not meet the side AP in triangle APQ. Therefore (by II 4) it meets the side PQ. 32. Convention. Using the notation of 31, we say: the points A, A' He on one and the same side of the plane, a, and the points A, B Me on different sides of the plane a. Ex. I. A straight cannot traverse more than 4 of the 7 regions of the plane determined by the straights of the sides of a triangle. Ex. 2. Four coplanar straights crossing two and two determine 6 points. Choosing 4 as vertices we can get two convex quadrilaterals, one of which has its sides on the straights. Ex. 3. Each vertex of an w-gon determines with the others (» — i) straights. So together they determine n{n — i)/2. Ex. 4. How many diagonals in a polygon of n sides. Ex. 5. What polygon has as many diagonals as sides? CHAPTER III. CONGRUENCE. III. The third group of assumptions: assumptions of congruence. 33. The assumptions of this group make precise the idea of congruence. 34. Convention. Sects stand in certain rela- tions to one another, for whose description the word congntcut especially serves us. Ill I. If A, B arc two points on a straight a, and A' a point oil the saute or anotlicr straight a', then we can find on tJic straight a' on a given ray from A' always one and only one point B' such that tJie sect AB is congruent to the sect A'B' . We write this in symbols AB = A'B'. Every sect iscongruent to itself, i.e., always ylS = ^45. The sect AB is always congruent to the sect BA, i.e., AB^BA. We also say more briefly, that every sect can be taken on a given side of a given point on a given straight in one and only one way. Ill 2. If a sect AB is congruent as ivcll to the sect A'B' as also to the sect A"B", then is also A'B' con- is i6 RATIONAL GEOMETRY. gruent to the sect A"B", i.e., if AB = A'B' and AB = A"B", then is also A'B' = A"B". Ill 3. On the straight a let AB and BC he two sects without common points, and furthermore A'B' and B'C two sects on the same or another straight, like- wise without common points; if then AB = A'B' and BC = B'C', so always also AC=A'C'. jL c' Fig. 10. 35. Definition. Let a be any plane and h, k any two distinct rays in a going out from a point 0, and pertaining to different straights. These two rays h, k we call an angle, and des- ignate it by^ {h, k) or 4- {k, h). The rays h and k, together with the point 0, separate the other points of the plane a into two regions of the following character : if A is a point of the one region and B of the other region, then every sect-train which joins A with B, goes either through or has with h or k sX least one point in common; on the contrary HA, A' are points of the same region, then there is always a sect-train which joins A with A' and neither goes through O nor through a point of the rays h, k. One of these two regions is distinguished from the other because each sect which joins any two points of this distinguished region always lies wholly Fig. II. CONGRUENCE. 17 in it; this distinguished region is called the interior of the angle {h, k) in contradistinction from the other region, which is called the exterior of the angle Qi, k). The interior of ^ {h, k) is wholly on the same side of the straight h as is the ray k, and altogether on the same side of the straight k as is the ray /;. The rays /;, k are called sides of the angle, and the point is called the vertex of the angle. Ill 4. Giveti any angle {li, k) in a plane a and a straight a' in a plane a' , also a determined side of a' on a'. Designate by W a ray of the straiglit a' start- ing from tlie point 0'; then tlierc is in the plane a' ONE AND ONLY ONE ray k' siicJi that the angle (h, k) is congruent to tlie angle Qi' , k'), and likewise all in- terior points of the angle {h' , k') lie on the given side of a'. ft^ ft' Fig. 12 In symbols: Every angle is congruent to itself, i.e., always ^ih,k)^^{h,k). The angle {h, k) is aht'ays congruent to the angle (k,h), i.e., 4{h,k)^4{k,'h). We say also briefly, that in a given plane CA-ery angle can be set off towards a given side against a l8 RylTIONAL GEOMETRY. given ray, but in a uniquely determined way. There is one and only one such angle congruent to a given angle. We say an angle so taken is uniquely determiijied. Ill 5. // an angle (h, k) is congruent as well to the angle Qi' , k') as also to the angle (h", k"), then is also the angle Qi' , ¥) congruent to the angle (h", k")', i.e., if ^ {h, k)=^ (h', k') and ^ {h, k)=^ {h", k"), then always i Qi' , k') = -^ Qi" , k"). 36. Convention. I^et ABC be any assigned tri- angle; we designate the two rays going out from A through B and C respectively by h and k. Then the angle {h, k) is called the angle of the triangle ABC included by the sides AB and AC or opposite the side BC. It contains in its interior all the inner points of the triangle ABC and is designated by ^BAC or ^A. Ill 6. // for two triangles ABC and A'B'C we have the congruences AB = A'B', AC = A'C', ^BAC^ ^B'A'C, then always are^ fulfilled the congruences ^ ABC ^^ A'B'C and ^ACB^^A'CB'. Deductions from the assumptions of congruence. 37. Convention. Suppose the sect AB congruent to the sect A'B'. Since, by assumption III t, also the sect AB is congruent to AB, so follows from III 2 that A'B' is congruent to AB; we say: the two sects AB and A'B' are congruent to one another. 38. Convention. Suppose ^ (h, ^) = ^ Qi', k'). CONGRUENCE. 19 Since (by III 4) ^ (h, k)^ :!f{h, k), therefore (by III s) 4 Oh' k') ^ ^ (/;, k). We say then: the two angles ^ (h, k) and ^ {W , k') are congruent to one another. 39. Definition. Two angles having the same ver- tex and one side in common, while the sides not common form a straight, are called adjacent angles. 40. Definition. Two angles with a common ver- tex and whose sides form two straights are called vertical angles. 41. Definition. Any angle which is congruent to one of its adjacent angles is called a riglit angle. Two straights which make a right angle are said to be perpendicular to one another. 42. Convention. Two triangles ABC and A'B'C are called eo)igrnetit to one another, if all the con- gruences AB^A'B', AC^A'C, BC^B'C, 4A^4A', ^B^^B', ^C^^C are fulfilled. 43. (First congruence theorem for triangles.) Triangles are congruent if they have two sides a)td the included angle congruent. In the triangles ABC and A'B'C take AB = A'B', AC^A'C, :^A=^A'. 20 RATIONAL GEOMETRY. To prove aABC= aA'B'C . Proof. By assumption III 6 the congruences ^B= 4B' and ^C= ^C"' are fulfilled, and so we have only to show that the sides BC and B'C are congruent to one another. Suppose now, on the contrary, that BC were not congruent to B'C, and take on ray B'C (by III i) the point D', such that BC = B'D'. Then the two triangles ABC and A'B'D' will have, since ^5 = ^B', two sides and the included angle respectively congruent ; therefore, by assumption III 6, ^ BAC = ^B'A'D'. Hence, by assumption III 5, ^B'A'C = ^B'A'D'. This is impossible, since, by. assump- tion III 4, against a given ray toward a given side in a given plane there is only one angle congruent to a given angle. So the theorem is completely established. 44. (Second congruence theorem for triangles.) Triangles are congruent if they have two angles and the included side congruent. Fig. 14. In the triangles ABC and A'B'C take AC'^A'C, ^A^iA', ^C^^C. To prove t^ABC= a A' B'C. CONGRUENCE. 21 Proof. Suppose now, on the contrary, AB is not = A'B', and take on ray A'B' the point D' , such that AB = A'D'. By III 6, -^-ACB^ ^A'C'D', but by hypotliesis ^ACB=^A'C'B'. Therefore (by Ills) ^ ^'C'B' ^ 4 A' CD'. But this is impossible, since (by III 4) in a given plane against a given ray toward a given side there is only one angle con- gruent to a given angle. Consequently our supposition, AB not = A'B', is false, and so AB = A'B'. Now follows (by 43) that aABC= aA'B'C. 45. Theorem. // tivo angles arc coigniciit, so are also tlicir adjacent angles. Take ^ABC= ^A'B'C. To prove 4CBD= ^C'B'D'. Proof. Choose the points .4', C, D' on the sides from B' so that A'B' = AB, C'B' = CB, DB = D'B'. In the two triangles ABC and A'B'C then the sides AB and CB are congruent respectively to the sides A'B' and CB', and since moreover the angles included by these sides are congruent by hypothesis, so follows (by 43) the congruence of those triangles, that is, we have the congruences AC = A'C and iBAC^^B'A'C. 2 2 RATIONAL GEOMETRY. Now since (by III 3) sect AD=A'D', so follows (again by 43) the congruence of the triangles CAD and C'A'D', that is, we have the congruences CD = C'D' and ^ADC^:^A'D'C', and hence in the triangles BCD and B'C'D' (by III 6),^i-CBD^ :f C'B'D'. 46. Theorem. Vertical angles are congruent. Proof. By III 4, -iABC^^CBA. Therefore, by 45, their adjacent angles are congruent, ^ CBD = i-ABF. Fig. 16. 47. Theorem. Through a point A, not on a straight a, there is one and only one perpendicular to a. Proof. Take any two points P, Q on a. Take from P against the side PQ of '4-APQ, and on the non-^ side of a, i-BPQ= ^APQ. Take PB = PA. Since A and B lie on different sides of a, there must be a point of sect AB on a. Then AOB is perpen- dicular to a. For (by 43) aBPO= aAPO, so ifBOP=^AOP. But these are adjacent. Therefore, by definition 41, AOP is a right angle. Moreover this. is the only perpendicular from A to a. For suppose any straight AO' perpendicular to a at 0', and take on this straight on the non-^ CO)^GPUBNCB. *3 side of a the sect O'B' ^O'A. Then from hypothe- sis ^PO'B' ^^.PO'A and so (by 43) aPO'B' = aPO'A. Therefore ^ B'PO' ^ ^ APO' and B' P ^ Fig. 17. AP. Therefore (by III 5 and III 2), ^B'P'0 = ^ BPO and B'P = BP Hence the points B and B' are not different. Therefore no second perpendicu- lar from .4 to a can exist. *48. Theorem. Let the angle {h, k) in the plane a be congruent to the angle Qi', k') in the plane a', and further let / be a ray of the plane a, which goes out from the vertex of the angle {h, k) and lies in the interior of this angle ; then there is always a ray I' in the plane a', which goes out from the vertex of the angle {h' , k') and lies in the interior of this angle, such that i (h, D^yf Qi', V) and i {k, 1)=^ (k', I'). Proof. Designate the vertex of ^ {h, k) by 0, and the vertex of 4 U'-'y k') by 0', and then determine on the sides h, k, h', k', the points A, B, A', B', so that we have the congruences OA = 0'A' and OB = 0'B'. * 48 may be omitted if 49 be taken as a redundant assimiption. 24 RATIONAL GEOMETRY. Because of the congruence of the triangles OAB and O'A'B' (by 43) AB = A'B', i-OAB^i-O'A'B', iOBA^^O'B'A'. The straight AB (by 30) cuts /, say in C; then we determine on the sect A'B' the point C, such that A'C' = AC, then is O'C the ray sought, V. Fig. 18. For, from AC=A'C' and AB=A'B', we may, by means of III 3, deduce the congruence BC = B'C'. Therefore (by III 6) i-AOC^i-A'O'C and 4BOC = iB'0'C'. • 49. Theorem. Let h, k, I on the one hand and h', k', I' on the other each be three rays going out from a point and lying in a plane ; if then we have the congruences ^ {h, /) = ^ Qi', V) and ^ (k, I) = ^ {k', I'), then also is always ^(h,k)^i.ih',k'). CONGRUENCE. 25 Proof. The rays are supposed such that either no point is interior to ^{h, I) and ^(k, I), or to ^(/i', I') and ^{k', I'), or else that if one of these angles be within a second, then the angle congruent to the first is within the fourth. I. In the first case, if / be supposed ivitJiin ^ Qi, k), take against h' toward k' , :^ (h' , k") = ^ {h, k). By Fig. 19. 48, take in angle {h', k") ray /" such that i- (h', I") ^ ^ {h, I) and ^ (/", k") ^ 7f (l, k). But by hy- pothesis ^{W, l')=^Qi,l). Therefore, by III 5, ^ {h', I') = ^ {h',l"), and so, by III 4, ray I" is iden- tical with ray /'. Then ^ {k", I") = ^ {k", V) = ^ {k, 1)=^ {¥, I'). So ^ {k", V) ^ i. (k\ I'), and, by III 4, ray k" is identical with ray k'. But 7f{h',k")=^{h,k). Therefore ^(h,k) = i{h',k'). If, however, I be supposed not ivithin ^ Qi, k), then it will lie in ^ {h", k") vertical to ^ {h, k). For it cannot lie in ^ {h, k") adjacent to ^ (/;, k), since then ^ (/, k) would contain ^ {h, /), contradicting the hypothesis in this case of no point interior to these two given angles. For like reason it cannot lie in 26 RATIONAL GEOMETRY. i-ih", k) adjacent to ^{h, k), since then ^ (/j, would contain ^ (/, k). Thus the ray m costraight with I is within ^ {h, k) , and m' costraight with I' is within ^ (h', k'). Fig. 20. Then (by 45) ^ (/j, m)= -4- (h', m') and ^ (^, w) = ifik', w!) [^'s adjacent to congruent ^ 's are con- gruent], and so this sub- case is reduced to the pre- ceding. II. The remaining case, where one angle ^{h, I) is within another, ^ (^, /), follows at once from 48. *5o. Theorem. All right angles are congruent. Let angle BAD be congruent to its adjacent angle CAD, and likewise let the angle B'A'D' be con- gruent to its adjacent angle C'A'D' . then are ^BAD, 4 CAD, 4 B'A'D', 4 C'A'D' all right angles. To prove iBAD= ^ B'A'D'. Proof. Suppose, contrary to our proposition, the right angle B'A'D' were not congruent to the right angle BAD, and then set off ^ B'A'D' against ray * 50 may be taken as a redundant assumption. CONGRUENCE. 27 AB so that the resulting side AD" falls either in the interior of the angle BAD or of the angle CAD ; sup- pose we have the first of these cases. Because ^B'A'D' ^ ^BAD", therefore, by 45 , ^C'A 'D' = ^ CAD" ; and since by hypothesis :fB'A'D'^ ^C'A'D', therefore, by Ills, ^ BAD" ^^ CAD". Since further ^ BAD is congruent to if CAD, so there is (by 48) within the angle CAD a ray AD"' such that /• c Fig. 21. B' D' i- BAD" =^CAD"' and also iDAD"=^DAD"'. But we had 4.BAD"= i-CAD", and therefore we must (by Ills) also have ^C.AD" = ^CAD"' . This is impossible, since (by III 4) every angle can be set off against a gi\-en ray toward a given side in a given plane only in oic way. Here^^^th is the proof for the congruence of right angles completed. 51. Corollary to so. At a point .4 of a straight a there is in a given plane containing o one and only one perpendicular to a. At .4 (by III 4) take a ray h making ^4- if,, b) = 4POA found in 47. 52. Definition. ^Mien any two angles are con- gruent to two adjacent angles, each is said to be the supplement of the other. 28 RATIONAL GEOMETRY. 53. Definition. If any angle can be set off against one of the rays of a right angle so that its second side lies within the right angle, it is called an acute angle. 54. Definition. Any angle neither right nor acute is called an obtuse angle. 55. Definition. A triangle with two sides con- gruent is called an isosceles triangle. 56. Theorem. The angles opposite the congruent sides of an isosceles triangle are congruent. Let ABC be an isosceles triangle, hsiv'mgAB^BC. To prove ^A=i.C. Proof. Since in the triangles ABC and CBA we have the con- gruences AB = CB, BC = BA, if ABC =4 CBA , therefore (by III 6) ^CAB^^ACB. 57. (Third congruence theorem for triangles.) Two triangles are congruent if the three sides of the one are congruent, respectively, to the three sides of the other. Fig. 22. Fig. 23. In the triangles ABC and A'B'C take AB^A'B', AC = A'C', BC = B'C'. CONGRUENCE. 29 To prove aABC^aA'B'C Proof. In the plane of ABC toward the side of the straight AC not containing B against the ray AC take the angle CAB"^ C'A'B'. Take the &ectAB" = A'B'. Then (by 43) aAB"C=aA'B'C' . Therefore B"C = BC, and A BCB" is isosceles ; there- fore (by 56) 4 CBB" = ^ CB"B. So also is a BAB" isosceles and .-. ^ABB"=^AB"B. Therefore (by 49) the angle ABC=i.AB"C. But :^AB"C = ^A'B'C'. .-. (by43) aASC= A^'5'C'. 58. If ^, J5, C be any three points not costraight, then (by the method used in 57) we can construct a point B" such that AB" = AB and CB" = CB. Therefore a point D such that no other point whatsoever, say D" , gives AD" = AD and CD" = CD, must be costraight with AC. 59- The following have been given as definitions: If A and B are two distinct points, the straight AB is, the aggregate of points P for none of which is there any point such that QA = PA and QB = PB. If ^, B, C are distinct points not costraight, the plane ABC is the aggregate of points P for none of which is there any point Q such that QA^PA, QB^PB, and QC^PC. 60. Convention. Any finite ntimber of points is called a figure; if all points of the figure lie in a plane, it is called a plane figure. 61. Convention. Two figures are called congruent if their points can be so mated that the sects and angles in this way coupled are all congruent. Congruent figures have the following properties: 3° RATIONAL GEOMETRY. If three points be costraight in any one figure their mated points are also, in every congruent figure, costraight. The distribution of points in corre- sponding planes in relation to corresponding straights is in congruent figures the same ; the like holds for the order of succession of corresponding points in corresponding straights. 62. The most general theorem of congruence for the plane and in general is expressed as follows : If {A, B, C, . . .) and (A', B' , C , . . .) are con- gruent plane figures and P denotes a point in the plane of the first, then we can always find in the plane of the second figure a point P' such that (A, B,C, . . ., P) and (A', B', C , . . . , P') are again congruent figures. If each of the figures contains at least three non- costraight points, then is the construction of P' only possible in one way. If {A, B, C,.. .) and (A', B', C , . . .) are con- gruent figures and P any point whatsoever, then we can always find a point P', such that the figures {A, B,C, . . . ,P) and (A', B', C, . . . , P') are congruent. If the figure {A, B, C, . . .) contains at least four non-coplanar points, then the construction of P' is only possible in one way. This theorem contains the weighty result, that all facts of congruence are exclusively conse- quences (in association with the assimiption-groups I and II) of the six assumptions of congruence already above set forth. This theorem expresses the existence of a cer- tain reversible unique transformation of the aggre- CONGRUENCE. 31 gate of all points into itself with which we are familiar under the name of motion or displacement. We have here founded the idea of motion upon the congruence assumptions. Thereby we have based the idea of motion on the congruence idea. The inverse way, to try to prove the congruence assumptions and theorems with help of the motion idea, is false and fallacious, since the intuition of rigid motion involves, contains, and uses the con- gruence idea. 63. Exercises. Ex. 6. Show a number of cases where two straights determine a point. Show cases where two straights do not determine a point. Are any of these latter pairs coplanar? Ex. 7 . Show cases where three coplanar straights deter- mine 3 points; 2 points; i point. Are there cases where they determine no point? Ex. 8. How many straights are, in general, deter- mined by 3 points? by 4 coplanar points? AVhat special cases occur? Ex. 9. Any part of a triangle together with the two adjoining parts determine the 3 other parts. Explain. Ex. 10. Try to state the first two congruence theorems for triangles so that either can be obtained from the other by simply interchanging the words side and angle. Ex. II. Principle of Duality in the Plane. In theorems of configuration and determination we may interchange point and straight, sect and angle. Try to write down a theorem of which the dual is true; is false. Ex. 12. If two angles of a triangle are congruent it is isosceles. Ex. 13. If the sides of a A axe =, so are the ^s. Dual? 32 RATIONAL GEOMETRY. / Ex. 14. In an isosceles A, sects to the sides from the ends of the base making with it = ^s are s. Ex. 15. If any two sects from the ends of a side of a A to the other sides making = ^s are = , the a is isosceles. 64. Definition. Two parallels are coplanar straights with no common point. 65. No assumption about parallels is necessary for the establishment of the facts of congruence or motion. 66. Theorem. Through a point A without a straight a there is always one parallel to a. Proof. Take the ray from the given point A through any point B of the straight a. Let C be any other point of the straight a. Then take in the plane ABC an angle congruent to '4- ABC against AB at the point A toward that side not containing C. The straight so obtained through A does not Fig. 24. meet straight a. If we supposed it to cut a in the point D, and that, say, B lay between C and D, then we could take on a a point D' , such that B lay between D and D' , and moreover AD = BD'. Be- cause of the congruence of the triangles BAD and ABD' (by 43), therefore ^ABD^^BAD'; and since the angles ABD' and ABD are adjacent angles, so must then, having regard to 45, also the angles congruent to them, BAD and BAD' , be ad- jacent angles. But because of 6, this is impossible. CONGRUENCE. 33 67. Definition. A straight cutting across other straights is called a transversal. 68. Definition. If, in a plane, two straights are cut in two distinct points A, B hy a. transversal, at each of these points four angles are made. Of these eight, four, having each the sect AB on a side [e.g., 3, 4, i', 2'], are called interior angles. The other four are called exterior angles. Pairs of angles, one at each point, which lie on the same side of the transversal, the one exterior and the other interior, are called corresponding angles [e.g., I and i']. Two non-adjacent angles on opposite sides of the ''^' ^^" transversal, and both interior or both exterior, are called alternate angles [e.g.; 3 and i']. Two angles on the same side of the transversal, and both interior or both exterior, are called con- jugate angles [e.g., 4 and i']. 69. Theorem. Two coplanar straights are parallel if a transversal makes congruent alternate angles. [Proved in 66.] 70. Theorem. If two straights cut by a trans- versal have corresponding angles congruent they are parallel. Proof. The angle vertical to one is alternate to the other. Ex. 16. If two corresponding or two alternate angles are congruent, or if two interior or two exterior angles on the same side of the transversal are supplemental, 34 . RATIONAL GEOMETRY. then every angle is congruent to its corresponding and to its alternate angle, and is supplemental to the angle on the same side of the transversal which is interior or exterior according as the first is interior or exterior. Ex. 17. If two interior or two exterior angles on the same side of the transversal are supplemental, the straights are parallel. Ex. 18. Two straights perpendicular to the same straight are parallel. Ex. 19. Construct a right angle. Ex. 20. On each ray from the vertex of a triangle co- straight with a side take a sect congruent to that side. The two new end-points determine a straight parallel to the triangle's third side. Ex. 21. On one side of any yf with vertex A take any two sects AB, AC and on the other side take congruent to these AB', AC Prove that BC and B'C intersect, say at D. Prove BC =B'C, ABCD = i^B'CD, -t^BAD^ ■4B'AD. Ex. 22. From two given points on the same side of a given st' find st's crossing on that given st', and making congruent ^ 's with it. Ex. 23. Construct a triangle, given the base, an angle at the base, and the sum of the other two sides [A from ffi, b, a-\-c]. Ex. 24. If the pairs of sides of a quadrilateral not con- secutive are congruent, they are ||. Ex. 25. On a given sect as base construct an isosceles a. Ex. 26. If on the sides AB, BC, CA of an equilateral A, AD =BE=CF, then ADEF is equilateral, as is A made by AE, BF, CD. CHAPTER IV. PARALLELS. IV. Assumption of Parallels (Euclid's Postulate). IV. Through a given point tliere is not more than one parallel to a given straight. 'ji. The introduction of this asstiniption greatly simplifies the foundation and facilitates the con- struction of geometry. 72. Theorem. Two straights parallel to a third are parallel. Proof. Were i and 2 not parallel, then there would be through their intersection point two par- allels to 3, which is in contradiction to IV. 73. Theorem. // a transversal cuts two parallels, the alternate angles are congruent. Proof. Were say ^BAD not = ^ABC, then we could through A (by III 4) take a straight making >fBAD'=^ABC [D' and D on same side oi AB], and a/ »' so we would have (by 69) through A two parallels to a, in contradiction to IV. 74. Corollary to 73. A perpendicular to one of two Fio- 26- parallels is perpendicular to the other also. 35 36 R/ITIONAL GEOMETRY. 75. Theorem. If a transversal cuts two parallels, the corresponding angles are congruent. Proof. The angle vertical to one is alternate to the other. Ex. 27. A straight meeting one of two parallels meets the other also. Ex. 28. A straight cutting two parallels makes con- jugate angles supplemental. Ex. 29. If alternate or corresponding angles are un- equal or if conjugate angles are not supplemental, then the straights meet. On which side of the transversal? 76. Theorem. A perpendicular to one of two parallels is parallel to a perpendicular to the other. Proof. Either of the two given parallels makes (by 74) right angles with both perpendiculars, which therefore are parallel by 69. 77. Corollary to 76. Two straights respectively perpendicular to two intersecting straights cannot be parallel. Proof. For if they were parallel, then (by 76) the intersecting straights would also be parallel. 78. Convention. When two angles are set off from the vertex of a third against its sides so that no point is interior to two, if the two sides not common are costraight, the three angles are said together to form two right angles. 79. The angles of a triangle together form, two right angles. Proof. Take alternate i.CBF=4-C and -^ABD = ^A ; then (by 69) can neither BF nor BD cut AC. By the parallel postulate ^^' ^^' IV, then is FBD a straight. PARALLELS. 37 80. Theorem. If two angles of one triangle are congruent to two of another, then the third angles are congruent. Proof. Gixen tA=i-A' Sind^B = 4.B'. Take CP parallel (||) to AB and C'P' \\ to A'B'. Then ta=-4A and ^p=:^B, ^a'^^A' and ^/?' = ^B'. .-.(by 49) tBCD = ^B'C'D'. /.(by 45) the adjacent angles ^ACB^:^A'C'B'. 81. Theorem. Two triangles are congrnait if they have a side, the opposite angle and another angle re- spectively congruent. Proof. By 80 and 44. Ex. 30. Ever}' triangle has at least two acute angles. Ex. 31. If the rajrs of one angle axe parallel or per- pendicular to those of another, the angles are congruent or supplemental. Ex. 32. In a right-angled triangle [a triangle one of whose angles is a right angle] the t'n-o acute angles are complemental (calling two angles coiiiptcvictits which together form a right angle). 82. Theorem. In any sect AB there is always one a)id oidy one point C such that AC = BC. Proof. Take an}' angle BAD at .4 against AB, and the angle congruent to it at B against BA and on the opposite side of a in the plane BAD; and take an-s' sect AD on the free ray from A, and one 38 RATIONAL GEOMETRY. BF congruent to it on the free ray from B. The sect DF must cut a, say in C, since D and F are on opposite sides of a. Moreover, C is between A and B. Otherwise one of them, say A , would be between B and C. But then DA would have a point A on F Fig. 29. BC, a side of triangle FBC, and so (by II 4) must meet another side. But this is' impossible, since it meets FC produced at D and is parallel to BF. Since thus ^A=i^B, and ^ACD^^BCF [ver- tical], therefore (by 81) aACD=aBCF. Therefore AC = BC. If we suppose a second such point C, then on ray DC take C'F' = DC'. Therefore (by 43) ^ C'BF' ^■4.DAC=^ABF, and BF' = AD = BF. Therefore F' is F and C is C. 83. Convention. The point C of the sect AB such that AC = BC may be called the bisection-point of AB, and to bisect AB shall mean to take this point C. Ex. 3^. Parallels through the end-points of a sect intercept on any straight through its bisection-point a sect this p't bisects. Ex. 34. In a right-angled triangle the bisection-point of the hypothenuse (the side opposite the r't^) makes equal sects with the three vertices. Hint. Take one acute 2jl in the r't 2Sl. PAR/ILLELS. 39 84. Theorem. Within any ^ Qi, k) there is always one and only one ray, I, such that ^ {h, l)= ^ {k, I). Proof. From the vertex take OA = OA'. By 82 take C, the bisection-point of A A'. Then aAOC=a A'OC [As with 3 sides = are=]. .-. ^A0C=4A'0C. If we suppose a second such ray OC, then aAOC'=a A'OC [as with 2 sides and the included ^ =are = ]. .-.AC'^A'C. .-. (by 82) C is C. 85. Convention. The ray Z of :^ (h, k) such that ■^ (h, /) = ^ (/, k) may be called the bisection ray or angle-bisector of ^ {h, k), and to bisect ^ (/i, ^) shall mean to take this ray I. Ex. 35. An angle may be separated into 2, 4, 8, 16, ... , 2" congruent angles. Symmetry. 86. Definition. Two points are said to be sym- metrical with regard to a straight, when it bisects at right angles their sect. The straight is called their axis of symmetry. Two points have always one, and only one, symmetry axis. A point has, with regard to a given axis of symmetry, always one, and only one, sym- FiG. 31. metrical point, namely, the one which ends the sect from the given point perpendicular to the axis and bisected by the axis. 40 RATIONAL GEOMETRY. rX ~L Fig. 33- 87. Definition. Two figures have an axis of symmetry when, with regard to this straight, every point of each has its sym- metrical point on the other. One figure has an Fig. 32. axis of symmetry when, with regard to this straight, every point of the figure has its symmetrical point on the figure. One figure is called symmetrical when it has an axis of symmetry. Any figure has, with regard to any given straight as axis, always one, and only one, symmetrical figure. 88. Theorem. An angle is symmetrical with re- gard to its bisector and the end-points of congruent sects from the vertex are symmetrical. Proof. Their sect is bisected at right angles by the angle-bisector. 89. Definition. A sym- metrical quadrilateral with a diagonal as axis is called a deltoid. 90. Definition. A sect whose end- points are the bi- section-points of opposite sides of a quadrilateral is called a median. So is the sect from a vertex of a triangle to the bisec- tion-point of the opposite side. Fig. 34. . ! . ■ .2 C |B Fig. 36. Fig. 35. PARALLELS. 41 91. Definition. A sjTmmetrical quadrilateral with a median as axis is called a symtra. Ex. 36. In a r't A if to set off one acute ^, a, in the other, ^, bisects, so is it with /J's sides. Ex. 37. The st' through the bisection-point of the base of a -I- A and the opposite vertex is ± to the base and bisects if . Ex. 38. The r't bi' of the base of a -I- A bisects ^ at the vertex. Ex. 39. The X from vertex bisects base and ^ in a •I- A. Ex. 40. The bisector of ^ at vertex of a •!■ a is r't bi' of base. Ex. 41. If a r't bi' of a side contains a vertex, the A is -I". Ex. 42. The bisector of an exterior if at vertex of -I- a is II to base, and inversely. Ex. 4 ^ The end of sect from intersection of congjuent sides of a -1- A costraiqht and s to one determines with I'nd of other a ± to base. Ex. 44. To erect a X at the end-point of a sect without producing the sect. Ex. 45. A II to one side of an :^ makes with its bisector and other side a -I- A. Ex. 46. The bisectors of the = ^s of a-l- A are s. Ex. 47. Every symmetrical quadrilateral not a del- toid is a symtra. Ex. 48. The intersection point of two s},Tnmetrical straights is on the axis. Ex. 49. The bisector of an angle is syminetrical to the bisector of the symmetrical angle. Ex. 50. A figure made up of a straight and a point is s^•mmetrical. Ex. 51. In any deltoid [i] One diagonal (the axis) is the perpendicular bisector of the other. [2] One diagonal (the axis) bisects the angles at its two vertices. [3] Sides which meet on one diagonal (the axis) are con- 4* RATIONAL GEOMETRY. gruent; so each side equals one of its adjacent sides. [4] One diagonal (not the axis) joins the vertices of congruent angles and makes congruent angles with the congruent sides. [5] The triangles made by one diagonal (the axis) are congruent. [6] One diagonal (not the axis) makes two isosceles triangles. Ex. 52. Any quadrilateral which has one of the six preceding pairs of properties (Ex. 51) is a deltoid. Ex. 53. A quadrilateral with a diagonal which bisects the angle made by two sides, and is less than each of the other two sides, and these sides congruent, is a deltoid with this diagonal as axis. Ex. 54. A quadrilateral with a side meeting a con- gruent side in a greater diagonal which is opposite con- gruent angles is a deltoid with that diagonal as axis. Ex. 55. In any symtra [i] Two opposite sides are parallel, and have a common perpendicular bisector. [2] The other two sides are congruent and make con- gruent angleii with the parallel sides. [3] Each angle is congruent to one and supplemental to the other of tlie ^wo not opposite it. [4] The diagonals are congruent and their parts adja- cent to the saine parallel are congruent. [5] One median bisects the angle between the two diagonals, and also ohe angle between the non-parallel sides (produced). Ex. 56. Any quadrilateral which has one of the pre- ceding five pairs of properties (Ex. 53) is a symtra. 92. Definition. A trapezoid is a quadrilateral with two sides parallel. 93. Definition. A parallelogram is a quadrilateral with each side parallel to another (its opposite). 94. Definition. A parallelogram with one angle right is called a rectangle. A parallelogram with two consecutive sides congruent is called a rhombus. A rectangle which is a rhombus is called a square. PARALLELS. 43 95. Theorem. The opposite sides and angles of a Fig. 37. parallelogram arc congruent, and its diagonals bisect cacJi other. Proof. aABC = A ADC [2 ^ 's and included side = ]. .-. BC=AD. .-. (as in 82) AF=FC and BF^FD. 96. Theorem. // three parallels make congruent sects on one transversal, they do on every transversal. Given a || 6 1| c, also AB = BC. To prove FG = GH. Proof. Take FL\\GM\\ AB. Then FL = AB = BC ^GM [95, opposite sides of allgmare^]. :.aFLG^ Fig. 38. aGMH[2 -4^ 's and included side =]. .*. FG=GH. 97. Corollary to 96. A straight through the bi- section-point of one side of a triangle and parallel to a second side bisects the third side. [In figure let F coincide \\-ith .^4.] 98. Inverse of 97. The straight through the bi- section-points of any two sides of a triangle is parallel to the third side. [For, by 97, it is iden- tical with the II to the third side through either bisection- point.] 44 RATIONAL GEOMETRY. Theorem. The sect whose end-points are the bisection-points of two sides of a triangle is congruent to each sect made in bisecting the third side. Proof. By97, 6^/^11 BCbi- ^ ^ sects AC. Since (by 98) Fig. 39. FG\\CH, .-. (hy 95) FG^CH. 100. Theorem. // two sides of a quadrilateral are congruent and parallel it is a parallelogram. Given AS = and \\CD. Proof. aABC=aADC. Fig. 40. .-.^ACB^^CAD. .-.CBWAD. Ex. 57. Every straight through the intersection of its diagonals cuts any parallelogram into congruent trapezoids. Ex. 58. A quadrilateral with each side equal to its opposite is a parallelogram. Ex. 59. A quadrilateral with a pair of opposite sides equal, and each greater than a diagonal making equal alternate angles with the other sides, is a parallelogram. Ex. 60. A quadrilateral with a side equal to its oppo- site, and less than a diagonal opposite equal angles, is a parallelogram . Ex. 61. A quadrilateral with each angle equal to its opposite is a parallelogram. Ex. 6z. A quadrilateral whose diagonals bisect each other is a parallelogram. PARALLELS. 45 loi. Theorem. In any sect AB there are always two, andonly two, points, C, D, such that AC = CD = DB. Fig. 41. Proof. Take on any ray from A, any sect AF, and a sect FG = AF, and a sect GH = FG. Take FC II GD II HB. Then, by 96, AC^CD = DB. Suppose two other such points C, D' . Then, by 98, C'F II D'G. Now HB' \\ GD' (by 96) makes D'B' = D'C' . .'. from our hypothesis and III i, B' is identical with B. /. since GD\\HB (by IV) D' is identical with D. .'. since FC\\GD (by IV) C is identical with C. 102. The two points, C, D, of the sect AB such that AC = CD = DB may be called the trisection- poiiits of AB. 103. Theorem. T/ii? f/zrct? 'medians of a triangle are copiinctal in that triscctioii-point of each remote from its vertex. Proof. Any median AG must meet any other CF, since A and G are on dif- ferent sides of the straight CF, and so the cross of st' AG with st' CF is on sect AG, and similarly it is on sect CF. If P, 0, are bisection-points of OC and OA, then (by 98 and Fig. 42. 99) PO II and =GF. .'. by 100 PQFG is a || gm and (by 9S) PF and QG bisect each other. A 46 RATIONAL GEOMETRY. 104. Definition. The cointersection-point of its medians is called the triangle's centroid. 105. Definition. A perpendicular from a vertex to the straight of the opposite side is called an altitude of the triangle. This opposite side is then called the base. The perpendicular from a vertex of a parallelogram to the straight of a side not through this vertex is called the altitude of the parallelogram with reference to this side as base. Ex. 63. The bisectors of the four angles which two intersecting straights make with each other form two straights perpendicular to each other. Ex. 64. If four coinitial rays make the first angle con- gruent to the third, and the second congruent to the fourth, they form two straights. Ex. 65. How many congruent sects from a given point to a given straight? Ex. 66. Does the bisector of an angle of a triangle bisect the opposite side? Ex. 67. The bisectors of vertical angles are costraight. Ex. 68. If two isosceles triangles be on the same base the straight determined by their vertices bisects the base at right angles. Ex. 69. Suppose a A to be 3 bars freely jointed at the vertices. Is it rigid? Are the ^s fixad and the joints of no avail? Of what theorem is this a consequence? How is it with a jointed quadrilateral? Why? Ex. 70. Joining the bisection-points of the sides of aA cuts it into 4 = As. Ex. 71. Joining the bisection-points of the consecutive sides of a quadrilateral makes a || g'm. Ex. 72. The medians of a quadrilateral and the sect joining the bisection-points of its diagonals are all three bisected by the same point. Ex. 73. If the bisection-points of two opposite sides PARALLELS. 47 of a II g'm ore joined to the vertices the diagonals are tri- sected. Ex. 74. The J.S from any point in the base of an •!• A to the sides are together an altitude. Ex. 75. The diagonals of a rectangle are =, oi a. rhom- bus are ±. Ex. 76. If 2 II s are cut by a transversal, the bisectors of the interior :^ s make a rectangle. Ex. 77. The angle-bisectors of a rectangle make a square. Ex. 78. If the ^ s adjoining one of the || sides of a trape- zoid are =, so are the others. Ex. 79. The bisectors of the interior ^ s of a trapezoid malce a quad' with 2 r't if s. Ex. 80. The bisection-point of one sect between jjs bisects any through it. Ex. Si. The = altitudes in •!■ A make with the base :^ s = to those made in bisecting the other if . Ex. 82. Through a given point within an if draw a sect terminated by the sides and bisected by the point. Ex. Sj. Sects from the vertex to the trisection-points of the base of -I- A are =. Ex. 84. If the if s made by producing a side of a A are = , so are the other sides. Ex. 85. If a quad' has 2 pairs of congruent consecutive sides, the other if s are = . Ex. 86. Two AS are = if two sides and one's median are respectively s Ex. 87. Two ■!• As are s if one if and altitude are = to the corresponding. Ex. 88. Two AS are s if a side, its altitude and an adjoining if are respectively = . Ex. 89. If 2 altitudes are = the A is •!• . Ex. 90. Two As are = if two sides and one's altitude are = to the corresponding. Ex. 91. Two AS are = if a side, its altitude and median are respectively = . Ex. 92. Two AS are = if a side and the other 2 altitudes are respectively s . 48 RATIONAL GEOMETRY. Ex. 93. Two AS are ^ if a side, an adjoii^ng ^ and its bisector are respectively =. Ex. 94. Two equilateral as are = if an altitude is =. Ex. 95. The bisector is within the ^ made by altitude and median. Ex. 96. In a right A one bisector also bisects ^ be- tween its altitude and median. Ex. 97. Two sects from the vertices of a A to the oppo- site sides cannot bisect each other. Ex. 98. The _Ls from 2 vertices of a A upon the median from the third are =. Ex. 99. Two II g'ms having an ^ and the including sides s are =. 106. Theorem. A point is on the perpendicular bisector of a sect whose end points make with it congruent sects. Take bisection point M of AB. Then (by 57) OMl AB. Fig. 43. Fig. 44. 107. Theorem. The three perpendicular bisectors of the sides of a triangle are copunctal (in its ' circum- center'). Proof. Bisect (by 82 and 51) AB at D and BC at F by perpendiculars. These perpendiculars (by 77) meet, say at 0. Therefore (by 43) AO =BO=CO. Therefore (by 106) is on ±bi' of AC. THE CIRCLE. 49 1 08. Theorem. The three altitudes of a triangle are copitnctal (in its 'orthocenter'). Given the ^ABC. To prove that the straights through .4, B, C perpendicular to the straights a, b, c respectively, are copunctal. ,-- / °/ >< / " /> c: \ / V /y ^\ / ^ /x \^ / v^ ^/ \' / \ / V B F to. 45. Proof. By 66, through A, B, C take B'C, A'C, A'B'WBC, AC, AB respectively. .-. AAB'C^ aABC = aABC'. [2:^s and included side =]. .-. AB'^AC, and AD is the J.bi' of B'C [± to ist of 2 ||s is J. to 2nd.] Similarly BE^.hi' of A'C; and CFLW of A'B'. .'. AD, BE, CF are copunctal by 107. CHAPTER V. PROBLEMS OF CONSTRUCTION. Existence theorems on the basis of assumptions I- IV, and the visual representation of such theorems by graphic constructions. Graphic solutions of the geometric problems by means of ruler and sect-carrier. 109. Convention. What are called problems of construction have a double import. Theoretically they are really theorems declaring that the exist- ence of certain points, sects, straights, angles, etc., follows logically by rigorous deductions from the existences postulated in our assumptions. Thus the possibility of solving such problems by elementary geometry is a matter absolutely essential in the logical sequence of our theorems. So, for example, we have shown (in loi) that a sect has always trisection points, and this may be expressed by saying we have solved the problem to trisect a sect. Now it happens that a solution of the problem to trisect any angle is impossible with only our assumptions. Thus any reference to re- sults following from the trisection of the angle would be equivalent to the introduction of additional assumptions. so PROBLEMS OF CONSTRUCTION. 51 But problems of construction, on the other hand, may have a reference to practical operations, usually for drawing on a plane a picture which shall serve as an approximate graphic repre- sentation of the data and results of the existential theorem. Our Assumptions I postulate the existence of a straight as the result of the existence of two points. This may be taken as authorizing the graphic desig- nation of given points and the graphic operation to join t^^•o designated points by a straight, and as guaranteeing that this operation can always be effected. Confining ourselves to plane geometry, on the basis of the same Assumptions I, we authorize the graphic operation to find the intersection-point of two coplanar non-parallel straights, and guaran- tee that this may alwaj's be accomplished. To practically perform these graphic operations, that is for the actual drawing of pictures which shall represent straights with their intersections, we grant the use of a physical instrument whose edge is by hypothesis straight, namely, the straight-edge or ruler. Thus Assumptions I give us as assumed con- structions, or as solved, the fundamental problem of plane geometry: Problem i. (a) To designate a given point of the plane; (b) to draw the straight determined by two points; to find tlic intersection of two non-parallel straights. no. Oiir Assumptions III postulate the existence on any given straight from any given point of it to- ward a given side, of a sect congruent to a given sect. 52 RATIONAL GEOMETRY. This may be taken as authorizing and guarantee- ing the graphic operation involved in what may- be called Problem 2. To set off a given sect on a given straight from a given point toward a given side. A physical instrument for actual performance of this construction in drawing might be called a sect- carrier. Our straight-edge will also serve as sect- carrier if we presume that the given sect may be marked off on it, and it then made to coincide with the given straight with one of the marked points in coincidence with the given point of the straight. Notice that in these graphic interpretations we freely use the terminology of motion, while the real existential theorems themselves are independent of motion, underlie motion, and explain motion. We assume that the motion of our physical instruments is rigid. 111. We now announce the important theorem that in our geometry all graphic problems can be solved, all graphic constructions effected, merely by using problems i and 2. Theorem. Those geometric construction problems {existential theorems) solvable by employing exclu- sively Assumptions I-IV are necessarily graphically solvable by means of ruler and sect-carrier. The demonstration will consist in solving with problems i and 2 the three following problems : 112. Problem 3. Through a given point to draw a parallel to a given straight. Given the straight AB and the point P. PROBLEMS OF CONSTRUCTION. S3 Construction. Join P with any point A of AB by Prob. I. On the straight PA beyond A take (by Prob. 2) AC^^P. JoinC with any other point B of AB. On the straight CB beyond B take BQ^BC. PQ is the parallel sought. Proof. By 98. 113. Problem 4. To draw a perpendicular to a given straight. Construction. Let A be any point of the given straight. Set off from A on this straight toward both sides two congruent sects, AB and AC, and then determine on any two other straights through A the points E and D, on the same side of AB^ and such that AB = AD = AE. Since ^ABD and -^ACE are angles at the bases of isosceles triangles, .". they are acute, .•. the straights BD and CE meet in F, and also the straights BE and CD in H. Then FH is the perpendicular sought. Proof. i-BDC and ^BEC are (by 56 and 79) right. Since (by 108) the altitudes of ^BCF are copunctal, .". FH is _L to BC. 114. Problem 5, To set off a given angle against a given straight, or to construct a straight cutting a given straight under a given angle. 54 RATIONAL GEOMETRY. Given /3 the angle to be set off, and A its vertex. Construction. We draw, by Prob. 3, the straight I through A \\ to the given straight against which the given angle /? is to be set off. By Prob. 4 draw a straight _L to / and a straight _L to one side of /?. Through any point B of the other side of /5 draw, by Prob. 3, ||s to these ±s. Call their feet C and D. Then (by Probs. 4 and 3) draw from A a. st' 1. to CD. Call its foot E. Then ^CAE=^. So EA will cut the given straight || to / under the given ^ p. Proof. Since -^-ACB and ^ADB are right, so (by 146) the four points A, B, C, D are concyclic. Consequently ^ACD= ^ABD (by 133) being in- scribed angles on the same chord AD and on the same side of it. Therefore their complements ■2^CAE= iBAD. 115. This completely demonstrates our theorem, 151, since the existential theorems in Assumptions II guarantee the solution of problems requiring no new graphic operations, such as to find a point within and a point without a given sect, and certain other problems of arrangement. 116. In our geometry, though constantly using graphic figures, we must never rely or depend upon them for any part of our proof. We must always PROBLEMS OF CONSTRUCTION. 55 take care that the operations undertaken on a figure also retain a purely logical validity. 117. This cannot be sufficiently stressed. In the right use of figures lies a chief difficulty of our in- vestigation. The graphic figure is only an approximate sug- gestive representation of the data. We cannot rely upon what we suppose to be our immediate per- ception of the relations in even the most accurate obtainable figure. In rigorous demonstration, the figure can be only a symbol of the conceptual content co^•ered by its underlying assumptions. The logical coherence should not be dependent upon anything supposed to be gotten merely from perception of the figure. No statement or step can rest simply on what appears to be so in a figm-e. Every statement or step must be based upon an assumption, a definition, a convention, or a preced- ing theorem. Yet the aid from figures, from sensuous intuition, is so inexpressibly precious, that any attempt even to minimize it would be a mistake. That treat- ment of a proposition is best which connects it most closely with a visualization of the figure, while yet not using, as if given by the figure, concepts not contained in the postulates and preceding propo- sitions. 118. As an immediate result of Prob. 5, the proofs in Chapters I-V of om- existential theo. ems give ruler and sect-carrier solutions of the coiTesponding prob- lems. We will now give some alternative solutions. 56 RATIONAL GEOMETRY. 119. Problem 6. At a given point A to make a right angle. Solution. Draw through A any straight AD, and through D any other straight BC, and make AD^BD^CD. Then is ^ SAC right. 49. ^ *> 120. Problem 7. From a given point A to drop a perpendicular upon a given straight BC. Solution. By Prob. 6, at A construct a rt. :^ BAC. Make BD^BA. Draw DE \\AC. Make BF^BE. Then is AF±BC. Proof. /^ABF = ADBE. [2 sides and inc. ^ = .J 121. Problem 8. At any point A on a straight BC to erect the perpendicular. Solution. By Prob. 7, from any point without the straight drop to it a perpendicular. By Prob. 3, ■ draw a parallel to this through A. 122. Problem 9. To bisect a given sect AB. Construction. Draw through B any other straight BC. Make on it BC^CD=DE. Produce AE=EF. Draw FDG. Then is AG ^GB. Proof. D is the centroid of A ASF. Fig. 50. PROBLEMS OF CONSTRUCTION. 57 123. Problem 10. To bisect a given angle. Construction. On one side of the given ^A take any two points B, C. On the other side takeAB'^AB, and AC ^AC. The sects BC and B'C intersect, say at D sector. B Fig. 52. AD is the desired bi- Cx. 124. Problem 11. (Without the Parallel Postu- late.) To draw a perpendicular to a given straight. Construction. Let AB be the given st'. Take another st' AC. On AB take AD =AC. Take (by 123) the bisector of ^CAD meeting CD in G. On AB take AF^AG. On Then is HF ± to AB. A A F Fig. 53. 'X» ray AG take AH=AD. Proof. aACG^a ADG ^ a AHF. 125. Problem 12. At a given point A on a given straight BC to erect the per- '° pendicular. Construction. Draw (by _ 124) YOZ ± to BC, meeting it^ at 0. Take or = 0Z. Pro- duce YA to A'. The bisector of i-ZAX is ± to BC. ■^^ y Y V / Fig. 54. 126. Problem 5. At a given point in a given straight to make an angle congruent to a given angle. Required, against the given ray AB of a, and A SS RATIONAL GEOMETRY. toward a given side of a, the C-side, to make an angles ^D (given). Construction. To one side of the given acute angle erect (by 124) FH ± DF, meeting the other side at H. Take AB^ ■pio"ss. " DF and BCIAB and BC^FH. .-. (by III 6) ^BAC^^FDH. 127. Problem. To draw a perpendicular to a given- straight from a given point without it. Let AB be the st' and C the p't. Draw (by 125) OZ _L to AB from a p't on it so as to meet AC, say in Z. Produce ZO to Y, making OZ = OY. Produce AY to F, making AF=AC. Then CF is ± to AB. 128. Definition. The cointersection-point of the three bisectors of the internal angles of a triangle, /, is called the triangle's in-center. Ex. 100. A right angle can be trisected. Ex. loi. To construct a triangle, given two sides and the included angle. Ex. 102. To construct a triangle, given two angles and the included side. Bx. 103. To construct a triangle, given two angles and a side opposite one of them. Ex. 104. To describe a parallelogram, given two sides and the included angle. Ex. 105. To construct an isosceles triangle, having given the base and the angle at the vertex. Ex. io6. To erect a perpendicular to a sect at its end- point, without producing the sect or using parallels. Hint. At this end-point against the given sect make PROBLEMS OF CONSTRUCTION. 59 any acute angle. At any other point of the sect make toward this a congruent angle. Beyond the intersection-point of the rays, make on this second ray a sect congruent to a side of this isosceles triangle. Its end is a point of the required perpendicular Ex. 107. To draw an angle-bisector without using the vertex. Ex. 108. Through a given point to draw a straight which shall ■ make congruent angles with two given straights. Ex. 109. In a straight find a point with which two given points give equal sects. Ex. no. From two given points on the same side of a straight to draw two straights intersecting on it and making congruent angles with it. Ex. III. To draw a straight through a given point between two given straights such that they intercept on it a sect bisected by the given point. Ex. 112. Through a given point to draw a st' making = ^s with the sides of a given ^. Ex. 113. Construct I- a from b and hb', from a and b; from P and a +b; from /3 and hb ; from b and /3 ; from p and hb; from p and a. Ex. 114. Construct r't a from a and he', from a and c; from a and a+b. Ex. 115. Construct A from p, a, and ha', from p, a, and /3; from its pedal; from h, a+c, a; from a, hb, p; from Ji, /z, 1 3. Ex. 116. Without prolonging two sects, to find the bisector of the ^ they would make. Ex. 117. From one end of the hypothenuse lay off a sect on it congruent to the J. from the end of this sect to the other side. Ex. 118. From I- A cut a trapezoid with 3 sides =. Ex. 119. To inscribe a sq. in a given r't •!• A. Ex. 120. Find point in side of -I- A where ± erected and produced to other side is = to base. 6o RATIONAL GEOMETRY. Ex. 121. Construct a from a, a, and that ta trisects a; from a and orthocenter ; from a and centroid. Ex. 122. Construct A from feet of medians; of alti- tudes. Ex. 123. (Brahmagupta.) If the diagonals of a quad' with opposite 4C's supplemental are ±, the st' through their intersection ± to any side bisects the opposite side. CHAPTER VI. SIDES AND ANGLES. 129. Convention. When a sect congruent to CD is taken on sect AB from A and its second end- point falls between .4 and B, then AB is said to be greater than CD; (AB> CD). When an angle congruent to 4 {h, k) is set off from vertex against one of the rays of i^AOB toward the other ray, if its second side falls within ^AOB, then -ifAOB is said to be greater than ^ {h, k). In symbols, '4-A0B> 4(h,k). 130. Theorem. // tlic first side of a triangle be greater than a second, tlien the angle opposite the first must be greater than- tJie angle opposite tlic second. Given BA > BC. 1 To prove 4C> ^.4. Proof. From B toward A take BD = BC. The end-point D of this sect then, because BA > BC, is between A and B, that is within ^ACB. Call F the point where the bisector of ^B meets the side AC. Then 4C=4FDB. But (by 79) yfFDByyfA. 6i 62 RATIONAL GEOMETRY. 131. Inverse. If ^Ay^-B, :. a>b. Proof. [From 56 and 130.] 132. Definition. Except the perpendicular, any sect from a point to a straight is called an oblique. 133. Theorem. From a point to a straight any oblique is greater than the perpendicular. Proof. Since ^ CAB is r't, .". (by 79) Fig. 57- ^A>^5. .". (byi3i)a>fe. 134. Theorem. Any two sides of a triangle are together greater than the third side. Proof. On St' BC, be- yond C, take CD = CA . .-. {hy s(>) 4 CDA^i^ CAD. But ^C is within i-DAB, .-. 4DAB>i.DAC^4D. .-. (by 131) BD>AB. 135. Theorem. (The ambiguous case.) If two tri- angles have two sides of the one congruent respectively to two sides of the other and the angles opposite one pair of congruent sides congruent, then the angles oppo- site the other pair are either congruent or supplemental. Fig. 58. C AH Fig. 59. Hypothesis. , A ABC and ^FGH with i-A = 4 F,AB ^FG, and BC^GH. SIDES AND ANGLES. 63 Conclusion. :^C =^H, or ^ C supplement of Proof. At B against BA take, on the side toward C, the ^ABC = 4G. If ray BC falls on ray BC, then (by 80) 7fC=^H. If not on BC, suppose C between C and A. Then (by 44) i-BC'A = 4H, and BC'^GH^BC. .-. (by 56) iBC'C^i-C. 136. Corollary to 135. Two triangles are con- gruent if they have two sides and the angle opposite the greater respectively congruent. 137. Definition. A triangle one of whose angles is a right angle is called a right-angled triangle, or more briefly a right triangle. The side opposite the right angle is called the liypothcnitse. 138. Corollary to 136. Two right-angled trian- gles are congruent if the hypothenuse and one side are respectively congruent. Ex. 124. If two triangles have two sides of the one re- spectively congruent to two sides of the other, and the angles opposite one pair of congruent sides congruent, then if these angles be not acute the triangles are congruent. Ex. 125. If two triangles have two sides of the one respectively congruent to two sides of the other, and the angles opposite one pair of congruent sides congruent, then if one of the angles opposite the other pair of con- gruent sides is a right angle the triangles are congruent. Ex. 126. If two triangles have two sides of the one respectively congruent to two sides of the other, and the angles opposite one pair of congruent sides congruent, then if the side opposite the given angle is congruent to or greater than the other given side the triangles are congruent. Ex. 127. If any triangle has one of the following proper- ties it has all: 64 RATIONAL GEOMETRY. I. Symmetry. ^. Two congruent sides. 3. Two congruent angles. 4. A median which is an altitude. 5. A median which is an angle-bisector. 6. An altitude which is an angle-bisector, 7. A perpendicular side-bisector which contains a vertex. 8. Two congruent angle-bisectors. Ex. 128. The difference of any two sides of a triangle is less than the third side. Ex. 129. From the ends of a side of a triangle the two sects to a point within the triangle are together less than the other two sides of the triangle, but make a greater angle. Ex. 130. Two obliques from a point making congruent sects from the perpendicular are congruent, and make congruent angles with the straight. Ex. 131. Of any two obliques between a given point and straight that which makes the greater sect from the foot of the perpendicular is the greater. Ex. 132. Of sects joining two symmetrical points to a third, that cutting the axis is the greater. 139. Theorem. // two triangles have two sides of the one respectively congruent to two sides of the other, then that third side is the greater which is opposite the greater angle. Proof. Take the triangles with one pair of congruent sides m common, BC, and on Fig. 60. ^j^g same side of BC the other air of congruent sides, BA, BA'. The bisector of :^ABA', being within ^ABC, meets ^C at a ,.o''n: G. Then (by 43) aABG^aA'BG. .-. AG SIDES AND ANGLES. 65 = A'G. But (by 134) A'G and GC are together greater than A'C. 140. Inverse of 139. If two triangles have two sides of the one respectively congruent to two sides of the other, then, of the angles opposite their third sides, that is the greater which is opposite the greater third side. Ex. 133. Two right triangles are congruent if the hy- pothenuse and an acute angle are congruent, or if a per- pendicular and an acute angle are congruent to a per- pendicular and the corresponding acute angle. Ex. 134. Given y4B a sect, Cits bisection-point, PA sPB. Prove PCXAB. Ex. 135. Inverse. GivenCP±bi' of AS. ProvePA=PB. Ex.136. GivenPMl.AMsPNJ.yl7V. Prove^PAMs ^ PAN or its complement. Ex.137. Inverse. Given :^ P^M = :^ PAAA. Prove PMi.AM^PN S.AN. V. The Archimedes Assumption.* V. Let A 1 be any point on a straight between any given points A and B\ take then the points A 2, Az, Ai, . . . , such that Ai lies between A and ^2, further- more A 2 between >liand Az, further ^3 between A2 and Ai, and so on, and also such that the sects AA-i, .41^2, A^Az, AzAi, . . . , are congruent; then in the series of points A2, Az, A4, . . , there is always such a point A„, that B lies between A and A„. 141. This postulate makes possible the introduc- tion of the continuity idea into geometry. We have not used it, and will not, since the whole of elemen- tary rectilineal geometry can be constructed with only Assumptions I-IV. *Archimeclis Opera, lec. Heiberg, vol. I, 1880, p. 11. CHAPTER VII. A SECT CALCULUS. 142. On the basis of assumptions I 1-3, and II-IV, that is, in the plane and without the Archimedes assumption, we will establish a sect calculus or geometric algorithm for sects, where all the oper- ations are identical with those for real numbers. The following proof is due to F. Schur. 143. (Pascal.) Let A, B,C and A', B' , C be two triplets of points situated respectively on two per- pendiculars and distinct from their intersection point 0'. If AB' is par- allel to A'B and BC parallel to B'C, then is also AC parallel to A'C. Proof. Call D' the point where the perpen- dicular from B upon the straight A'C meets the straight B'A'. Then C is the orthocenter of the triangle BA'D' ; therefore D'CLA'B and .-. JL^B'. Fig. 61. Gonsequently C is also the orthocenter of the tri- 66 A SECT CALCULUS 67 angle AB'D' ; .-. AD'IB'C and .-. also ±BC'. Con- sequently B is the orthocenter of triangle AC'D'; .-. AC'±D'B and .-. AC'\\A'C. 144. Instead of the word "congruent" and the sign =, we use, in this sect calculus, the word "equal" and the sign =. 145. We begin by showing how from any two sects to find unequivocally a third by an operation we will call addition. 146. li A, B, C are three costraight points, and B lies between A and C, then we designate c=AC as the sum of the two sects a = AB and h=BC, and write to express this c = a-\-b. B I, n. y a ^ 6 j 1« c=a+6 -^ Fig. 62. To add the two sects a and 6 in a determined order, we start from any point A, and take the point B such that AB = , that is = a. Then on the straight AB beyond B we take the point C such that BC = b. Then the sect AC is what we have designated as the sum of the two sects a =AB and b =BC in the order a + b. 147. From III 3 follows immediately that this sum is independent of the choice of the point A, and independent of the choice of the straight AB. By III I, it is independent of the order in which the sects are added. Therefore a + b = b + a. 148. This is the couuuiitative law for addition. Thus the commutative law for addition holds good, 68 RATIONAL GEOMETRY. is valid, for our summation of sects. But this law is not at all self-evident, and expresses no general magnitude relation, but a wholly definite geometric fact ; for a, b are throughout not numbers, but only- symbols for certain geometric entities, for sects. 149. The sects a and b are called less than c; in symbols: aa, c>b. 150. To add to a +b a further sect c, take on straight AB beyond C the sect CD = c. Then the sect AD = (a +b) +c. But this same sect AD is, by the given definition of sum, also the sum of the sects AB^ and BD, that is of the sects a and (6 +c). Thus a +{b +c) = (a +b) +c, and so is verified and valid what is called the associative law for addition. 151. To define geometrically the product of a sect a by a sect b, we employ the following con- struction. We choose first an arbitrary sect, which remains the same for this whole theory, and designate it by i. Set off on one side of a right angle, starting from the vertex 0, first the sect i and then, likewise from the vertex O, the sect b. Then set off on the other side the sect a. Join the end-points of the sects i and a, and draw a parallel to this straight through the end-point of the sect b. The sect which this parallel determines on the other side is the product ab. 152. To prove for our so defined multiplication A SECT CALCULUS. 69 of sects the commutative law for multiplication, ab = ba, first construct in the defined way the sect ab. Then take on the first (the horizontal) side of the right "■^l angle the sect a, and on the second (the vertical) side the sect b. Join the end of i with the end of b on the vertical side and draw a par- allel to this join through the end of a on the horizontal side : this, by definition, cuts off on the vertical side the sect ba, which (by Pascal) is identical with the previously constructed ab, because of the parallelism of the dotted lines. 153. We emphasize that this definition is purely geometric; ab is not at all the product of two num- bers. 154. To prove for our multiplication the asso- ciative law for multi- FlG. 64. 6c) a V plication a(bc) = (ab)c we construct first the e=ba Va\ sect d = bc, then da, d=bc "^vVNk further the sect e = ba, b ^W^ and finally ec. The ii^ end-points of da and 1 c a ec coincide (by Pas- Fig. 65. cal), and by the com- mutative law follows the above formula foi the as- sociative law of sect multiplication. 155. Finally is valid in our sect-calculus also the distributive law a(b--\-c) =ab + aL. 70 RATI0N/1L GEOMETRY. o(b+c) To demonstrate it we construct the sects ab, ac, and a(b + c), and draw through the end-point of the sect c (see Fig. 66) a parallel to the other side of the right angle. The congruence of the two right-angled tri- angles shaded in the _, figure and the appli- ^*'' cation of the theorem of the equality of opposite sides in the parallelogram give then the desired proof. 156. If b and c are any two sects, there is always one and only one sect a such that c = ab; this sect c a is designated by the notation -r, and is called the quotient of c by b. Ex. 138. In r't A,/ij makes ^ s =to a and ^. Ex. 139. Always Wc <§(a +6). Ex. 140. Prom point without acute ^ a, ±s to sides make 1^ =a. Ex. 141. The sect joining the bisection-points of the non-|| sides of a trapezoid is || to the || sides and half their sum. Ex. 142. The trisection-points of the sides of an equi- lateral A form a regular hexagon. Ex. 143. The ±s from A and B upon m-c are =. Ex. 144. Construct ■!■ A from 6 and a+hb; from /3 (or hb) and perimeter. Ex. 145. The sum of the three sects from any point A SECT CALCULUS. 7^ within a A to the vertices is < the sum and > i sum of the 3 sides. Ex. 146. Construct ir't A from b +c. Ex. 147. In a given st' find a point to which sects from 2 given points have the least sum. Ex. 148. The sum of the medians in A is < the sum and > J sum of sides. Ex. 149. Construct A from o - ^ and c; A from a, /?, r; A from a, /9, R. Ex. 150. If hb meets b at D, construct a from ht, a -AD, c-CD. Ex. 151. A quad' is a trapezoid if an opposite pair of the 4 AS made by the diagonals are =. CHAPTER VIII. PROPORTION AND THE THEOREMS OF SIMILITUDE. 157. With help of the just-given sect-cacuas EucHd's theory of proportion can in the following manner be established free from objection and with- out the Archimedes assumption. 158. Convention. If a, b, a', b' are any four sects, then the proportion a:b=a':b' shall mean nothing but the sect equation ab' = ba'. 159. Definition. Two triangles are called similar if their angles are respectively congruent. Sides between vertices of congruent angles are called cor- responding. 160. Theorem. In similar triangles the sides are proportional. Given a, b and a', b' corresponding sides in two similar triangles. To prove the proportion a:b=a':b'. Proof. We consider first the special case, where the angles included by a, b and a', b' in the two tri- angles are right, and sup- We 72 Fig. 67. pose both triangles on the same right angle. PROPORTION AND THE THEOREMS OF SIMILITUDE. 73 then set off from the vertex on one side the sect i, and take through the end-point of sect i the parallel to the two hypothenuses. This parallel determines on the other side the sect e. Then is, by our defini- tion of the sect-product, b=ca, b'=ea'. Conse- quently we have ab'=ba', that is, a:b=a': b'. We pass now to the general case. Construct in each of the two similar triangles the in-center I, respectively /', and drop from these the three per- pendiculars r, respect- ively r', on the triangle's sides. Designate the re- spective sects so deter- mined on the sides of the triangles by at, a^, b^, ba, Fig- 68. Ca, ct, respectively ui, a/, b/, ba, cj, cj'. The just- proven special case of our theorem gives then the proportions ab:r=ab : r b^:r = br: Oc'. r=ac' -.r', ba:r = ba.':r'. From these we conclude by the distributive law a:r=a':r', b:r=b':r', and consequently, in virtue of the commutative law of multiplication, a:a'=r:r'=b:b', and a:b = a':b'. i6i. From the just-proven the'irem (i6o) we get easily the fundamental theorem of the theory of proportion, which is as follows : 74 RATIONAL GEOMETRY. Theorem. If two parallels cut off on the sides of any angle the sects a, b, respectively a', b', then holds good the proportion a:b=a':b'. Inversely, when four sects a, b, a', b' fulfill this proportion, if the pairs a, a' and b, b' are set off upon the respective sides of any angle, then the straight joining the end- points of a and b is parallel to that joining the end- points of a' and b' . Proof. First, since parallels make with the sides of the given angle similar triangles, therefore (by i6o) a:b = a':b'. Second, for the inverse. Through the end-point of a' draw a parallel to the straight joining the end- points of a and b, and call the sect it determines on the other side b". Fio.bg. rj.^^^ ^y p.^g^ a:b=:a':b". But by hypothesis a:6=a': 6'. .". b"=b'. 162. Thus we have fotinded with complete rigor the theory of proportion on the basis of the Assumption-groups I-IV. 163. Corollary to 161. If straights are cut by any number of parallels the corresponding inter- cepts are proportional. 164. Corollary to 160. Parallels are divided proportionally by any three copunctal transversals. 165. Corollary to 161. Two triangles are similar if they have two sides proportional and the in- cluded angles congruent. 166. Definition. A point P, costraight with AB but without the sect AB, is said to divide the sect AB externally into the sects PA, PB. PROPORTION AND THE THEOREMS OF SIMILITUDE. 75 167. Corollary to 161. A sect can be divided nternally or externally in proportion to any two unequal given sects. The point of internal divi- sion is unique; likewise the point of external divi- sion. 168. Theorem. The bisector of any angle of a triangle or of its adjacent angle divides the opposite side in proportion to the other two sides. [Proof. Take AF \\ to bisector BD. Then BF = c.-\ 169. Definition. A sect divided internally and externally in proportion is said to be divided har- monically, and the four points are called a harmonic range. 170. Theorem. A perpendicular frcnn the riglit angle to the hypothenuse divides a right-angled tri- angle into two oiliers similar to it, and is tlic mean proportional between the parts of the hypothenuse. Each side is the mean propor- tional between the hypothenuse and its adjoining part. Proof. The r't A ABC ~ r't "^ ° aABD, since 4 A is common. '°' '^' 76 RATIONAL GEOMETRY. 171. Theorem. The square of the hypothenuse equals the sum of the squares of the two sides. Proof. AC:AB = AB:AD,fha.%{s,AB-^=AC-AD. Same way BC^ =AC- DC. Now add . .-. AB^ +BC^=AC{AD +DC) =AC^. 172. Theorem. Triangles having their sides taken in order respectively proportional are similar. A' Fig. 72. In the triangles ABC and A'B'C let AB:A'B' = AC:A'C' = BC:B'C'. To prove that the triangles ABC and A'B'C axe similar ('~). Proof. Upon AB take AF^A'B', and upon AC take AH^A'C. Then AB:AF =AC:AH. .-. (by 165) aASC ~ to AAFH. .-. AB:AF = BC:FH. But by hypothesis AB:AF = BC:B'C'. .■.FH = B'C'. .-. AAFH ^ A A'B'C [3 sides^]. .-. AABC~AA'B'C'. 173. Definition. Similar polygons are those of which the angles taken in order are respectively equal [i.e., congruent], and the sides between the equal angles proportional. 174. Theorem. Two similar polygons can be divided into the same number of triangles respect- ively similar. PROPORTION y4ND THE THEOREMS OF SIMILITUDE. 77 175- In our geometrical constructions we may for the scLt-carrier substitute the unit-sect carrier or set sect, an instrument for setting off one single fixed sect, say the unit. 176. Problem 3. Through a given point to draw d pdralld to a given straiglit. Construction. Join the given point P with any D /\, ' IS Fig. 73. point ,4 of the given straight a, and beginning at .4 set off twice in succession on a the unit, say to B and C. Let D be any point on AP, and E the cross of PC and BD, and finally F the cross of AE and CD. Then is PF\\a. Proof. Suppose the parallel to a through P meets BD at K and CD at F'. Then aDKP- A DBA, and aDKF'^^DBC. .-. (by 160) DK-.KP =DB:BA, and DK:KF'=DB:BC. .-. KP=KF'. Now aKE'F'-^aBE'A, and aKEP-aBEC, :.KF':BA=KE':BE', and KP:BC = KE -.BE. .-. KE':BE'=KE:BE. Fig, 74. 78 R/ITION/iL GEOMETRY. So E' IS, E. .'. F' is F, and PF the parallel to a through P. 177. Problem 2. To set off a given sect on a given straight from a given point toward a given side. Construction. Let AB be the sect to be set off and P the given point on /, the given straight. Draw through P the parallel to AB and set off on it the unit PC, and on / the iinit PD. The parallel to AP through B cuts PC, say in Q. The parallel to CD through Q cuts I in F. Then PF = AB. Ex. 152. If AB is divided harmonically by P, P', then PP' is divided harmonically by A, B. Ex. 153. If two triangles have the sides of one respect- ively parallel or perpendicular to the sides of the other they are similar. Ex. 154. The corresponding altitudes of two similar tri- angles are proportional to any two corresponding sides. Ex. 155. To divide a sect into parts proportional to given sects. Ex. 156. A sect can be divided into any number of equal parts. Ex. 157. To find the fourth proportional to three given sects. Ex. 158. To find the third proportional to two given sects. Ex. 159. If three non-parallel straights intercept propor- tional sects on two parallels they are copunctal. Ex. 160. One side of a A is to either part cut off by a st' II to the base as the other side is to the corresponding part . Ex. i6i. If a straight divides two sides of a A propor- tionally, it is II to the third side. PROPORTION AND THE THEOREMS OF SIMILITUDE. 79 Ex. 162. The bisectors of an interior and an exterior ^ at one vertex of a A divide the opposite side harmonically. Ex. 163. The perimeters of two ~ polygons are pro- portional to any two corresponding sides. Ex. 164. A median and two sides of a trapezoid are copunctal. Ex. 165. The hypothenuse is divided harmonically by any pair of st's through the vertex of the r't ^ , making = -}^s, with one of its sides. Ex. 166. The bisection-point of the base of a A and any point on a || to the base through the vertex make a sect cut harmonically by a side and the other side produced. Ex. 167. I divides h as 6 to a +c. Ex. 168. Sects from the ends of the base of a A to the intersections of a || to base with the sides intersect on a median. Ex. 169. A from /3, a/c, R. Ex. 170. The sides of the pedal [A whose vertices are the feet of the altitudes] cut off As ~ to the original. Ex. 171. Three points being given, to determine an- other, through which if any st" be drawn, J.s upon it from, two of the former, shall together be equal to the J. from the third. Ex. 172. From two given sects to cut off two propor- tional to a second given pair so as to leave remainders proportional to another given pair. Ex. 173. In A , if sects from the ends of the base to the opposite sides intersect on the altitude, the joins of its foot to their ends will make equal angles with the base. Ex. 174. Construct A from h, §, and that tb makes segments as )" to n. Ex. 175. A from /3, vib, and ^ between b and im. Ex. 176. A from h and ±s on it from A and C. Ex. 177. A from /3, a-c, and difference of segments made by hb. Ex. 178. R't A from a +t, and b +c. Ex. 179. A from a - ^, a:b= in:u, and a third propor- 8o RyfTIONAL GEOMETRY. tional to the difference of segments made by he and the lesser side. Ex. i8o. A from, a, a +h, a +c. Ex. i8i. A from a, R, and b:c = m:n. Ex. 182. A from a, b, a—hb. Ex. 183. Divide a given sect harmonically as m to n. Ex. 184. In ~ As, a:a' =ha-h'a' =ma-m'a' = ta:fa' =r:r = R:R'. Ex. 185. Two r't As are ~ if hypothenuse and a X are proportional. Ex. 186. R't A from a and the non-adjacent segment made by he. Ex. 187. If 2 As have two sides of the one proportional to two sides of the other, and ^ s, one in each, opposite one corresponding pair of these sides = , the ^ s opposite the other pair are either = or supplemental. Ex. 188. The altitude to hypothenuse is a fourth pro- portional to it and the sides. Ex. 189. To inscribe a sq. in a given A. CHAPTER IX. EQUIVALENCE. The theory of equivalence in the plane. 178. We take as basis for the investigations in the present chapter the Assumptions I, 1-3, and II-IV. We exclude the Archimedes assumption. Our theory of proportion and sect-calculus put us in position to fotind the Euclidean theory of equiva- lence by means of the assumptions named, that is, in the plane and independent of the Archimedes assumption. 179. Convention. If we join two points of a polygon P by any sect-train which runs wholly in the interior of the polygon we obtain two new poly- gons, Pi and P2, whose inner points all lie in the interior of P. We say: P is separated or cut into Pi and P2', Pi and P2 together compose P. 180. Definition. Two polygons are called equiv- alent if thej^ can be cut into a finite number of triangles congruent in pairs. 181. Definition. Two potygons are said to be equivalent by completion if it is possible so to annex 81 82 RATIONAL GEOMETRY. equivalent polygons to them that the two polygons so composed are equivalent. 182. We will use the sign of equality ( = ) between polygons to denote "equivalent by completion." 183. From these definitions follows immediately: By imiting equivalent polygons we get again equiv- alent polygons. If we take away equivalent poly- gons from equivalent polygons the remaining poly- gons are equivalent by completion. Furthermore, we have the following propositions : 184. Theorem. Two polygons Pj and P2 equiv- alent to a third P3 are equivalent. Two polygons equivalent by completion to a third ' are equivalent by completion. Proof. By hypothesis there is as well for P^ as for P2 an assignable partition into triangles such that each of these two partitions corresponds re- spectively to a partition of the polygon P3 into congruent triangles. / \ A/ A \ / X yb'+c\ .-. :^^>r't :if. Iia'15 = minor arc A'B'. Proof. Since (by 43) aACB= aA'C'B'; .■.AB = A'B'. Moreover, if D is any point within arc AB then ray CD is within ^ACB. Hence (by 48) there is within ^ A'C'B' a ray CD' meeting arc A'B' in D', which makes i-A'C'D'=^ACD and ^B'C'D' = tBCD. :. (by 43) A'D'^AD and B'D'^BD. Also any point D" of the minor arc A'B' such that A'D" = AD would (by 58) be on the ray making ^A'C'D"= ^ACD= 4A'C'D', and hence identical with D'. THE CIRCLE. "3 273. Corollary. Any arc may be bisected. 274. Theorem. Any two congruent arcs have con- gruent radii. Given arc .4il/5 = arc A'M'B'. To prove CA = C'A'. Fig. 107. Proof. The bisector of 4 ACB cuts arc AMB in a point il/ such that, (by 43) d.ACM=ABCM. .-. AM=BM and ^ BMC= i AMC. From hypoth- esis there is a point il/' of arc A'M'B' such that aA'M'B'=aAMB. .-.A'M'^B'M'. .-. (by 58) aA'M'C'^^B'M'C. :. ^A'AL'C'^iB'M'C. :. (by 48 and 84) i.A'M'C'= i-AMC. .• . (by 44) the two isosceles triangles aAMC = aA'M'C. . AC = A'C'. 275. Inverse of 273 Congruent minor arcs are intercepted by congruent angles at the center. Proof. Since from hypothesis chord AB = c\iord A'B', .-. (by 274 and 58) aACB= aA'C'B'. .-. i-ACB^^A'C'B'. 276. Theorem. In a circle or in circles ivitJi con- griicni radii, congruent chords have congriioit minor arcs. 114 RATIONAL GEOMETRY. For the angles at the center on the congruent chords are congruent (by 58) [As with 3 sides =]. .'. (by 272) the minor arcs they intercept are con- gruent. 277. Theorem. Given a minor arc and a circle of congruent radius. There are on the circle two and only two arcs with a given end-point, congruent to the given arc. Proof. An angle at the center which intercepts the given arc can be set off (by III 4) once and only once on each side of the radius to the given point. 278. Theorem. From any point of a circle there are not more than two congruent chords, and the chords are congruent in pairs, one on each side of the diameter from that point. Proof. If AB is any chord, take at center C on the other side of AC, the ^ACB'=4ACB; :. by 43, i^ACB'= aACB. :.AB' = AB. Moreover, were B" the end- point of a third chord from A p^^ ^^g congruent to AB and to AB' , then B, B' , B" would be at once on (DC{CA) and OA{AB), which, by 122, is impossible. 279. Definition. If all the points of one arc are points of a second, but the second has also points not on the first, then the second is said to be greater than the first and any arc congruent to the second is said to be greater than any arc congruent to the first. THE CIRCLE. iiS 280. Theorem. In a circle or in circles with con- gritcnt radii, of two angles at the center, the greater intercepts tJw greater arc and chord. Hypothesis. CA=C'A'. iACD> i-A'C'B'. Conclusion. Arc AD > arc A' B'. & Fig. 109. Proof. From C against CA toward D, (by III 4) take i.ACB^:ifA'C'B'. Then from hjrpothesis ray CB is within ^ ACD. : . B is within arc AD. :. (by 192) arc ^Z?>arc AB. But (by 185) arc A'B' = axc AB. .'. (by 192) arc AD> arc A'B'. Moreover hA'C'B' has two sides CA', C'B'^CA, CD of A ACD, but -^-ACDy ^A'C'B', :. (by 179) ADyA'B'. 281. Inverse of 280. In a circle or in circles with congruent radii the greater chord has the greater angle at the center and the greater minor arc. For (by 140) it has the greater angle at the cen- ter, and .". , by 193, the greater minor arc. 282. Inverse of 280. In a circle or in circles with congruent radii, the greater minor arc has "6 RATION /I L GEOMETRY. the greater angle at the center and the greater chord. 283. Theorem. In a circle or in circles with con- gruent radii, congruent chords have congruent perpen- diculars from the center, and the lesser chord has the greater perpendicular. Proof. Of two chords from A on the same side of the diameter AC, one, say AD, is (by III 4) without the angle CAB made by the other, and hence its end-point Z?ison the minor arc A5. Hence (by 282) ^ACB> ■4.ACD and AB>AD. Moreover, the sect from the center to the bisection-point of AD, since D and so every point of AD is on the opposite side of AB from C, crosses the straight AB and .'. is > the perpendicular from C to AB. Moreover, congruent chords anywhere have con- gruent perpendiculars (by 138). 284. Inverse of 283. In a circle or in circles with congruent radii, chords having congruent perpen- diculars from the center are congruent, and the chord with the greater perpendicular is the lesser. For (by 283) it cannot be greater nor congruent. Two Circles. 285. A figure formed by two circles is symmetrical with regard to their center-straight as axis. IHE CIRCLE. 117 Every chord perpendicular to this axis is bisected by it. If the circles have a common point on this straight they cannot have any other point in common, for any point in each has in that its symmetrical point with regard to this axis, and circles with three points in common are identical. 286. Two circles with one and only one point in common are called tangent, are said to touch, and the common point is called the point of tangency or contact. 287. If two circles touch, then, since there is only one common point, this point of contact is on the center-straight, and a perpendicular to the center- straight through the point of contact is a common tangent to the two circles. Ex. 238. Two circles cannot mutually bisect. Ex. 239. The chord of half a minor arc is greater than half the chord of the arc. Ex. 240. In a circle, two chords which are not both diameters do not mutually bisect each other. Ex. 241. All points in a chord are within the circle. Ex. 242. Through a given point within a circle draw the smallest chord. Ex. 243. Rays from center to intersection points of a tangent with || tangents are ±. Ex. 244. A circle on one side of a triangle as diameter passes through the feet of two of its altitudes. Ex. 245. In I- j^ABCiiD on ^S =BC, prove CD>AD. Ex. 246. A circumscribed parallelogram is a rhombus. Ex. 247. In aABC, having AB>BC, the median BD makes ifBDA obtuse. Ex. 248. If AB, a side of a reg^ar A, be produced to D, then AD>CD> BC. Ex. 249. If SZ? is bisector fe, and A5>SC", then SO CZ?. ii8 RATIONAL GEOMETRY. Ex. 250. How must a straight through one of the common points of two intersecting circles be drawn in order that the two circles may intercept congruent chords on it ? Ex. 251. Through one of the points of intersection of two circles draw the straight on which the two circles intercept the greatest sect. Ex. 252. If any two straights be drawn through the point of contact of two circles, the chords joining their second intersections with each circle will be on parallels. Ex. 253. To describe a © which shall pass through a given point, and touch a given © in another given point. Ex. 254. To describe a © which shall touch a given 0, and touch a given st' [or another given ©] at a given point. Ex. 255. The foot of an altitude bisects a sect from orthocenter to circum-©. Ex. 256. If from the end-points of any diameter of a given © Xs be drawn to any secant, their feet give with the center = sects. Ex. 257. A, B, I, Ic axe concylic. Ex. 258. If tb meets circum-© in D, then DA ^DC =DI. Ex. 259. The Xs at the extremities of any chord make = sects on any diameter. Ex. 260. If in any 2 given tangent ©s be taken any 2 II diameters, an extremity of each diameter, and the point of contact shall be costraight. Ex. 261. If 2 s touch internally, on any chord of one tangent to the other the point of contact makes sects which subtend = :^s at the point of tangency of the ©s. Ex. 262. 2ma> = -aFAD, .-. BC-AD =FD-AC. .-. AB-CD+BC-AD=BF-AC+FD-AC>BD-AC. 309. Corollary to 308 (Ptolemy). The product of the diagonals of a cyclic quadrilateral equals the sum of the products of the opposite sides. (For then F falls on BD.) 310. Theorem. If a cyclic polygon be equilateral it is regular. Ex. 277. Every equiangular polygon circumscribed about a circle is regular. Ex. 278. Every equilateral polygon circumscribed about a circle is regular if it has an odd number of sides. Ex. 279. Every equiangular polygon inscribed in a circle is regular if it has an odd number of sides. Ex. 280. The chords on a sf through a contact-point of two Os are proportional to their diameters; and a common tangent is a mean proportional between their diameters. Ex. 281. The sum of the squares of the segments of 2 X chords equals the sq' of the diameter. Ex. 282. On the piece of a tangent between two || tan- THE CIRCLE. 127 gents the contact-point makes segments whose product is the square of the radius. Ex. 283. To inscribe in and circumscribe about a given O a A ~ to a given A . Ex. 284. Rr\ac=b:2{a+b+c). Ex. 285. In a r't A the ± sides are as the in-radii of As made by he. Ex. 286. A quad' is cycHc if diagonals cut so that product of segments of one equals product of segments of the other. Ex. 287. If one chord bisect another, and tangents from the extremities of each meet, the st' of their intersection points is II to the bisected chord. Ex. 288. The diagonals of a regular pentagon cut one another in the golden section, and the larger segments equal the sides. Ex. 289. From the vertex of an inscribed a a sect to the base || to a tangent at either end of the base is a fourth proportional to the base and two sides. Ex. 290. Straights from the vertices of any a to the contact-points of the in-O are copunctal. Ex. 291. If a chord is bisected by another, either seg- ment of the first is a mean proportional between the seg- ments of the other. Ex. 292. The diameter of a is a mean proportional between the sides of the circumscribed regular a and hexagon. Ex. 293. From the center of a given © to draw a st' cutting off from a given tangent a sect any multiple of the segment between © and tangent. Ex. 294. The vertices of all as on the same base with sides proportional are on a o with center costraight with base and radius a mean proportional between the sects from its center to the ends of base. CHAPTER XI. LENGTH AND CONTENT OF THE CIRCLE, 311. Theorem. The product of two sides of a triangle equals the product of two sects from thai vertex making equal angles with the two sides and extending, one to the base, the other to the circle cir- cumscribing the triangle. Proof. aCBD~aABE. 312. Corollary I to 311. If i^f,BD and BE coincide they bisect the angle B ; .■.AB-BC=BD-BE = BD{BD +DE) = BD' +BD-DE =BD' +CD -DA (hy 248). Fig no. Therefore the square of a bisector together with the product of the sects it makes on a side equals the product of the other two sides. 313. Corollary II to 311. If BD be an altitude, BE is a diameter, for then i^BAE is r't; there- fore in any triangle the product of two sides equah the product of the diameter of the circumscribed circle by the altitude upon the third side. Ex. 296. To find the bisectors of the angles of a tri- angle, given the sides. 4= -[abs{s —6)1^. a +0 28 LENGTH AND CONTENT OF THE CIRCLE. 129 Ex. 297. To find the radius of the O circumscribing a tri- angle. Rule: Divide the product of the three sides by four times the area of the triangle. R=abc/4A. Ex. 298. The in-radius. equals area over half sum of sides. [r=A/s]. Ex. 299. The side of an equilateral triangle is 6=i?(3)* = 2r(3)*- Ex. 300. The radius of circle circumscribing triangle 7, 15, 20, is 12J. The in-radius is 2. Ex. 301. To find the radius of an escribed circle. Rule: Divide the area of the triangle bv the difference between half the sum of its sides and the tangent side. [r, = A /(s -a)]. Ex. 302. A =(rrir2rs)i. Ex.303. i/ri+i/r2 + i/rs = i/r. Ex. 304. The sum of the four squares on the four sides of any. quadrilateral is greater than the sum of the squares on the diagonals by four times the square on the sect joining the mid-points of the diagonals. Ex. 305. The sum of the squares on the four sides of a parallelogram is equal to the sum of the squares on the diagonals. Ex. 306. The product of the external segments (sects) made on one side by the bisector of an external angle of a triangle equals the square of the bisector together with the product of the other two sides. Ex. 307. Find r,, r^, r,, when a = 7, 6 = 15, c = 2o. 314. Here for the first time the Archimedes Assumption is admitted, in so far as this chapter needs continuity. 313. Definition. A polygon whose sides are chords is 'inscribed,' whose sides are tangents is 'circumscribed.' 316. We assume that with every arc is connected 130 RATIONAL GEOMETRY. one, and only one, sect greater than the chord, and, if the arc be minor, less than the siim of the ' sects on the tangents from the extremities of the arc to their intersection, and such that if an arc be cut into two arcs, this sect is the sum of their sects. This sect we call the length of the arc. There is one, and only one, sect greater than the perimeter of any inscribed polygon and less than the perimeter of any circumscribed polygon, namely, the length of the circle. 317. In practical science, every sect is expressed by the unit sect preceded by a number. From our knowledge of the number and the imit sect it multiplies, we get knowledge of the sect to be expressed, and we can always construct this expression. For science, the unit sect is the centi- meter ["""J, which is the hundredth part of the sect called a meter, two marked points on a special bar of platinum at Paris, when the bar is at the tem- perature of melting ice. If an angle of an equilateral triangle be taken at the center of a circle, the chord it intercepts equals the radius. Therefore the length of a semi- circle is greater than three times its radius. Again, taking any diameter, then the diameter perpendicular to this, then perpendiculars at the four extremities of these, we have a square of tan- gents equal to 8r. Therefore the length of a semicircle is less than four times its radius. LENGTH AND CONTENT OF THE CIRCLE. 131 318. Theorem. The lengths of circles are propor- tional to their radii. Let c and c' be the lengths of the circles O0(r), O0'(/). To prove c:c' =r:r'. Proof. Let / be the fourth proportional such that r-y =c:f. Inscribe in O (}'(/) any polygon of perimeter p', and in O0(r) the similar polygon of perimeter p. Then p:p'=r:r'. [The perimeters of two similar inscribed polygons are as the radii.] .-. c-.f^p-.p'. :. c:p=f:p'. ButO^, .-. f>p'. Hence / is greater than the perimeter of any polygon inscribed in O0'(/). In the same way / is less than the perimeter of any polygon circumscribed to OO'(r'). .•./=c'. 319. Corollary to 318. The lengths of two arcs of two different circles, but intercepted by equal central angles, are as the radii. 320. Since c:r=c':r', .'. the number of its radii in the length of any circle is always the same constant, designated by 2n. Hence the length of any circle is 27cr. I3» RATlON/tL GEOMETRY. Historical Note on it. 320a. We have proved that n is greater than 3 and less than 4, but the Talmud says: "What is three handbreadths around is one handbreadth through," and oitr Bible also gives this value 3. [I Kings vii. 23; II Chronicles iv. 2.] Ahmes (about 1700 B.C.) gave [4/3]* = 3 -16. Archimedes placed it between 34-f and 3I. Ptolemy- used 3tVV- The Hindoo Aryabhatta (b. 476) gave 3-1416; the Arab Alkhovarizmi (flourished 813-833) gave 3-1416; Adriaan Anthoniszoon, father of Adriaan Metius [in 1585] gave 355/113=3-1415929; Ludolf van Ceulen [i 540-1610] gave the equivalent of over 30 decimal places [tt = 3-141592653589793238462643383279 + ] (the decimal fraction was not yet invented), and wished it cut on his tomb at Ley den. Vega gave 140 decimal places; Dase, 200; Richter, 500. In 1873 Wm. Shanks gave 707 places of decimals. The symbol Tt is first used for this number in Jones's "Synopsis Palmariorum Matheseos," Lon- don, 1706. In 1770 Lambert proved n irrational, that is, inexpressible as. a fraction. In 1882 Lindemann proved n transcendent, that is, not a root of any algebraic equation with rational coefficients, arid hence geometrically inconstructable. LENGTH /iND CONTENT OF THE CIRCLE. 133 Kochansky (1685) gave the following simple construction for the length of the semicircle : Fig. 120. At the end-point A of the diameter BA draw the tangent to the circle OC{CA). Take ^ACE = half the angle of an equilateral triangle =4 r't ^ . On the tangent, take EF = 2,r. Then BF is with great exactitude the length of the semicircle. In fact BF =r[i2,\ — 2(3)*]* = 3-141 5^- 321. Definition. The circle with the unit sect for radius is called the unit circle. The length of the arc of unit circle intercepted by an angle with vertex at center is called the size of the angle. The angle whose size is the unit sect is called a radian. 322. Theorem. A radian intercepts on any circle an arc whose length is that circle's radius. If /( denote the number of radians in an angle and I the length of the arc it intercepts on circle of radius r, then 323. Definition. An arc with the radii to its end-points is called a sector. 324. In content, an inscribed regular polygon of 211 sides equals a triangle with altitude the radius 134 RATIONAL GEOMETRY. and base the perimeter of an inscribed regular polygon of n sides; these sides being diagonals of deltoids with the radius for the other diagonal.] In content, a circumscribed regular polygon of n sides equals a triangle with altitude r and base the polygon's perimeter. There is one, and only one, triangle intermediate in content between the series of inscribed regular polygons and the series of circumscribed regular polygons, namely, that with altitude r and base the circle's length. The content of this triangle is called the content or surface of the circle, and its area, \rc=r'^T:, the area of the circle. So the areas of circles are pro- portional to the areas of squares on their radii. 325. Definition. From analogous considerations, the surface of a sector is defined as the content of a triangle with the arc's radius for altitude and the arc's length for base. 326. The area of a sector is the product of the length of its arc, by half the radius. Ex. 308. The areas of similar polygons are as the squares of corresponding sides. Ex. 309. Find the length of the circle when r = 14 units. Ex. 310. Find the diameter of a wheel which in a street 19,635 meters long makes 3125 revolutions. Ex. 311. Find the length and area of a circle when r = -]. Ex. 312. If we call one-ninetieth of a quadrant a degree of arc and its angle at the center a degree of angle, find the size of this ^ (size of 2(^1°). Ex. 313. How many degrees in a radian? Ex. 314. The angles of a triangle are as 1:2:3. Find the size of each. Find the number of degrees in each. Ex. 315. The angles of a quadrilateral are as 2:3:4:7. Find each in degrees and ra LENGTH /tND CONTENT OF THE CIRCLE. 13S Ex, 316. In what regular polygon is every angle i68j°? Ex. 317. If a r't^ be divided into — congruent parts, how many of them would a radian contain? Ex. 318. Find the length of the arc pertaining to a central angle of 78° when r=i-5 meters. Ex. 319. Find an arc of 112° which is 4 meters longer than its radius. Ex. 320. Calling n=~i^-, find r when 64° are 70-4 meters. Ex. 321. Find the inscribed angle cutting out one-tenth of the circle. Ex. 322. An angle made by two tangents is the differ- ence between 180° and the smaller intercepted arc. Make this statement exact. Ex. 323. Find the size of half a right angle. Ex. 324. Find the size of 30°; 45°; 60°. Ex. 325. How many radians in 7r°? in 240°? Ex. 326. Express the size of seven-sixteenths of a right angle. Ex. 327. How many radians in the angle made by the hands of a watch at 5: 15 o'clock? at quarter to 8? at 3: 30? at 6: 05? Ex. 328. The length of half a quadrant in one circle equals that of two-thirds of a quadrant in another. Find how many radians would be subtended at the center of the first by an arc of it equal in length to the radius of the second. Ex. 329. Find the number of degrees in an angle whose size is i; is f ; is |; is i}; is fjr. 2 • e Ex. 330. The size of the sum of two angles is — -Tt and their difference is 17°; find the angles. Ex. 331. How many times is the angle of an isosceles triangle which is half each angle at the base contained in a radian? Ex. 332. Two wheels with fixed centers roll upon each other, and the size of the angle through which one turns gives the number of degrees through which the other 136 RATIONAL GEOMETRY. turns in the same time. In what proportion are the radii of the wheels? Ex. 333. The length of an arc of 60° is 36!; find the radius. Ex. 334. Find the circle where ^30° is subtended by arc 4 meters long. Ex. 335. If be area of circle, prove I /[0/i(n)> + 1 /[0/.(r,)]*+ 1 /[O/sWJi = I /[0/(r)]i. Ex. 336. The perimeters of an equilateral triangle, a square, and a circle are each of them 12 meters. Find the area of each of these figures to the nearest hundredth. Ex. 337. An equilateral triangle and a regular hexagon have the same perimeter; show that the areas of their inscribed circles are as 4 to 9. Ex. 338. Find the number of degrees in the arc of a sector whose area equals the square of its radius. Ex. 339. Find area of sector whose radius equals 25 and the size of whose angle is f . Ex. 340. The length of the arc of a sector is 16 meters, the angle is -J- of a r't ^ . Find area of sector. Ex. 341. If 2 As have a common base, their areas are as the segments into which the join of the vertices is di- vided by the common base. Ex. 342. The area of a circum-polygon is half perimeter by in-radius \_\pr]. Ex. 343. The area of a rhombus is half the product of its diagonals. Ex. 344. If we magnify a quad' until a diagonal is tripled, what of its area? Ex. 345. If the sum of the squares on the three sides of a A = 8 times the square of a median the A is r't-angled. Ex. 346. Lengthening through A the side 6 of a A by c and c by 6, they become diagonals of a symtra whose area is to that of the A as {h + cY to be. Ex. 347. If upon the three sides of a r't A as corre- sponding sides similar polygons are constructed that on the hypothenuse =the sum of those on the Is. LENGTH AND CONTENT OF THE CIRCLE. 137 Ex. 348. The area of any r'tA =the sum of the areas of the two lunes or crescent-shaped figures made by de- scribing semi- OS outwardly on the ±s and a semi-o on the hypothenuse through the vertex of the r't ^ [called the lunes of Hippocrates of Chios (about 450 b.c.)]- Ex. 349. (Pappus.) Any two ||g'ms on two sides of a A are together =to a ||g'm on the third side whose consecutive side is = and |1 to the sect joining the common vertex of the other ||g'ms to the intersection of their sides || to those of the A (produced) . Ex. 350. If all the sides of a quad' are unequal, it is impossible to divide it into = as by straights from a point' within to its vertices. Ex. 351. The joins of the centroid and vertices of a a trisect it. Ex. 352. Make a symtra triple a given symtra. Ex. 353. On each side of a quad' describe a sq' out- wardly. Of the four as made by joining their neighbor- ing comers, two opposite =the other two and = the quad'. Ex. 354. If from an ^ a we cut two = as, one i , the sq' of one of the = sides of the -I- a equals the product of the sides of the other a on the arms of the ^ a. Ex. 355. If any point within a Ijg'm be joined to the four vertices, one pair of as with || bases = the other. Ex. 356. One median of a trapezoid cuts it into =parts. Ex. 357. Transform a given a into an = | a. Ex. 358. Transform a given -I- a into a regular a. Ex. 359. Construct a polygon ~ to two given ~ poly- gons and = to their sum. Ex, 360. If a vertex of a a moves on a X to the oppo- site side, the difference of the squares of the other sides is constant. Ex. 361. The ^ bi's of a rectangle make a sq', which is half the sq' on the difference of the sides of the rectangle. Ex. 362. The bisectors of the exterior ^ s of a rectangle make a sq' which is half the sq' of the sum of the sides of the rectangle. Ex. 363. The sum of the squares made by the bisectors 138 RATIONAL GEOMETRY. of the interior and exterior ^ s of a rectangle equals the sq' of its diagonal; their difference is double the rect- angle. Ex. 364. If on the hypothenuse we lay off from each end its consecutive side, the sq' of the mid sect is double the product of the others. Ex.365. In aABC, BD-a = BF-c. Ex. 366. In a trapezoid, the sum of the sq's on the diagonals equals the sum of the sq's on the non-|| sides plus twice the product of the || sides. Ex. 367. Prove ri+r2-\-r, = r+/^R. Ex. 368. To bisect a a by a st' through a given point in a side; by a st' || to a side; _L to a side. Ex. 369. Trisect a ■]■ A by ||s. Ex. 370. A quad' equals a a with its diagonals and their ^ as sides and included ^ . Ex. 371. The areas of as inscribed in a © are as the products of their sides. Ex. 372. Construct an equilateral a, given the altitude. Ex. 373. a from ^s and area. Ex. 374. Triple the squares of the sides of a A is quad- ruple the sq's of the medians. Ex. 375. Any quad' is divided by its diagonals into four AS whose areas form a proportion. Ex. 376. AH-HD =BH-HE. Ex. 377. The area of a -I- r't a is -J-c^. Ex. 378. Construct -I- a =given a with same b and hb. Ex. 379. Bisect any quad' by a st' from any vertex ; from any point in a side. Ex. 380. Any st' through the bisection-point of a diag- onal bisects the ||gm. Ex. 381. ^ia' + b' + c') =4(ma'' + inb' + mc'). Ex. 382. Upon any st' the sum of the ±s from the vertices of a a is thrice the i from its centroid. Ex.383. In r't A, 50^ = 4{ma' + nic') . Ex. 384. Trisect a quad'. Ex. 385. Find A =aABC, but with sides m, n; with side m and adjoining ^ S; and opposite ^ d. LENGTH AND CONTENT OF THE CIRCLE. I3Q Ex.386. Find ■!• A = aABC", but with base m; with side in. Ex. 387. Find a =given polygon. Ex. 388. From any point in an equilateral a the three ±s on the sides together ==the altitude. Ex. 389. Sects from the bisection-point of a non-|| side of a trape^.oid to oppsite vertices bisect it. Ex. 390. If the products of the segments of two inter- secting sects are=, their ends are concyclic. Ex. 391. Area of r't A = product of the segments of the hypothenuse made by X from I. Ex. 392. In r't A, areas of as made by he are propor- tional to areas of their in-©s. Ex.393. I /ha+i/hb+i/hc = i/r. Ex.394. hahbhc=ia+b + c)^r'/abc. Ex. 395. If ha', hb', he' be the perpendiculars from any point within a a, upon the sides, prove ha /ha+hb'/hb + hc'/he-l. Ex.396, r = \AI -BI -Clia+h + c) /abc. Ex. 397. abc =a{Aiy + h{Biy + c{CI)\ Ex. 3 98 . {Aiy-\- (BI) ' + (CI) '= ab + ac + bc- babe I {a + h^c). Ex. 399. i?-fr = ±s from O on sides. Ex. 400. In -I- A, if 6 =hb, then |fo =R. Ex. 401. R=2R of hDEF Ex. 402. Area of a J,, /;, I^^ahc/2r. Ex. 403. If ga, qb, qe be the sides of the 3 sq's inscribed in a A, then i/gu = i/Aa+ i/a; i/gt = i//f6+i/6; i/gc = I /lie + I /c. Ex. 404. i/r= i/ha+'i./hb+i/hr\ i/r, = — i//t P> ^ are coplanar. Fig. 133. 350. Corollary to 349 and 342. Two perpen- diculars to a plane are parallel. 351. Inverse of 350. If the first of two parallels is perpendicular to a plane, the second is also per- pendicular to that plane. For the perpendicular erected to the plane from the foot of the second is (by 350) parallel to the first and so (by IV) identical with the second. 352. Theorem. // one plane he perpendicular to one of two intersecting straights, and a second plane perpendicular to the second, they meet and their meet is perpendicular to the plane of the two straights. Hypothesis. Let a be _[jCA at A and ^ be ±CB at B. Proof. The meet AD of a with plane ACB is (by 334) \_AC, and likewise BD±BC; .'. (by 77) GEOMETRY OF PLANES. 151 AD meets BD. Thus the planes a and /?, having D in common, meet in DP, and (by 351) their meet PD is _L to a straight through D \\ to AC, and also _L to a straight through D || to BC. Fig. 134. 353. Theorem. Two straights, each parallel to till- same straight, are parallel to one aiiotlicr, even thoiigJi tlic tJircc be }iot coplanar. For a plane J_ to the third will (by 351) be J. to each of the others; /. (by 350) they are ||. Ex. 418. Are st's || to the same plane ||? Are planes || to the same st' ||? Ex. 419. A plane || to the meet of two planes meets them in ||s. 354. Definition. The projection of a point upon a plane is the foot of the perpendicular from the point to the plane. The projection of a straigJit upon a plane is the assemblage of the projections of all points of the straight. 355. Theorem. The projection of a straiglit on a plane is tlie straight throiigit the projections of any tico of its points. IS2 RATIONAL GEOMETRY. Given A', P', B' , the projections of ^, P, B, points of the straight AB, on the plane a. To prove P' in the straight A'B'. Proof. A, A', B, B' are (by 349) coplanar. P is (by I 5) in this same plane, .". (by 350 and 327) so is PP' ; .■. (by 9) A\ P', B' are costraight. Fig. 135. 356. Corollary to 355. A straight and its pro- jection on a given plane are coplanar. If a straight intersects a plane, its projection passes through the point of intersection. A straight parallel to a plane is parallel to its projection on that plane. 357. Theorem. A straight makes with its own Fig. 136. projection upon a plane a less angle than with any other straight in the plane. GEOMETRY OF PLANES. 153 Hypothesis. Let A' and BA' be the projection of A and BA on a, and BC any other straight in a, through B. Conclusion. t^ABA' < 2^ ABC. Proof. Take BC = BA'. Then AA' its inclination. Ex. 422. Equal obliques from a point to a plane are equally inclined to it. 359. Definition. Parallel planes are such as nowhere meet. 360. Theorem. Planes perpendicular to the same straiglit are parallel. Proof. They cannot (by 345 and 347) have a point in common. Ex. 423. A St' and a plane ± to the same st' are ||. 361. Theorem. Every plan-e through only one of two parallels is parallel to the other. Given AB || CD in a, and /? another plane through AB. To prove CD \\ ^. 154 RATIONAL GEOMETRY. Since ^S is in a and in ^, it contains (by 9) every point common to the two planes. Fig. 137. But CD is wholly in a. So to meet /? it must have a point in common with a and /?, that is, it must meet AB. But by hypothesis AB^CD. Ex. 424. Through a given point to draw a st' |1 to two given planes. Ex. 425. If a II a, and b the meet of a with p, ^ on a, then a \\ b. Ex. 426. Through A determine a to cut b and c. 362. Problem. Through either of two straights not coplanar to pass a plane parallel to the other. Pig. 138. If AB and CD are the given straights, take CF\\AB. Then (by 361) DCF\\AB. GEOMETRY OF PL/tNES. iSS Determination. There is only one such plane. For, through DC any plane || AB meets plane ABC in the parallel to AB through C, .". is identical with CDF. Ex. 427. Through a point without a plane pass any number of st's || to that plane. Ex. 428. Through a point without a st' pass any number of planes || to that st'. Ex. 429. Planes on a 1| a meet a in ||s. Ex. 430. If ffl II a and a || p, then a || to the meet a/3. 363. Problem. Through any given point P to pass a plane parallel to any two given straights, Fig. 139. a, b. [The plane determined by the parallel to a through P, and the parallel to b through P.] [There is only one such plane.] Ex. 431. Through two non-coplanar straights one and only one pair of jj planes can be passed. 364. Theorem. The intersections of two parallel planes with a third plane are parallel. Proof. They cannot meet, being in two parallel planes; yet they are coplanar, being in the third plane. 156 RATIONAL GEOMETRY. 365. Corollary to 364. Parallel sects included between parallel planes are equal. Ex. 432. If two II planes meet two || planes, the four meets are ||. Ex. 433. If o II to the meet ap, then a || a and a || /?. Ex. 434. If a II p, aA-a is ±/9. Ex. 435. Through A draw a || p. [Solution unique.] Ex. 436. If, in a, a cross a', in /9, b cross b', and a || b, a' II b', then a \\ /?. Ex. 437. Through A all st's || a are coplanar. Ex. 438. Two planes || to a third are ||. Ex. 439. The intercepts on ||s between a and o || a are =. Ex. 440. If AB II oc and BC \\ a, then plane ABC \\ a. Ex. 441. If three sects .are = and ||, the as of their adjoining ends are = and ||. Ex. 442. If ^ in a II a, AB || a is in a. 366. Theorem. If two angles have their sides respectively parallel and on the same side of the straight through their vertices, they are equal. Fig. 140. Hypothesis. AB\\A'B' with A and A' on the same side of BB' ; also CB \\ C'B' with C and C on same side of BB'. GEOMETRY OF PL/iNES. 157 Proof. From A take (by 66) AA'\\BB'; : . (by 95) AA' = BB' and A'B'=AB. In same way CC \\ and = BB' and B'C = BC. But then AA' = CC' and (by 353) AA'WCC; :. (by 100) AC=A'C'; .-. (by 58) A ABC = A A' B'C. 367. Corollary to 366. Parallels intersecting the same plane are equally inclined to it. 368. Definition. Let two planes, a, /?, intersect in the straight a. Let A and A' be points on a. Erect now at A and A' perpendiculars to a in one hemiplane a' of «, and also in hemiplane /?' of ^. Then (by 366) the angle of the perpendiculars at A is equal to the angle of the perpendiculars at A'. Fig. 141. We call this angle ilic iiicliiutioii of the hvo hcmi- plciiics a' and ,3'. When the inclination is a right angle the planes are said to be perpendicular to each other. 369. Theorem. If a straight is perpendicular to a given plane, any plane containi)ig this straight is perpoidijcnlar to the given pla>ic. Proof. At the foot of the given perpendicular erect in the given plane a perpendicular to the 158 RATIONAL GEOMETRY. meet of the planes. From the definition of a perpendicular to a plane (334) the given perpen- dicular makes with this a r't ^ ; but this angle is (by 368) the inclination of the planes. Ex. 443. A plane ± the meet of two planes is JL to each; and inversely. Ex. 444. Through a in a draw /3± a. 369 (6). Corollary to 369. A straight and its pro- jection on a determine a plane perpendicular to a. 370. Theorem. // two planes are perpendicular to each other, any straight in one, perpendicular to their ineet, is perpendicular to the other. 371. Corollary to 370. If two planes are per- pendicular to each other, a straight from any point in their meet, perpendicular to either, lies in the other. For the perpendicular to their meet in one is perpendicular to the other, and (by 340) there is only one perpendicular to a plane at a point. Ex. 445. If A in ffXA from A, a±/? is in a. Ex. 446. If a st' be || to a plane, a plane J. to the st' is X. to the plane. 372. Corollary to 371. If each of two inter- secting planes is perpendicular to a given plane, their meet is perpendicular to that plane. Proof. The perpendicular to this third plane from the foot of the meet of the others is (by 371) in both of them. Ex. 447. Through a st' || a to pass p || a. Ex. 448. Through a draw a± /?. Ex. 449. Through A draw aX/? and x- GEOMETRY OF PLANES. 159 373. Theorem. If two straights be cut by three parallel planes the corresponding sects are proportional. Fig. 142. Let AB, CD be cut by the parallel planes a, /?, r in A, E, B and C, F, D. To prove AE:EB=CF: FD. Proof. If AD cut ^ in G, then (by 364) EG \\ BD and AC II GF. .-. (by 235) AE:EB=AG:GD and AG:GD= CF:FD. .■.AE:EB=CF:FD. Ex. 450. Investigate the inverse of 373. 374. Theorem. Two straights not coplanar have one, and only one, common perpendicular. Given a and b not coplanar. To prove there is one, and only one, straight perpendicular to both. Proof. Through any point A oi a take c \\ b. Then (by 361) the plane ac or a \\ b. The projection i6o RATIONAL GEOMETRY. b' of 6 on a is (by 356) || b and cuts a, say in B' ; else were a || h' \\ b. Then, in plane b'b, B'B drawn Lb' is Fig. 143. (by 370) _La and (by 74) also Ub. It is the only common perpendicular to a and b. For any common perpendicular meeting a at B" is Ub" through B" II h, which is in a, and .'. ±a, hence B" is a point of b' the projection of 6 on a; .". identical with B', the cross of b' with a. 375. Corollary to 374. Their common perpen- dicular is the smallest sect between two straights not coplanar. For (by 142) BB' =B,B/ =-(S + 35). 4 Fig. 150. Proof. Any prismatoid may be divided into tetrahedra, all of the same altitude as the pris- matoid; some, as CFGO, having their apex ia the top of the prismatoid and their base T\ithin its base; some, as OABC, ha\4ng three stimmits within the top and the fotuth in the base of the Fig. 151. 1 78 RATIONAL GEOMETRY. prismatoid, thus having for base a point and for top a triangle; and the others, as ACOG, having for base and top a pair of opposite edges, a sect in the plane of the base and a sect in the plane of the top of the prismatoid, as OG and AC. Therefore if the formula holds good for tetra- hedra in these three positions, it holds for the prismatoid, their sum. In (i) call 5i the section at two-thirds the alti- tude from the base 5, ; then 5j is Ja from the apex. Therefore (by 397) the areas :. A =^(5, + 35.) =^(5,4-^5,) ^\aB„ which (by 382) equals Y^, the volume of this tetra- hedron. In (2) the base B^ is a point, and S^ is fa from this point, which is the apex of an inverted pyramid. •■• (by 397) the areas52:T2 = (fa)2:o='; :.S., = \T2\ 4 4 In (3) let KLMN be the section S^. Now the areas AANK:AAGO=AN^^AG^ = {lay:a^ = i:<); AGNM:AGAC = GN':GA' = i^ay:a^ = 4:g. POLYHEDRONS AND yOLUMES. 179 But the whole tetrahedron D^ and the pyramid CANK may be considered as having their bases in the same plane, AGO, and the same altitude, a perpendicular from C: .■.CANK:D^= A ANK: A AGO = i:g; .-.CANK^^D,. In the same way OGXM : D, = A GNM : AG AC = 4:9; :.OGNM=iD,; .-. CANK + OGNM ^^D, ; .• CKLMX + OKLMiX = ^D,. But by (398) CKLMX + OKL]\IN = ^-iaSs + 4 4 since here the area 53 = 0. 400. Corollary to 399. Since, in a frustum of a pyramid, B and S are similar; .•. if fc and 5 be corre- sponding edges, B:S = b^:s'; .-. the voltime F^-B(i + ^j. 401. Definition. A prism is a prismoid whose base and top are congruent. A right prism is one whose lateral edges are perpendicular to its base. i8o RATIONAL GEOMETRY. A parallelepiped is a prism whose bases are prallelograms. A cuboid is a parallelopiped whose six faces are rectangles. A cube is a cuboid whose six faces are squares. 402. Corollary to 400. To find the volume of any prism. Rule. Multiply its altitude by the area of its base. Formula, V{P)=a-B. .•. To find the volume of a cuboid. Rule. Multiply together any three copunctal edges, that is, its length, breadth, and thickness. 403. A cube whose edge is the unit sect has for volume this unit sect, since 1X1X1 = 1. Any polyhedron has for volume as many such unit sects as the polyhedron contains such cubes on the unit sect. The number expressing the volume of a poly- hedron will thus be the same in terms of our unit sect, or in terms of a cube on this sect, considered as a new kind of imit, a unit solid. Such units, though traditional, are tmnecessary. Ex. 525. If the altitude of the highest Egyptian pyramid is 138 meters, and a side of its square base 228 meters, find its volume. Ex. 526. The pyramid of Memphis has an altitude of 73 Toises; the base is a square whose side is 116 Toises. If a Toise is 1-95 meters, find the volume of this pyramid. Ex. 527. A pyramid of volume 15 has an altitude of 9 units. Find the area of its base. Ex. 528. Find the volume of a rectangular prismoid of 12 meters altitude whose section at two-thirds the POLYHEDRONS AND VOLUMES. i8i attitude is s meters long and 2 meters broad, and base 7 meters long and 4 meters broad. Ex. 529. In a prismoid 15 meters tall, whose base is 36 square meters, each basal edge is to the top edge as 3 to 2. Find the volume. Ex. S30. Every regular octahedron is a prismatoid whose bases and lateral faces are all congruent equilateral tri- angles. Ex. S3 1. The bases of a prismatoid are congruent squares of side h, whose sides are not parallel. What are the lateral faces? Ex. 532. If from a regular icosahedron we take off two five-sided pyramids whose vertices are opposite summits, there remains a prismatoid bounded by two con- gruent regular pentagons and ten equilateral triangles. Ex. 533 . Both faces of a prismatoid of altitude a are squares; the lateral faces isosceles triangles. The sides of the upper base are parallel to the diagonals of the lower base, and half as long as these diagonals; and b is a side of the lower base. Find the volume. Ans. \ab^. Ex. S34- The top of a prismatoid of altitude a =6 is a square of side b =10; the section at one-third the altitude from base is 10. Find volume. Ex. 535. Every prismatoid is equal in volume to three pyramids of the same altitude with it, of which one has for base half the sum of the prismatoid's bases, and each of the others its mid-cross section; D^la i^^^ + 2il/) = MB, +4M +B,). Ex. S36. If a prismatoid have bases with angles re- spectively equal and their sides parallel, in volume it equals a prism plus a pyramid, both of the same altitude with it, whose bases have the same angles as the bases of the prismatoid, but the basal edges of the prism are half the sum, and of the pyramid half the difference, of the corresponding sides of both the prismatoid's bases. Ex. S3 7. If the bases of a prismatoid are trapezoids whose mid-sects are b^ and b^, and whose altitudes are 1 82 RATIONAL GEOMETRY. 1 .1 1 /cii +cf2 61+62 I a, —a, 6, — 6,\ a, and a,, the volume = o(-^ ^—^ ^ +—^ ^■- ^"l . ' ' \2 2322/ Ex. 538. A side of the base of a frustum of a square pyramid is 25 meters, a side of the top is 9 meters, and the height is 240 meters. Required the volume of the frustum. Ex. 539. The sides of the square bases of a frustum are 50 and 40 centimeters. Each lateral edge is 30 centi- meters. How many liters would it contain? Ex. 540. In the frustum of a pyramid whose base is 50 and altitude 6, the basal edge is to the corresponding top edge as 5 to 3. Find volume. Ex. 541. Near Memphis stands a frustum whose height is 142-85 meters, and bases are squares on edges of 185-5 and 3-714 meters. Find its volume. Ex. 542. In the frustum of a regular tetrahedron, given a basal edge, a top edge, and the volume. Find the altitude. Ex. 543. A wedge of 10 centimeters altitude, 4 centi- meters edge, has a square base of 36 centimeters perimeter. Find volume. Ex. 544. The diagonal of a cube is n. Find its volume. Ex. 545. The edge of a cube is n. Approximate to the edge of a cube twice as large. Ex. 546. Find the cube whose volume equals its super- ficial area. Ex. 547. Find the edge of a cube equal to three whose edges are a, b, I. Ex. 548. If a cubical block of marble, of which the edge is i meter, costs one dollar, what costs a cubical block whose edge is equal to the diagonal of the first block? Ex. 549. If the altitude, breadth, and length of a cuboid be a, 6, I, and its volume V, (i) Given a, 6, and superficial area; find V. (2) Given a, b, V; find I. (3) Given V, (ab), (bl); find I and b. (4) Given V, l-r) , (-r ) ; find a and 6. POLYHEDRONS AND t^OLUMES. 183 (5) Given (ab), (al), (bl); find a and b. Ex. 550. If 97 centimeters is the diagonal of a cuboid with square base of 43 centimeters side, find its volume. Ex. 551. The volume of a cuboid whose basal edges are 12 and 4 meters is equal to the superficial area. Find its altitude. Ex. 552. In a cuboid of 360 superficial area, the base is a square of edge 6. Find the volume. Ex. 553. A cuboid of volume 864 has a square base equal in area to the area of two adjacent sides. Find its three dimensions. Ex. 554. In a cuboid of altitude 8 and superficial area 160 the base is square. Find the volume. Ex. 555. The volume of a cuboid is 144, its diagonal 13, the diagonal of its base 5. Find its three dimensions. Ex. 556. In a cuboid of surface 108, the base, a square, equals in area the area of the fotir sides. Find volume. Ex. 557. What is the area of the sheet of metal re- quired to construct a rectangular tank (open at top) 12 meters long, 10 meters broad, and 8 meters deep? Ex. 558. The base of a prism 10 meters tall is an isosceles right triangle of 6 meters hjrpothenuse. Find volume. Ex. 559. In a prism the area of whose base is 210 the three sides are rectangles of area 336, 300, 204. Find volume. Ex. 560. A right prism whose volume is 480 stands upon an isosceles triangle whose base is 10 and side 13. Find altitude. Ex. 561. In a right prism whose volume is 54. the lat- eral area is four times the area of the base, an equilateral triangle. Find basal edge. Ex. 562. The vertical ends of a hollow trough are parallel equilateral triangles with i meter in each side, a pair of sides being horizontal. If the length between the triangular ends be 6 meters, find the volume of water the trough will contain. CHAPTER XIV. TRIDIMENSIONAL SPHERICS. 404. Definition. If C is any given point, then the aggregate of all points A for which the sects CA are congruent to one another is called a sphere. C is called the center of the sphere, and CA the radius. Every point B, such that CA > CB is said to be within the sphere. If CA < CD, then D is without the sphere. 405. Theorem. Any ray from the center of a sphere cuts the sphere in one, and only one, point. 406. Theorem. Any straight through its center cuts the sphere in two, and only two, points. 407. Definition. A sect whose end-points are on the sphere is called a chord. 408. Definition. Any chord through the center is called a diameter. Its end-points are called opposite points of the sphere. 409. Theorem. Every diameter is bisected by the center. 410. Corollary to 106 and 409. A plane through its center meets the sphere in a circle with radius equal to that of the sphere. Such a circle is called a great circle of the sphere. 184 TRIDIMENSIONAL SPHERICS. 185 411. Corollary to 410. All great circles of the sphere are congruent, since each has for its radius the radius of the sphere. 412. Theorem. Any two great circles of a sphere bisect each otlicr. Proof. Since the planes of these circles both pass through the center of the sphere, therefore on their intersection is a diameter of the sphere which is a diameter of each circle. 413. Theorem. If any number of great circles pass through a given point, they will also pass through the opposite point. Proof. The given point and the center of the sphere determine the same diameter for each of the circles. 414. Corollary to 413. Through opposite points an indefmite number of great circles can be passed. 415. Theorem. Through any two non-opposite points on a sphere, one, and only one, great circle can be passed. Proof. For the two given points and the center of the sphere determine its plane. 416. Definition. A straight or plane is called tangent to a sphere when it has one point, and only one, in common \\4th the sphere. Two spheres are called tangent to each other when they have one point, and only one, in com- mon. 417. Theorem. .4 straigJit or plane liaving the foot of the perpendicular to it from the center in com- mon with tlie sphere is tangent. Proof. This perpendicular, a radius, is (by 142) i86 RATIONAL GEOMETRY. less than any other sect from the center to this straight or plane. Therefore every point of the straight or plane is without the sphere except the foot of this radius. 418. Theorem. If a straight has a given point not the foot of the perpendicular to it from the center in common with a sphere, it has a second point on the sphere. Proof. This is the other end-point of the sect from the given point bisected by this perpen- dicular. 419. Theorem. If a plane has a point not the foot of the perpendicular to it from the center in common with a sphere, it cuts the sphere in a circle. Proof. If A be the common point and C the foot of the perpendicular, the circle OC{CA) Fig. 152. is on the sphere. 420. Corollary to 419. The straight through the center of any circle of a sphere perpendicular to its plane passes through the center of the sphere. 421. Definition. The two opposite points in which the perpendicular to its plane, through the center of a circle of the sphere, meets the sphere, are called the poles of that circle, and the diameter between them its axis. 422. Theorem. Any three points on a sphere deter- mine a circle on the sphere (I 3 and 419). 423. Theorem. The radius of any circle of the sphere whose plane does not contain the center TRIDIMENSIONAL SPHERICS. 187 of the sphere is less than the radius of a great circle. Proof. The hj^othenuse (by 142) is > a side. 424. Definition. A circle on the sphere whose plane does not contain the center of the sphere is called a small circle of the sphere. 425. Inverse of 417. Every straight or plane tangent to the sphere is perpendicular to the radius at the point of contact. For if not it would have (by 418) another point on it. 426. Theorem. // tico spheres have two points in common they cut in a circle ivJiose center is in their center-straight and wliose plane is perpendicular to that straight. Hypothesis. Let C and be the centers of the spheres having the points .4 and B in common. Conclusion. They have in common all points, and only those, on a circle mth center on OC and plane J_ to OC. Fig. 153. Proof. Since, by 58, i\ACO=^BCO, .-.perpen- diculars from -4 and B upon OC are equal and meet OC at the same point, D. Thus all, but only, points like .4 and B, in a plane _L to OC, and points of qD{DA), are on both spheres. RylTIONAL GEOMETRY. 427. Corollary to 426. If two spheres are tan- gent, either internally or externally, their centers and point of contact are costraight. 428. Theorem. Four points, not coplanar, deter- mine a sphere. Proof. Let A, B, C, D be the four given points. Then (by 352) the plane a ± to and bisecting AB meets /? ± to and bisecting BC in EH ± ABC, and meets d± to and bisecting BD in FO±ABD. :.EHl_EG, and FOlFG, and (by 77) EH and FO meet, say, at 0; .'. (by 346) is one, and the only center of a sphere containing A, B, C, D. 429. Corollary to 428. The four perpendiculars to the faces of a tetrahedron through their circum- centers, and the six planes bisecting at right angles the edges, are copunctal in its circumcenter. 430. Problem. To inscribe a sphere in a given tetrahedron. Construction. Through any edge and any point from which perpendiculars to its two faces are equal, take a plane. Likewise with the other edges in the same face. The cointersection of these three planes is the incenter required. 431. Theorem. The sects joining its pole to points on any circle of the sphere are equal. Proof (343). Pig. 154. TRIDIMENSIONAL SPHERICS. 189 Since equal chords have the great-circle-arcs join- on its c Hence sphere of all the 43'2. Corollary to 431. congi'uent minor arcs, .' ing a pole to points circle are congruent, if C is any point in a n, then the aggregate points A in a, for which great-circle-arcs CA are con- gruent to one another is a circle. Fig. 155. 433. Theorem. The great-circle-arc joining any point ill a great circle ivith its pole is a quadrant. Proof The angle at the center is right. 434. Theorem. // .4, B are iion-oppositc, tlic point P is a pole of their great circle iclicn the arcs PA, PB are both great- circle -quadrants. For each of the angles POA, POB is right and .-. POl^OAB. 435. Definition. The angle between two great-circle-arcs on a sphere, called a spherical angle, is the angle between tangents to those arcs at their point of meeting. 436. A spherical angle is the inclination of the two hemiplanes containing the arcs. 437. Theorem. Any great circle tli rough a pole of a given great circle is perpendicular to the given great circle. Proof. Their planes (by 369) are at right angles. 438. Inverse of 437. Any great circle perpen- FiG. 156. 190 RATIONAL GEOMETRY. dicular to a given great circle will pass through its poles. 439. Theorem. // a sphere be tangent to the par- allel planes containing opposite edges of a tetrahe- dron, and sections made in the sphere and tetra- hedron by one plane parallel to these are of equal area, so are sections made by any parallel plane. i^/ Cr' •' "^ Pig. 157. Hypothesis. Let KJ be the sect J. to the edges EF and GE. in the || tangent planes. Then KJ = DT, the diameter. Let Ol{IP) =M0, sections made by the plane ±DT at / and ±KJ at R, where KR =DI. Let ABCLSN be any parallel plane through a point A of the sphere. Conculsion. LN = QCiCA). Proof. Since A LEU- A MEW, and A LHV~ A MHZ, .-. MW:LU =EM -.EL =JR :JS (by 373). MZ -.LV = HM :HL=KR:KS. But (by 366) ^ZMW = ^ VLU. .: (by 299) area MO -.area. LN = WM-MZ:UL-LV =JR-RK:JS-SK. TRIDIMENSIONAL SPHERICS. 191 But (by 325) area o/(/P) :area (DC{CA) =PI^:AC^ = TI-ID:TC-CD (hy24rS); .•. area LA/' = area oC{CA). 440. Cavalieri's asstimption. If the two sec- tions made in two solids between two parallel planes by any parallel plane are of equal area, then the solids are of equal volume. 441. Theorem. The volume of a sphere of radius r is f Trr*. Proof. A tetrahedi-on on edge, and a sphere with this tetrahedron's altitude for diameter, have (by 439) all their corresponding sections of equal area, if any one pair are of equal area. Hence (by 440) they are of equal volume. •'• (by 399) vol. sphere = faS. But a = 2r, and (by 245) S = -^r ■ ^r ■ -a. : Vol. sphere = ^J • 2;- • lr-\r ■ 7v = ^nr^. 442. Definition. The area of a sphere is the quotient of its volume by one-third its radius. Area of sphere = 4-?'-. 443. Corollary to 324. The area of a sphere is quadruple the area of its great circle. 444. Definition. A spherical segment is the piece of a sphere between two parallel planes. If one of the parallel planes is tangent to the sphere, the segment is called a segment of one base. 445. Corollary to 439 and 440. The volume of a spherical segment is --0\' + 3r/), where r^ is the 192 RATIONAL GEOMBTRY. radius of the section two-thirds the altitude from the base whose radius is r-^. If the segment is of one base its volume is lanr^^] which in terms of r, the radius of the sphere, is na'^i.r — j, and equals J7rafr2^ +— j. If we eliminate r^hy intro- ducing rj, the radius of the top, the volume of the segment is InalTiir^^ + r^^) -\-a^]. 447- Corollary to 305. ^^ = 2DP ' ^^^^'^' ^^^T- Ex. 563. A circle on a sphere of 10 centimeters radius has its center 8 centimeters from the center of the sphere. Find its radius. Ex. 564. The sects from the centers of circles of equal area on a sphere to the center of the sphere are equal. Ex. 565. Where are the centers of spheres through three given points? Ex. 566. Find the volume of a sphere whose area is 20. Ex. 567. Find the radius of a globe equal to the sum of two globes whose radii are 3 and 6 centimeters. Ex. 568. A section parallel to the base of a hemisphere, radius i, bisects its altitude. Find the volume of each part. Ex. 569. The areas of the parts into which a sphere is cut by a plane are as 5 to 7. To what numbers are the volumes of these parts proportional? Ex. 570. The volume of a spherical segment of one base and height 8 is 1200. Find radius of the sphere. Ex. 571. Find the volume of a segment of 12 centi- meters altitude, the radius of whose single base is 24 centimeters. TRIDIMENSIONAL SPHERICS. I93 Ex, 572. In terms of sphere radius, find the altitude of a spherical segment n times its base. Ex. 573. Find volume of a spherical segment of one base whose area is 15 and base 2 from sphere center. Ex. 574. In a sphere of lo centimeters radius find the radii r^ and r, of the base and top of a segment whose altitude is 6 centimeters and base 2 centimeters from the sphere center. Ex. 575. Out of a sphere of 12 centimeters radius is cut a segment whose volume is one-third that of the sphere and whose bases are congruent. Find the radius of the bases. Ex. 576. Find the radius of a sphere whose area equals the length of a great circle. Ex. S77- Find the volume of a sphere the length of whose great circle is n. Ex. 578. Find the radius of a sphere whose volume equals the length of a great circle. Ex. 579. The volume of a sphere is to that of the cir- cumscribing cube as ;t to 6. Ex. 580. Find altitude of a spherical segment of one base if its area is -4 and the volume of the sphere V. Ex. 581. The radii of the bases of a spherical segment are 5 and 4; its altitude 3. Find volume. CHAPTER XV. CONE AND CYLINDER. 448. Definition. The aggregate of straights de- termined by pairing the points of a circle each with the same point not in their plane is called a circular cone of two nappes. This point is called the apex of the cone. Each straight is an element. The straight determined by the apex and the center is called the axis of the cone. The rays of the cone on the same side of a plane through the apex perpendicular to the axis are one nappe of the cone. The sects from the apex to the circle are often called the cone, and are meant when we speak of the area or the volume of the cone. The altitude is the perpendicular from the apex to the plane of the circle, the 'base.' When each element makes the same angle with the axis, the cone is called a right cone. In a right cone all sects from apex to circle are equal, and each is called the slant height. 449. Theorem. Every section of a circular cone by a plane parallel to the base is a circle. Let the section D'H'B'F' of the circular cone A-DHBF be parallel to the base. To prove D'H'B'F' a circle. 194 CONE AND CYLINDER. I9S Proof. Let C be the center of the base, and C the point of the axis AC in the plane D'H'B'. The plane through AC and any element AB gives radii CB, CD, and parallel to them the sects C'B', CD'. .-.(by 7s)aASC~a^5'C' and aACD-aAC'D'. .-. (by 234) C'B' : CB=AC': Fig. 158. AC = C'D':CD. But CB=CD. .■.C'B'=C'D'. 450. Corollary to 449. The axis of a circular cone passes through the center of every section parallel to the base. 451. Theorem. // a circular cone and a tetra- hedron have equal altitudes and bases of equal area and in the same plane, sections by a plane parallel to the bases are of equal area. Fig. 159. Proof. BC : B'C =AC : AC =AL : AL' = = 1T : YT =VH : VH' =GH : CH\ .-. BC^ : WC'^ = GH- : CH" But (by 325) area qC{CB) : area qC{C'B') ip^ RATIONAL GEOMETRY. =BC^ : WC'^^GH^ : G^" = area aFGH : area aF'G'H' (by 300). But by hypothesis area OC{CB) =area aFGH. .-. Area oC'CC'S') -area aF'G'H'. 452. CoroUary to 451. Volume of circular cone is (by 440) = volume of tetrahedron of equal altitude and base =■|a;rr^ 453. Theorem. The lateral area of a right cir- cular cone is half the product of the slant height by the length of the circle taken as base. Proof. Its lateral surface is intermediate be- tween those of the inscribed and those of the cir- cumscribed pyramids. .'. K =^ch=nrh. 454. Definition. A truncated pyramid or cone is the portion included between the base and a plane meeting all the elements. A frustum of a cone is the portion included be- tween the base and a plane parallel to the base. 455. Theorem. The lateral area of a frustum of a right circular cone is half the product of its slant height by the sum of the lengths of its bases. .■.F = ih{c,+c,) = Tzh{r^+r^). 456. Corollary to 451 and 399. The volume of the frustum of a circular cone, V-F^\a7z{r,' + ^r^), where r^ is the radius of 5. 457. Definition. A circular cylinder is the assem- CONE AND CYLINDER. 197 blage of straights each through a point of a given circle but not in its plane, and all parallel. The portion of this assemblage included between, two planes parallel to the circle is also called a circular cylinder. The sects the planes cut out are called the elements of the cylinder. The two circles in these planes are called the bases of the cylinder. The sect joining their centers is called the axis. A sect perpendicular to the two planes is the altitude of the cylinder. If the elements are perpendicular to the planes, it is a right cylinder; otherwise an oblique cylinder. A section whose plane is perpendicular to the axis is called a rigl'.t section of the cylinder. Any two elements, being equal and parallel, are opposite sides of a parallelogram; hence the bases and all sections parallel to them are equal circles. A truncated cylinder is the portion between a base and a non-parallel section. 458. Theorem. Tlie volume of a circular cylin- der is the product of its base by its altitude. Proof. If a prism and cylinder have equal altitudes and bases of equal area, any sections parallel to the bases are of equal area. .•. (by 402) V-C = a7:r\ 459. The lateral area of a circular cylinder is the product of an element by the length of a right sec- tion: C = 2r.ra. Proof. Its lateral surface is intermediate be- ,^-x-c/ 7^: Fig. i6o. 198 RATIONAL GEOMETRY. tween those of the inscribed and those of the cir- ctunscribed prisms. 460. Corollary to 459. The lateral area of a truncated circular cylinder is the prod- uct of the intercepted axis by the length of a right section. Proof. For substituting an oblique section for the right section through the same point of the axis changes neither the area nor the volume, since the portion between the sections is the same above as below either. 461. Corollary to 460. The volume of a trun- cated circular cylinder is the product of the inter- cepted axis by the area of the right section. 462. Archimedes' Theorem. The volume of a sphere equals two-thirds the volume of the circum- scribed cylinder. Proof. The volume of the circumscribed cylin- Ex. 582. In a right circular cylinder of altitude a, call the lateral area C and the area of the base B. (i) Given a and C; find r. (2) Given B and C; find a. (3) Given C and a = 2r; find C + 2B. (4) Given C +2B and a=^r; find C. (5) Given a and B +C; find r. Ex. 583. The lateral area of a right circular cylinder is equal to the area of a circle whose radius is a mean proportional between the altitude of the cylinder and the diameter of its base. Ex. 584. In area, the bases of a right circular cylinder together are to the lateral surface as radius to altitude. Ex. 585. If the altitude of a right circular cylinder CONE y4ND CYLINDER. 199 is equal to the diameter of its base, the lateral area is four times that of the base. Ex. 586. How much must the altitude of a right cir- cular cylinder be prolonged to inci-ease its lateral area by the area of a base? Ex. 587. The lateral area of a right circular cone is t\\'ice the area of the base; find the vertical angle. Ex. 588. Call the lateral area of a right circular cone K, its altitude a, the basal radius r, the slant height /;. (i) Given a and r; find K. (2) Given a and Ii; find 7v. (3) Given K and /; ; find r. Ex. 589. How much canvas is required to make a conical tent 20 meters in diameter and 12 meters high? Ex. 590. How far from the vertex is the cross-section which halves the lateral area of a right circular cone? Ex. 591. Given the volume and lateral area of a right circular cylinder; find radius. Ex. 592. Given lateral area and altitude of a right circular c3-linder; find volume. Ex. 593. A right cylinder of volume 50 has a circum- ference of 9; find lateral area. Ex. 594. In a right circular c^-linder of volume 8, the lateral ai-ea equals the sum of the bases; find altitude. Ex. 595. If in three cylinders of the same height one radius is the sum of the other two, then one lateral area is the simi of the others, but contains a greater volume. Ex. 596. What is the relation between the volumes of two cylidei-s when the radius of one equals the alti- tude of the other? CHAPTER XVI. PURE SPHERICS. 463. If, instead of the plane and straight, we take the sphere and its great circle, that is, its geodesic or straightest, then much of our plane geometry holds good as spherics, and can be read off as spherics. Deducing spherics from a set of assumptions which give no parallels, no similar fig- ures, we get a two-dimensional non-Euclidean geome- try, yet one whose results are also part of three- dimensional Euclidean. I. Assumptions of association on the sphere. I i'. For every point of the sphere there is always one and only one other point which with the first does not determine a straightest. This second point we will call the opposite of the first. Two points, not each the other's opposite, always deter- mine a straightest. Such points are said to be on or of the straightest, and the straightest is said to be through them. I 2'. Every straightest through a point is also through its opposite. Fig. 161. PURE SPHERICS. Fig. i6i. 1 3'. Any two points of a straightest, not each the other's opposite, determine this siraigJitcst; and on every straightest there are at least two points not opposites. 1 4' There are at least three points not on the same straightest. 464. Theorem. // 0' is the opposite of 0, tJien O is the opposite of 0'. Proof. If is not the opposite of 0', they deter- mine a straightest. There is a point P not on this straightest (by 1 4'), and this point is not the opposite of O, since it is not 0' . .'. 0, P deter- mine (by I i') a straightest which (by I 2') goes through O'. .". 0, 0' do not determine a straightest. 465. Theorem. Two distinct straightests can- not have three points in common. [Proved as ia 6.] II. Assumption of betweenness on the sphere. These assumptions specify how "between" may be used of points in a straightest on a sphere. II i'. No point, is between two opposites. II 2'. Between any two points not opposites there is always a third point. II 3'. If B is between .4 and C, then B is also between C and A, and is neither C nor .4. II 4'. If -4 and B are not op- posites, then.there is always a point C sucJ> that B is between A and C. Fig. 163. 202 RATIONAL GEOMETRY. II 5'. Of any three points, not more than one can be between the other two. II 6'. If 5 is between A and C, and C is between A and D, then B is between A and D. II 7'. Between no two points are there two opposites. 466. Theorem. No point is between its opposite and any third point. Proof. If A' is the opposite of A, then it is not between B and A. For since A and B are not opposites, there is (by II 4') a point C such that A is between B and C. But then we would have (by II 6') A' between B and C. But A and A'cannot (by II 7O both be between the same two points. 467. Definition. Two points A and B, not oppo- sites, upon a straightest a, we call a sect and desig- nate it with AB or BA. The points between A and -B are said to be points of the sect AB or also situated within the sect AB. The remaining points of the straightest a are said to be situated without the sect AB. The points A, B are called end-points of the sect AB. II 8'. (Pasch's assumption.) On the sphere, ^ ^^,^ let A, B, C he three points, /^ \. not all on a straightest, and /^^' ~~"""A ^'^ ^^° opposites, and let a be a I A \ straightest on which are none Kv^^^ / L3-W °^ ^^^ points A, B, C; if then \ ^~"^^~~~y / the straightest a goes through \,^__^ ty'^ ^ point within the sect AB, Pj(, jg^ it must always go either through a point of the sect BC or through a point of the sect AC; but it can- not go through both. PURE SPHERICS. 203 Fig. 165. 468. Theorem. Every straightest a separates the other points of the sphere into two regions, of the following character: every point .4 of the one re- gion determines with every point B of the other region, not its opposite, a sect .46 within which lies a point of the straightest a ; on the contrary, any two points. A and A' of one and the same region al- ways determine a sect AA' which contains no point of a. [Proved as in 22.5 Points in the same one of these two regions are said to be on the same sid^ of a. 469. Theorem. The points of a straightest a other than two opposites, 0, 0', are separated by 0, O' into two classes such that or 0' is between any point of the one and any non-opposite point of the other, but neither nor 0' is between two of the same class. Proof. Take any other straightest h through and .-. through 0'. It (by 468) cuts the sphere into two regions. Now (by II 4') a is not wholly in either of these regions; but all its points other than and 0' are in these regions. Two in the same region have no point of b between them. But and 0' are points of b. Two Fig. 166. 204 R/ITlON/tL GEOMETRY. not opposites in different regions have a point of b between them; .'. either or 0'. 470. Definition. The parts of ^a straightest determined by a point of it (with its opposite 0') are called rays from 0. and 0' are called end-points of the rays. II 9'. If C is a point of ray PP', every other point of the ray is between C and P or C* and P'. 471. Theorem. Two opposites cannot both be on the same ray. Proof. II 2', II 9' and 466. 472. Theorem. Every straightest has a point in common with any other. Proof. If not, consider the straightest deter- mined by any point of the one and a point of the other. This would have on one ra,y a pair of opposites, contrary to 471. 473. Definition. On the sphere, a system of sects, AB, BC, CD, . . . KL is called a sect- train, which joins the points A and L with one another. The points within these sects together with their end-points are all to- gether called the points of the sect-train. In particular, if the point L is identical with the point A, then the sect-train is called a spherical polygon. The sects are called the sides of the 'P'^^-'^^i- spherical polygon;, their end- points its vertices. PURE SPHERICS. 205 Polygons with three vertices are called spherical triangles. Fig. 168. Fig. 169. 474. Theorem. Every spherical triangle sepa- rates the points of the sphere not pertaining to its sect-train into two regions, an inner and an outer. [As in 29.] 475. Convention. On a given straightest OA, the two rays 00', from to its opposite 0', are distinguished as of opposite sense. This distinc- tion may be indicated by a qualitative use of the signs + and — (plus and minus), as in writing positive and negative numbers. Any sect PO' or ray from P through 0', or any sect PB where B is between P and 0', has the sense of that ray 00' on which is P. Then also BP is of sense opposite that of PB. III. Assumptions of congruence on the sphere. Ill i'. If A, B are two points, not opposite, on a straightest a, and A' a point on the same or another straightest a', then we can find on a given ray of the straightest a' from A' always one and 2o6 RATIONAL GEOMETRY. only one point B', such that the sect AB is con- gruent to the sect A'B'. Always AB = AB = BA. Ill 2'. If AB = A'B' and AB^A"B", then is also A'B' = A"B". Ill 3'. On the straightest a let AS and BC be two sects without common points, and fiuther- more A'B' and B'C two sects on the same or another straightest, likewise without common points; if then AB = A'B' and BC = B'C', then is also AC = A'C'. 476. Definition. On the sphere, let h, k be any two distinct rays from a point 0, which pertain to different straightests. These two rays h, k from we call a spherical angle, and desig- nate it by ^ {h, k) or :^ {k, h). The rays /j, k, together with the point separate the other points of the sphere into two re- gions, the interior of the angle Fig. 170. and the exterior. |A.s in 35.] The rays h, k are called sides of the angle, and the point the vertex. Ill 4'. On the sphere, given a spherical angle -^ Qi, k), and a straightest a', also a de- termined side of a'. Designate by h' a ray of the straightest a' starting from the point 0' : then is there one and only one ray k' such that the Fig. 171. PURE SPHERICS. 207 ^(ft, k) is congruent to the angle '^ih', k'), and likewise all inner points of the angle ■4. (h', k') lie on the given side of a'. Always ? {h, k)^^ {h, k)^^ {k, h). ^ III 5'. If ? {h, k)^4 ih', k') and ? Qi, k)^ ? ill", k"), then is also '? {h', k') = ? ih", k"). 477- Convention. On the sphere let ABC be an assigned spherical triangle; we designate the two rays going out from A through B and C by Ji and k respect- ively. Then the angle ^ {h, k) is called the angle of the triangle ABC included by the sides AB and AC, or opposite the side BC. ' '^*" It contains in its interior all the inner points of the spherical triangle ABC and is designated by ^BAC or ^A. Ill 6'. On the sphere, if for two triangles ABC and A'B'C we have the congruences AB^A'B', AC^A'C, '^BAC = ^ B'A'C, then also always are fulfilled the congruences 'i ABC = ^ A'B'C and ^ ACB =^ A' C B' . 478. Convention. When the sect AB is set off on a ray starting from A, if the point B falls within the sect AC, then the sect AB\^ said to be less than the sect AC. Fig. 173- 2o8 RATIONAL GEOMETRY. Fig. 174. In symbols, ABAB. AB > CD when E between A and B gives AE = CD or BE = CD, using = for s . Fig. 175. 479. Convention. When ^AOB is set off from vertex 0' against one of the rays of ^A'O'C toward the other ray, if its second side falls within 4^A'0'C, then the '^AOB is said to be less than the ^A'O'C. In symbols, '^AOBK^A'O'C. Then also :!fA'0'C is said to be greater than '^AOB. In symbols, '^A'0'C>^AOB. 480. Definition. Two spher- ical angles, which have the vertex and one side in common and whose not-common sides make a straightest are called adjacent angles. Fig. 176. PURE SPHERICS. 20g 481. Definition. Two spher- ical angles with a common vertex and whose sides make two straightests are called verti- cal angles. Fig. 178. Pig. 177. 482. Definition. A spherical angle which is congruent to one of its adjacent angles is called a right angle. Two straightests which make a right angle are said to be perpendicular to one another. 483. Convention. If ^, 5 are points which deter- mine a straightest, then we may designate one of the regions or hemispheres it makes as right from the straightest AB taken in the sense of the sect AB (and the same hemisphere as left from BA taken in the sense from B to A). If now C is any point in the right hemisphere from AB, then we designate that hemisphere of AC in which B lies as the left hemisphere of AC. So we can finally fix for each straightest which hemisphere is right from this straightest taken in a given sense. Of the sides of any angle, that is designated as the right which Kes on the right hemisphere of 2IO RATION/I L GEOMETRY. that straightest which is determined (also, in sense) by the other side^ while the left side is that lying on the left of the straightest which is determined (also in sense) by the other side. Two spherical triangles with all their sides and angles respectively congruent are called congruent if the right side of one angle is congruent to the right side of the congruent angle, and its left side to that angle's left; but if the right side of one angle be congruent to the left side of the con- gruent angle, and its left side to that angle's right, the triangles are called symmetric. 484. Theorem. Two spheri- cal triangles are either con- gruent or symmetric if they have two sides and the in- cluded angle congruent. [Proved as in 43.] Fig. 179. 485. Theorem. Two spheri- cal triangles are either con- gruent or symmetric if a side and the two adjoining angles are respectively congruent. [Proved as in 44.] Fig. 1 80. PURE SPHERICS. 211 486. Theorem. If two spherical angies are con- gruent, so are also their adjacent angles. [Proved as in 45.] 487. Theorem. Vertical spherical angles are congruent. [Proved as in 46.] 488. Theorem. All right an- gles are congruent. [Proved as in 50.] Fig. 181. 489. Theorem. At a point A of a straightest a there is not more than one perpendicular to a. [Proved as in 52.] 490. Definition. When any two spherical angles are congruent to two adjacent spherical angles each is said to be the sitppleiuoit of the other. 491. Definition. If a spher- ical angle can be set off against one of the rays of a right angle so that its second side lies within the right angle, it is called an acute angle. Fig. 182. 492. Definition. A spherical angle neither right nor acute is called an obtuse angle. RATIONAL GEOMETRY. 493. Definition. A spheri- cal triangle with two sides congruent is called isosceles. Fig. 183. 494. Theorem. The angles opposite the congru- ent sides, of an isosceles triangle are congruent. [Proved as in 57.] 495. Theorem. If two angles of a spherical tri- angle be congruent, it is isosceles. [Proved as in 485.] 496. Theorem. Two spher- ical triangles are either con- gruent or symmetric if the three sides of the one are con- gruent, respectively, to the three sides of the other. gProved as in 58.1] Fig. 184. 497. Theorem. Any two straightests perpendicular to a given straightest intersect in a point from which all sects to the given straightest are perpendic- ular to it and congruent. Fig. 185. PURE SPHERICS. 213 Given r't ^A=:^C. To prove PA = PC = PD, and ^D rt. Proof. By 495, P^= PC and (by 485) PA = P'A. .-. (by 484) ^PDA^^P'DA; .-. (by 488) ^PDA= '^PAD; .-.(by 495) PD^PA. Fig. 186. 498. Definition. The two opposite points at which two perpendiculars to a given straightest intersect are called its poles, and it the polar of either pole. A sect from a pole to its polar is called a quadrant. 499. Theorem. All quadrants are congruent. ^ Let AB and A'B' be two quadrants. To prove AB = A'B'. Proof. At A take a r't? BAC, and also at A'. On AC take a sect AC, and on A'C take A'C = AC. At C" and C Fig. 187. take straightests 1.AC and These contain B and . (by 485) AB = A'B'. Theorem. A point ivhich is a quadrant from two points of a straightest not oppo- sites is its pole. Let PA and PC be two quadrants. Proof. At A and C erect Fig. 188. A'C. B'. . 500 214 R/ITIONAL GEOMETRY. perpendiculars intersecting at P'. Then (by 499 and 496) '^PAC = 'iP'AC. 501. Theorem. // three sects front a point to a straightest be equal, they are quadrants. Proof. They are sides of two adjacent isosceles triangles, and hence perpendiculars. 502. Contranominal of 501. If three equal sects from a point be not quadrants, their three other end- points are not on a straightest. 503. Theorem. Through a point A, no' on a straightest a, there is to a always a perpendicular straightest which, if A be not a pole of a, is unique. Proof. Take any sect QR on a. Take on the other side of a from A, '^BQR^'^AQR, and QB = QA. Then (by 484) AB± a at 5. Moreover, if there were a second straightest perpendic- ular to a from A, then A would (by 4 )8) be a pole of a. 504. Definition. A point B of a given ray 00' such that BO = BO' will be called the bisection-point of the ray. A point B between A and C such that AB = BC is called the bisection-point of the sect AC. 505. Problem. To bisect a given ray 00'. Construction. At two points of the given ray not Fig. 189. PURE SPHERICS. 2IS Fig. 190. both end-points erect perpen- diculars [take (by III 4') an- gles = to ^5 in 503], inter- secting at p. Take another ray from 0, not on the same straightest as the given ray, and at two points of it not both end-points erect perpen- diculars intersecting at 0. The straightest PO bisects the given ray 00'. Proof. Since P and Q are poles, .'. ^B = r't 4 = '^D. .-. (by 485) OB = BO'. 506. Theorem. // and Q' are opposiics, then ivith vertex 4- Qi, k) = ^ {h, k) ivith vertex 0' . Proof. Bisect (by 505) ray h at A and ray k at C. Then (by 496) ?AOC= ^AO'C. 507. Definition. From the vertex 0, a ray h with- in ^{h, k) making ^ (/?, b)= ^{b, k) will be called the bisector of ^ {h, k). 308. Problem. To bisect a given spherical angle. •^ Construction. By 505, bi- sect the rays of the angle ^5 at H and F Take A between H and B and from F on FB' take FC^HA. Then AC intersects HF at D, and BD bisects ^HBF. Proof. By 496, ?.4rB^ '^ CAB' ; by 485 . HD^FD; .-. by 496, ^ HBD = 4 FBD. Fig. 191- 2l6 R/tTION/tL GEOMETRY, 509. Problem. To bisect a given sect. Construction. At the end- points erect perpendiculars [by taking (by III 4') angles = ?5 in 503]. ^ Bisect (by 508) the ^ be- tween them. Proof. By 497 and 484. F'^. 192. 510. Corollary. In an isosceles triangle the bi- sector of the angle between equal sides bisects at right angles the third side. 511. Theorem. If two spher- ical triangles have two sides of the one equal respectively to two sides of the other, and the angles opposite one pair of equal sides equal, then the angles opposite the other pair are either equal or supple- FiG. 193. mental. [Proved as in 175.] 512. Definition. In any spherical triangle the sect having as end-points a vertex and the bisec- tion-point of the opposite side is called a median. 513. Theorem. An angle adjacent to an angle of a spherical triangle is greater than, equal to, or less than either of the interior non-adjacent angles, accord- ing as the median from the other interior non-adjacent angle is less than, equal to, or greater than a quadrant. And inversely. Proof. Let ^ACD be an angle adjacent to '^ACB PURE SPHERICS. 217 Fig. 194. of aACB. Bisect AC at F. On straightest BF beyond F take FH = FB. . . (by 484) '^BAF= ^HCF. If now the median BF be a quadrant BFH is a ray and H is on BCD. If the median 5F be less than a quadrant, H' is within ^ACZ?. .■.^H'CA<^DCA. .-.^DCAy^BAC. If 5C be greater than a quadrant, H" is without ^ACD. .■.iH"CF>'^DCF. .-.^DCAk^BAC. 514. Definition. Two sects respectively congru- ent to two made by a point on a ray with its end- points are called supplcincntal. 515. Theorem. The supplements of congruent sects are congruent. Proof. They are sums or differences of quadrants and congruent sects less than quadrants; and (by 499) all quadrants are congruent. 516. Theorem. // a median be a quadrant, it is an angle- bisector, and the sides of the bisected angle are supplcniental. Let median BD in /\ABC be a quadrant. Proof. In ^ABD and 2CB'D (by 484) AB = CB' and ^ABD^^CB'D^ ^CBD (by 506). Fig. 195. 2t8 RATIONAL GEOMETRY. 517. Inverse. If two sides of a triangle are sup- plemental, the median is a quadrant. Proof. ^ ABC = or •!• 2AB'C (by 496). .-. 2 ABD = or -I- 2CB'D (by 484). .-. BD=DB'. 518. Corollary. If two sides of a triangle are sup- plemental, the opposite angles are supplemental. 519. Theorem. Two spherical triangles are either congruent or symmetric if they have two angles of the one respectively equal to two of the other, the sides oppo- site one pair equal, and those opposite the other pair not supplemental. Given '^B=^E; '^C^'^F; AB = DE.; AC not supplemental to FD. Proof. On ray BC take BG = EF. G must be C, else would we have a 'aACG with adjacent ^^GB = :^ACG in- terior non-adjacent and .•. with median a quadrant (by 513) and .-. (by 516) with AC supplemental to AG, that is, to FD. 520. Theorem. Two spheri- cal triangles are either congru- ent or symmetric if they have in each one, and only one, right angle, equal hypothenuses and another side or angle con- gruent. Given ^C^'^H=r\ ?, and c = h. li a=f, then if AC>g, takeCD=g. .-. BD=h = c, and (by sio)the Fig. 196. Fig. 197. PURE SPHERICS. 219 Fig. bisector of 4 DBA is ± to CDA. .-. i^by 498) B is pole to CDA. .-.'^A is also r't. ^ If ?A = ?F, then if 4 ABC >?(7, take ^ABD = 'i-G- .■.{hy4Ss)'^BDA = ^H=^C = T't^. .-.B is pole to CDA . 521. Theorem. The straight- est ihroiigli tlie poles of two straiglitcsis is tlic polar of their intersect ion- points. Let A and B be poles of a and h, which intersect in P To prove AB the polar of P. Proof. AP and EP are quadrants. 522. Corollary to 521. The straightest through the poles of two straightests is perpendicular to both. 523. Corollary to 521. If three straightests are copunctal, their poles are on a straightest. 524. Definition. If A, B, C are the vertices and a, b, c the opposite sides of a spherical triangle, and A' that pole of a on the same side of a as ^4, B' of b as B, C of c as C, then A'B'C is called the polar triangle of ABC. 525. Definition. Of a spherical triangle A, B, C, the polar triangle is A', B', C where A' is that pole of BC or a on the same side oi a as A, B' of b as B, C of c as C. 526. Theorem. 7/ of two spherical triangles the second is the polar of the first, then the first is the polar cf the second. Let ABC be the polar of A'B'C. RATIONAL GEOMETRY. To prove A'B'C the polar of ABC. Proof. Since B is pole of A'C, .-. BA' is a quad- rant; and since C is pole of A'B', .-. CA' is a quadrant; • '• (by 500) A' is pole of BC. In like manner, B' is pole of ^C", and C of ^5. More- over, since by hypothesis A and A' are on the same side of B'C and A is pole of S'C, .■. sect AA' is less than a quad- are on the same side of BC, of And so for B' and C . Fig. igg. rant. .'.A and ^ which A' is pole. 527. Theorem. In a pair of polar triangles, any angle of either intercepts, on the side of the other which lies opposite it, a sect which is the supplement of that side. Let ABC and A'B'C be two polar triangles. Proof. Call D and E re- spectively the points where ray A'B' and ray A'C meet BC. Since B is pole of A'C, . . BE is a quadrant, and since C is pole of A'B', .■ CD is a quadrant. Fig. 200. But BE + CD=BC + CE + CB + BD = BC+iEC + CB + BD) =BC + DE. 528. Theorem. Two spherical triangles are either congruent or symmetric if they have three angles of the one respectively equal to three angles of the other. PURE SPHERICS. 221 Proof. Since the gi\'en triangles are respectively equiangular, their polars are respectively equilateral. For (by 484) equal angles at the poles of straightests intercept equal sects on those straightests; and these equal sects are the supplements of correspond- ing sides of the polars Hence these polars, having three sides respectively equal, are respectively equiangular. Therefore the original triangles are respectively equilateral, which was to be proved. 529. Corollary to 511. Two spherical triangles are either congruent or symmetric if they have two sides of the one respecti\'ely equal to two of the other, the angles opposite one pair equal, and those opposite the other pair not supplemental. 530. Theorem. // hvo sides of a splicrical triangle arc each less than a quadrant, any sect from the third side to the opposite vertex is less than a quadrant. Let AB and BC be each less than a quadrant. To prove SZ? < quadrant. Proof. Let FG, the polar of B, meet BD at H. If H were between B and D, then GHF would (by II 9') meet CA, and so have a point on each of the three sides of aAB'C, which (by II 9') is impossi- ble. Hence D is between B ^ riG. 20T and H. That is 51) < quad- rant. 531. Corollary to 530 and 513. If two sides of a spherical triangle are each less than a quadrant, the 222 RATIONAL GEOMETRY. angle opposite either is less than the supplement of the angle opposite the other. 532. Theorem. // two sides of a spherical triangle he each less than a quadrant, as the third side is greater or less than one of these, so is it with the opposite angles. And inversely. Let in "aABC, BC and another side, AB, be each less than a quadrant, and ACyAB. To prove ^ABO'^ACB. Proof. Within .4C take D making AD=AB. Then (by i;^o) DB is less than a quad- FiG. 202. ^^ ^ „ . - .^„ rant. .-. (by 531) ^ADB> ? C. But 2 ABC > '^ABD = ^ADB > ?C". 533. Theorem. // the three sides of a spherical tri- angle are each less than a quadrant, any two are together greater than the third. [Proved as in 174.] 534. Definition. On the sphere, the assemblage of points which with a given point give congruent sects is called a circle. The given point is called a pole of the circle. Any one of the congruent sects is called a spherical radius of the circle. Thus a straightest is a circle with a quadrant for spherical radius. But henceforth, for convenience, by circle we will mean a circle with a radius not a quadrant. A sect whose end-points are on a circle is called a spherical chord, or simply a chord. A chord containing a pole is called a diameter. Since the supplements of congruent sects are (by PURE SPHERICS. 223 515) congruent, therefore every circle has two poles which are opposite points, and its spherical radius to one pole is the supplement of that to the other. Always one spherical radius is less than a quadrant. Call its pole the g-pole, and it the g-radius. 535. Theorem. Any splicri- cal chord is bisected by the perpoidicidar from a pole. Proof. AD =^BD {hy S2o). 536. Corollary. A straight- est perpendicular to a diam- eter at an end-point has only this point in common with the circle. ' ^°^" 537. Definition. A straightest with one and only one point in common with a circle is called a tangent to the circle. 538. Theorem. // an oblique from a point to a straigJitcst he less than a quadrant, tlicn tJicre is one and only one perpendicular sect from iJie point to the straightest 'which meets it at less than a quadrant from the foot of the oblique and this is less than a quadrant. Let BA be obhque to CA and < q, and BC ± CA . Proof. Then CA cannot = g, else would BA =q. Hence CI may be taken '^PCA^2P^C. .-.^PAD acute. .-. (by 532) PD'^PAC >r't '^PAB. .-. ^PCA ad- jacent to obtuse -^PCF is acute. But it is also obtuse, being (by 494) Fig. 206. PURE SPHERICS. 225 = -i^PAC. This is impossible, .-.BA not LAP; .'. (by 540) it has a second point on the circle. 542. Corollary. A tangent has no point within the circle. 543« Theorem. // less than a quadrant, the per- pendicular is the least sect from a point to a straightest. Proof. If any other sect from P to AC were less than the perpendicular PA, then AC would have a point within the circle with q-po\e P and g-radius PA, and . (by 541) a second point on this circle, which (by 536) is impossible. 544. Convention. In general a stun of sects is a number of quadrants plus a sect. 545. Theorem. Any two sides of a spherical tri- angle are together greater than the third. Proof. Since each side is less than two quadrants, we have only to prove AB + BC'>AC when ABBD. AB+BC>DB+BC= DC =AC. Fig. 2og. (3) If C^ >q, take on it CZ? = g and make CBF^q. Then sects >15 and DF cross at X', for F is on the non-C-side of AB, while Z? is on the C-side of AB. .-.DF must have a point on straight AB. But all points of sect DF are inte- rior to ^ C", .•. this intersection point is on sect AB, which is all of straight AB within ^ C. Then (by 543) BFCB + BC' = CA + AC'. .■.CB + BA>CA. 546. Definition. A convex spherical polygon is one no points of which are on different sides of the straightest of any of its sides. 547. Theorem. A convex spherical polygon is less than one containing it. Fig. 2x0. PURE SPHERICS. 227 Fig. 548. Theorem. The sum of the sides of a convex spherical polygon is less than four quadrants. Proof. It is within, hence less than, any one of its angles. 549. Theorem. // one angle of a spherical triangle be greater tJuin a second, the side opposite the first must be greater tlian the side opposite the second; and inversely. Given ^C>^B. Proof. Take ?DCS=?B. Then(by49s)LiC = L'B. But (hy 545) DC + DA> AC. 550. Theorem. In a cyclic quadrilateral, the sum of one pair of opposite angles equals the sum of the other pair. Proof. By isosceles triangles. 551. Theorem. Of sects join- ing tiuo symmetrical points to a third, tliat cutting the axis is the greater. Proof. BA=BD + DA =BD + DA'>BA'. Fig. 2x2. 552. Theorem. // two spherical triangles have two sides of tJu- one equal to tivo sides of the other, but the included a)ig!cs unequal, then that third side is the greater ivliicli is opposite the greater angle; and in- versely. 228 RATIONAL GEOMETRY. Proof. Against one of the equal sides of one tri- angle construct a triangle with elements equal to those in the other. Bisect the angle made by the pair of equal sides. This axis cuts the third side, which is opposite the greater angle. 553. Theorem. // each of the two sides about a right angle is less than a quad- rant, then the hypothenuse is less than a quadrant. Proof. Extend the two sides BA, BC, taking BF = BD = quadrant. Then (by 500) B is pole of DF. .'. (by 498and497) ^Fisr't. .'.(by Fig- 213- 495) L'F is a quadrant, .'.(by 500) DA is a quadrant. .". (by 530) AC" < quadrant. 554. Inverse of s S3. If the hypothenuse and a side are each less than a quadrant, then the other side is less than a quad- rant. Proof. If B is r't (Fig. 222), and AB and AC each q. 555. Theorem. The straightest bisecting two sides of a triangle meets the third side at a quadrant from its bisection-point. Let the straightest through A', B', the bisection- points of two sides BC, CA , meet the third side pro- duced at D and D' . PURE SPHERICS. 229 Proof. Take (by 538) AL, BM, CN J. A'B' and such that each is 2 r't t's and < 6 r't ^ 's. 568. Corollary II to 566. Every ^ of a A is >^e. 569. Corollary III to 566. A spherical polygon is equivalent to half the content of its spherical excess. Ex. 597. If a spherical angle adjacent to one angle of a spherical triangle is equal to a second angle of the triangle, the sides opposite these are together a ray. Ex. 598. In a spherical triangle and the spherical triangle determined by the opposites of its vertices the sides and angles are respectively congruent. Ex. 599. Where are the vertices of spherical triangles on a given base the sum of whose other sides is a ray ? Ex. 600. Does a triangle ever coincide with its polar? Ex. 601. The difference of any two angles of a spherical triangle cannot exceed the supplement of the third. Ex. 602. The bisector of an angle passes through the pole of the bisector of the supplemental adjacent angle. Ex. 603. If two straightests make equal angles with a third, the sects from their poles to its are equal. Ex. 604. If a straightest be through the pole of a second, so is the second through a pole of the first. Ex. 605. If two circles be tangent, the point of contact is on their center-straightest. Ex. 606. The common secant of 2 intersecting Os bisects a common tangent. Ex. 607. The three common secants of 3 ©s which intersect each other are copunctal. Ex. 608. If a quad' can have a o inscribed in it, the sums of the opposite sides are = . Ex. 609. If two equal ©s intersect, each contains the orthocenters of as inscribed in the other on the common chord as base. Ex. 610. Three equal ©s intersect at a point H, their other points of intersection being A, B, C. Show that PURE SPHERICS. 833' H is the orthocenter of a ABC; and that the a formed by the centers of the circles is sto h.ABC. Ex. 611. The feet of Xs from A of i^ABC on the ex- ternal and internal bi's of ^ s B and C are co-st' with the bisection-points of h and c. Does this hold for "a? Ex. 612. (Bordage.) The centroids of the 4 'as de- termined by four concyclic points are concyclic. Ex. 613. The orthocenters of the 4 'as determined by four concyclic points, A,B,C, D, are the vertices of a quad' = to A BCD. The incenters are vertices of an equian- gTilar quad'. Ex. 614. (Brahmegupta.) If the diagonals of a cyclic quad' are ±, the -L from their cross on one side bisects the opposite side. Ex. 615. If the diagonals of a cyclic quad' are I., the feet of the Xs from their cross on the sides and the bisec- tion-points of the sides are concyclic. Ex. 616. If an inscribed equiangular polygon have an odd number of sides, it is equilateral. Ex. 617. If a circumscribed equilateral polygon have an odd number of sides, it is equiangular. Ex. 61S. If one of two equal chords of a O bisects ihe other, then each bisects the other. Ex. 619. The tri-rectangular a is its o'wm polar. Ex. 620. All = AS on the same side of the same base have their two sides bisected by the same straightest. Ex. 621. If the base of a a be given, and the vertex variable, the straightests through the bisection-points of the two sides always pass through two fixed points. Ex. 622. If .1 and A' be opposites, then as ABC, A'BC are called cohtuar. A pole of the straightest bisecting AB and AC is also pole of the circum-O of the colunar i^A'BC. Ex. 623. Given b and a:+ j-— /3 to construct q-pole and radius of circum-O- Ex. 624. If a+ ^ = x, the q-pole of circum-o is bisection- point of c. 234 R/tTlONAL GEOMETRY. Ex. 625. Two AS with one ^ the same and the opposite escribed Os=,have equal perinaeters. Ex. 626 The tangent at A to the circum-© of aABC makes with AB and AC'^s whose difference =/? — r- Ex. 627. The q-pole of the circum- o of a a coincides with that of the in- © of the polar a ; and the spherical radii of the 2 ©s are complementary. Ex. 628. From each 2^ of a a a X is drawn to the straightest through the bisection-points of ihe adjacent sides. Prove these ±s copunctal. Ex. 629. Through each y; of a a a straightest is drawn to make the same ^ with one side as the i- on the base makes with the other side. Prove these copunctal. Ex. 630. Two birectangular "as are = if the oblique ^s are = , or if the sides not quadrants are = . Ex. 631. In A, if c is fixed and a+^^iz, then C is on a fixed straightest. Ex. 632. (Joachimsthal.) If two diagonals of a com- plete spherical quadrilateral are quadrants, so is the third. Ex. 633. (i) A quad' whose diagonals bisect each other (a cenquad) has its opposite sides = ; (2) and in- versely. (3) Also its opposite x; s = ; (4) and inversely. (5) Every straightest through this .bisection-point (spherical center) cuts the quad' into =halves. (6) Its opposite sides make = alternate -^Cs with a diagonal. (7) Inversely, a quad' with a diagonal making with each side a ^ = to its alternate is a cenquad. (8) So is a quad with a pair of opposite sides = and making = alternate ^ s with a diagonal. (9) Also a quad' with a pair of opposite sides =, and a diagonal making = . alternate -^ s with the other sides and opposite "^s not supplemental. (10) From the spherical center ±s on a pair of opposite sides are =. PURE SPHERICS. 235 (11) If two consecutive ^s of a cenquad are =,it has a circuni-0. (12) If two consecutive sides of a c:nquad are=,it has an in- O . ^13) The polar of a cenquad is a concentric cenquad. (14) A pair of opposite sides of a cenquad intersect on the polar of its spherical center. (15) Any two consecutive vertices of a cenquad and the opposites of the other two are concyclic. (16) If ABCD be a cenquad, then .4, B, C, D' and A', B', C, D ai^e on = Os with q-poles opposites. Ex. 634. The sides of a a intersect the corresponding sides of its polar on the polar of thei- orthocenter. Ex. 635. The sect which a ^ intercepts on the polar of its vertex equals a sect between poles of its sides. Ex. 636. If a spherical quad' is inscribed, and another circumscribed touching at the vertices of the first, the crosses of the opposite sides of these quad's are on a straightest . Ex. 637. The crosses of the sides of an inscribed a with the tangents at the opposite vertices are on a straightest. CHAPTER XVII. ANGLOIDS OR POLYHEDRAL ANGLES 570. Theorem. The area of a spherical angle, L, is 2r^u. Proof. For we have the proportion, area of ^: area of J sphere = size of ^ : size of r't ^ = size of ^ at center : size of r't ^ ; that is, L : rV=-M : ^tt. .". L = 2r'u. 571. Corollary to 570 and 566. The area of a spherical triangle is the size of its spherical excess multiplied by its squared radius. If e' is the u of e, A =e'r^. 572. Corollary to 571. To find the area of a spherical polygon, multiply its spherical excess in radians by the squared radius. 573. Definition. Three or more rays, a, b, c, from the same point, V, taken in a certain order and such that no three consecutive are coplanar, determine a figure called a polyhedral angle or an angloid. The common point V is the vertex, the rays a, b, c, . . . are edges, the angles ^ab, ^bc, . . . are faces, 236 ^NGLOIDS OR POLYHEDRAL ANGLES. 237 and the pairs of consecutive faces are the diJicdrals of the angloid. According to the number of the rays, 3, 4, S, . . . the angloid is called trihedral, tetrahedi-al, penta- hedral, . , and in general polyhedral. 574. If a unit sphere be taken with the vertex of the angloid as center, this determines a spherical polygon vhose angles are of the same size as the inclinations of the angloid's dihedrals, ^Yhile the length of each side of the polygon is the size of the coiTesponding face-angle of the angloid. Hence from any property of spherical polygons we may infer an analogous property of angloids. For example, the following properties of trihe- drals have been proved in our treatment of spheri- cal triangles: I. Trihedrals are either congruent or symmetrical which have the followmg parts congruent : (i) Two face-angles and the included dihedral. (2) Two dihedrals and the included face-angle. (3) Three face-angles. (4) Three dihedrals. (3) Tu'o pairs of dihedrals and the face-angles opposite one pair equal, opposite the other pair not supplemental. (6) Two pairs of face-angles and the dihedrals opposite one pair equal, opposite the other pair not supplemental. IL As one of the face-angles of a trihedral is gi-eater than or equal to a second, the dihedral oppo- s-te the first is greater than or equal to that opposite the second, and inversely. 238 RATIONAL GEOMETRY. III. Symmetrical trihedrals are equivalent or eqtiivalent by completion. IV. Any two face-angles of a trihedral are together greater than the third. V. In two trihedrals having two face-angles respec- tively congruent, if the third is greater in the first, so is the opposite dihedral, and inversely. VI. In any trihedral the sum of the three face- angles is less than four right angles. VII. In any trihedral, the sum of the three dihe- drals is greater than two and less than six right angles. In the same way, defining a polyhedral as convex when any polygon formed by a plane cutting every face is convex, we have : VIII. In any convex polyhedral any face-angle is less than the sum of all the other face-angles. Proof. Divide into trihedrals and apply IV re- peatedly. IX. In any convex polyhedral the sum of the face-angles is less than four right angles. X. The three planes which bisect the dihedrals of a trihedral are costraight. XL The three planes through the edges and the bisectors of the opposite face-angles of a tri- hedral are costraight. XII. The three planes through the bisectors of the face angles of a trihedral, and perpendicular to these faces, respectively, are costraight. XIII. The three planes through the edges of a trihedral, and perpendicular to the opposite faces, respectively, are costraight. /INGLOIDS OR POLYHEDRAL ANGLES. 239 XIV. If two face-angles of a trihedral are right, the dihedrals opposite are right. Ex. 638. The face angles of any trihedral are propor- tional to the sides of its X on any sphere. Ex. 639. The area of a a is to that of the sphere as its spherical excess is to 8 r't ^s {c'-.^it). Ex. 640. Find the angles and sides of an equilateral a whose area is \ the sphere. Ex. 641. The angle-sum in a r't a is < 4 r't ^s. Ex. 642. If one of the sects which join the bisection- points of the sides of a a be a quadrant, the other two ai"e quadrants. Ex. 643. Cut a tetrahedral by a plane so that the sec- tion is a llgm. Ex. 644. To cut by a plane a trirectangular trihedral so that the section may equal any given A. Ex. 645. The base AC and the area of a a being given, the vertex B is concyclic ^^-ith -4' and C Ex. 646. Given a trihedral; to each face from the vertex erect a perpendicular ray on the same side as the third edge; the trihedral they form is called the polar of the given one. If one trihedral is the polar of a second, then the second is also the polar of the first. Ex. 647. If two trihedrals are polars, the face angles of the one are supplemental to the inclinations of the corre- sponding dihedrals of the other. Ex. 648. If two angles of a a be r't, its area varies as the third ^ . Ex. 649. If t', one minute, is one sixtieth of a degree, and 1", one second, is one sixtieth of a minute, find the area of a a from the radius ;•, and the angles a =20° 9' 30", ,8=55° 53' 3-'". r=ii4° 20' 14". Ans. o.iSisr'. Ex. 650. All trihedrals having two edges common, and, on the same side of these, their third edges prolongations of elements of a right cone containing the two common edges, are equivalent. 24° RATIONAL GEOMETRY. Ex. 651. Equivalent Xs on the same side of the same base are between copolar = ©s. Ex. 652. Find the spherical excess of a 'a in degrees from its area and the radius. Ex. 653. If any angloid whose size is i, that is, any angloid which determines on the unit sphere a spherical polygon whose area is i, be called a steradian, and all the angloids about a point be together called a steregon, then a steregon contains 47c steradians. APPENDIX I. THE PROOFS OF THE TWO BETWEENNESS THEOREMS 16 AND 17, TAKEN FOR GRANTED IN THE TEXT.* 57S. Theorem 1. If B is bchcccn A and C, and C is bctivccn A and D, then C is hchcccn B aitd D. Proof. Let A, B, C, Dhe on a. Through C take a straight c other than a. On c take a point E other than C. On the straight BE between B and E take F. Thus between B and F is no point of c. Now between A and F there can be no point of c, else c would (by II 4) have a point between A and B, since, by the construction of F, c cannot have a point between B and F. Thus C would be between A and B, contrary to our hypothesis that B is between A and C. Thus since c cannot have a point between A and F, it must (by II 4) have a point between F and D. Thus we have the three non-co-straight points F, B, D, and c with a point between F and D, and, by construction, none between F and B. Therefore it must (by II 4) have a point between B and D. So C is between B and D. * These proofs are due to my pupil, R. L. Moore, to whom I have been exceptionally indebted throughout the making of this book. 241 242 RATIONAL GEOMETRY. 576. Theorem II. If B is between A and C, and C is between A and D, then B is between A and D. Proof. Let A, B, C, Dhe on a. Through B take a straight b other than a. On b take a point E other than B. On the straight CE between C and E take F. Thus between C and F is no point of b. Then since by hypothesis B is between A and C, therefore b must (by II 4) Pig. *io. have a point between A and F. Thus we have the three non-co-straight points A, F, D, and b with a point between A and F. There- fore b must have (by II 4) a point between ^ and D, or between F and D. But it cannot have a point between F and D, for then it must (by II 4) have a point either between F and C, contrary to our con- struction, or else between C and D, contrary to Theorem I, by which C is between B and D. There- fore it has a point between A and D. So B is be- tween A and D. 577. Theorem. Ill Any four points of a straight can always be so lettered, ABCD, that B is between A and C and also between A and D, and furthermore C is between A and D and also between B and D. Proof. We may (by II 3) letter three of our points B, C, D, with C between B and D. Now as regards B and D, and our fourth point A, either A is between B and D, or B is between A and D,ov D is between A and B. If B is between A and D, we have fulfilled the hypothesis of Theorems I and II. APPENDIX I. 243 If D is between A and B, then interchanging the lettering for B and D, that is calHng B, D and D, B, we have the hypothesis of Theorems I and II. There only remains to consider the case where A is be- tween B and D. If now C is between D and A , we have fulfilled the hjrpo thesis of Theorems I and II, by calling D, A, and C, B, and A, C, and B, D. If, however, A were between C and D we would have fulfilled the hypothesis of Theorems I and II by writing for A, B, for D, A, and for B, D. We have left only one sub-case to consider, that where D is between A and C. This sub-case is impossible. Suppose ABCD on a. Through C take a straight c other than a. On c take a point E other than C. On the straight DE between D and E take F. Thus between D and F is no point of c. P'°- *"■ Then since by hypothesis C is between B and D, therefore c must (by II 4) have a point between B and F. Therefore we have the three non-costraight points B, F, A , and c with a point between B and F. Therefore c has (by II 4) either a point between B and A, or a point between F and A . But it cannot have a point between F and A, else it would (by II 4) either have a point between F and D, contrary to our construction, or else between D and A, giving C between D and .4, contrary to our hypothesis D between A and C. 244 RATIONAL GEOMETRY. So C would be between B and A, but this with D between A and C gives (by Theorem I) D between A and B, contrary to our hypothesis A between B and D. Thus there is always such a lettering that B is between A and C, and C between A and D, whence (by Theorem I) C is between B and D, and (by Theorem II) B is between A and D. 578. Theorem, ylr^ A, B,C, D points of a straight, such that C lies between A and D and B between A and C, then lies also B between A and D, but not between C and D. A B c D Fig. 222. Proof. The points ABCD, in accordance with 577, have an order in which two are each between the remaining pair and of this remaining pair neither is between two others. But here by hypothesis C and B axe between others. So we reach the follow- ing arrangements ACBD, DBCA, ABCD, DCBA. Of these arrangements, however, the first two do not satisfy the hypothesis. For in both arrang- ments C lies between A and B, which (by II 3) con- tradicts the hypothesis "-B between A and C" In the third and fourth arrangement appears, by 577, that C lies between B and D, therefore, by II 3, B cannot lie between C and D. 579. Theorem. Between any two points of a straight there are always indefinitely many points. Proof. By II 2, there is between A and B at least one point C; likewise there is between A and /IPPENDIX I. 245 C at least one point C Further, there is within AC at least one point C" , which likewise is within ,4 B but not within C'B ; therefore since C lies within C'B, C" cannot be identical with C. In this way ^ji 91 £! 2 S_ Fig. 223. we get e^•cr new points of AB without ever coming to an end. 580. Theorem. If ABCD is an arrangement of four points con-esponding to 577, then there is besides this aiTangement only still the inverse which fulfills 577. [The proof is essentially already given in pro^dng 57S.] 581. Theorem. If any finite number of points of a straight are g'i^'en, then they can always be ar- ranged in a succession .4, B, C, D, E, . . . , K, such that B lies bet\\-cen .4 on the one hand and C, D, E K on the other, further C between .4, B on one side and D, E K on the other, then D between A, B, C on the one side and E, . . , K on the other, and so on. Besides this distribution there is only one other, the re\-ersed arrangement, which is of the same character. [This theorem is a generalization of 577.] Fig. 224. Proof. Our theorem holds for four points by 577 and 5 So. 246 RATIONAL GEOMETRY. We may show that the theorem remains valid for w + I points if it holds for n points. Let A^A2A3 . . . An be the desired arrangement for n points. If further we take an additional point then there are forthwith three cases possible: (i) A I lies between X and An; (2) An lies between X and A^; (3) X lies between A^ and An- In the third case we prove further, that there is one and only one number m, such that X lies between A m and A m+i- Finally we show that the following arrangements in the three cases have the desired properties: (i) XA,A-A, . . . An-, (2) A,A,A,...AnX; (3) Aj^A^Ag . . . AmXAm+t • • • A„; and that they with their inversions are the only ones which possess those properties. APPENDIX II. THE COMPASSES. 582. Euclid's third postulate is: About any cen- ter with any radius one and only one circle may he taken. This has been understood in ordinary geom- etries as authorizing the use of a physical instrument, the compasses, for drawing a circle with any center and any radius. But this is only made fruitful, beyond the sect- carrier, in problem solving, by a new assumption: Assumption of the Compasses. Assumption Yl. If a straight have a point within a Circle, it lias two points on the circle. Corollary. // a circle have a point within and a point ivithont another circle, it has two points on this other. 583. Problem. From a given point without the circle to \ draw a tangent to the circle. Construction. Join the given point A with the cen- ter C, meeting the circle in B. Erect BD± to CB, and (by VI i) cutting in 247 ■-.- ^' H Fig. 225. 248 RATIONAL GEOMETRY. D the OC(CA). Join DC, meeting eC(CB) in F. Then AF is tangent to OCiCB). Proof. Radius CA, ± to chord HD, bisects arc HD; .'.if we rotate the figure until H comes upon the trace of A, then A is on the trace of D; .'. tan- gent HB on trace of AF. Determination. Always two and only two tan- gents. 584. Problem. To construct a triangle of which the sides shall be equal to three given sects, given that any two whatever of these sects are together greater than the third. Fig. 226. Given the three sects a, b, c, any two whatever together greater than the third. Construction. On a straight OF from take OG = b. Take O0(a), and QG{c). Since a + c>&, these (by VI 2) intersect, say at K. i^^OGK is the triangle required. 585. Problem. To construct a triangle, given two sides and the angle opposite one of them. Given a, c, and C. APPENDIX II. 249 Case i. If ac. [^C acute.] I. li c = p, the perpendicular from B on CA, there is only one triangle. II. If c>p, then A^ and A' are on the same side of C and there are two different b triangles which fulfill the condi- tions, namely, A'BC and A^BC (Fig. 238). This is called the ambiguous case. III. If c<^, no triangle. Fig. 229. APPENDIX III. THE SOLUTION OF PROBLEMS. 586. A problem in geometry is a proposition ask- ing for the graphic construction of a figure which shall satisfy or fulfill certain given conditions or requirements. It has been customary to use the ruler and compasses; that is, to allow our assump- tions I-V and also VI (Appendix II), but no others. Of these, assumptions V-VI have usually been superfluous and unnecessary, the problems treated not requiring the compasses, but only rtiler and set- sect. 587. When we know how to solve a problem, the treatment consists of (i) Construction: Indicating how the ruler and sect-carrier or ruler and compasses are to be used in effecting what is required. (2) Proof: Showing that the construction gives a figure fulfilling all the requirements. (3) Determination: Considering the possibility of the solution, and fixing whether there is only a single solution or suitable result of the indicated procedure, or more than one, and discussing the limitations which sometimes exist, within which alone the solution is possible. 250 APPENDIX III. 251 588. The first step toward finding a desired solu- tion is usually what is called Geometrical Analysis. This consists in supposing drawn a figure like the one desired, also containing the things given, and then analyzing the relations of the given things among themselves and to the things or figure sought, or the elements necessary for attaining such figure. 589. Methods of procedure in problem-solving. I. Siicccsshie SiibsHtiifions. — We may substitute for the required construction another from which it would follow, and for this another, perhaps simpler, until one is reached which we know how to accom- plish. Just so, in attempting to find a demonstration for a new theorem, we may freely deduce from the de- sired proposition by use of invertible theorems, and if thus we reach a known proposition, the inversion of the process will give the demonstration sought. Fig. 230. I Example i. Theorem. If from any point P on the circumcircle of the triangle ABC be drawn PA', 252 RATIONAL GEOMETRY. PY, PZ perpendicular to the sides, the points X, Y, Z will be costraight, the Simson's st' of A for P. Analysis. We will have proven X, Y, Z costraight if on joining XY, XZ we show ^PXY supplement- ary to :if PXZ. But again this is proven if we show :^PXZ supplementary to i-ABP and i^PXY = ^ABP. But ^PXZ is supplement of i^ABP since P, X, Z, B are concyclic. And since P, Y, C, X are con- cyclic, ^PXY is supplement of ^PCY, as is also ^ABP, since P, C, A, B are concyclic. I Example 2. Problem. Construct a A, given an angle, the side opposite and two sects proportional to the other two sides, [a from a, a, b/c] Analysis. By a and a is (by 165) the circumcircle given. Bisect ^BAC by AD and prolong AD to meet the circle again (by 138) in E. Then (by 242) CD/BD = b/c. So (by 241) the point D is known. Moreover, since arc BE = arc CE, the point E is (by 225) known. Therefore, taking BC = a, the points D and E can be constructed, and thus A by the prolongation of ED to the circle. Analogous Problems: Ex. i. Given a, R, h/c. Ex. 2. Given a, R, b/c. Ex. 3. Given a, BD, CD. II. Data. — The explicit giving of certain things may involve the implicit giving of others more im- mediately available. Fig. 231. APPENDIX III. 253 Such an implicitly given thing has been called a datum. For example, if a straight and a point without it be given, the perpendicular from the point to the straight is a datum. If an angle and a point within it be given, then the sect from the point to the vertex, the sect from the point to one side drawn parallel to the other side, and the sect this cuts off from the side are data. With an angle a are given the constructible parts \a, \a, etc., but not ^a, \a, etc.; also the supple- ment and complement. If the sum and difference of two magnitudes are given so are the magnitudes. If in a triangle of the three things, a side, the oppo- site angle, the circumradius, two are given, so is the third. So also with base, altitude, area. II Example i. To construct a triangle from one side, the opposite angle, and the difference of the other two angles, (a from a, a, /?—;-.) Analysis. . Since a is known, so is also ^-|- ;- as its supplement. .'. /? and j are known, and we have a side and the two adjoining angles. III. Translation. — Again new auxiliary parts may advantageously be introduced. Certain procedures are found particularly fertile. In any triangle ABC, transporting AC parallel to itself into BD, and extending BA equal to itself to E, we have that the sides of EDC are double the medians of ABC and parallel to them. 254 RATIONAL GEOMETRY. The sides of ABC are two-thirds the medians of EDC, and A is its centroid. Two of the altitudes of the triangles AED, ARC, ADC are equal to two altitudes of ABC and the content of EDC is triple that of ABC. If we have given such elements of ABC as render possible, through these properties, the determina- tion of one of the triangles EDC, AED, ADC, AEC, then the triangle ABC will always be constructible. Ill Example i. Problem. Construct a triangle, given one median m^, the angle between the others, ■m^ and m^, and two sects proportional to them : [A from nil, ^of m^ and m^, m^/m^. Analysis. In the aEDC we know DC = 2FC = 2Wi ; also ^E and DE/EC. Therefore the problem is reduced to I Example 2. IV. Symmetry. — ^Add to the tentatively con- structed figure the figure symmetrical to it or to a part of it, with respect to a chosen straight as axis. Particularly adapted for axis is an angle-bisector or a perpendicular. Especially does this show differences of sects or angles. APPENDIX in. 255 IV Example i. Problem. To construct a tri- angle, given two sides and the difference of the opposite an- gles: [A from a, b, a — /?]. Analysis. Take CE -I- CA with respect to perpendicular CD. Then ^AEC=^ a. But ^ ARC = ^B+ :^BCE = /?+fAPC^^BPC. Ex. 666. Any 2X chords intersect in a given point of a given- ©. Find the locus of the bisection-point of a chord joining their ends. Ex. 667. The locus of a point, the sum of the squares of whose sects from the vertices of a given equilateral a equals twice the square on one of the sides, is the circum-© . Ex. 668. The locus of the end of a given sect from the point of contact and on the tangent is a concentric ©. Ex. 669. Find the locus of the foot of the X from P on a st' through B. Ex. 670. Find the locus of the end of sect from P cut by st' a into parts as m to n. Ex. 671. Find locus of end of sect from st' a cut by P into parts as m to n. Ex. 672. Sects II and with ends in the sides of ^ a are cut into parts as m to n. Find the locus of the cutting points. Ex. 673. Find the locus of a point P if PAiPB =m:n. Ex. 674. The locus of the cross of two tangents to <3C{r), the st' of whose chord of contact rotates about a fixed point P is a st' p±CP. P is called the pole of p, and p the polar of P, with respect to the given O. Ex. 675. If A (given) is on the polar of X (variable), find the locus of X. APPENDIX HI. 26s Ex. 676. To find the locus of a point from which rays through the ends of a given sect make a given :^ . Ex. 677. Find the locus of the vertex B of As, given b and a.c=m: 11. Ex. 67S. [Circle of ApoUonius.] If a sect is cut into parts as di to n, and the interior and exterior points of division are talcen as ends of a diameter, this O contains the vertices of all as on the given sect, whose other two sides are as 11: to n. Ex. 679. Find the locus of those points in a plane a from which ra}-s to the ends of a given sect not on a are ± . Ex. 680. Find the locus of P if PA =PB =PC. Ex. 6S1. If h' is the projection of b on a\\b, and b' ±a in «, find the locus of the bisection-point of a given sect AB if A on a and B on b. Ex. 682. A variable st' is || to a given plane and meets tx^-o non-coplanar st's. Find the locus of a point which cuts the intercepted sect into parts as m to 11. Ex. 683. Find the locus of the point from which -Ls to three coplanar st's are=. Ex. 684. Find the locus of the point having one or two of the foUo'tt'ing: 1 1) Equal sects to two given points; (II) Equal J_s to two given intersecting st's; (III) Equal ±s o two given planes. Ex. 6S5. Find the locus of the poles of great circles maldng a given angle with a given great circle. Ex. 686. Calling 2 ^s c mplciiiciiial wh n their sum is a r't "4- , what is the locus of the intersection of ravs from .4 and B making '4- with AB the complement of 4- with BAl Ex. 68 7. Calling a chord the chord of contact of the point of interesection of tangents at its extremities, what is the locus of points whose chords of contact in oC(r) equal r? Ex. 688. The locus of vertex of a =s', on given b, is St' II b at altitude hb, where blw = 23^- 266 RATIONAL GEOMETRY. Ex. 689. The locus of vertex of A on given h, and with a^-c^= 425 Sy mtra 41 Talmud 132 Tanchord angle 107 Tangent 106, 223 circles 117 1 plane 185 spheres 185 straight 57, 185 straightest 223 Tetrahedral 237 Tetrahedron 165 circumcenter of 188 edges of 1 65 faces of 165 summits of 165 volume of 169 Theorem 3 of Archimedes 198 of Bordage,. 233 of Brahmagupta 60, 233 of Euler 166 of Heron 98 of Hippocrates of Chios ... . 137 of Joachimsthal 234 of Lexell 229 of Moore 241 of Pappus 137 of Pascal 6g of Ptolemy . i2g of Pythagoras 8 . Top of prismatoid i7» INDEX. 273 PAGE Top of segment 191 Trace 90 Translation 122, 253 Transversal 33 partition 87, 169 plane 169 Trapezoid 42 Triangle 11 altitudes of 46 angle bisectors of 109 angle of 18, 207 area of 87 base of 46 equilateral 96 isosceles 28, 212 medians of 40 polar 219 right 37, 63 spherical 205 Triangles, congruent 19 similar 7^ symmetric 210 Trihedral 237 Trisection-points 44 Truncated cone 196 cylinder 197 Truncated pyramid 196 PAGB Unit 68, 94, 130, 180 circle 133 Vega 132 Veronese 170 Vertex of angle 17 Vertical angles 19, 209 Vertices of polygon 10 Volume 168 of cone 194 of cuboid 180 of cylinder 197 of polyhedron 174 of prism 180 of prismatoid 176 of pyramid 176 of sphere 191 of spherical segment 191 of tetrahedron 169 of truncated cylinder 198 Within 6 the sphere 184 tetrahedron 165 Without 6 the sphere 184 tetrahedron 165 SHORT-TITLE CATALOGUE OF THE PUBLICATIONS OP JOHN WILEY & SONS, New York. London: CHAPMAN & HALL, Limited. ARRANGED UNDER SUBJECTS. 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