IN MEMORIAM FLOR1AN CAJORI *"'• 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. Raticnal Geometry. A Text-book for the Science of Space. Based on Hilbert's Foundations, nmo, viii -f- 285 pages, 247 figures. Cloth, $1.75. RATIONAL GEOMETRY A TEXT-BOOK FOR THE SCIENCE OF SPACE BASED ON HILBERT'S FOUNDATIONS BY GEORGE BRUCE HALSTED A.B. and A.M., Princeton; Ph.D., Johns Hopkins FIRST EDITION' FIRST THOUSAND NEW YORK 'jOHN WILEY & SONS London : CHAPMAN & HALL, Limited 1904 &Ar46'3 Copyright, 1904, BY GEORGE BRUCE HALSTED. CAJORl ROBERT DRUMMOND, PRINTER, NHW YORK. fcUf PREFACE. Writing to Professor Hilbert my desire to base a text-book on his foundations, he answered : M Ueber Ihre Idee aus meinen Grundlagen eine Schul-Geo- metrie zu machen, bin ich sehr erfreut. Ich glaube auch, dass dieselben sich sehr gut dazu eignen wer- den." Geometry at last made rigorous is also thereby made more simple. George Bruce Halsted. Kenyon College, Gambier, Ohio. iii 9181iM CONTENTS. CHAPTER I. FAGE Association I CHAPTER II. Betweenness 5 CHAPTER III. Congruence 15 CHAPTER IV. Parallels 35 f CHAPTER V. ' The Circle 49 } CHAPTER VI. Problems of Construction 62 CHAPTER VII. Sides, Angles, and lArcs A 73 CHAPTER VIII. A Sect Calculus 87 CHAPTER IX. Proportion and the Theorems of Similitude 98 VI CONTENTS. CHAPTER X. PAGB Equivalence 109 CHAPTER XL Geometry of Planes 139 CHAPTER XII. Polyhedrons and Volumes 163 CHAPTER XIII. Tridimensional Spherics 194 CHAPTER XIV. Cone and Cylinder 205 CHAPTER XV. Pure Spherics 212 CHAPTER XVI. Angloids or Polyhedral Angles 248 APPENDIX I. Proofs of Betweenness Theorems 253 APPENDIX II. The Compasses 259 APPENDIX III. The Solution of Problems 262 TABLE OF SYMBOLS. We denote triangle by A ; the vertices by A, B, C; the angles at A, B, C by a, {3, y; the opposite sides by a, b, c; the altitudes from A, B, C by h a , hb, h c ; the bisectors of a, /?, y by t a , tb, U\ the medians to a, b, c by m a , nib, m c ; the feet of ha, hb, he by 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, b, c by l lt I t , I 3 ; their ex-radii by r„ r„ r s ; the circumcenter by O; the circumradius by R; angle by ^; angles by ^s; angle made by the rays B A and BC by ^,4i?C; angle made by the rays a and b both from the point O by ^ (a, b) or 2£a&; bisector by bi'; circle by O; circles by Os; circle with center C and radius r by ©C(r); congruent by = ; equal or equivalent by completion by = ; for example [exempli gratia] by e.v. • greater than by > ; less than by < ; minus by — ; parallel by ||; parallels by ||s; vil viii TABLE OF SYMBOLS. parallelogram by \\g'm; perimeter [sum of sides] by p; perpendicular by j_ ; perpendiculars by Is; plus by + ; quadrilateral by quad'; right by r't; spherical angle by ^ ; spherical triangle by 'a ; similar by ~ ; symmetrical by •!• ; therefore by ,\ RATIONAL GEOMETRY. CHAPTER I. ASSOCIATION. The Geometric Elements. i. 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. We think three different sorts of things. The things of the first kind we call points, and designate them by A, B, C, . . . ; the things of the second system we call straights, and designate them by a, b, c, ... ; the things of the third set we call planes, and designate them by a, /?, y, . . . . 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," " congruent." The exact and complete description of these rela- 2 ^ RATIONAL GEOMETRY. tiqns r is accomplished by means of the assumptions |©f •gqcimetry: 4. The assumptions of geometry separate into five groups. Each of these groups expresses certain con- nected fundamental postulates of our intuition. 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 1. 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 we use also the expression : ' ' the straights " a " and ' ' b "have the point A in common." I 2. Any two distinct points of a straight determine this straight; and on every straight there are at least two points. That is, if AB determine a and AC determine a, and B is not C, then also B and C determine a. I 3. Three points, A, B, C, not costraight, always determine a plane a. ASSOCIATION. 3 We use also the expressions: A, B, C "lie in" a, A, B, C, "are points of" a, etc. I 4. Any three non-costraight points A, B, C of a plane a determine this plane a. I 5. // 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. 16. If two planes a, /? have a point A in common, then they have besides at least another point B in common. I 7. In every plane there are at least three non- costraight points. There are at least four non-co- straight 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 6) they have a second point in common, and therefore (by I 5) each has in it the straight which (by I 1) 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 5) 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 3) a plane, in which (by I 5) they and the given straight lie. Any plane on this point and straight would be on the three points already used, hence (by I 4) 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 3) a plane in which (by I 5) each of the two straights lies. Any plane on these straights would be on the three points already used, hence (by I 4) 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 1. If A, B, C are points of a straight, and B lies between A and C, then B also lies between C and A, and is neither C nor A. 4 § ?■ Fig. II 2. If A and C are two points of a straight, then there is always at least one point B, which lies between Fig. A and C, and at least one point D, such that C lies between A and D. 5 6 RATIONAL 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 or 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 are 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, E, . . . , AT on the other, further C between A, B on the one hand and D, E t . . . , 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.] 21. 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 within 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. Fig. 4. 22. Theorem. Every straight a, which lies in a plane a, separates the other points of this plane a into two regions, of the following character: every RATIONAL GEOMETRY. point A of the one region determines with every point B of the other region a sect AB, within which lies a point of the straight a; on the contrary, any two Fig. 5. points A, A f of one and the same region always deter - mine a sect A A' which contains no point of a. Proof. Let A be 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 : (1) On PP' lies no point of a. (2) On QQ' lies no point of a. (3) On PQ lies always a point of a. BETU/EENNESS. 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 AG' \ therefore (by 21) none within QQ'. (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. 23. Convention. If A, A' } } B are four costraight points such that is between A and B but not between A and ^4'; then we say: the points A, A f Fig. 7. lie w //t£ 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. 24. Definition. The assemblage, aggregate, or to- tality of all points of the straight a situated on one and the same side of O is called a ray starting from 0. Consequently every point of a straight is the ori- gin of two rays. 25. Convention. Using the notation of 22, we say: the points P, P' lie in the plane a on one and the same side of the straight a and the points P, Q lie in the plane a on different sides of the straight a. 26. 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. io RATIONAL GEOMETRY. Proof. Let be their common point and A another point on b, 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 22) can contain no point either of a or of b. So also if both were Fig. 8. on the A side of a and the non-5 side of b ; or both on the non-A side of a and the B side of b ; or both on the non-.4 side of a and the non-B side of b. 27. 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. 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. A sect not a side but whose end-points are ver- tices is called a diagonal of the polygon. BETIVEENNESS. n Polygons with 3, 4, 5, . . . , n vertices are called respectively triangles, quadrilaterals, pentagons, . . . , n-gons. 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 a, 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 A is a point of the inner {interior point) and B a point of the outer < {exterior point), then every sect-train which joins A with B has at least one point in common with the ^3 polygon; on the contrary if A, A' are two points bf of the inner and B, B' two points of the outer, then ^ there are always sect-trains, which join A with A'^ and B with B' and have no point in common with the polygon. There are straights in a which lie wholly outside the polygon; on the contrary no such straights which lie wholly within the polygon. Proof. Any simple polygon by joining its ver- tices gives a number of triangles. For a triangle ABC there is (by 26) a region with points on the 12 RATIONAL GEOMETRY. A side of BC, the B side of CA, and the C side of A B, i.e., an inner region. Moreover, the straight determined by a point on b and a point on c both in non-A lies wholly without the region ABC, since it cannot again meet b or c and so cannot (by II 4) Fig. 9. 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 straight determined by the point within and any point on a side has (by II 4) a point on another side, thus making another tri- angle, 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 tri- angle. 30. Corollary to 29. A straight through a ver- tex and a point within a triangle has a point within the 'opposite' side. 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 A B, within which lies a point of a\ on the contrary any two points A and A' of one and the same region always determine a sect A A', which contains no point of a. Proof. Let A be a point which does not lie 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 lies a point of a. Now is to be shown: (1) On PP' lies no point of a. (2) On QQ' lies no point of a. (3) On PQ lies always a point of a. (1) From hypothesis neither within AP nor AP f lies a point of a. Suppose now a point of a 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 & point of a, likewise within AQ'. The intersection straight of the planes a and AQQ' therefore meets two sides of the triangle A QQ' ; 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 RATIONAL 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' lie on one and the same side of the plane a, and the points A, B lie 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 n-gon determines with th : others (n — 1) straights. So together they determine n(n — 1)/2. Ex. 4. How many diagonals in a polygon of n sides. Ex. 5. What polygon has as many diagonals as sides? - r 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 congruent especially serves us. Ill 1. If A, B are two points on a straight a, and A' a point on the same or another straight a' , then we can -find on the straight a! on a given ray from A' always one and only one point B' such that the sect AB is congruent to the sect A'B'. We write this in symbols AB = A'B'. Every sect is congruent to itself, i.e. , always A B = AB. 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 well to the sect A'B' as also to the sect A"B", then is also A'B' con- is 1 6 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 be 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 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 Fig. 11. ......... every sect-tram which joins A with B, goes either through or has with h or k at least one point in common; on the contrary if A, A' are points of the same region, then there is always a sect-train which joins A with A' and neither goes through 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 CONGRUENCE. 1 7 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 (h, 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 h. The rays h, k are called sides of the angle, and the point is called the vertex of the angle. Ill 4. Given any angle (h, k) in a plane a and a straight a' in a plane a! , also a determined side of a' on a' . Designate by h' a ray of the straight a' start- ing from the point 0' ; then there is in the plane a f one and only one ray k' such that the angle (h, k) is congruent to the angle (h', k') y and likewise all in- terior points of the angle {h\ k') lie on the given side of a'. k In symbols: Every angle is congruent to itself, i.e., always The angle (h, k) is always congruent to the angle (AU), i.e., $<&*)■*(*.*). We say also briefly, that in a given plane every angle can be set off towards a given side against a 1 8 RATIONAL 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 determined. Ill 5. // an angle (h, k) is congruent as well to the angle (h\ k') as also to the angle (h" } k"), then is also the angle (h' t k') congruent to the angle (h", k") ; i.e., if * (h, k)=if (h' t V) and * (h, k)^4 (h", k"), then always 4 (h\ k') = %. (h" t k"). 36. Convention. Let 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 A B 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 4 A. Ill 6. If for two triangles ABC and A'B'O we have the congruences AB^A'B', AC^A'C, 4 BAC m 4 B'A'C, then always are fulfilled the congruences 4 ABC m 4 A'B'C and * ACB s 4 A'C'B'. 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 a£ (h, k) m -4. (h' t k'). CONGRUENCE. 1 9 Since (by III 4) 4 (h, k)=4 (h, k), therefore (by III 5) * (h/ k') = 7f (h, k). We say then: the two angles 4 Qh k) an d ^ (#, &') 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 right 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 congruent to one another, if all the con- gruences AB = A'B\ AC**A'C t BC = B'C, lA=lA' t IB^IB', IC^tC are fulfilled. 43. (First congruence theorem for triangles.) Triangles are congruent if they have two sides and the included angle congruent. In the triangles ABC and A'B'C take AB = A'B\ AC = A'C\ 4A=^-A'. 20 RATIONAL GEOMETRY. To prove aABC = aA'B'C. Proof. By assumption III 6 the congruences ifB= ~^B' and ^C= ~4-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' t such that BC = B'U. Then the two triangles ABC and A'B'T)' will have, since ^-B = ^-B', two sides and the included angle respectively congruent; by assumption III 6, consequently, are in particular the two angles B AC and B'A'D* con- gruent to one another. By assumption III 5, con- sequently, must therefore also the two angles B' A'C ! and B' A'T)' be congruent to one another. This is impossible, since, by assumption 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.) Two triangles are congruent if a side and the two adjoining angles are respectively congruent. Fig. 14. In the triangles ABC and A' B'C take AC = A'C' % ^A^^A' % i-C^i-C. To prove aABC^ aA'B'C . . CONGRUENCE. 21 Proof. Suppose now, on the contrary, AB is not = A'B', and take on ray A'B' the point D' t such that AB^A'U. By III 6, *ACB = ^A'CD', but by hypothesis ^ACB^ ^A'CB'. Therefore (by III 5) 1A'C'B'= I 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. // two angles are congruent, so are also their adjacent angles. Take ^ABC= TfA'B'C. To prove ifCBD^ ^CB'D'. Proof. Choose the points A', C, D' on the sides from B f so that A'B' = AB, CB' = CB, DB = D'B'. Jn the two triangles ABC and A'B'C then the sides AB and CB are congruent respectively to the sides A'B 1 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 i-BAC^iB' A'C . 22 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 = CD' and $ ADC & if A' PC', and hence follows, through consideration of the triangles BCD and B'C'D' (by III 6), the congruence of the angles CBD and C'B'U. 46. Theorem. Vertical angles are congruent. Proof. By III 4, ^ABC^^CBA. Therefore, by 45, their adjacent angles are congruent, ^.CBD = 4ABF. 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 ifAPQ, and on the non-A side of a, *fBPQ= iAPQ. Take PB = PA, Since A and B lie on different sides of a, there must be a point O of sect A B on a. Then AOB is perpen- dicular to a. For (by 43) aBPO= aAPO, so ^BOP^^fAOP. But these are adjacent. Therefore, by definition 41, A OP is a right angle. Moreover this perpendicular is unique. For sup- pose any. straight AO' perpendicular to a at 0', and CONGRUENCE. 23 take on this straight on the non-^4 side of a the sect 0'B' = 0'A. Then from hypothesis i{PO'B' = IPO* A and so (by 43) aPO'B'= aPO'A. There- Fig. 17. fore 4 B'PO' =^APO' and B'P ^AP. Therefore (by III 5 and III 2), ^B'PO'= * BPO and B'P = BP. Hence the points B and B' are not different. There- fore no second perpendicular from A to a can exist. 48. Theorem. Let the angle (h, k) in the plane a be congruent to the angle (/*', k') in the plane «', 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 /' 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, l)=^f (h', V) and * (k, l)=^f (&', /')• Proof. Designate the vertex of if (h, k) by 0, and the vertex of 4 W* k') by 0', and then determine on the sides h, k, h\ k f , the points A, B, A\ B', so that we have the congruences OA^O'A' and 0B = 0'B'. 24 RATIONAL GEOMETRY. Because of the congruence of the triangles OAB and O'A'B' (by 43) AB = A'B', ^OAB^^fO'A'B', iOBA= ^fO'B'A'. The straight AB (by 30) cuts /, say in C\ then we determine on the sect &!B' the point C, such that A'C'sAC, then is O'C the ray sought, V . In fact, 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) %. AOC= * A' O'C' and ^BOC ^^B'O'C. 49. Theorem. Let h, k, I on the one hand and h', k', V on the other each be three rays going out from a point and lying in a plane ; if then we have the congruences ^ (h, I) = ^ (h', /') and ^ (k, I) m 4 (k', /')> then also is always CONGRUENCE. 25 Proof. The rays are supposed such that either no point is interior to ^-{h, I) and ?£(k, /), or to "4-{h\ I') and 4 (&', /'), 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 within 4 (h, k), take against h' toward k' t 4 (/*', k") a 4 (h, k). By Fig. 19. 48, take in angle (h' t k") ray I" such that 4 (h' t Z") = l(h,l) and *(/",£")=*(/,&). But by hy- pothesis i(h\ V)mt(h,f). Therefore, by III 5, t(. (/*', V) = 4 (fc',/"), and so, by III 4, ray I" is iden- tical with ray /'. Then 4 (k", /") m 4 (£", V) = 4 (&, = * (&', l')- So 4 (k", V) s 4 (k' } /'), and, by III 4, ray k" is identical with ray k' '. But 4 (h\ k") 3 4 (h, k) . Therefore 4 (h, k) = 4(h',k'). If, however, / be supposed not within ^(/j, &), then it will lie in 4 i)i" ', k") vertical to 4 (h, k). For it cannot lie in 4-{h,k") adjacent to ^ (/*, &), since then 4 (I, k) would contain 4 (h* 0> 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£(h", k) adjacent to £(/*, k), since then 4 (h, 1) would contain %. (/, fe). Thus the ray m costraight with I is within i£(h, k), and m' costraight with V is within 4 Q 1 ', k'). Fig. 20. Then (by 45) ^ (&, w) = ^ (/*', m') and ^ (&, w) = ^-(k',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, s£ (k, I), follows at once from 48. 51. Theorem. All right angles are congruent. Let angle BAD be congruent to its adjacent angle CAD, and likewise let the angle B'A'D r be con- gruent to its adjacent angle C'A'D'\ then are ■4. BAD, 4 CAD, 4-B'A'D', ^CA'D' all right angles. To prove 4BAD= fB'A'U. Proof. Suppose, contrary to our proposition, the right angle B'A'D' were not congruent to the right angle BAD, and then set off 4-B r A'D t against ray 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' '= 4 BAD" \ therefore, by 45, * C'A'D' = i CAD" ; and since by hypothesis ^B'A'D' = 4 C'A'D', therefore, by III 5, * BAD" = * CAD". Since further 4 BAD is congruent to ^CVIZ}, so there is (by 48) within the angle CAD a ray AD'" such that Df ° ,D* C Fig. si, $BAD"=$CAD'" and also ^fDAD"^ 4 DAD'". But we had ^BAD"= 4 CAD", and therefore we must (by III 5) also have t{.CAD"= if CAD'". This is impossible, since (by III 4) every angle can be set off against a given ray toward a given side in a given plane only in one way. Herewith is the proof for the congruence of right angles completed. 52. Corollary to 51. At a point A of a straight a there is not more than one perpendicular to a. 53. Definition. When any two angles are con- gruent to two adjacent angles, each is said to be the supplement of the other. 54. Definition. If any angle can be set off against one of the rays of a right angle so that its second 28 RATIONAL GEOMETRY. side lies within the right angle, it is called an acute angle. 55. Definition. Any angle neither right nor acute is called an obtuse angle. 56. Definition. A triangle with two sides con- gruent is called an isosceles triangle. 57. Theorem. The angles opposite the congruent sides of an isosceles triangle are congruent. Let ABC be an isosceles triangle, having AB^BC. To prove ^ A == if.C. Proof. Since in the triangles ABC and CBA we have the con- gruences AB = CB, BC = BA, ^ABC=^f CBA , therefore (by III 6) $CAB=ifACB. 58. (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. Fig. 23. In the triangles ABC and A'B'C take ABsA'B', ACmA'C, BCmB'C. CONGRUENCE. 29 To prove a ABC = a A'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"= CA'B'. Take the sect AB" = A' B'. Then (by 43) &AB"C = *A'B'C. Therefore B"C = BC, and A BCB" is isosceles ; there- fore (by 5 7) 7f CBB" = t(. CB"B. So also is a BAB" isosceles and ,\ * ABB" = if AB"B. Therefore (by 49) the angle ABC=^AB"C. But ^AB"C = 4. A'B'C. :. (by 43) aABC^ a A'B'C. 59. If A, B, C be any three points not costraight, then (by the method used in 58) 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. 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 Q such that QA a PA and QB = PB. If A, 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 t QB^PB, and QC^PC. 60. Convention. Any finite number of points is called a figure; if all points of the figure lie in 3 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' t 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- cost raight 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' t such that the figures (A, B, C, . . . , P) and (A', B\ C, . . . , P f ) 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 assumption-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. 3 1 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; 1 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? What 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. 11. 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 are ^, so are the a£s. Dual? 32 RATIONAL GEOMETRY. Ex. 14. In an isosceles A, sects to the sides from the ends of the base making with it m sfs are = . 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 ^fABC against AB at the point A toward that side not containing (7. The straight so obtained through A does not ^L B C D 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 ABD and BAD' (by 43), therefore 4 ABD = ^BAU \ and since the angles ABD' and ABD are adjacent angles, so must then, having regard to 45, also the angles BAD and BAD' be adjacent angles. But because of 6, this is not the case. CONGRUENCE. S3 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 by a transversal, at each of these points four angles are made. Of these eight, four, having each the sect A B on a side [e.g., 3, 4, 1', 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., 1 and 1']. Two non-adjacent angles on opposite sides of the IG ' 25 * transversal, and both interior or both exterior, are called alternate angles [e.g., 3 and 1']. 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 the 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 if 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^aB'CD, if BAD = ifB'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 if '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 a, 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 there is not more than one parallel to a given straight. 71. The introduction of this assumption greatly simplifies the foundation and facilitates the con- struction of geometry. 72. Theorem. Two straights parallel to a third arc parallel. Proof. Were 1 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 ^.BAU^i-ABC [D' and / D on same side of AB], and _d 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 Fig. 26. parallels is perpendicular to the other also. 35 3 6 RATIONAL 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 t 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 tCBF=*C and $ABD m gA ; then (by 69) can neither BF nor BD cut AC. By the parallel postulate IV, then is FBD a straight. Fig. PARALLELS. 37 80. Theorem. If two angles of one triangle are congruent to two of another, then the third angles are congruent. Proof. Given tA=^A f and 4-B^4.B'. Take CP parallel (||) to AB and C'P' || to A'B'. Then ta^i-A and ^.p = ^B y ^.a'^^A' and ^/?' = *B'. .'.(by 49) *BCD^*B'C'D'. /.(by 45) the adjacent angles Z ACB = $ A'C'B'. 81. Theorem. Two triangles are congruent if they have a side, an adjoining and the opposite angle re- spectively congruent. Proof. By 80 and 44. Ex. 30. Every triangle has at least two acute angles. Ex. 31. If the rays of one angle are parallel or per- pendicular to chose 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 two acute angles are complemental (calling two angles complements which together form a right angle). 82. Theorem. In any sect AB there is always one and only one point C such that AC = BC. Proof. Take any angle BAD at A 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 any 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 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=^B, and i-ACD^^BCp [ver- tical], therefore (by 81) aACD=aBCF. Therefore AC = BC. If we suppose a second such point C, then on ray DC take CF' = DC. Therefore (by 43) i CBF f =^DAC=4ABF, and BF' = AD = BF. Therefore F f is F and C is C. 83. Convention. The point C of the sect A B such that i4C = BC may be called the bisection-point of A J5, and to fo*s£c/ A 5 shall mean to take this point C. Ex. ^t,. Parallels through the end-points of a sect intercept congruent sects on any straight through its bisection-point. 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 2^ in the r't ^. PARALLELS. 39 84. Theorem. Within any •£ (h, k) there is always one and only one ray, /, such that •£ (h, /) = ^ (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=]. .\ ^AOC=^fA'OC. If we suppose a second such ray OC, then a ,4 0(7= a A'OC [as with 2 sides FlG and the included a£ =are = ]. i.ACmA'C :. (by 82) CisC. 85. Convention. The ray /of ?{ (h, k) such that ^ (/t, /) = ^ (/, &) may be called the bisection ray or bisector of ^ (/t, &), and to bisect a£ (/*, &) shall mean to take this ray /. Ex. 35. An angle may be separated into 2, 4, 8, 1 6, ... v 2 M 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. 4o RATIONAL GEOMETRY. Z, Fig. S3. 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. Fig. 35. PARALLELS. 41 91. Definition. A symmetrical quadrilateral with a median as axis is called a symira. Ex. 36. In a r't A if to set off one acute £, a, in the other, /?, bisects, so is it with /?'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 + A bisects if at the vertex. Ex. 39. The _L from vertex bisects base and ^ in a •I- A. Ex. 40. The bisector of ^ at vertex of a -I- 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 ^ at vertex of + a is || to base, and inversely. Ex. 43. The end of sect from intersection of congruent sides of a + A costraight and ■ to one determines with end of other a ± to base. Ex. 44. To erect a i. at the end-point of a sect without producing the sect. Ex. 45. A || to one side of an ^ makes with its bisector and other side a -I- A. Ex. 46. The bisectors of the ■ ?f s of a-l- A are =. Ex. 47. Every symmetrical quadrilateral not a del- toid is a symtra. Ex. 48. The intersection point of two symmetrical straights is on the axis. Ex. 49. The bisector of an angle is symmetrical to the bisector of the symmetrical angle. Ex. 50. A figure made up of a straight and a point is symmetrical. Ex. 51. In any deltoid [1] 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- 42 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 [1] Two opposite sides are parallel, and have a common perpendicular bisector. [2] The other two sides are congruent and make con- gruent angles with the parallel sides. [3] Each angle is congruent to one and supplemental to the other of the two not opposite it. [4] The diagonals are congruent and their parts adja- cent to the same parallel are congruent. [5] One median bisects the angle between the two diagonals, and also the angle between the non-parallel sides (produced). Ex. 56. Any quadrilateral which has one of the pre- ceding five pairs of properties (Ex. 55) 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 are congruent, and its diagonals bisect each other. Proof. aABC= a ADC [side and 2 adjoining *ss]. :.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||c, also AB = BC. To prove FG = GH. Proof. Take FL\\GM\\ AB. Then FL^AB^BC = GM [95, opposite sides of a||gm are = ]. .*. aFLG = aGMH [side and 2 adjoining i-s = ]. .'. 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 with A.] 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 || to the third side through either bisection- point.] Fig. 38. 44 RATIONAL GEOMETRY. 99. 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. By 97 GH \\ BC bi- sects AC. Since (by 98) Fig. 39. FG\\CH, :. (by 95) FG^CH. 100. Theorem. If two sides of a quadrilateral ore congruent and parallel it is a parallelogram. Given AB = and \\CD. Proof, a ABC a a ADC. Fig. 40. /. ^fACB= if CAD. :. CB\\AD. 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. 62. A quadrilateral whose diagonals bisect each other is a parallelogram. PARALLELS. 45 10 1. Theorem. In any sect AB there are always two, and only 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\\GD\\HB. Then, by 96, AC^CD^DB. Suppose two other such points C \ D' '. Then, by 98, C'F || D'G. Now HB' \\ GD' (by 96) makes D'B' s-D'C. .'. from our hypothesis and III 1, B' is identical with B. .'. since GD\\HB (by IV) D' is identical with D. .'. since FCJGT? (by IV) C is identical with C. 102. The two points, C, D, of the sect AB such that AG = CD = DB may be called the trisection- points oi AB. 103. Theorem. 77t£ three medians of a triangle are copunctal in that trisection-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, Q, are bisection-points of OC and OA, then (by 98 and Fig. 42. 99) PQ || and =GF. .'. by 100 PQFG is a || gm and (by 95) PF and QG bisect each other. 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 fixed 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 a A 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 || g'm are joined to the vertices the diagonals are tri- sected. Ex. 74. The J_s from any point in the base of an f A to the sides are together an altitude. Ex. 75. The diagonals of a rectangle are =, of a rhom- bus are ±. Ex. 76. If 2 || 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 make a quad' with 2 r't ^s. Ex. 80. The bisection-point of one sect between ||s bisects any through it. Ex. 81. The = altitudes in -I- A make with the base if s ss to those made in bisecting the other ^ . Ex. 82. Through a given point within an ^ draw a sect terminated by the sides and bisected by the point. Ex. 83. Sects from the vertex to the trisection-points of the base of + A are =. Ex. 84. If the ^ s made by producing a side of a A are s, so are the other sides. Ex. 85. If a quad' has 2 pairs of congruent consecutive sides, the other ^s are =. Ex. 86. Two As" are = if two sides and one's median are respectively = . Ex. 87. Two + As are = if one ^ and altitude are ■ to the corresponding. Ex. 88. Two as are = if a side, its altitude and an adjoining ^ are respectively = . Ex. 89. If 2 altitudes are = the A is + . Ex. 90. Two As are = if two sides and one's altitude are s 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 ■ . 48 RATIONAL GEOMETRY. Ex. 93. Two as are m if a side, an adjoining ^ and its bisector are respectively = . Ex. 94. Two equilateral as are ■ if an altitude is =. Ex. 95. The bisector is within the if made by altitude and median. Ex. 96. In a right A one bisector also bisects if 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 ±s from 2 vertices of a A upon the median from the third are =. Ex. 99. Two || g'ms having an if and the including sides = are = . Ex. 100. The ± from the circumcenter to a side is half the sect from the opposite vertex to the orthocenter. Ex. 101. The centroid is the trisection point of the sect from orthocenter to circumcenter remote from the ortho- center CHAPTER V. THE CIRCLE. 106. Definition. If C is any point in a plane a, then the aggregate of all points A in a, for which the sects CA are congruent to one another, is called a circle. [qC(CA).] C is called the center of the circle, and CA the radius. 107. Theorem. Any ray from the center of a circle and in its plane a cuts the circle in one, and only one, point. 108. Theorem. Any straight through its center and in its plane a cuts the circle in two, and only two, points, and these are on opposite sides of its center. Proof. On each of the two rays determined in this straight by the center there is (by III 1) one, and only one, sect congruent to the radius of the circle. 109. Definition. A sect whose end-points are on the circle is called a chord. no. Definition. Any chord through the center is called a diameter. in. Theorem. Every diameter is bisected by the center of the circle. 49 5° RATIONAL GEOMETRY. Fig. 43- ii2. Theorem. No circle can have more than one center. Proof. If it had two, the diameter through them would have two bisection -points, which (by 82) is impossible. 113. Theorem. The straight through the bisection- point of a chord, and the center of the circle, is perpen- dicular to the chord. Proof. A ACO = A BCO, [ a s with 3 sides = are =]; .*. if ACO = if BCO. But they are adjacent; .'. by definition COL to AB. 114. Corollary to 113. The cir- cle is symmetrical with regard to any one of its diameters as axis. 115. Corollary to 1 1 3 . If with the end-points of a sect each of two points gives congruent sects the two determine its perpen- dicular bisector. 116. Corollary to 1 1 5 . If two circles have two points in com- mon their center-straight is the perpendicular bisector of their .common chord. 117. Theorem. The perpendicular bisecting any chord contains the center. The perpendicular from the center to a chord bisects it. Proof. By 113 the three properties pertain to one straight. But any two suffice to determine that straight. 118. Corollary to 117. Every point which taken THE CIRCLE. 5 1 with two points gives congruent sects is on the per- pendicular bisector of their sect. 119. Theorem. Every point on the perpendicular bisector of a sect taken with its end- points gives con- gruent sects. 120. Theorem. A straight cannot have more than two points in common with a circle. Proof. If it had a third, then, "since (by 117) the perpendicular bisecting any chord contains the cen- ter, there would be two perpendiculars from the center to the same straight, which (by 47) is im- possible. 121. Theorem. Chords which mutually bisect are diameters. Proof. The perpendicular bisector of each con- tains the center. 122. Theorem. Circles with three points in com- mon are identical. Proof. The center is on the perpendicular bi- sectors of the chords. 123. Theorem. Any three points not costraight de- termine a circle. Proof. If A , B y C be not costraight, bisect (by 82) AB at D and BC at F by perpendiculars. [Take (by III 4) angles = to ^C in 84.] These perpendiculars (by 77) meet, say at 0. Therefore (by 119) A0 = BO = CO. Therefore A, B, Fig. 45- C are on the circle with center and radius AO. By 122 the three are on no other circle. 52 RATIONAL GEOMETRY. 124. Corollary to 123 and 120. [Points on the same circle are called concyclic.'] Every three points are costraight or concyclic. No three points are costraight and concyclic. 125. Definition. The circle through the vertices of a triangle is called its circumcircle, O0(R), and the center of the circumcircle is called the circum- center of the triangle ; its radius, R, the circum-radius. 126. Corollary to 123. The three perpendicular bisectors of the sides of a triangle are copunctal in its circumcenter. 127. Theorem. The three altitudes of a triangle are copunctal. Given the a ABC. To prove that the straights through A, B, C perpendicular to the straights a, b, c respectively, are copunctal. Fig. 46. Proof By 66, through A, B, C take B'C, A'C, A'B'\BC, AC, AB respectively. /. aAB'C = aABC= aABC [as with a side and 2 adjoining ^s= are =]. .' AB'^AC, and AD is the ±bi' of B'C [_L to 1 st of 2 || s is 1 to 2nd]. Similarly BEA.W of A'C'\ and CFJ_bi' of A'B'. :. AD, BE, CF axe copunctal by 126. THE CIRCLE 53 128. Definition. The point of cointersection of the three altitudes is called the orthocenter of the triangle. 129. Theorem. If any ray, /, be taken -within a given angle, 4 (K &)> the bisectors m, n of the two angles so made form an angle congruent to each of the angles made in bisecting the given angle. Proof. On the other side of h from k take *(*,#)»*(*,*). Then since * (h, n) = $ (n, h) and *(n f fc) s *(*,#), .\ (by 49) *(M)^ *(»,#). Fig. 47. But since * (n, /) ■ * (fc, «), ,\ * (n, /) = ^ (£', A). But also af (/, tn) = £(h, m). .*. (by 49) ^(w, m) = £(£', w), .*. w bisects ^(w, £'). .*. (by 48) £(w, w) = £(&, /*) where 6 bisects £(/*, fe). 130. Theorem. The bisectors of adjacent angles make a right angle. frtt Fig. 48. Proof. Extend one of the bisectors, as /, through the vertex 0. Then *(/', *)«*(/, h') [vertical 54 RATIONAL GEOMETRY. ^s are ■]. .*. (by III 5) *(l',h)=t(l,k). But by hypothesis, £(h, m) = £(k, m). .'.(by 49) £(/, m) = 4-V , m). .'.by definition £(l, m) is right. Ex. 102. If a straight satisfy any two of the following conditions it also satisfies the others: 1. Passing through the center. 2. Perpendicular to the chord. 3. Bisecting the chord. 4. Bisecting the angle at the center. Ex. 103. Every axis of symmetry for a circle contains the center. Ex. 104. Where are the bisection- points of a set of parallel chords? Ex. 105. Where are the bisection-points of a set of equal chords? Ex. 106. If from any point three sects drawn to a circle are congruent, that point is the center. 131. Theorem. If any ray / be taken without a given angle, 4- (K k) , and of the two angles so formed, one, £(k, I), be within the other, if(k, /), then the angle formed by their bisectors, £(b, n) t is congruent to each of the angles as 4- (k> m) made in bisecting the given angle. o Fig. 49. Proof. By 129, since k is within £(/, h), .*. tim, n) =4-(b, I). But by hypothesis £(n,k)=- *(/, »), .*. (by 49) £(k, m) = Z(n, b). THE CIRCLE. 55 132. Definition. An angle whose vertex is on a circle of which its sides contain chords is called an inscribed angle, and said to be upon the chord be- tween its sides. 133. Theorem. Inscribed angles upon the same chord and the same side of it are congruent. Proof. 1 st. If the chord BC be a diameter the straight through the vertex A of any inscribed angle and the cen- ter makes two isosceles tri- angles. .'. bisecting £BOF we b| get ZDOFmtBAO. In same way ZHOF^ZCAO. :. (by 49) *BAC = *DOH 9 which (by 130) is right. 2d. If the vertex A be on the same side of the straight BC as the center 0, then sect OA cannot cut BC, and (by III 1) the center is between A and the other point A' of the circle on the straight AO. If now ray OA be costraight with a side of ifBOC, then aAOC being isos- Fig. 50. Fig. 51. celes, the bisector OF of ^ BOC makes $F0C = ZBAC. Again if ray OA' is within ■4 BOC, then the bisectors OF and OH make lA'OF^ifOAB and 4A'0n=i.0AC. ;. (by 49) %-BAC =^F0H t which Fig. 52. 5 6 RATIONAL GEOMETRY. (by 129) is congruent to each angle made in bisect ing BOC. If, however, ray OA' is without ?{BOC, then the bisector OD of 4- A'OB makes ^AVD=^OAB, and the bi- sector OE of ^-A'OC makes tA'OEs i-OAC. .'. (by 49) ^BAC = 4D0E, which (by 131) is congruent to each angle made Fig. 53. in bisecting BOC. 3d. If the vertex A and the center be on opposite sides of the straight BC. Let A f be the other point of the circle on the straight AO. Then the six angles of the two triangles ABC, A'BC to- gether form four right angles. But by case 1st, the two angles at C form a right angle, likewise the two at B. .'. ^BAC is the supplement of ~4-BA'C. 134. Corollary to 133. The inscribed angle upon a diameter is right. 135. Definition. A polygon whose sides are congruent and Fig. 54? whose angles are congruent is called regular. 136. Definition. A polygon whose vertices are concyclic is called cyclic. 137. Corollary to 133. In a cyclic quadrilateral the opposite angles are supplemental. 138. Theorem. If a straight have one point in common with a circle and be not perpendicular to the radius to that point, it has also a second point in com- mon with the circle. THE CIRCLE. 51 Let the straight a have the point P in com- mon with the OC[CP] and be not _L to CP. From C drop CM _L to a. From M on a set off MP' mMP. .-.CP'mCP. .'.P'ison OC[CP]. Fig. 55. 139. Definition. A straight which has two points in common with a circle is called a secant. 140. Theorem. A straight perpendicular to a diam- eter at an end-point has only this end-point in common with the circle. Proof. Any chord is (by 117) bisected by the perpendicular from the center. 141. Definition. A straight which has only one point in common with a circle is called a tangent to the circle, and the point is called the point of contact. 142. Theorem. If BC be perpendicular to AB f iC and D any point on the straight A B other than B, and on ray CD we take sect CF = CB, then F is within sect CD. Proof. Otherwise since A CBF is isosceles, two an- gles of a triangle would each be right or each obtuse, which (by 79) is impossible. 143. Theorem. If the rays of one angle are within another the angles are not congruent. Fig. 56. 58 RATIONAL GEOMETRY. Fig. 57. 144. Theorem. Proof. For suppose £ (h, m) m % (k, I) and k, I within 4 (h, m). On the other side of m from the points of h there is a ray n such that ~4- (m, n)~ 7f(h t k). ;. (by 49) *(*,») = ^.(h, m). .'. from our hypothesis ^ (k, n) = ^ (k, l) 7 which (by III 4) is impos- sible. If P be a point within the triangle ABC, then angle APC is not congruent to angle ABC. Proof. The ray BP must (by 30} have on it a point D within the sect AC. :.PD is within ifAPC. From P take PF || BC. It makes 4FPD = ^CBD. From P take P^H^E. It makes .-. (by 49) ^-FPG = i-CBA. If now we supposed 4 CBA = ^ CPA we should have ^FPG = ifCPA, which (by 143) is impossible. 145. Theorem. // two triangles have a side in com- mon and the angles opposite it congruent, and with vertices on the same side of it, the four ver- tices are coney die. Proof. If the circle through A, B, C did not contain D, then (by 138) it would have a second Fig. 59. point in common with the D A Fig. 58. ■4GPD=^ABD. THE CIRCLE. 59 straight AD or CD, or else we would have OCljCD and OALAD. But in this last case the whole circle except points A and C would be within ^ADC. For the center O would be within ifADC, being then on the bisector of if ADC since, a ADC being isos- celes, aAOD=aCOD [3 sides = ]. Hence the point B would be within a ADC, which (by 144) is impossi- ble. But it is just as im- possible that AD or CD (besides A or C) should have a point the circle other than D. For then We Fig. 60. on U would impos- have ^.ADC^^AD'C, which (by 79) is sible. 146. Theorem. // two opposite angles of a quad- rilateral are supplemental it is cyclic. Proof. Given the if CD A is the supplement of ^ B. On the circle determined by A , B, C take a point D' on the same side of AC as D. Then ^D'^^D, being each the supplement of ifB. .'. (by 145) D is concylic with A CD\ that is, with ABC. 147. Corollary to 146. A quadrilateral is cyclic if an angle is congruent to the angle adjacent to its opposite. Fig. 61. 60 RATIONAL GEOMETRY. Ex. 107. Defining a tanchord angle as one between a tangent to a circle and a chord from the point of contact, prove it congruent to an inscribed angle on this chord. Ex. 108. An angle made by two chords is how related to the angles at the center on chords joining the end- points of the given chords? • Ex. 109. The vertices of all right-angled triangles on the same hypothenuse are coney clic. Ex. no. Tangents to a circle from the same external point are congruent, and make congruent angles with the straight through that point and the center. Ex. in. Two congruent coinitial chords are symmetric with respect to the coinitial diameter. Ex. 112. If triangles on the same base and on the same side of it have the angles opposite it equal, the bisectors of these angles are copunctal. Ex. 113. The end-points of two congruent chords of a circle are the vertices of a symmetrical trapezoid. Ex. 114. The chord which joins the points of contact of parallel tangents to a circle is a diameter. Ex. 115. A parallelogram inscribed in a circle must have diameters for diagonals. Ex. 116. Of the vertices of a triangle and its ortho- center, each is the orthocenter of the other three. Ex. 117. At every point on the circle can be taken one, and only one, tangent, namely, the perpendicular to the radius at the point. Ex. 118. The perpendicular to a tangent from the center of the circle cuts it in the point of contact. Ex. 119. The perpendicular to a tangent at the point of contact contains the center. Ex. 120. The radius to the point of contact is perpen- dicular to the tangent. Ex. 121. An inscribed ||g'm is a rectangle. Ex. 122. The bisector of any ^ of an inscribed quad' intersects the bisector of the opposite exterior if on the 0. Ex. 123. The O with one of the m sides of a -I- A as diameter bisects the base. THE CIRCLE. 61 Ex. 124. The radius is = to the side of a regular in- scribed hexagon. Ex. 125. If the opposite sides of an inscribed quad' be produced to meet, the bisectors of the ^s so formed are ±. Ex. 126. The circles on 2 sides of a A as diameters in- tersect on the third side (in the foot of its altitude). Ex. 127. The altitudes of a A are the ^ bisectors of its pedal A (the feet of the altitudes). Ex. 128. AB a diameter; AC any chord; CD tangent; BD±CD, meets AC on QB(BA). Ex. 129. Find a point from which the three rays to three given points make = ^ s. Ex. 130. The circumOs of 3 As made by 3 points on the sides of a A, 2 with their vertex, are copunctal. V. The Archimedes Assumption.* V. Let A x be any point on a straight between any given points A and B\ take then the points A 2 , A 3> A 4 , . . . , such that A x lies between A and A 2> further- more A 2 between A x and A 3 , further A 3 between A 2 and A 4 , and so on, and also such that the sects AA lf AiA 2 , A 2 A 31 A 3 A 4 , . . . , are congruent; then in the series of points A 2t A 3 , A At . . . , there is always such a point A n , that B lies between A and A n . 148. 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 the ordi- nary school-geometry can be constructed with only Assumptions I-IV. * Archimedis Opera, rec. Heiberg, vol. I, 1880, p. 11. CHAPTER VI. PROBLEMS OF CONSTRUCTION. Existence theorems on the basis of assumptions I-V, and the visual representation of such theorems by graphic constructions. Graphic solutions of the geometric problems by means of ruler and sect-carrier. 149. 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, circles, etc., follows logically by rigorous deduction 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 101) 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. 62 PROBLEMS OF CONSTRUCTION. 63 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 two 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 always 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 1. (a) To designate a given point of the plane; (b) to draw the straight determined by two points; to -find the intersection of two non-parallel straights. 150. Our 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. 64 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. 151. We now announce the important theorem that in our geometry all graphic problems can be solved, all graphic constructions effected, merely by using problems 1 and 2. Theorem. Those geometric construction problems {existential theorems) solvable by employing exclu- sively Assumptions I-V are necessarily graphically solvable by means of ruler and sect-carrier. The demonstration will consist in solving with problems 1 and 2 the three following problems : 152. 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. 65 Construction. Join P with any point A of AB by Prob. 1. On the straight PA beyond A take (by Prob. 2) AC^AP. 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. 153. 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 4 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. -4-BDC and ^BEC, as inscribed angles on the diameter BC, are (by 134) right. Since (by 127) the altitudes of &BCF are copunctal, .\ FH is 1 to BC. 154. 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. Fig. 63. 66 RATIONAL GEOMETRY. Given /? the angle to be set off > and A its vertex. Construction. We draw, by Prob. 3, the straight / through A || to the given straight against which the given angle fi is to be set off. By Prob. 4 draw a straight ± to / and a straight J_ to one side of /?. Through any point B of the other side of ft draw, by Prob. 3, ||s to these J_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 ^fCAE= t 9. So EA will cut the given straight || to I under the given ^/?. Proof. Since ^-ACB and ifADB are right, so (by 146) the four points A, B, C, D are concyclic. Consequently ^fACD= ifABD (by 133) being in- scribed angles on the same^ chord AD and on the same side of it. Therefore their complements 4CAE= 4 BAD. 155. 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; while Assumption V would simply guarantee the finding of a point with- out a given sect by repeating a certain specific ap- plication of our Prob. 2. 156. In our geometry, though constantly using PROBLEMS OF CONSTRUCTION. 67 graphic figures, we must never rely or depend upon them for any part of our proof. We must always take care that the operations undertaken on a figure also retain a purely logical validity. 157. 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 covered 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 figure. 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. 158. As an immediate result of Prob. 5, the proofs in Chapters I-V of our existential theorems give ruler 68 RATIONAL GEOMETRY. Fig. 65. and sect-carrier solutions of the corresponding prob- lems. We will now give some alternative solutions. 159. 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 ifBAC right. [Inscribed angle on a diameter.] 160. 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. i-BAC. Make BD=BA. Draw DE \\AC. Make BF=BE. Then is AF± BC. Proof. ^ABF= aDBE. [2 sides and inc. %■ = .] 161. 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 . 162. Problem 9. To bisect a given sect AB. Construction. Draw through B any other straight BC. Make onitBC=CD = DE. Produce AE^EF. Draw FDG. Then is AG=GB. Proof. D is the centroid of aABF. Fig. 66. PROBLEMS OF CONSTRUCTION. 69 163. Problem 5. At a given point in a given straight to make an angle congruent to a given angle. Required, against the given ray A B of a, and toward a given side of a, the C-side, to make an angle = ^ D (given) . Construction. To one side of the given acute angle erect (by 161) FH _L DF, meeting the other side at H. Take AB=DF and BCLAB and BC^FH. .-. (by III 6) t£BAC= ^-FDH. 164. Problem 10. To bisect a given angle. Construction. On one side of the given ~4-A take any two points B, C. On the other side take AB f = AB, and AC 6 9- = AC. The sects BC and B'C intersect, say at D. AD is the desired bi- sector. 165. Problem. To join two points by an arc con- taining a given angle. Let A, B be the two points, a the given angle. Make an angle BAC supplemental to a. Erect the perpendicular to AC at A, and to AB at the bi- section-point. Their point of in- tersection is the center of the required circle. ^ AFB = a. Proof. Their supplements Fig. tAOD^lBAC (complements of ^OAD). 7© RATIONAL GEOMETRY. 1 66. Problem. To describe a circle touching three given intersecting but not copunctal straights. Construction. At the points of intersection draw the angle-bisectors. From the cross of any two of these bisectors, the perpen- dicular upon either of the three straights is the radius of a circle touching all three. 167. Definition. The cointersection-point of the three bisectors of the internal angles of a triangle, /, is called the triangle's in-center [r, the in-radius]; Ol(r) the in-circle. 168. Definition. A circle touching one side of a triangle and the other two sides produced is called an escribed circle, or ex-Q. The three centers I l9 I 2y I 3 of the escribed circles Oli(r g ), Ol 2 {r 2 ) t OJ 3 (r 3 ) of a triangle are called its ex-centers. Ex. 131. A right angle can be trisected. Ex. 132. To construct a triangle, given two sides and the included angle. Ex. 133. To construct a triangle, given two angles and the included side. Ex. 134. To construct a triangle, given two angles and a side opposite one of them. Ex. 135. To describe a parallelogram, given two sides and the included angle. Ex. 136. To construct an isosceles triangle, having given the base and the angle at the vertex. Ex. 137. 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. 71 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. 138. Construct a circle containing two given points with center on a given straight. Ex. 139. To draw an angle-bisector without using the vertex. Ex. 140. Through a given point to draw a straight which shall make congruent angles with two given straights. Ex. 141. In a straight find a point with which two given points give equal sects. Ex. 142. 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. 143. 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. 144. Through a given point to draw a st' making = ^s with the sides of a given •£. Ex. 145. Construct + A from b and hb', from a and b; from /? and a -\-b; from fi and hb; from b and /?; from p and hb] from p and a; from 6 and r. Ex. 146. Construct r't A from a and h c ; from a and c; from c and r; from a and r; from /? and r; from a and a + b; from R and r. Ex. 147. Construct A from p, a, and h a ; from p, a, and /?; from its pedal; from b, a + c, a; from a, hb, p\ from /„ I 2t I 3 . Ex. 148. Without prolonging two sects, to find the bisector of the ^ they would make. Ex. 149. Describe O through two given points with center on given st' ; with given radius. Ex. 150. From one end of the hypothenuse lay off a sect on it congruent to the ± from the end of this sect to the other side. 72 RATIONAL GEOMETRY. Ex. 151. From -I- A cut a trapezoid with 3 sides =. Ex. 152. To inscribe a sq. in a given r't-l- A. Ex. 153. Find point in side of -I- A where J- erected and produced to other side is = to base. Ex. 154. To describe a O which shall pass through a given point and touch a given st' at a given point. Ex. 155. AB, AC, BD, CE, are chords. BD || AC, CE || AB. Then AF \\ DE is a tangent. Ex. 156. To describe a O whose center shall be in one ± side of a r't A while the O goes through the vertex of the r't ^ and touches the hypothenuse. Ex. 157. To describe a O of given radius with center in one side of a given ^ and tangent to the other side. Ex. 158. Construct A from a, a, and that to. trisects a; from a and orthocenter; from a and centroid. Ex. 159. Construct A from a, /?, R; from feet of medians; of altitudes. Ex. 160. (Brahmagupta.) If the diagonals of an in- scribed quad' are ±, the st' through their intersection _L to any side bisects the opposite side. CHAPTER VII. SIDES, ANGLES, AND ARCS. 169. Convention. When a sect congruent to CD is taken on sect A B from A and its second end- point falls between A and B, then A B is said to be greater than CD; (AB>CD). When an angle congruent to 4 (h f k\ is set off from vertex against one of the rays of ^.AOB toward the other ray, if its second side falls within 4-AOB, then ^.AOB is said to be greater than ^ (h, k). In symbols, ?fAOB> 170. Theorem. // the first side of a triangle be greater than a second, then the angle opposite the first must be greater than the angle opposite the second. Given BA>BC. To prove ^C> ^A. Proof. From B toward A take BD = BC. The end- point D of r IG. 72. this sect then, because BA > BC, is between A and B, that is within ^-ACB, as is therefore also CD. Then is aBDC isosceles, .-. tCDB= i-DCB. ;. ^ACB> 4 BCD or ^.CDB. But (by 79) 4-CDB> 4 A. 73 74 RATIONAL GEOMETRY, 171. Inverse. If ^A> 7{B, .'.a>b. Proof. [From 57 and 170.] 172. Definition. Except the perpendicular, any sect from a point to a straight is called an oblique. 173. Theorem. From a point to a straight any oblique is greater than the perpendicular. Proof. Since ^-CAB is r't, .'. (by 79) Fig- 73. -4A> ^B. .*. (by 171) a>b. 174. Theorem. Any two sides of a triangle are together greater than the third side. Proof. On st' BC, be- yond C, take CD^CA. •'• (by 57) i-CDA^^CAD. But AC is within * DAB, :. ^-DAB> ^DAC=ifD. .'. (by 171) BD>AB. 175. Theorem. (The ambiguous case.) // 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. 74. c A H Fig. 75. Hypothesis, a ABC and a FGH with ^ A = 4 F, AB^FG, and BC^GH. SIDES, ANGLES, AND ARCS. 75 Conclusion. ~4-C= ?fH } or ^.C supplement of *H. Proof. At B against BA take, on the side toward C, the ^ABC' = ^G. If ray BC falls on ray BC, then (by 80) ^fC= %-H. If not on BC, suppose C between C and A. Then (by 44) ^fBCA^^fH, and BC'mGHmBC. .'. (by 57) ^BCC= *C. 176. Corollary to 175. Two triangles are con- gruent if they have two sides and the angle opposite the greater respectively congruent. 177. 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 hypothcnuse. 178. Corollary to 176. Two right-angled trian- gles are congruent if the hypothenuse and one side are respectively congruent. Ex. 161. If two triangles have two sides of the one respec- tively 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. 162. 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. 163. 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. 164. If any triangle has one of the following proper- ties it has all: 76 RATIONAL GEOMETRY. i. Symmetry. 2. 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. 165. The difference of any two sides of a triangle is less than the third side. Ex. 166. 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. 167. Two obliques from a point making congruent sects from the perpendicular are congruent, and make congruent angles with the straight. Ex. 168. 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. 169. Of sects joining two symmetrical points to a third, that cutting the axis is the greater. 179. 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 in common, BC, and on the same side of BC the other pair of congruent sides, BA, BA' '. The bisector of i-ABA', being within ^fABC, meets AC at a points. Then (by 43) aABG = aA'BG. .'.AG SIDES, ANGLES, AND ARCS. 77 = j4'G\ But (by 174) A'G and GC are together greater than A'C . 179''. Inverse of 179. 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. 170. 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. 171. Given AB a sect, C its bisection-point, PA m PB. Prove PC LAB. Ex. 172. Inverse. Given CPJLbi' of AB. Prove PA=PB. Ex.173. Given PM±AM=PN±AN. Prove^.PAM = ^ PAN or its complement. Ex. 174. Inverse. Given ifPAM = ^PAN. Prove PM ± AM mPNL AN. 180. Definition. If AB is a diameter of a circle with center C, then the two points of the circle on any other diameter, being on opposite sides of C, are (by 25) on opposite sides of the straight AB. Hence the points of the circle other than the points A } B, are separated by AB into two classes of points uniquely paired. One of these classes together with the point A is called a semicircle. The other, with B } is the associated semicircle; A and B are called end-points of each semicircle. 181. Definition. If A, D are two points on the circle with center C, then, since (by 142) the end of 78 RATIONAL GEOMETRY. the perpendicular from C to the straight AD falls within a radius, therefore the points of the circle are not all on the same side of the secant AD. Hence the points of the circle other than the points A, D are separated by AD into two classes of points. One of these classes, together with the point A, is called an arc. The other, with the point D, is called the associated or explemental arc. A and D are called end-points of each arc. Of these two arcs the arc on the side oi AD re- mote from the center is called the minor arc. The arc on the same side of AD as the center is called the major arc. The chord AD is said to be the chord of each of the two arcs. Thus to every arc pertains a chord, and to every chord pertain two arcs. Fig. 78. 182. Definition. Two arcs AB, A'B', are called congruent when, the end-points being mated, to every point C of the first arc corresponds one, and only one, point C of the second, such that AC=A'C and BC=B'C. SIDES, ANGLES, AND ARCS. 79 183. Corollary to 182. Congruent arcs have con- gruent chords. 184. Definition. An angle having its vertex at the center of the circle is called an angle at the center, and is said to intercept the arc and chord, whose end points are on the angle's sides and whose other points are within the angle. 185. Theorem. In a circle or in circles with con- gruent radii, congruent angles at the center intercept congruent arcs. Fig. 79. Given tACBmtA'C'B'. To prove the minor arc AB = minor arc A 'B'. Proof. Since (by 43) *ACBm 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 9 in D' , which makes * A'CUm * ACD and i£B'CD' = *BCD. :. (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 4A'CD" = 4 ACD = tA'C'D', and hence identical with D'. 8o RATIONAL GEOMETRY. 1 86. Corollary 187. Theorem. gruent radii. Given arc AMB = arc A'M'B To prove CA=C'A'. Any arc may be bisected. Any two congruent arcs have con- Fig. 80. Proof. The bisector of 4 ACB cuts arc AMB in a point M such that, (by 43) aACM=aBCM. .'. AM^BM and if BMC= 4 AMC. From hypoth- esis there is a point M' of arc A'M'B' such that a A'M'B' m a AMB. .' . A'M' m B'M'. .-. (by 58) aA'M'C'^aB'M'C. .'. 4A'M'C'^4B'M'C. .'. (by 48 and 84) ifA'M'C' = 4 AMC. .'. (by 44) the two isosceles triangles aAMC = aA'M'C. .'.AC^A'C. 188. Inverse of 185. Congruent minor arcs are intercepted by congruent angles at the center. Proof. , Since from hypothesis chord AB = chord A'B' } .-. (by 187 and 58) aACB^ aA'C'B'. .'. ifACB^ifA'C'B'. 189. Theorem. In a circle or in circles with con- gruent radii, congruent chords have congruent minor arcs. SIDES, ANGLES, AND ARCS. 81 For the angles at the center on the congruent chords are congruent (by 58) [As with 3 sides = ]. .'. (by 185) the minor arcs they intercept are con- gruent. 190. 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. 191. 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 f =^fACB, .-.by 43, ^ACB'= &ACB. :.AB f mAB. Moreover, were B" the end- point of a third chord from A congruent to AB and to AB f , then B, B' , B" would be at once on OC(CA) and OA(AB), which, by 122, is impossible. 192. 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. 82 RATIONAL GEOMETRY. 193. Theorem. In a circle or in circles with con- gruent radii, of two angles at the center, the greater intercepts the greater arc and chord. Hypothesis. CA^CA'. 4ACD> 4 A'CB'. Conclusion. Arc AD> arc A'B'. Fig. 82. Proof. From C against CA toward D, (by III 4) take ifACB= ^-A'C'B'. Then from hypothesis ray CB is within ^fACD. .*. B is within arc AD. .-. (by 192) arc AD > arc AB. But (by 185) arc A f B' = avc AB. .'. (by 192) arc AD> arc A'B'. Moreover a A'CB' has two sides CA', CB' = CA, CD of aACD, but ^fACD> * A'CB', ,\ (by 179) ADyA'B'. 194. Inverse of 193. 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 i79 & ) it has the greater angle at the cen- ter, and .*. , by 193, the greater minor arc. 195. Inverse of 193. In a circle or in circles with congruent radii, the greater minor arc has SIDES, ANGLES, AND ARCS. 83 the greater angle at the center and the greater chord. 196. 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 D is on the minor arc A B. Hence (by 195) ^-ACB> ^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 .'. (by 142) is > the perpen- dicular from C to AB. Moreover, congruent chords anywhere have con- gruent perpendiculars (by 178). 197. Inverse of 196. 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 196) it cannot be greater nor congruent. Two Circles. 198. A figure formed by two circles is symmetrical with regard to their center-straight as axis. 84 RATIONAL GEOMETRY. 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. 199. 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. 200. 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. 175. Two circles cannot mutually bisect. Ex. 176. The chord of half a minor arc is greater than half the chord of the arc. Ex. 177. In a circle, two chords which are not both diameters do not mutually bisect each other. Ex. 178. All points in a chord are within the circle. Ex. 179. Through a given point within a circle draw the smallest chord. Ex. 180. Rays from center to intersection points of a tangent with || tangents are JL. Ex. 181. A circle on one side of a triangle as diameter passes through the feet of two of its altitudes. Ex. 182. In + a ABC HD on AB^BC, prove CD>AD. Ex. 183. A circumscribed parallelogram is a rhombus. Ex. 184. In &ABC, having AB> BC, the median BD makes ^ BDA obtuse. Ex. 185. If AB, a side of a regular A, be produced to D, then AD> CD> BC. Ex. 186. If BD is bisector fe, and AB> BC, then BO CD. SIDES, ANGLES, AND ARCS. 85 Ex. 187. How must a straight through one of the common points of two intersecting circles be drawn in order tha>t the two circles may intercept congruent chords on it? Ex. 188. Through one of the points of intersection of two circles draw the straight on which the two circles intercept the greatest sect. Ex. 189. 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. 190. To describe a O which shall pass through a given point, and touch a given O in another given point. Ex. 191. To describe a O which shall touch a given O, and touch a given st' [or another given O] at a given point. Ex. 192. The foot of an altitude bisects a sect from orthocenter to circum-O. Ex. 193. If from the end-points of any diameter of a given O J_s be drawn to any secant their feet give with the center ■ sects. Ex. 194. A, B, I, h are concylic. Ex. 195. If tb meets circum-O in D, then DA =DC =DI. Ex. 196. The _l_s at the extremities of any chord make = sects on any diameter. Ex. 197. If in any 2 given tangent Os be taken any 2 || diameters, an extremity of each diameter, and the point of contact shall be costraight. Ex. 198. If 2 Os 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 Os. Ex.199. 2m > = a, c>b. 209. To add to a + b a further sect c, take on straight A B 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 (b + c). Thus a + (b + c) = (a + 6) + c, and so is verified and valid what is called the associative law for addition. 210. To define geometrically the product of a sect a by a sect 6, we employ the following construction. We choose first an arbitrary sect, which remains the same for this whole theory, and designate it by 1. This we set off from their inter- section point on one of two per- pendicular straights. On the other we set off on one ray a, on the other b. The circle through the free end-points of 1, a, b de- termines on the fourth ray a sect c. Then we name this sect c the product of the sect a by the sect b; and we write c = ab. By s As and 133, ab=ba. This is the commuta- tive law for multiplication. 211. Considering the triangle of the end-points of 9 o RATIONAL GEOMETRY. i and a, it is equiangular to that of the end- points of b and c. This gives as an easy construction for our sect product the following: Set off on one side of a right angle, starting from the vertex 0, first the sect i and then, likewise from the vertex 0, 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 FlG * 87 parallel determines on the other side is the product ab ; or we may call it ba, since, as we have already seen, ab=ba, which is also given by the fact that the triangle of the end- points of i and b is equiangular to that of the end-points of a and c. 212. We emphasize that this definition is purely geometric; ab is not at all the product of two num- bers. 213. To prove for our multiplication the asso- da=(bc) a A SECT CALCULUS. 91 dative law for multiplication a (be) = (ab)c we con- struct first the sect d = bc } then da, further the sect e=ba, and finally ec. The end-points of da and ec coincide (by Pascal), and by the commuta- tive law follows the above formula for the associa- tive law of sect multiplication. 214. Finally is valid in our sect-calculus also the distributive law a(b + c) =ab + ac. 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. 89) a parallel to the other side of the right angle. The congruence of the two right-angled triangles shaded in the figure and the application of the theorem of the equality of Xb+c) b+c opposite sides in the parallelogram give then the desired proof. 215. 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 j- t and is called the quotient of c by b. 9 2 RATIONAL GEOMETRY. The Sum of Arcs. 216. Definition. If a and b are two arcs with equal radii, their sum, a + b, is the arc obtained by taking together as one arc the arc a and an arc congruent to b having as one of its end- points an end- point of a and its points taken as outside of a. 217. Theorem. In the same circle or in circles with equal radii, if minor arc a m minor arc a! and Fig. 90. minor arc 6 = minor arc b' , then arc (a + 6)= arc (a' + 6'). Let minor arc AB = a and minor arc BD=b f minor arc A'B' =a! and minor arc B'D' =b r . To prove arc ABD= arc A'B'D'. Proof. '*CBF=&C'B'F' (two sides and in- cluded 4) .\*CBF=tC'B'F'. In same way 4CBH^ ^C'B'H'. .'. (by 49) ^HBF= ^H'B'F'. ;. (two sides and included ^) chord A D= chord A'U . .'. (by 189) minor arc AD = minor arc A'D'\ and if a -j- b is minor, so (by inverse of 133) is a' + 6'. But if a + b be not a minor arc, then if P be any point on the semicircle or major arc a + 6, take *D'C'P'=*DCP with P' on the same side of A SECT CALCULUS. 93 D'O with reference to A' as P of DC with reference to A. Thus D'P'=DP t also ^DPA^^D'P'A' Fig. 91. and t(DAP = $D'A'P', since ,4 and A' are on the same side of DP and D'P'. :. aAPD= aA'P'D'. 218. Definition. If an angle at the center is right the arc it intercepts is called a quadrant. 219. Corollary to 217. In a circle or in circles with equal radii the sum of any two quadrants is congruent to the sum of any other two, and all semicircles are congruent. A circle is the sum of two semicircles or four quadrants. Congruent major arcs are the sums of con- gruent semicircles and congruent minor arcs. 220. Convention. We may look upon a semi- circle as an arc whose chord is a diameter, and we may look upon a whole circle as a major arc whose two end-points coincide. The explemental minor arc will then be one single point. We may even think of arcs on a circle greater than the whole circle. In such a case certain points on the circle are considered more than once. 221. Any arc may now be expressed as a sum of a number of quadrants and a minor arc. 94 RATIONAL GEOMETRY. The Sum of Angles. 222. Definition. The sum of two acute angles or of a right angle and an acute angle is the angle obtained by setting off one against the other from its vertex with no interior point in common, and then omitting the common ray. The sum of any two or more angles is an aggre- gate of right angles and one acute angle such that, taken as angles at the center of any one circle, the sum of the intercepted quadrants and the arc inter- cepted by the acute angle equals the sum of the arcs intercepted by the angles to be added together. 223. Corollary to 79. The sum of the three angles of any rectilineal triangle is two right angles. The sum of two supplemental angles is two right angles. 224. In the familiar terminology of motion circles with equal radii are called congruent, and we say they can be made to coincide if the center of one be placed on the center of the other. Since, in their congruence, any one given point of the one can be mated with any point of the other, we say, after coincidence the second circle may be turned about its center, and still coincide with the first. Hence also a circle can be made to slide along itself by being turned about its center. This expresses a fundamental characteristic of the circle. It allows us to turn any figure connected with the circle about the center without changing its relation to the circle. Such displacement is called a rotation. A displacement of a figure connected with a A SECT CALCULUS. 95 straight, in which the straight slides on its trace, is called translation. That translation can be effected without rotation is an assumption about equivalent to the parallel Assumption IV. 225. Theorem. The diameter perpendicular to a chord bisects the angle at the center, and the two arcs, minor and major, made by the chord. 226. Convention. Parallel secants or parallel chords are said to intercept the two arcs whose points are between the parallels. 227. Theorem. Parallel chords intercept congruent arcs. Given AB\\A'B'. To prove minor AA r = minor arc BB'. Proof. If CD J_ AB then also (by 74) CD±A'B'. Then (by 117 and 58) ^ACD^^BCD and t(.A'CD = ^B'CD. ;. (by 49) ^ACA'^^BCB'. :. (by 185) minor arc AA f = minor arc BB' . 228. Theorem. If a simple plane polygon be cut into triangles by diagonals within the polygon the sum of their angles, together with four right angles, equals twice as many right angles as the polygon has sides. Proof. By a diagonal within the polygon cut off a triangle. This diminishes the number of sides by one and the sum of the angles by two right angles. So reduce the sides Fig. 93. 96 RATIONAL GEOMETRY. to three. We have left two more sides than pairs of right angles. 229. Definition. The exterior angle at any ver- tex of a polygon is the angle between a side and the ray made by producing the other side through the vertex. 230. Theorem. In any convex plane polygon the sum of the exterior angles, one at each vertex, is four right angles. Proof. The exterior angle is the supplement of the adjacent angle in the polygon. This pair gives a pair of right angles for every side. But (by 228) the angles of the polygon give a pair of right angles for every side except two. Ex. 205. In r'tA, he makes £s = to a and /?. Ex. 206. Always m c <%(a-\-b). Ex. 207. From point without acute #! a, is to sides make ^ = a. Ex. 208. The sect joining the bisection-points of the non-|| sides of a trapezoid is | to the || sides and half their sum. Ex. 209. How many sides has a polygon, the sum of whose interior ^s is double the sum of its exterior ^s? Ex. 210. How many sides has a regular polygon, four of whose ^s are together 7 r't ^s? Ex. 211. The trisection-points of the sides of an equi- lateral A form a regular hexagon. Ex. 212. The±s from A and B upon m c are =. Ex. 213. Find the sum of 3 non-consecutive £s of an inscribed hexagon. Ex. 214. Construct •!• A from b and a+hb', from fi (or hb) and perimeter. Ex. 215. The sum of the three sects from any point within a A to the vertices is < the sum and > J sum of the 3 sides. A SECT CALCULUS. 97 Ex. 216. Construct -l-r't A from b + c. Ex. 217. In a given st' find a point to which sects from 2 given points have the least sum. Ex. 218. The sum of the medians in A is < the sum and > £ sum of sides. Ex.219. Construct A from a — /? and c. a from a, /?, r; A from a, /?, R. Ex. 220. The sum of 2 opposite sides of a circumscribed quad' is half the perimeter. The sum of the £s they sub- tend at the center is 2 r't ^s. Ex. 22i. In r't A, a + b =c+2r =2R + 2r. Ex. 222. From the vertices of A as centers find 3 radii which give Os tangent, two and two. Ex. 223. If H is orthocenter, the 4 circum-Os of A, B, C, H are =. Ex. 224. Of /, /„ 7 2 » ^3» each is the orthocenter of the other 3 , and the 4 circum- s are m . Ex. 225. If, of a pentagon, the sides produced meet, the sum of the ^s formed is 2 r't ^s. Fx. 226. If hb meets b at D, construct A from ho, a— AD, c-CD. Ex. 227. A quad' is a trapezoid if an opposite pair of the 4 As made by the diagonals are ■. CHAPTER IX. PROPORTION AND THE THEOREMS OF SIMILITUDE. 231. With help of the just-given sect-calculus Euclid's theory of proportion can in the following manner be established free from objection and with- out the Archimedes assumption. 232. Convention. If a, b, a', b f are any four sects, then the proportion a:b=a' : b' shall mean nothing but the sect equation ab' =ba'. 233. Definition. Two triangles are called similar if their angles are respectively congruent. Sides between vertices of congruent angles are called cor- responding. 234. Theorem. In similar triangles the sides are proportional. 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- Fig. 94. angles are right, and sup- pose both triangles on the same right angle. We Given PROPORTION AND THE THEOREMS OF SIMILITUDE. 99 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=ea, 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 /, 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 a b} a c , b Ci b a , Fig. 95. c b , respectively a b , a/, b/ t b c r ' rJ The just- proven special case of our theorem gives then the proportions a b :r = a b : r', b c :r=b/:r', a c :r=a c ': r', b a :r Wir*. From these we conclude by the distributive law a\r = a':r' y 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'. 235. From the just-proven theorem (234) we get easily the fundamental theorem of the theory of proportion, which is as follows: ioo 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 f . Proof. First, since parallels make with the sides of the given angle similar triangles, therefore (by 234) a: b=a' : b f . 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" . Then by First a:b=a': b" . But by hypothesis a\b=a' :b f . .'. b" =b'. 236. Thus we have founded with complete rigor the theory of proportion on the basis of the Assumption-groups I-IV. 237. Corollary to 235. If straights are cut by any number of parallels the corresponding inter- cepts are proportional. 238. Corollary to 234. Parallels are divided proportionally by any three copunctal transversals. 239. Corollary to 235. Two triangles are similar if they have two sides proportional and the in- cluded angles congruent. 240. Definition. A point P, costraight with A B, but without the sect AB, is said to divide the sect AB externally into the sects PA, PB. 7 \b' V Fig. 96. PROPORTION AND THE Tk£GR£M$ 'OF ZIMJUT-'JDE. 101 241. Corollary to 235. 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. 242. Theorem. The bisector of any angle of a Fig. 97. 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] 243. 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. 244. Theorem. A perpendicular from the right angle to the hypothenuse divides a right-angled tri- angle into two others similar to it, and is the mean proportional between the parts of the hypothenuse. Each side is the mean propor- tional between the hypothenuse and its adjoming part. Proof. The r't a ABC ~ r't c &ABD, since ^A is common. Fig. 98. IQ 2 gJTIQNAl GEOMETRY. ©245. Corollary to 244. The perpendicular from any point a in a circle to the diameter is the mean proportional between the parts of the diameter. Fig. 99. 246. Theorem. The square of the hypothenuse equals the sum of the squares of the two sides. Proof. AC : AB =AB :AD, that is, AB 2 =AC-AD. Same way BC 2 =AC-DC. Now add. .'. AB* + BC 2 =AC(AD + DC)=AC 2 . 247. Theorem. Triangles having their sides taken in order respectively proportional are similar. Fig. 100. 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 are similar (~). Proof. Upon AB take AF=A'B\ and upon AC take AH=A'C. Then AB:AF=AC:AH. .'.(by 239) a ABC ~ to aAFH. .\AB:AF = BC:FH. But by hypothesis AB'.AF = BC:B'C. :.FH=B'C. :. aAFH=aA'B'C [3 sides s]. .-. a ABC- a A'B'C. PROPORTION AND THE THEOREMS OF SIMILITUDE. 103 248. Theorem. The product of the sects into which a given point divides chords of a given circle is constant. Fig. ioi. Hypothesis. Let chords AB and CD intersect in P. Conclusion. AP-PB=CP-PD. Proof. lPAC=tPDB (by 133), and * tAPC=^BPD\ \\ aAPC-aBPD. 249. Corollary to 248. From a point taken on a tangent the square on the sect to the point of con- tact equals the product of the sects made on any secant. The Golden Section. 250. Problem. To divide a sect so that the product of the whole and one part equals the square of the other part. Required on AB to find P such that AB-PB = AP\ Construction. Draw BC LB A and =\AB On the straight AC take D between A and C, and E beyond C such that CD^CB= CE. Take AP=AD io4 RATIONAL GEOMETRY. and AP'=AE. P and P' divide AB internally and externally in the golden section. Proof. By 249, AB 2 =AD-AE=AP(AP + AB) = AP 2 + AP-AB. .-. AB(AB -AP) =AP 2 . .\ AB-PB =AP\ Again, AB 2 = AE-AD=P'A(AE-DE) = P'A{P , A-AB)=P'A 2 -AB-P'A. .\AB(AB + P'A)=P'A 2 . ;.AB-P'B = P'A 2 . p' Fig. 102 251. Corollary to 250. If a is any sect divided in the golden section, its greater part #=-f(5)* — I J For (by 246) AC 2 = AB 2 +BC 2 =a 2 +- = 1 v J ' 4 \ 2 .-. AP=AD = AC-CD=ia(5)*--- 252. Theorem. The products of opposite sides A of a non-cyclic quadrilateral are together greater than the product of its diagonals. Proof. Make ifBAF^ •4 CAD, and ^ABF^ 4-ACD. Join FD. Then aABF- aACD, :.BA:AC=FA-AD. Fig. 103. PROPORTION AND THE THEOREMS OF SIMILITUDE. 105 But this shows (since 4BAC=^FAD), aBAC^aFAD. From aABF-aACD, .'. AB-CD = BF -AC. From aBAC-aFAD, .-. BC-AD =FD-AC. ?. AB-CD + BC-AD=BF'AC + FD-AC>BD-AC. 253. Corollary to 252 (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.) 254. 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. 255. Theorem. Two similar polygons can be divided into the same number of triangles respect- ively similar. 256. Theorem. If a cyclic polygon be equilateral it is regular. Ex. 228. If AB is divided harmonically by P, P', then PP' is divided harmonically by A, B. Ex. 229. If two triangles have the sides of one respect- ively parallel or perpendicular to the sides of the other they are similar. Ex. 230. The corresponding altitudes of two similar tri- angles are proportional to any two corresponding sides. Ex. 231. To divide a sect into parts proportional to given sects. Ex. 232. A sect can be divided into any number of equal parts. Ex. 233. To find the fourth proportional to three given sects. Ex. 234. To find the third proportional to two given sects. Ex. 235. If three non-parallel straights intercept pro- portional sects on two parallels they are copunctal. 106 RATIONAL GEOMETRY. Ex. 236. Every equiangular polygon circumscribed about a circle is regular. Ex. 237. Every equilateral polygon circumscribed about a circle is regular if it has an odd number of sides. Ex. 238. Every equiangular polygon inscribed in a circle is regular if it has an odd number of sides. Ex. 239. One side of a A is to either part cut off by a st' || to the base as the other side is to the corresponding part. Ex. 240. If a straight divides two sides of a A propor- tionally, it is || to the third side. Ex. 241. The bisectors of an interior and an exterior 4- at one vertex of a A divide the opposite side harmonically. Ex. 242. The perimeters of two ~ polygons are pro- portional to any two corresponding sides. Ex. 243. A median and two sides of a trapezoid are copunctal. Ex. 244. The chords on a st' through a contact-point of two Os are proportional to their diameters; and a common tangent is a mean proportional between their diameters. Ex. 245. The sum of the squares of the segments of 2 J_ chords equals the sq' of the diameter. Ex. 246. On the piece of a tangent between two || tan- gents the contact-point makes segments whose product is the square of the radius. Ex. 247. To inscribe in and circumscribe about a given O a A we draw from one vertex A of the triangle A through each dividing-point of the par- tition, that is, through each vertex of the triangles A *, a transversal ; by these transversals the triangle a is cut into certain triangles A,. Each of these triangles A t is cut by the dividing-sects of the given partition into certain triangles and quadrilaterals. If in each quadrilateral we draw a diagonal then each triangle A t is cut into certain triangles A ts . We will now show that the partition into trian- gles Afc, as well for the triangles a, as also for the triangles A k , is a chain of transversal partitions. Fig. hi. In fact, first is clear, that every partition of a tri- angle into part-triangles can always be effected by a series of transversal partitions, if, in the partition, no dividing-points lie in the interior of the triangle, and besides at least one side of the triangle remains free from dividing-points. Now these conditions are evident for the trian- gles a t from the circumstance that for each of them EQUIVALENCE. 1 1 7 the interior and one side, that opposite the point A, are free from dividing-points. But also for every a^ is the partition into a /s reducible to transversal partitions. In fact, if we consider a triangle a k , then there is, among the transversals from A in the triangle a a certain transversal which either cuts the triangle a k into two triangles, or else upon which a side of a k falls. For we recall that within no At are there dividing- points. By construction, through every vertex of Ajt goes a transversal from A ; and there is always one vertex of & k for which this transversal has a second point not in the region exterior to A * ; it therefore either goes through the interior of A k or upon it is a side of A*. In this latter case this side of the triangle A k remains altogether free from further dividing-points in the partition into triangles a /5 . In the other case the sect of that transversal within the triangle A k is for both the triangles thus arising a side which in the partition into triangles a ts remains surely free from further dividing-points. From the considerations at the beginning of this demonstration the area A (a) of the triangle a equals the sum of all areas A (a { ) of the triangles A t, and this sum is equal to the sum of all areas A(A ts ). On the other hand is also the sum of the areas A(A k ) of all triangles a k equal to the sum of all areas A(a is ). Hence, finally, the area A (a) is also equal to the sum of all areas A(A k ). So the theorem is completely proven. « n8 RATIONAL GEOMETRY. 281. Definition. If we define the area A(JP) of a polygon as the sum of the areas of all triangles into which it is cut in a certain partition, then the area of a polygon is independent of the way it is cut into triangles, and consequently determined uniquely simply by the polygon itself. Proof. Suppose a c to be the triangles of a cer- tain partition, and A k those of any other partition. Considering these two partitions simultaneously, in general is every triangle A c cut into polygons by sects pertaining to a*. Now we introduce sects sufficient to cut these polygons themselves into tri- angles A s . Then the triangles a c have (by 280) for the sum of their areas the sum of the areas of a 5 . But so also have the triangles A &. [This fact, that the sum named is independent of the way of cutting up the polygon, is the kernel, the essence of this whole investigation.] 282. Corollary to 281. Equivalent polygons have equal area. 283. Moreover, if P and Q are two polygons equiv- alent by completion, then there must be, from the definition, two equivalent polygons P' and Q', such that the polygon compounded of P and P 1 is equiv- alent to the polygon compounded of Q and Q' . From the two equations i4(P + P')~.4(Q + g'), A(P')=A(Q')> we deduce at once A(P) =A(Q), that is, polygons equivalent by completion have equal area. 284. From this latter fact we get immediately the proof of the theorem of 271 (Eu. I, 39). For, EQUIVALENCE. HO designating the equal bases of the two triangle?, by b, the corresponding altitudes by h and h\ we then conclude from the assumed equivalence by completion of the two triangles that they must also have equal area; that is, it follows \bh = \bh\ and, consequently, after division by \b, h=h'\ which was to be proved. Area and Equivalence-by-completion. 285. In what precedes we have found that poly- gons equivalent-by-completion have always equal area. The inverse is also true. 286. To prove the inverse, we consider first two triangles ABC and AB'C with a common right angle at A. The areas of these two triangles are expresesd by the formulas A(ABC)=iAB.AC, A(AB'C'=iAB'-AC. If we assume that these two areas are equal, we have AB.AC=AB'-AC, or AB:AB'=AC':AC, and from this it follows (by 235) that the two straights BC and B'C are parallel, and then we recognize (from 266) that the two triangles BOB' and BC'C are equivalent-by-completion. By an- nexing the triangle ABC it follows that the two triangles ABC and AB'C are equivalent-by-com- pletion. Thus we have proved that two right- FlG. 112. 120 RATIONAL GEOMETRY. angled triangles with equal area are also always equivalent-by-completion . 287. Take now any triangle, with base b and altitude h, then this is (by 266) equivalent-by- completion to a right-angled triangle with the two perpendicular sides b and h\ and since the original triangle evidently has the same area as the right- angled triangle, so it follows that in the preced- ing article the limitation to right-angled triangles was not necessary. Thus we have shown that any two triangles with equal area are also always equivalent-by -completion. 288. Now let P be any polygon with area b. Let P be cut into n triangles with the respective areas b lf b 2 , . . . b n \ then is b=b 1 + b 2 + . . . + b n . Construct now a triangle ABC with the base A B =b and the altitude h = i and mark on the base the points A lt A 2 , . . . A n , such that b l =AA l , b 2 =A 1 A 2 , . . . b n _ x =i„_ 2 A„.„ b n =A n _ 1 B. Since &n b Fig. 113. tne triangles within the polygon P have respectively the same area as the triangles AA l C i A ± A 2 C, . . . • A n _ 2 A n _ x C, A n _ t BC, so they are, by what has just been proven, equivalent-by-completion to these. EQUIVALENCE. 1 2 1 Consequently the polygon P is equivalent-by- completion to a triangle with the base b and the altitude h = i. Hence follows, with help of theorem 287, that two polygons of equal area are always equivalent- by-completion. 289. We may combine the two results found in this article 288 and in 283 into the following theorem: Two polygons equivalent-by-completion have always the same area; and inversely, two polygons with equal area are always equivalent-by- completion. 290. In particular two rectangles equivalent-by- completion which have one side in common must also have their other sides congruent. 291. Also follows the theorem: // we cut a rectangle by straights into several triangles and leave out even one of these triangles, then we cannot with the remaining triangles fill out the rectangle. In what precedes is shown that this theorem is completely independent of the iVrchimedes assump- tion. Moreover, without the application of the Archimedes assumption, this theorem 291 does not suffice of itself for demonstrating Eu. I, 39. 292. Definition. Of two polygons P and Q, we call P of lesser content (respectively, of equal, of greater content) than Q f according as the area A(P) is less (equal, greater) than A(Q). 293. From what precedes it is clear that the concepts of equal content, of lesser content, of greater content are mutually exclusive. 294. Further, we see that a polygon which lies 122 RATIONAL GEOMETRY. wholly within another polygon must always be of lesser content than this latter. 295. Herewith we have established the essential theorems of the theory of superficial content, wholly upon considerations of the congruence of sects and angles, and without assuming superficial content to be a magnitude. 296. Theorem. The area of any parallelogram is the product of the base by the altitude. 297. Corollary to 296. The area of any rec- tangle or square is the product of two consecutive sides. 298. A square whose side is the unit sect has for area this unit sect, since 1X1=1. Any polygon has for area as many such unit sects as the polygon contains such squares on the unit sect. The number expressing the area of a polygon will thus be the same in terms of our unit sect or in terms of a square on this sect considered as a new kind of unit, a unit surface, or unit of content. Such units, though traditional, are unnecessary and sometimes exceedingly awkward, as, for ex- ample, the acre. Ex. 284. If twice the number expressing the area of a triangle be divided by the number expressing the base, the quotient is the number expressing the altitude. Ex. 285. One side of a triangle is 35-74, and the alti- tude on it is 6-3. Find the area. 299. Theorem. // two triangles (or parallelograms) EQUIVALENCE. i 2 3 have one angle of the one congruent to one angle of the other, their areas are proportional to the products of their sides about the congruent angles. then Let 4Cm$C % Area a ABC \AC-BB AC-BD Area &A'B'C~ \A'C -B'U ~ A'C -B'U' R71 BC But in • ~ a 's BCD and B'C'D', -§rjy==g,£r. Area a ABC AC-BC " Area aA'B'C A'C -B'C 300. Corollary to 299. The areas of similar tri- angles are proportional to the squares of correspond- ing sides. 301. Problem. To construct a rectangle, given two consecutive sides. Construction. Take a straight and a perpendic- ular to it. From the vertex of ? — --- — -§ the right angle take one given sect on the straight, the other _ on the perpendicular. Through their second end-points draw perpendiculars. These (by 77) meet. They inter- sect in the fourth vertex of the rectangle required. Proof. By construction the figure is a parallelo- gram with one right angle; .'.a rectangle. Fig. 115. 124 RATIONAL GEOMETRY. 302. Corollary to 301. So we may construct a square on any given sect. 303. Theorem. The square on the hypothenuse of any right-angled triangle is equivalent to the sum of the squares on the other two sides. Hypothesis, a ABC, r't- angled at C. Conclusion. Square on A B is equivalent to sq. on AC + sq. on BC. Proof. On hypothenuse AB, on side opposite C, con- ABDF. From its vertices , and /. || BC. Drop BK, A Il6. Fig. struct (by 302) the sq. D, Fdrop DH, FG±CA FL±DH, and ?.. (complements of (complements of (complements of aDBK=aFDL = || AC. Then $ABC = tDBK 4ABK). Also ^BDK^iDFL Also -4-DFL = ^AFG ;. (by 44), aABC = .-. BCHK is sq. on BC, sq. on AB = AFLKB + ^.LDF) •4.AFL). a FAG. and FGHL =sq. on AC. . 2 A ABC = sq. on BC + sq. on AC. 304. Problem. To construct an equilateral triangle on a given' sect. Construction. On the st. AB from A take the given sect AB. At B erect to AB a perpendicular. On this perpen- dicular, from B take BC, the given sect. Join AC. At C erect to the straight AC the perpendicular CD. On CD from C take CD, the given sect. Join AD. Bisect AD at E, "7- and AB at F. Erect at F to AB EQUIVALENCE. 125 the perpendicular FG. From F take FG=AE. ABG is the required equilateral triangle. __Proof. AG 2 =AF 2 + FG 2 = (iABy + ({ 3 (AB) 2 = AB\ 305. Theorem. In any triangle y the square of a side opposite any acute angle is less than the sum of the squares A^^ of the other two sides by twice y j* ^s^ the product of either of those sides C Z — ! i _^ A and a sect from the foot of that side's altitude to the vertex of the acute angle. Proof. Let a, 6, c denote the sides, and h denote 6's altitude, and / the sect from its foot to the acute angle A . a 2 -h 2 = (b-j} 2 = b 2 -2bj + j 2 = b 2 -2bj + c 2 -h 2 ; .'.a 2 = b 2 -2bj + c 2 . 306. Theorem. In an obtuse-angled triangle the square of the side opposite the obtuse angle is greater than the sum of the squares of the other two sides by twice the product of either of those sides and a sect from the foot of that side's altitude to the vertex of the obtuse angle. Ex. 286. Find the area of an isosceles triangle whose base is 60 and each of the equal sides 50. Ex. 287. If two triangles (or parallelograms) have an angle of one supplemental to an angle of the other, their areas are as the products of the sides including the supple- mentary angles. Ex. 288. The area of any circumscribed polygon is half the product of its perimeter by the radius of the inscribed circle. 126 RATIONAL GEOMETRY. Ex. 289. To find the area of a trapezoid. Rule: Multiply the sum of the parallel sides (its bases) by half their common perpendicular (its altitude). Ex. 290. The area of a trapezoid equals the product of its altitude by its median (the sect joining the bisection- points of the non-parallel sides). 307. To find the altitudes of a triangle in terms of its sides. B Either ^ A or ^ C is acute. Suppose ^ C acute, c 2 = a 2 + b 2 — 2bj (by 305). a 2 + ^2_ c 2 h Fig. 119. ,\ ; hb 2 =a 2 -j 2 =a 26 * (a 2 + b 2 -c 2 ) 2 4 a 2 b 2 -(a 2 + b 2 -c 2 ) 2 4b 2 4b 2 (2ab + a 2 + b 2 -c 2 )( 2 ab -a 2 -b 2 + c 2 ) 4b 2 [(a + b) 2 -c 2 ][c 2 -(a-b) 2 ] 4b 2 (a + b + c)(a + b-c)(c + a-b)(c-a + b) 4b 2 Put (a + b + c)=2S. Then a + b-c = 2S-2C 25 2(5 —c)2(s —6)2(5 — a) .\W:- 4b'' ;.h h =l[s(s-a)(s-b)(s-c)]K 308. (Heron.) To find the area of a triangle in terms of its sides. b 2 $ A= %bk b = -.-[s(s-a)(s-b)(s-c)] ; ,\ A =[5(5 -a) (s -b) (s-c)f. EQUIVALENCE. 127 Ex.291. If a 2 =b 2 +c 2 , ;.^.A=r\^. lia 2 >b 2 +c 2 , .'. £A>r , t £. Iia 2 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 straight are parallel. Proof. They cannot (by 345 and 347) have a point in common. Ex. 423. A st' and a plane JL to the same st' are ||. 361. Theorem. Every plane through only one of two parallels is parallel to the other. Given AB\\ CD in a, and /? another plane through AB. To prove CD || /?. *5 RATIONAL GEOMETRY. Since AB is in a and in /?, it contains (by 9) every point common to the two planes. Fig. r37. But CD is wholly in a. So to meet fi it must have a point in common with a and /?, that is, it must meet A B. But by hypothesis AB \\ CD. Ex. 424. Through a given point to draw a st' || to two given planes. Ex. 425. If a || a, and b the meet of a with /?, /? on a then a \\ b. Ex. 426. Through A determine a to cut b and c. 362. Problem. Through either of two straights not co planar to pass a plane parallel to the other. Fig. 138. If AB and CD are the given straights, take CF || AB. Then (by 361) DCF \\ AB. GEOMETRY OF PLANES *53 Determination. There is only one such plane. For, through DC any plane || A B 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 \\ a meet « in ||s. Ex. 430. If a I| a and a || p, then a || to the meet a/?. 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 || 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. J 54 RATIONAL GEOMETRY. 365. Corollary to 364. Parallel sects included between parallel planes are equal. Ex. 432. If two || planes meet two || planes, the four meets are ||. Ex. 433. If a || to the meet aft, then a \\ a and a \\ ft. Ex. 434. If a || ft, aLa is ±ft. Ex. 435. Through A draw a \\ ft. [Solution unique.] Ex. 436. If, in a, a cross a' , in ft, b cross b' , and a || b, a' || b', then a || ft. 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 a \\ a are -», Ex. 440. If AB |1 a and BC || «, then plane A BC \\ a. Ex. 441. If three sects are = and ||, the as of their adjoining ends are = and ||. Ex. 442. If A in a || a, AB \\ a is in a. 366. Theorem. // two angles have their sides respectively parallel and on the same side of the straight through their vertices y they are equal. Fig. 140. Hypothesis. AB\\A'B f with A and A' on the same side of BB f ; also CB \\ C'B' with C and C on same side of BB'. GEOMETRY OF PLANES. 155 Proof. From A take (by 66) A A 1 1| BB' '; .\ (by 95) AA' = 5£' and A '5' =A£. In same way CC'H and - BB' and B'C = BC. But then A A' - CC and (by 353) AA'\\CC'\ :. (by 100) AC-A'C; /. (by 58) aabcsaa'S'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 a, and also in hemiplane /?' of p. 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 the inclination of the two hemi- planes a' and /?'. When the inclination is a right angle the planes are said to be perpendicular to each other. 369. Theorem. // a straight is perpendicular to a given plane, any plane containing this straight is perpendicular to the given plane. Proof. At the foot of the given perpendicular erect in the given plane a perpendicular to the 156 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 a£ ; but this angle is (by 368) the inclination of the planes. Ex. 443. A plane _L the meet of two planes is JL to each; and inversely. Ex. 444. Through a in a draw fi±a. 369 (b). 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 meet, 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 a_L/?, from A, a±fi is in a. Ex. 446. If a st' be || to a plane, a plane _L to the st' is 1 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 /? || a. Ex. 448. Through a draw «JL/9. Ex. 449. Through A draw a±p and j. GEOMETRY OF PLANES. 57 373. Theorem. // two straights be cut by three parallel planes the corresponding sects are proportional. Fig. 142. Let A B, 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 t then (by 364) EG \\ BD and AC\\GF. :. (by 235) AE:EB=AG:GD and ^(7:6^= CF:FL>. .'. AE:EB=CF:FD. Ex. 450. Investigate the inverse of 373. 374. Theorem. Two straights not co planar 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 of a take c\\b. Then (by 361) the plane ac or a || b. The projection 158 RATIONAL GEOMETRY. b' of b on a cuts a, say in B' ; else were a || b f \\ b. Then, in plane b% B'B drawn ±b' is (by 370) ±a Fig. 143- and (by 74) also ±b, and is the only common per- pendicular to a and b. For any common per- pendicular meeting a at B" is J_ b" through B" || 6, which is in a, and .\ ±«, hence -B'' is a point of 6' the projection of b on a; .'. identical with B f , the cross of b f with a. 375. Corollary to 374. Their common perpen- dicular is the smallest sect between two straights not coplanar. For (by 142) BB' ")- Adding these equations to one another and notic- ing that (by 3866) on the one side V{OT x )+ V(OT 2 ) + . . . + V{OT n ) = V(OBCD), on the other side V(OT h )+V(OT l2 ) + . . . + V(OT In ) = V(OABC), POLYHEDRONS AND VOLUMES. 177 we find, using I to mean the sum from w = i up to n=n, V(OBCD) = ViOABC) + T[V(t n ') + V(t n ") + ^(C")J- n<= 1 Furthermore, the division of the tetrahedron OBCD by the plane ABC into the tetrahedra OABC and ABCD is a transversal partition, and so (by 384) V(OBCD) - V(OA£C) + F(yl5CL>). The last two equations give finally V (ABCD) = "T\y(tJ) + V(t n ") + V(t H '")] ; . n = i therefore also for this case our theorem is proven. Case 3 . lies without the tetrahedron on one of its boundary planes, that is in the plane of one of its faces; for example, on the plane ABD (Fig. 152). Fig. 152. Then the projections A', B' , D' are costraight; therefore one of them, say A' f lies between the other two. 178 RATIONAL GEOMETRY B' , C, D' make a triangle which is cut by the sect A'C into two part-triangles A'B'C and A'C'D'. Correspondingly the plane OA'C cuts the tetra- hedron ABCD into two part-tetrahedra AA t BC, and AAfiD\ and since this plane goes through the edge AC, therefore it makes a transversal partition, and we have (by 384) V(ABCD) = V(AA X BC) + V(AAfD). Since, however, lies on the prolongation of the edge A L A, therefore (by Case 2) V{AA l BC)=IV{t b ) and V{AA,CD)=lV{t d ), where t b are all the tetrahedra into which in accord- ance with the above method the tetrahedron AA X BC is divided [and t d those in AA L CD]. These last three equations give now V(ABCD) = IV(t), where t are all the part-tetrahedra of ABCD. Case 4. If lies on no one of the boundary planes of the tetrahedron ABCD, then no three of the pro- jection-points A', B', C, D' are ccstraight. Consider first the case in which one of these points (say A f ) falls within the triangle B'C'D' made by the other three (Fig. 153). The joining sects A'B', A'C', A'D' cut the tri- angle B'C'D' into three part-triangles, and corre- spondingly the tetrahedron ABCD is divided into three part-tetrahedra AAJBC. AAJ3D, AA.CD, POLYHEDRONS AND VOLUMES. 179 giving a case of Partition method I ; therefore (by 386c) V(ABCD) - V(AA X BC) + V(AA X BD) + V(AA X CD). If we now, as above, divide the triangles A'B'C\ A'B'D' , A' CD' into part-triangles, project these Fig. 153. back upon the corresponding part-tetrahedra, and designate the part-tetrahedra obtained in the above given way of the tetrahedra AA X BC, AA X BD, AA X CD with t d , t c , t b respectively, then we obtain the three equations V{AA X BC) -IV(t 4 ) ; V{AA X BD) = IV(t c ) ; V(AA x CD)=ZV(t b ), since lies on the ray A X A, that is we have here each time Case 2. These last four equations give now V(ABCD) ■■= IV (t), where / are all the part-tetrahedra of ABCD. iSo RATIONAL GEOMETRY. Case 5. If, finally, again lies in no one of the boundary planes of the tetrahedron A BCD, but each of the projections A', B r , C, D' falls without the triangle made by the remaining projection-points, then the projections of the edges of the tetrahedron make a convex quadrilateral with its two diagonals. This quadrilateral is divided by its diagonals into four triangles, M'A'C, M'A'D' % M'B'C\ M'B'D' (Fig. 154). Fig. 154. The plane OA'B' divides the tetrahedron by a transversal partition into two parts, so that we have (by 384) V(ABCD) - V(AMBC) + V(AMBD) 9 where M is the point corresponding to the inter- section point M' of the diagonals A'B r and CD'. POLYHEDRONS AND VOLUMES. 181 If we now divide each triangle M'A'C, M'A'D', M'B'C, M'B'D' into part-triangles and project back upon the faces of the tetrahedron A BCD, then the tetrahedra AMBC and AMBD are divided each into part-tetrahedra, which, in general, may be designated respectively t c and t d . Since now, moreover, lies in the plane AMB, therefore (by Case 3) we have the equations V{AMBC)=lV(t c ) and V(AMBD) = IV(t d ). Consequently is also in this case V(ABCD) = IV(t). This completes the proof of the theorem that in every partition of a tetrahedron by central projec- tion the sum of the volumes of the part-tetrahedra equals the volume of the whole tetrahedron. 386/1. The most general partition can be built up from the four partition methods already given, and this proves the fundamental theorem 386. For let A BCD be a tetrahedron and P lt P 2 , . . . P k the part-tetrahedra which arise from any par- tition of it. If, now, we project all these tetrahedra from the point A upon the face BCD, then their projections, which necessarily all fall within the triangle BCD, overlie and overlap, in general, manifoldly, and cut one another into polygons. When we cut these polygons into triangles and join their vertices with A we divide each tetrahedron, P m (m = i, 2, . . . k), into a number of truncated tetrahedra (including perhaps pyramids of five summits and 1 82 RATIONAL GEOMETRY. complete tetrahedra) which in turn in the well- known way we divide into further part-tetrahedra. Since the so obtained partition of each part- tetrahedron P m into further part-tetrahedra, which may be designated t m ' t t m " y . . . *£«>, is accom- plished with the aid of central projection, for each projection-center lies without each tetrahedron, only that one with the summit A excepted, so is V{P m ) - V{t m ') + F(V') + . . . + V(^J), If we now give m successively the values i, 2, . . . k, we obtain k such equations, which added give the following: But, on the other hand, every tetrahedron AT, where T is a part-triangle of BCD, cuts out from the aggregate of part-tetrahedra t a, set, and each tetrahedron of this set appears once and only once in the above sum. At the same time all summits of these last part- tetrahedra lie on the three edges from A of the particular tetrahedron AT; that is, AT is divided by this set of tetrahedra according to Partition method II. Furthermore, the whole tetrahedron ABCD is divided into tetrahedra AT according to Partition method I, so that it is divided according to Parti- tion method III into part-tetrahedra t. Now this complex of tetrahedra / is identical POLYHEDRONS AND VOLUMES 183 with the complex t$ t where / = i, 2, . . . n m \ m = i, 2, . . . k. Consequently V(ABCD)-iV(Wiz;j: : *.)• which combined with the previous equation gives the desired proof of the fundamental theorem m = k V(ABCD) = 2 V(P m ). m=i 387. Theorem. If a polyhedron P is cut into tetrahedra in two different ways, then the sum of the volumes of the tetrahedra of the first partition equals that of the second. Proof. Suppose P divided into m tetrahedra hs t 2 , . . . t m , and again into n tetrahedra */, t 2 \ . . . U. Construct a tetrahedron T which contains the polyhedron P, and cut the polyhedron bounded by the surface of P, and that of T in any definite way into tetrahedra 3T/, TV, . . . Thus we obtain two partitions of the tetrahe- dron T and (by 386) the equations V(T)=V(t l ) + V(Q + ... + V(t m ) + V(T l ') + V(T,') + ... V(T)=V(t l ') + V(t 2 ') + ... + V(t n ') + V(T l ') + V(T 2 ') + ... whence V(t t ) + V(t 2 ) + ... + V(t m ) = V(t 1 ') + V(t 2 ') + ... + V(tJ). l8 4 RATIONAL GLOME TRY 388. Definition. The volume of a polyhedron is the sum of the volumes of any set of tetrahedra into which it is cut. 389. Definition. Two polyhedra P and Q of equal volume are said to be equal. P is called greater than Q if the volume of P is greater than the volume of Q ; less if volume is less. UV(P)=V(Q), we say P^Q. UV(P)>V(Q), we say P>Q. IiV(P) '• D 2 =-(B 2 + 3 S 2 ) =-(o + iT 2 ) =iaT 2 = Y 2 . 4 4 In (3) let KLMN be the section S 3 . Now the areas AANK:AAGO=AN 2 -^G 2 ^(^a) 2 :a 2 = i:g; AGNM:AGAC = GN 2 :GA 2 = (%a) 2 :a 2 == 4:9. But the whole tetrahedron D 3 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 3 = a ANK: a AGO = 1:9; :.CANK=\D 3 . In the same way OGNM :D 3 = aGNM : aGAC = 4:9; .• . CA iVX + O^A^M = {P, ; .• . CKLMN + OKLMN - *D 9 . But by (398) OKLMN + OKLMN = i-iaS t + i • § aS 3 = Ja5 3 ; / . ^L> 3 = JaS 3 ; /.£ 3 = ^3^(£ 3 + 3S 3 ), 4 4 since here the area B 3 = o. POLYHEDRONS AND VOLUMES. 189 400. Corollary to 399. Since, in a frustum of a pyramid, B and 5 are similar; .*. if b and s be corre- sponding edges, 2\ B:S = b 2 :s 2 ; /.the volume F=-b(i+^A. 401. Definition. A prism is a prismatoid whose base and top are congruent. A right prism is one whose lateral edges are perpendicular to its base. A par allelo piped 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. j 03. 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 unit, a unit solid. Such units, though traditional, are unnecessary. 190 RATIONAL GEOMETRY, 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 top is 5 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. 530. Every regular octahedron is a prismatoid whose bases and lateral faces are all congruent equilateral tri- angles. Find its volume in terms of an edge b. Ex. 531. The bases of a prismatoid are congruent squares of side b, whose sides are not parallel; the lateral faces are eight isosceles triangles. Find the volume. Ans. £a6 2 (2 + 2*). Ex. 532. If from a regular icosahedron we take off two five-sided pyramids whose vertices are opposite summits, there remains a solid bounded by two congruent regular pentagons and ten equilateral triangles. Find its volume from an edge b. Ans. %b\$+ 2(5)*]. Ex. 533. Both bases 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. £a& 2 . Ex. 534. The upper base of a prismatoid of altitude a = 6 is a square of side, 62 = 7-07107; the lower base is a square of side 6, = 10, with its diagonals parallel to sides of the upper base; the lateral faces are isosceles triangles. Find volume. Ex. 535. Every prismatoid is equal in volume to three POLYHEDRONS AMD VOLUMES. 19 1 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 = * a (~T^ +2M ) =MB 1 +4M+B 2 ). Ex. 536. 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. 537. If the bases of a prismatoid are trapezoids whose mid-sects are b x and b 2 , and whose altitudes are b l +b 2 1 a x —a 2 b x —b 2 \ a t and a 2 , the volume = a( ■ — ■ 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. 19 2 RATIONAL GEOMETRY. 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 1 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, b, I, and its volume V, (1) Given a, b, and superficial area; find V. (2) Given a, b, V; find /. (3) Given V, (ab), (bl); find I and b. (4) Given V, l-j-J , l-j J ; find a and b. (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 four 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 to meters tall is an isosceles right triangle of 6 meters hypothenuse. Find volume. POLYHEDRONS AND VOLUMES. 19 3 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 1 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 XIII. 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. 194 TRIDIMENSIONAL SPHERICS. 195 4i i. 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 other. 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. // 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 indefinite 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 with the sphere. Two spheres are called tangent to each other when they have one point, and only one, in com- mon. 417. Theorem. A straight or plane having the foot of the perpendicular to it from the center in com- mon with the sphere is tangent. Proof. This perpendicular, a radius, is (by 142) 196 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. // 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. 158. 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. 197 of the sphere is less than the radius of a great circle. Proof. The hypothenuse (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. // two spheres have two points in common they cut in a circle whose center is in their center-straight and whose plane is perpendicular to that straight. Hypothesis. Let C and be the centers of the spheres having the points A and B in common. Conclusion. They have in common all points, and only those, on a circle with center on OC and plane _L to OC. Fig. 159. Proof. Since, by 58, aACO=aBCO, /.perpen- diculars from A and B upon OC are equal and meet OC at the same point, D. Thus all, but only, points like A and B, in a plane J_ to OC, and points of OD(DA), are on both spheres. 198 RATIONAL 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 J_ to and bisecting A B meets /? ± to and bisecting BC in EH 1 ABC, and meets dl_ to and bisecting BD in FOLABD. .\EH±EG y and F01.FG, 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). Fig. 160. TRIDIMENSIONAL SPHERICS. r 99 432. Corollary to 431. Since equal chords have congruent minor arcs, .'. the great-circle-arcs join- ing a pole to points on its circle are congruent. Hence if C is any point in a sphere a, then the aggregate of all points A in a, for which the great-circle-arcs CA are con- gruent to one another is a circle. 1 Fig. 161. 433. Theorem. The great-circle-arc joining any point in a great circle with its pole is a quadrant. Proof The angle at the center is right. 434. Theorem. If A, B are non- opposite, the point P is a pole of their great circle when the arcs PA, PB are both great- circle -quadrants. For each of the angles POA, POB is right and /. PO±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 through a pole 0} 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. 162. 200 RATIONAL GEOMETRY. dicular to a given great circle will pass through its poles. 439. Theorem. If 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. D Vr: c "" !>-- r -&b 1 \ /*/ s / 1 \ ^V / / T \i ^/f Fig. 163. Hypothesis. Let KJ be the sect _L to the edges EF and GH in the || tangent planes. Then KJ = DT, the diameter. Let Ql(IP) = M0, sections made by the plane ±DT at I and IK J at R, where KR =DI. Let ABCLSN be any parallel plane through a point A of the sphere. Conculsion. LN = qC(CA). 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=ifVLU. ;. (by 299) area MO :area LN = WM-MZ :UL-LV = JR-RK:JS : SK. TRIDIMENSIONAL SPHERICS. 201 But (by 325) area el(IP) :area oC(CA) =PI 2 :AC 2 -77 -ID :TC -CD (by 245); .*. area LN = area OC(CVI). 440. Cavalieri's assumption. 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 nr 3 . Proof. A tetrahedron 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 = JaS. . But a = 2r, and (by 245) 5 = fr-|r-7r. .'. Vol. sphere = f •2r-§r-|r-7r=|^r 3 . 442. Definition. The area of a sphere is the quotient of its volume by one-third its radius. Area of sphere = 4^r 2 . 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 a spherical segment is -7r(^i 2 + 3^ 3 2 ), where r 3 is the 202 RATIONAL GEOMETRY. radius of the section two-thirds the altitude from the base whose radius is r v If the segment is of one base its volume is \anr^\ which in terms of r, the radius of the sphere, is na 2 lr — ], and equals %na(r 2 2 -\ — ). If we eliminate r 3 by intro- ducing r 2 , the radius of the top, the volume of the segment is inalsir^ + r^) + a 2 ]. 446. Problem. Given a portion of a sphere, find its radius. Construction. Take any three points of the part given, say A, B, C. The plane A, B, C (by 419) cuts the sphere in a circle. The straight at D, the center of this circle perpendicular to the plane ABC, contains the center of the sphere (by 420) and therefore meets the sphere, say at P. In the plane PAD draw AP'A_ to AP and meet- ing DO in P'. Bisect PP' in 0. Then is the center of the sphere and OP is the radius. Proof. is circumcenter of PAP f . :.OP = OA. But since OD is_LO(D)ZM at A ;.OA=OB = OC. 447. Corollary to 305. Fig. 164. OP DA 2 + DP : 2DP that is, R = r 2 +h'< 2h Ex. 563. A circle on a sphere of 10 centimeters radius TRIDIMENSIONAL SPHERICS. 203 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 1, 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. 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 10 centimeters radius find the radii r t and r 2 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. 577. Find the volume of a sphere the length of whose great circle is n. 204 RATIONAL GEOMETRY. 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 n to 6. Ex. 580. Find altitude of a spherical segment of one base if its area is A 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 XIV. 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. 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. 205 2o6 RATIONAL GEOMETRY. 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 CB\ CD'. .'.(by >]$)&ABC~&AB'C and aACD-^aACD'. ;. (by 234) CB' : CB=AC: Pig. 165, AC-CD* i CD. But CB= CD. .'.CB'=CD'. 450- Corollary to 449. The axis of a circular cone passes through the center of every section parallel to the base. 451. Theorem. If 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. 166. Proof. BC : B'C-AC : AC =AL : AD '.- - VT : VT - VH : VH' =GH : G'H'. .'.BC 2 :B'C 2 = GH 2 : CH' 2 But (by 325) area oC(CB) : area oC(CB') CONE AND CYLINDER. 207 = BC 2 : B'C' 2 =GH 2 : (TE' 2 = area &FGH : area aF'G'H' (by 300). But by hypothesis area OC(CB) =area aFGH. .-. Area oC'{C'B') -area aF'G'H' . 452. Corollary to 451. Volume of circular cone is (by 440) = volume of tetrahedron of equal altitude and base = -J ar.r 2 . 453. Theorem. The lateral area of a right cir- cular cone is half the product of the slant height by the length of the base. Proof. It has the same area as a sector of a circle with the slant height as radius and an arc equal in length to the length of the cone's base. • ". (by 323) 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. Proof. It is the difference of the areas of two sectors with a common angle, the lengths of the arcs of the sectors being equal to the lengths of bases of the frustum. .*. F = \h(c x + c 2 ) = 7ih(r l + r 2 ). Fig. 167. 456. Corollary to 451 and 399. 2o8 RATIONAL GEOMETRY. The volume of the frustum of a circular cone, T/.F = ia 7 r(r 1 2 + 3 r 3 2 ), where r 3 is the radius of 5. 457. Definition. A circular cylinder is the assem- 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 right 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. The 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 = a7ir 2 . CONE AND CYLINDER. 209 459. The lateral area of a circular cylinder is the product of an element by the length of a right section : C ' = 27tra. Fig. 168. Proof. It is equal to the area of a parallelo- gram with one side an element and the consecu- tive side equal to the length of a base. An altitude of this parallelogram equals the length of the right section. 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. Fig. 169. 210 RATIONAL GEOMETRY. Proof. The volume of the circumscribed cylin- der = 7rr 2 -2r = 2 nr 3 . Ex. 582. In a right circular cylinder of altitude a, call the lateral area C and the area of the base B. (1) Given a and C; find r. (2) Given B and C; find a. (3) Given C and a = 2r; find C + 25. (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 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 increase its lateral area by the area of a base? Ex. 587. The lateral area of a right circular cone is twice 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 h. (1) Given a and r; find K. (2) Given a and h; find K. (3) Given K and h; 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 cylinder; find volume. CONE AND CYLINDER. 21 1 Ex. 593. A right cylinder of volume 50 has a circum- ference of 9; find lateral area. \ Ex. 594. In a right circular cylinder of volume 8, the lateral area 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 sum of the others, but contains a greater volume. Ex. 596. What is the relation between the volumes of two cyliders when the radius of one equals the alti- tude of the other? CHAPTER XV. 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. 170. PURE SPHERICS. 1 3'. Any two points of a straightest, not each the other's opposite, determine this straightest; 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. If O f is the opposite of 0, 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 0, since it is not 0'. .*. 0, P deter- mine (by Ii') a straightest which (by I 2') goes through 0' . .'. O, O f do not determine a straightest. 465. Theorem. Two distinct straightests can- not have three points in common. [Proved as in 6J II. Assumptions of betweenness on the sphere. 466. These assumptions specify how " between" may be used of points in a straightest on a sphere. II 1'. No point is between two opposites. II 2'. No point is between its opposite and any third point. II 3'. Between any two points not opposites there is always a third point. II 4'. If B is between A and C, then B is also between C and A, and is neither C nor A. II 5'. If A and B are not opposites, then there 214 RATIONAL GEOMETRY. Fig. 172. is always a point C such that B is between A and C. II 6 r . Of any three points, not more than one can be between the other two. II 7'. If B is between A and C, and C is between A and D, then B is between ^4 and D. II 8'. Between no two points are there two opposites. ■ 467. Definition. Two points A and B, not oppo- sites, upon a straight est a, we call a s£c* and desig- nate it with AB or BA. The points between A and B are said to be points of the sect A B or also situated within the sect A 5. 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 9'. (Pasch's assumption.) let A, B, C be three points, not all on a straightest, and no two opposites, and let a be a straightest on which are none of the points A, B, C\ if then the straightest 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; but it can- not go through both. 468. Theorem. Every straightest a separates On the sphere, Fig. 173. PURE SPHERICS. 215 the other points of the sphere into two regions, of the following character: every point A of the one re- gion determines with every point B of the other region, not its opposite, a sect AB within which lies a point of the straightest a; on the contrary, anv two points A and A' of ; 4 ft -1 FlG - x 74- one and the same region al- ways determine a sect A A' which contains no point of a. [Proved as in 22.] Points in the same one of these two regions are said to be on the same side of a. 469. Theorem. The points of a straightest a other than two opposites, 0, 0', are separated by 0, 0' 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 f is between two of the same class. Proof. Take any other straightest b through and .'. through 0'. It (by 468) cuts the sphere into two regions. Now (by II 5 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 1 are points of b. Two Fig. 175. 21 6 RATIONAL 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 10'. 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 3', II 10' and II 2'. 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 ray a pair of opposites, contrary to 471. 473. Definition. On the sphere, a system of sects, A B, 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 o) 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 FlG - I 7 6 - spherical polygon; their end- points its vertices. PURE SPHERICS. 217 Polygons with three vertices are called spherical triangles. Fig. 177. Fig. 178. 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 1'. 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 r from A' always one and 2l8 RATIONAL GEOMETRY. only one point B' , such that the sect A B is con- gruent to the sect A'B f . Always AB = AB = BA. Ill 2'. If AB = A'B f and ,4B = A"£", then is also A'B'^A"B". Ill 3 r . On the straightest a let AB and BC be two sects without common points, and further- 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 -4. (h, k) or ^ (&, h). The rays /&, &, together with the point separate the other points of the sphere into two re- gions, the interior of the angle Fig. 179. anc [ the exterior. [As 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 ^ (h, 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 f such that the Fig. 180. PURE SPHERICS. 219 %. (h, k) is congruent to the angle ?£(h\ k'), and likewise all inner points of the angle ^ (h\ k') lie on the given side of a! . Always ? (h, k) = ^. (h, A?) = ? (k, h). Ill 5'. If ? (h, k)=$ {W, k') and ? (h, fc)s ? (&", &"), then is also * (/*', £') ^ ? (/*", fe"). 477. Convention. On the sphere let ABC be an assigned spherical triangle; we designate the tw r o rays going out from A through B and C by h and k respect- ively. Then the angle ^ (h t 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 f C> then also always are fulfilled the congruences t ABC ~t A'B'C and ?ACB^ *A'CB'. 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 A B is said to be less than the sect AC. Fig. 182, 220 RATIONAL GEOMETRY. Fig. 183. In symbols, ABAB. AB> CD when E between A and B gives A is = CD or 5£ = CD, using = f or m . 479. Convention. When 4-AOB is set off from vertex 0' against one of the rays of 'ifA'O'C toward the other ray, if its second side falls within i$A'&G % then the ^AOB is said to be less than the ^A'O'C. In symbols, ^-AOBK^A'O'C. Fig. 184. Then also ^A'O'C is said to be greater than 4A0B. In symbols, ^.A'O'Cy^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. 185. PURE SPHERICS, 221 481. Definition. Two spher- ical angles with a common vertex and whose sides make two straightests are called verti- cal angles. Fig. 186. Fig. 187. 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 A, B 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 lies on the right hemisphere of 222 RATIONAL 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. 188. 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. 189. PURE SPHERICS. 223 486. 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. 190. 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 supplement 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. 191. 492. Definition. A spherical angle neither right nor acute is called an obtuse angle. 224 RATIONAL GEOMETRY. 493. Definition. A spheri- cal triangle with two sides congruent is called isosceles. Fig. 192. 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. [Proved as in 58.] Fig. 193. 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. 194. PURE SPHERICS. 225 Given r't ^A= ifC. To prove PA=PC=PD, and ^Drt. Proof. By 495, PA = p C and (by 485) PA^P'A. ;. (by 484) 4PDA= ?P'DA; .'. (by 488) %PDA= ^.PAD; .'.(by 495) PDmPA. Fig. 195. 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 f . Proof. At A take a r't? £AC, and also at A'. On AC take a sect AC, and on AC take A'C 7 a AC. At C and C Fig. 196. take straightests ±AC and A r C r . These contain B and B'. :. (by 485) AB = A'B'. 500. Theorem. A ^oiw/ which 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 226 RATIONAL GEOMETRY. perpendiculars intersecting at P' . Then (by 499 and 496) IPAC^P'AC. 501. Theorem. // three sects from 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. straightest a, Through a point A, no' on 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= 4-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 498) be a pole of a. 504. Definition. A point B of a given ray 00' such that BO s 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 PURE SPHERICS. 227 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 Q. The straightest PQ bisects the given ray 00'. Proof. Since P and Q are poles, .'. ^.B = r't ?=?£>. A (by 485) OB = BO'. 506. Theorem. If and 0' are opposites, then with vertex ^ (h, k) = ^ (h, k) with 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 b with- in ?(/&, k) making ^ (h, b)= £(&, k) will be called the bisector of $ (h, k). 508. Problem. To bisect a given spherical angle. Construction. By 505, bi- sect the rays of the angle ?B at H and F. Take A between H and B and from F on FB' take FC^HA. Then ^C intersects HF at Z), and £D bisects ^HBF. Proof. By 496, %ACB= ? CAS';. -.by 485, HD~FD\ .\ by 496, * tf££ = ? F££>. 228 RATIONAL GEOMETRY, 509. Problem. To bisect a given sect. Construction. At the end- points erect perpendiculars [by taking (by III 4') angles = ?S in 503]. ^ Bisect (by 508) the £ be- tween them. Proof. By 497 and 484. FlG - 2DI - 510. Corollary. In an isosceles triangle the bi- sector of the angle between equal sides bisects at right angles the third side. 5 xx. 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. 202. 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. 229 Fig. 203. 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 £CX>. If the median BF be less than a quadrant, if' is within IfACD. .-. ^/fCA^JDCA .-. jDCA > ?£AC. If BC be greater than a quadrant, if" is without .-. 1H"CF>*DCF. .\ ?£>CV1g,takeCD=g. .\BD = Fig. 205. Fig. 206. h = c, and (by 510) the PURE SPHERICS. 231 bisector of $DBA is _L to CD A. .*. ^by 498) B is pole to CD A. .'. %-A is also r't. If %.A m ?F, then if $ABC >?£, take $ABD = *G. .'. (by 485) ^BDA = ?tf = ?C = r't ?. .-. £ is pole toCZM. 521. Theorem. The straight- est through the poles of two straightests is the polar of their intersection-points. Let A and B be poles of a and b, which intersect in P. To prove A B the polar of P. Proof. AP and £P are , Fig. 207. 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 f that pole of a on the same side of a as A, B' of b as B y 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 of a as A, B r of b as £, C of c as C 526. Theorem. If of two spherical triangles the second is the polar of the first, then the first is the polar of the second. Let ABC be the polar of A'B'C 232 RATIONAL GEOMETRY. Fig. 208. To prove A'B'C the polar of ABC. Proof. Since B is pole of A f 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 AC, and C of AB. More- over, since by hypothesis A and A' are on the same side of B'C and A is pole of B'C, .'. sect A A' is less than a quad- rant. .'. A and A 1 are on the same side of BC, of which A' is pole. And so for B' and O \ 527. Theorem. J« a /?) = BC + £>£. 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. 2 33 Proof. Since the given 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 respectively equal to two of the other, the angles opposite one pair equal, and those opposite the other pair not supplemental. 530. Theorem. // two sides of a spherical triangle are each less than a quadrant, any sect from the third side to the opposite vertex is less than a quadrant. Let A B and BC be each less than a quadrant. To prove BD < 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 t and so have a point on each of the three sides of a AB'C, which (by II 9' ) is impossi- ble. Hence D is between B and H. That is BD< quad- rant. 531. Corollary to 530 and 513. If two sides of a spherical triangle are each less than a quadrant, the Fig. 210. 234 RATIONAL GEOMETRY. Fig. 211. angle opposite either is less than the supplement of the angle opposite the other. 532. Theorem. If two sides of a spherical triangle be 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 2 ABC, BC and another side, AB, be each less than a quadrant, and AC>AB. To prove ^fABC>^ACB. Proof. Within AC take D making AD = AB. Then (by 530) DB is less than a quad- rant. .'. (by 531) 1ADB> ? C. But 2 ABC > ^ABD m %ADB > ?C. 533. Theorem. If 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. 2 35 Fig. 212. 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 ^-radius. 535. Theorem. Any spheri- cal chord is bisected by the perpendicular from a pole. Proof. AD=BD (by 520). 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 straightest be less than a quadrant, then there is one and only one perpendicular sect from the 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 oblique to CA and ^PCA = 2PAC. . \ %PAD acute, /.(by 532) PD1PAC >r't $PAB. .«. If PC A ad- jacent to obtuse ^PCF is FlG - 2I 5- acute. But it is also obtuse, being (by 494) PURE SPHERICS 2 37 = ifPAC. This is impossible, .-. BA not A_AP\ .'. (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-pole P and ^-radius PA, and .'. (by 541) a second point on this circle, which (by 536) is impossible. 544. Convention. In general a sum 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 ABy S43 )ADDB+BC= DC -AC. Fig. 218. (3) If CA > q } then in ABC all sides are less than quadrants. •*• (by 533) CB + BA + AC >CB + 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. PURE SPHERICS. 2 39 548. 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. If one angle of a spherical triangle be greater than a second, the side opposite the first must be greater than the side opposite the second ; and inversely. Given ?C> ?£. Proof. Take %DCB = ?£. Then(by495)ZX7 = L>J3. But (by 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 two symmetrical points to a third, that cutting the axis is the greater. Proof. BA=BD + DA = BD + DA'>BA'. Fig. Fig. 221, 552. Theorem. // two spherical triangles have two sides of the one equal to two sides of the other, but the included angles unequal, then that third side is the greater which is opposite the greater angle', and in- versely. 240 RATIONAL GEOMETRY. Fig. 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 498 and 497) ^Fisr't. .'.(by 495) DF is a quadrant. . \ (by 500) DA is a quadrant. /. (by 530) AC < quadrant. 554. Inverse of 553. 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 and BMD' (by 519) AD=BD'. /.if C be bi- section-point of AB, we have C , A+^L>=C , 5 + J B J D , =^. 556. Theorem. The end-points of any sect taken with any point on its perpendicular bisector give equal sects. 557. Corollary 556. Every point on the perpen- dicular bisector of a sect is pole of a circle through its end-points. 558. Corollary to 557. The perpendicular bisectors of the sides of a spherical triangle are copunctal (in its circumcenter) . 559. Corollary I to 555. The altitudes of a spherical triangle are copunctal (in its orthocenter. For, regarding A'B'C as the triangle, the perpen- dicular to DC at C is the polar of D, and .'. J_ to A'B'. Similarly, the perpendicular to BA' at A' is _L to B'C, etc. So the three altitudes of A'B'C are copunctal in the circumcenter of ABC. 560. Corollary II to 555. (Lexell). The vertices of spherical triangles of the same 242 RATIONAL GEOMETRY. angle-sum on the same base are on a circle copolar with the straightest bisecting their sides. For AO=BO, ^OAB =$OBA, ^.LAB = ^MBA = i[A+B + C]. Hence a A OB is fixed, and .'. OC [supplemental to OA]. 561. Theorem. The straight- ests through the corresponding vertices of a triangle and its polar are copunctal in the com- mon orthocenter. Proof. For AA' is J_ to BC and B'C\ since it passes through their poles. Fig. 224. Equivalence. 562. Theorem. Any angle made with a side of a spherical triangle by joining its end-point to the circumcenter, equals half the angle-sum less the opposite angle of the tri- angle. Proof. For ? A + ?B + ?C - 2 £ OCA + 2 ? OCB ± 2^.0AB. .'. %OCA = \&A + %B + $C] - [? OCB ± $OAB] = M $A + ? B + ?C] - ?£. 563. Corollary to 562. Symmetrical spherical tri- angles are equivalent or equivalent by completion. For the three pairs of isosceles triangles formed by joining the vertices to the' circumcenters, hav- Fig. 225. PURE SPHERICS. 243 Fig. 226. ing respectively a side and two adjoining angles congruent, are congruent. 564. Theorem. Of the triangles formed by three non-copunctal straightests, two containing vertical angles are together equivalent to that angle. To prove 2ABC + 2AB'C = ^.ABA'CA. Proof. B'C=BC, each being supplement of CB'. Again A C = A 'C (supple- ments of AC). Again AB' = A'B (supplements of AB). .-. (by 496) 2AB f C = 2BCA f . .-. 2 ABC + 2AB'C = a ABC+2BCA' ^ABA'CA. 565. The spherical excess, e, of a spherical triangle is the excess of the sum of its angles over two right angles. In general the spherical excess of a spherical poly- gon is the excess of the sum of its angles over twice as many right angles as it has sides less two. 566. Theorem. A spherical triangle is equivalent to half its spherical excess. Proof. Produce the sides of the a ABC until they meet again two and two at A', B f y C . The a ABC now appears in three angles, if A, ^.B, %.C. But (by 564) $A = 2aBC + 2AB'C. .-. ?A + ?£4 ?C FlGl 227 - =2 r't?s + 2^5C. 2AAJ3C = ?A + ?£+?C-2r't ?'s=e. 244 RATIONAL GEOMETRY. 567. Corollary I to 566. The sum of the angles of a a is > 2 r't f'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 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 straigh tests 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 0s 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 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. 245 H is the orthocenter of a ABC; and that the a formed by the centers of the circles is =to a ABC. Ex. 611. The feet of ±s from A of a ABC on the ex- ternal and internal bi's of ^ s B and C are co-st' with the bisection-points of b 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 ABCD. The incenters are vertices of an equian- gular quad'. Ex. 614. (Brahmegupta.) If the diagonals of a cyclic quad' are JL, the -L from their cross on one side bisects the opposite side. Ex. 615. If the diagonals of a cyclic quad' are 1 , the feet of the ±s 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. 618. If one of two equal chords of a O bisects he other, then each bisects the other. Ex. 619. The tri-rect angular a is its own 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 A and A' be opposites, then as ABC, A'BC are called colunar. A pole of the straightest bisecting A B and AC is also pole of the circum-O of the colunar a A'BC. Ex. 623. Given b and a + r— /? to construct q-pole and radius of circum-O- Ex. 624. If a+ /? = r, the q-pole of circum-o is bisection- point of c. 246 RATIONAL GEOMETRY. Ex. 625. Two as with one ^ the same and the opposite escribed Os=,have equal perimeters. Ex. 626 The tangent at A to the circum-O of a ABC 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-O of the polar *a ; and the spherical radii of the 2 O s are complementary. Ex. 628. From each / of a a a 1 is drawn to the straightest through the bisection-points of the adjacent sides. Prove these ±s copunctal. Ex. 629. Through each ^ of a a a straightest is drawn to make the same ^ with one side as the -L on the base makes with the other side. Prove these copunctal. Ex. 630. Two birect angular as are = if the oblique ^s are = , or if the sides not quadrants are = . Ex. 631. In a, if c is fixed and a+/9=w, 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. (1) A quad' whose diagonals bisect each other (a cenquad) has its opposite sides = ; (2) and in- versely. (3) Also its opposite ^s = ; (4) and inversely. (5) Every straightest through this bisection-point (spherical center) cuts the quad' into = halves. (6) Its opposite sides make = alternate ^s 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 Is on a pair of opposite sides are =. PURE SPHERICS. 247 (11) If two consecutive ^ s of a cenquad are =, it has a circum-o. (12) If two consecutive sides of a cenquad 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 coney clic. (16) If ABCD be a cenquad, then A, B, C, D' and A\ B' t C, D are on = Os with q-poles opposites. Ex. 634. The sides of a a intersect the corresponding sides of its polar on the polar of their 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 XVI. ANGLOIDS OR POLYHEDRAL ANGLES 570. Theorem. The area of a spherical angle, L, is 2r 2 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 : r 2 7i=u : \iz. .'. L = 2rht. 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, 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 tbc, . . . are faces, 248 ANGLOIDS OR POLYHEDRAL ANGLES. 249 and the pairs of consecutive faces are the dihedrals of the angloid. According to the number of the rays, 3, 4, 5, . . . the angloid is called trihedral, tetrahedral, 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 whose angles are of the same size as the inclinations of the angloid's dihedrals, while the length of each side of the polygon is the size of the corresponding 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 following parts congruent : (1) Two face-angles and the included dihedral. (2) Two dihedrals and the included face-angle. (3) Three face-angles. (4) Three dihedrals. (5) Two 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. II. As one of the face-angles of a trihedral is greater than or equal to a second, the dihedral oppo- site the first is greater than or equal to that opposite the second, and inversely. 250 RATIONAL GEOMETRY. III. Symmetrical trihedrals are equivalent or equivalent 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, defin'ng 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. XI. 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. ANGLOIDS OR POLYHEDRAL ANGLES, 251 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 a 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 ifs {e'-.^n). 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 are quadrants. Ex. 643. Cut a tetrahedral by a plane so that the sec- tion is a ||gm. 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 with A' 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 1', 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 r, and the angles a =20° 9' 30", /?=55° 53' 32", r=n4° 20' 14". Ans. o.i8i 3 r 2 . 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. 252 RATIONAL GEOMETRY. Ex. 651. Equivalent as on the same side of the same base are between copolar = Os. 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 1, that is, any angloid which determines on the unit sphere a spherical polygon whose area is 1, be called a steradian, and all the angloids about a point be together called a steregon, then a steregon contains 47: steradians. APPENDIX I. THE PROOFS OF THE TWO BETWEENNESS THEOREMS 16 AND 17, TAKEN FOR GRANTED IN THE TEXT.* 575. Theorem I. If B is between A and C, and C is between A and D, then C is between B and D. Proof. Let A, B, C, D be 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 y 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 t 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 5 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. 253 254 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, D be 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) Fig. 229. have a point between A and F. Thus we have the three non-co-straight points A, F, A and 6 with a point between A and F. There- fore b must have (by II 4) a point between A 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, A BCD, 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,or D is between A and B. If B is between A and D, we have fulfilled the hypothesis of Theorems I and IL APPENDIX I. 255 If D is between A and B, then interchanging the lettering for B and D, that is calling B, D and A 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 hypothesis of Theorems I and II, by calling D, A, and C, B, and A, C, and 5, D. If, however, A were between C and £> 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 A BCD on a. Through Ctake a straight c other than a. On c take a point £ other than C. On the straight DE between D and E take F. Thus between D and F is no point of c. FlG - 23 °- Then since by hypothesis C is between B and Z>, therefore c must (by II 4) have a point between 5 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 A, contrary to our hypothesis D between A and C, 256 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. ^4r# 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. Fig. 231. Proof. The points A BCD, 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 are between others. So we reach the follow- ing arrangements ACBD, BBC A, 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 67 ' 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 APPENDIX I. 257 C at least one point O . Further, there is within AC at least one point C", which likewise is within A B but not within OB ; therefore since C lies within OB, C cannot be identical with C. In this way liio. 232. we get ever new points of AB without ever coming to an end. 580. Theorem. If ABCD is an arrangement of four points corresponding to 577, then there is besides this arrangement only still the inverse which fulfills 577. [The proof is essentially already given in proving 578.] 581. Theorem. If any finite number of points of 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, E, . . . , K on the other, further C between A } 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 reversed arrangement, which is of the same character. [This theorem is a generalization of 577.] Fig. '■33- Proof. Our theorem holds for four points by 577 and 580, 258 RATIONAL GEOMETRY. We may show that the theorem remains valid for n + 1 points if it holds for n points. Let A 1 A 2 A 3 . . . A n be the desired arrangement for n points. If further we take an additional point then there are forthwith three cases possible : (1) A t lies between X and A n \ (2) A n lies between X and A 1 \ (3) X lies between A t and A n . 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+l . Finally we show that the following arrangements in the three cases have the desired properties: (1) XA,A 2 A 3 . ..A n ; (2) A^A^^AnX; (3) ^-1-^2^3 • • • A m XA m+l . . . A n \ 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 be 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 two new assumptions : Assumptions of the Compasses. Assumption VI 1 . If a straight have a point within a circle, it has two points on the circle. Assumption VI 2. If a circle have a point within and a point without another circle, it has two points on this other. 1 583. 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 1) cutting in 259 \ 260 RATIONAL GEOMETRY. D the OC(CA). Join DC, meeting OC(CB) in F. Then AF is tangent to OC(CB). Proof. Radius CA, _L 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 en 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. 235. 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 OG(c). Since a + c>b, these (by VI 2) intersect, say at K. a 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. 26 Case i. If ac. [^fC acute. J I. If c = p, the perpendicular from B on CA, there is only one triangle. II. If c>p, then A t and A f are on the same side of C and there are two different B triangles which fulfill the condi- >ions, namely, A'BC and A t BC (Fig. 238). This is called the ambiguous case. III. If c2I 6 Reckoning 272 Rectangle 42 Regions 7 Regular polygon 56 Rhombus 42 Richter 130 Right 221 angle 19 circular cone 205 circular cylinder 208 section of cylinder 208 triangle 37,75 Rotation 94 Ruler 63 Scalene 87 Schatunovsky 172 Schur 87 Secant 57 Second 251 Sect 6, 213 -calculus 87 -carrier 64 greater 73 Section 73 common 139 right 208 Sector 132 area of 132 Sect- train 10, 216 Segment, base of 201 of circle 201 of one base 201 of sphere 201 of straight 201 top of 202 volume of 201 Semicircle 77 Sense 217 Shanks 131 Side of angle 17 of plane 14 of point on straight. . . 9 of polygon 10 of straight 9 284 INDEX. PAGE Similar 98 polygons 105 Similarity 267 Simson 264 Size of angle 131 Skew 161 Slant height of cone 205 Small circle 197 Solution of problems .... 262 Sphere i94 area of 201 center of 194 circumscribed 198 diameter of 194 great circle of 1 94 inscribed 198 radius of 194 small circle of 197 tangent 195 volume of 201 Spherical angle 199 center 246 chord 234 excess 243 polygon 216 radius 234 segment 201 tangent 235 triangle 217 Spherics 212 Square 42 Steradian 252 Steregon 252 Straight 29 edge 63 Straights 1 Straightest 212 Successive substitutions.. 263 Sum of angles 94 of arcs 9 2 of sects 88, 237 Summits 163 Supplement. 27, 223 Supplemental sects 229 Surface 163 Symbols vii Symmetry 39, 266 Symmetrical points 39 Symtra 41 Talmud 130 Tanchord angle 60 PAGE Tangent 57, 235 circles 84 Tangent, plane 195 spheres 195 straight 57. 1 95 straightest 235 Tetrahedral 249 Tetrahedron 163 circumcenter of 198 edges of 163 faces of 163 summits of 163 volume of 167 Theorem 3 of Archimedes 209 of Bordage 245 of Brahmegupta. ... 72, 245 of Euler 164 of Heron 126 of Hippocrates of Chios. 135 of Joachimsthal 246 of Lexell 241 of Moore 253 of Pappus 135 of Pascal 87 of Ptolemy 105 of Pythagoras 112 Top of prismatoid 184 of segment 202 Trace 90 Translation 94, 265 Transversal 33 partition 115, 167 plane 167 Trapezoid 4 2 Triangle 11 altitudes of 46 angle bisectors of 70 angle of 18, 219 area of 115 base of 46 equilateral 123 isosceles 28, 224 medians of 40 polar 231 right 37.75 spherical 217 Triangles, congruent 19 similar 98 symmetric 222 Trihedral , , 249 INDEX. 285 PAGE Trisection-points 44 Truncated cone 207 cylinder 20S Truncated pyramid 207 tetrahedra 175 Unit 89, 122, 129, 189 circle 131 Vega 130 Veronese 168 Vertex of angle 17 Vertical angles 19 Vertices of polygon 10 Volume 166 of cone 205 of cuboid 189 PAGE Volume of cylinder 208 of polyhedron 184 of prism 189 of prismatoid 186 of pyramid 185 of sphere 201 of spherical segment. . . 201 of tetrahedron 167 of truncated cylinder. . 209 Within 6 the sphere 194 tetrahedron 163 Without 6 the sphere 1 94 tetrahedron 163 SHORT-TITLE CATALOGUE OF THE PUBLICATIONS OF JOHN WILEY & SONS, New York. 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(McMillan.) 8vo, 2 50 Sondericker's Graphic Statics, wuti Applications to Trusses, Beams, and Arches 8vo, 2 00 * Trautwine's Civil Engineer's Pocket-book i6mo, morocco, 5 00 Wait's Engineering and Architectural Jurisprudence * 8vo, 6 00 Sheep, 6 50 Law of Operations Preliminary to Construction in Engineering and Archi- tecture. 8vo, 5 00 Sheep, 5 50 Law of Contracts 8vo, 3 00 Warren's Stereotomy — Problems in Stone-cutting 8vo, 2 50 Webb's Problems in the U?e and Adjustment of Engineering Instruments. i6mo, morocco, 1 25 * Wheeler's Elementary Course of Civil Engineering 8vo, 4 00 Wilson's Topographic Surveying 8vo, 3 So BRIDGES AND ROOFS. Boiler's Practical Treatise on the Construction of Iron Highway Bridges . . 8vo, 2 00 * Thames River Bridge 4to, paper, 5 00 Burr's Course on the Stresses in Bridges and Roof Trusses, Arched Ribs, and Suspension Bridges 8vo, 3 5<> Du Bois's Mechanics of Engineering. VoL II Small 4to, 10 00 Foster's Treatise on Wooden Trestle Bridges 4to, 5 00 Fowler's Coffer-dam Process for Piers 8vo, 2 50 Greene's Roof Trusses .". 8vo, 1 25 Bridge Trusses 8vo, 2 so Arches in Wood, Iron, and Stone 8vo, 2 50 Howe's Treatise on Arches 8vo, 4 00 Design of Simple Roof-trusses in Wood and Steel 8vo, 2 00 J#hnson, Bryan, and Turneaure's Theory and Practice in the Designing of Modern Framed Structures Small 4to, 10 00 Merriman and Jacoby's Text-book on Roofs and Bridges: Parti. — Stresses in Simple Trusses 8vo, 2 -50 Part n. — Graphic Statics 8vo, 2 50 Part m.— Bridge Design. 4th Edition, Rewritten 8vo, 2 50 Part IV.— Higher Structures 8vo, 2 50 Morison's Memphis Bridge 4to, 10 00 Waddell's De Pontibus, a Pocket-book for Bridge Engineers. . . i6mo, morocco, 3 00 Specifications for Steel Bridges i2mo, 1 25 Wood's Treatise on the Theory of the Construction of Bridges and Roofs . 8vo, 2 00 Wright's Designing of Draw-spans: Part L —Plate-girder Draws 8vo, 2 50 Part II.— Riveted-truss and Pin-connected Long-span Draws 8vo, 2 50 Two parts in one volume **vo, 3 50 HYDRAULICS. Bazin's Experiments upon the Contraction of the Liquid Vein Issuing from an Orifice. (Trautwine.) 8vo, Bovey's Treatise on Hydraulics 8vo, Church's Mechanics of Engineering 8vo, Diagrams of Mean Velocity of Water in Open Channels paper, Coffin's Graphical Solution of Hydraulic Problems i6mo, morocco, Flather's Dynamometers, and the Measurement of Power nmo, FolwelTs Water-supply Engineering 8vo, Frizell's Water-power 8vo, Fuertes's Water and Public Health nmo, Water-filtration Works nmo, Ganguillet and Kutter's General Formula for the Uniform Flow of Water in Rivers and Other Channels. (Hering and Trautwine.) 8vo, Hazen's Filtration of Public Water-supply 8vo, Hazlehurst's Towers and Tanks for Water- works 8vo, Herschel's 115 Experiments on the Carrying Capacity of Large, Riveted, Metal Conduits 8vo, Mason's Water-supply. (Considered Principally from a Sanitary Stand- point.) 3d Edition, Rewritten 8vo, Merriman's Treatise on Hydraulics. 9th Edition, Rewritten 8vo, * Michie's Elements of Analytical Mechanics 8vo, Schuyler's Reservoirs for Irrigation, Water-power, and Domestic Water- supply Large 8vo, •* Thomas and Watt's Improvement of Riyers. (Post., 44 c. additional), 4to, Turneaure and Russell's Public Water-supplies 8vo, Wegmann's Desien and Construction of Dams 4to, Water-supply of the City of New York from 1658 to'1895 4to, Weisbach's Hydraulics and Hydraulic Motors. (Du Bois.) 8vo, Wilson's Manual of Irrigation Engineering ,Small 8vo. Wolff's Windmill as a Prime Mover 8vo, Wood's Turbines 8vo, Elements of Analytical Mechanics 8vo, MATERIALS OF ENGINEERING. Baker's Treatise on Masonry Construction 8vo, Roads and Pavements 8vo, Black's United States Public Works Oblong 4to, Bovey's Strength of Materials and Theory of Structures 8vo, Burr's Elasticity and Resistance of the Materials of Engineering. 6th Edi- tion, Rewritten 8vo, Byrne's Highway Construction 8vo, Inspection of the Materials and Workmanship Employed in Construction. i6mo, Church's Mechanics of Engineering 8vo, Du Bois's Mechanics of Engineering. Vol. I Small 4to, Johnson's Materials of Construction Large 8vo, Keep's Cast Iron 8vo, Lanza's Applied Mechanics 8vo, Martens's Handbook on Testing Materials. (Henning.) 2 vols 8vo, Merrill's Stones for Building and Decoration 8vo, Merriman's Text-book on the Mechanics of Materials 8vo, Strength of Materials nmo, Metcalf's Steel. A Manual for Steel-users nmo, Patton's Practical Treatise on Foundations 8vo, 7 2 OO 5 OO 6 OO 1 50 2 50 3 OO 4 OO 5 OO 1 50 2 50 4 OO 3 OO 2 50 4 00 5 00 4 OO 5 OO 6 OO 5 OO 5 OO to OO 5 OO 4 OO 3 OO 2 50 3 OO 5 00 5 00 5 OO 7 50 7 50 5 OO 3 OO 6 OO 7 50 6 00 2 50 7 50 7 50 5 OO 4 OO I OO 2 OO 5 00 T »5 I oo 3 50 2 00 2 00 8 00 2 00 3 50 2 50 5 00 4 00 3 00 I 25 2 00 3 00 4 00 Rockwell's Roads and Pavements in France i2mo, Smith's Materials of Machines i2mo, Snow's Principal Species of Wood 8vo, Spalding's Hydraulic Cement i2mo, Text-book on Roads and Pavements i2mo, Thurston's Materials of Engineering. 3 Parts 8vo, art I. — Non-metallic Materials of Engineering and Metallurgy 8vo, Part II. — Iron and Steel 8vo, Part III. — A Treatise on Brasses, Bronzes, and Other Alloys and their Constituents 8vo, Thurston's Text-book of the Materials of Construction 8vo, Tillson's Street Pavements and Paving Materials ; 8vo, Waddell's De Pontibus. (A Pocket-book for Bridge Engineers.) . . i6mo, mor. , Specifications for Steel Bridges i2mo, Wood's Treatise on the Resistance of Materials, and an Appendix on the Pres- ervation of Timber 8vo, Elements of Analytical Mechanics.. 8vo, Wood's Rustless Coatings: Corrosion and Electrolysis of Iron and Steel. . .8vo, RAILWAY ENGINEERING. Andrews's Handbook for Street Railway Engineers. 3X5 inches, morocco, 1 25 Berg's Buildings and Structures of American Railroads 4to, 5 00 Brooks's Handbook of Street Railroad Location i6mo. morocco, 1 50 Butts's Civil Engineer's Field-book i6mo, morocco, 2 50 Crandall's Transition Curve i6mo, morocco, 1 50 Railway and Other Earthwork Tables 8vo, 1 50 Dawson's "Engineering" and Electric Traction Pocket-book. i6mo, morocco, 5 00 Dredge's History of the Pennsylvania Railroad: (1879) Paper, 5 00 * Drinker's Tunneling, Explosive Compounds, and Rock Drills, 4to, half mor., 25 00 Fisher's Table of Cubic Yards Cardboard, 25 Godwin's Railroad Engineers* Field-book and Explorers' Guide i6mo, mor., 2 50 Howard's Transition Curve Field-book i6mo, morocco. 1 so Hudson's Tables for Calculating the Cubic Contents of Excavations and Em- bankments 8vo, 1 00 Molitor and Beard's Manual for Resident Engineers i6mo, 1 00 Nagle's Field Manual for Railroad Engineers i6mo morocco. 3 00 Philbrick's Field Manual for Engineers i6mo, morocco, 3 00 Searles s Field Engineering i6mo, morocco, 3 00 Railroad Spiral i6mo, morocco, 1 50 Taylor's Prismoidal Formulae and Earthwork 8vo, 1 50 * Trautwine's Method of Calculating the Cubic Contents of Excavations and Embankments by the Aid of Diagrams 8vo, 2 00 The Field Practice of [Laying Out Circular Curves for Railroads. i2mo, morocco, 2 50 Cross-section Sheet Paper, 25 Webb's Railroad Construction. 2d Edition, Rewritten i6roo. morocco, 5 00 Wellington's Economic Theory of the Location of Railways Small 8vo, 5 00 DRAWING. Barr's Kinematics of Machinery 8vo, 2 50 * Bartlett's Mechanical Drawing 8vo, 3 00 * " ' " Abridged Ed 8vo, 1 50 Coolidge's Manual of Drawing 8vo, paper, 1 00 Coolidge and Freeman's Elements of General Drafting for Mechanical Engi- neers. (In press.) Durley's Kinematics of Machines 8vo, 4 00 8 Hill's Text-book on Shades and Shadows, and Perspective 8vo, 2 00 Jamison's Elements of Mechanical Drawing. (7n press.) Jones's Machine Design: Part I. — Kinematics of Machinery 8vo, 1 50 Part n. — Form, Strength, and Proportions of Parts 8vo, 3 00 MacCord's Elements of Descriptive Geometry . , , 8vo, 3 00 Kinematics; or, Practical Mechanism , 8vo, 5 00 Mechanical Drawing , . . . , 4to, 4 00 Velocity Diagrams 8vo, 1 50 * Mahan's Descriptive Geometry and Stone-cutting , 8vo, 1 50 Industrial Drawing. (Thompson.) 8vo, 3 5© Reed's Topographical Drawing and Sketching 4to, 5 00 Reid's Course in Mechanical Drawing 8vo, 2 00 Text-book of Mechanical Drawing and Elementary Machine Design . . 8vo, 3 00 Robinson's Principles of Mechanism 8vo, 3 00 Smith's Manual of Topographical Drawing. (McMillan.) 8vo, 2 50 Warren's Elements of Plane and Solid Free-hand Geometrical Drawing. . i2mo, 1 00 Drafting Instruments and Operations i2mo, 1 25 Manual of Elementary Projection Drawing i2mo, 1 50 Manual of Elementary Problems in the Linear Perspective of Form and 5 Shadow i2mo, 1 00 Plane Problems in Elementary Geometry i2mo, 1 25 Primary Geometry i2mo, 75 Elements of Descriptive Geometry, Shadows, and Perspective 8vo, 3 50 General Problems of Shades and Shadows 8vo, 3 00 Elements of Machine Construction and Drawing 8vo, 7 50 Problems. Theorems, and Examples in Descriptive Geometrv 8vo, 2 50 Weisbach's Kinematics and the Power of Transmission. (Hermann and Klein.) 8vo, 5 00 Whelpley's Practical Instruction in the Art of Letter Engraving i2mo, 2 00 Wilson's Topographic Surveying 8vo, 3 50 Free-hand Perspective 8vo, 2 50 Free-hand Lettering 8vo. 1 00 Woolf's Elementary Course in Descriptive Geometry Large 8vo, 3 00 ELECTRICITY AND PHYSICS. Anthony and Brackett's Text-book of Physics. (Magie.) ... .Small 8vo, 3 °° Anthony's Lecture-notes on the Theory of Electrical Measurements i2mo, 1 00 Benjamin's History of Electricity 8vo. 3 00 Voltaic Cell. 8vo, 3 00 Classen's Quantitative Chemical Analysis by Electrolysis. (Boltwood.). ,8vo, 3 0° Crehore and Squier's Polarizing Photo-chronograph 8vo, 3 00 Dawson's "Engineering" and Electric Traction Pocket-book. . i6mo, morocco, 5 00 Dolezalek's Theory of the Lead Accumulator (Storage Battery). (Von Ende.) i2mo," 7 2 50 Duhem's Thermodynamics and Chemistry. (Burgess.) 8vo, 4 00 Flather's Dvnamo meters, and the Measurement of Power i2mo, 3 00 Gilbert's De Magnete. (Mottelay.) 8vo, 2 50 Hanchett's Alternating Currents Explained i2mo, 1 00 Hering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 2 50 Holman's Precision of Measurements 8vo, 2 00 Telescopic Mirror-scale Method, Adjustments, and Tests.. .. .Large 8vo, 75 Landauer's Spectrum Analysis. (Tingle.) 8vo, 3 00 Le ChateUer's High-temperature Measurements. (Boudouard — .Burgess. )i2mo, 3 00 Lob's Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz.) i2mo. 1 00 • Lyons's Treatise on Electromagnetic Phenomena. Vols. I. and II. 8vo, each, 6 00 * Michie. Elements of Wave Motion Relating to Sound and Light 8vo, 4 00 9 Niaudet's Elementary Treatise on Electric Batteries. (FishDack. ) lomo, 2 50 • Rosenberg's Electrical Engineering. (Haldane Gee — Kinzbrunner.). . . .8vo, 1 50 Ryan, Norris, and Hoxie's Electrical Machinery. VoL L 8vo, 2 50 Thurston's Stationary Steam-engines 8vo, 2 50 * Tillman's Elementary Lessons in Heat . 8vo, 1 30 Tory and Pitcher's Manual of Laboratory Physics Small 8vo, 2 00 Ulke's Modern Electrolytic Copper Refining 8vo, 3 00 LAW. * Davis's Elements of Law 8vo, 2 50 * Treatise on the Military Law of United States 8vo, 7 00 * Sheep, 7 50 Manual for Courts-martial i6mo, morocco, 1 50 Wait's Engineering and Architectural Jurisprudence 8vo, 6 00 Sheep,' 6 50 Law of Operations Preliminary to Construction in Engineering and Archi- tecture 8vo, 5 00 Sheep, s 5o Law of Contracts 8vo, 3 00 Winthrop's Abridgment of Military Law nmo, 2 50 MANUFACTURES. Bernadou's Smokeless Powder — Nitro-cellulose and Theory of the Cellulose Molecule i2mo, 2 50 Bolland's Iron Founder i2mo, 2 50 ** The Iron Founder," Supplement i2mo, 2 50 Encyclopedia of Founding and Dictionary of Foundry Terms Used in the Practice of Moulding i2mo, 3 00 Eissler's Modern High Explosives 8vo, 4 00 Effront's Enzymes and their Applications. (Prescott.) 8vo, 3 00 Fitzgerald's Boston Machinist i8mo, 1 00 Ford's Boiler Making for Boiler Makers i8mo, 1 00 Hopkins's Oil-chemists* Handbook 8vo, 3 00 Keep's Cast Iron 8vo, 2 50 Leach's The Inspection and Analysis of Food with Special Reference to State Control. (In preparation.) Metcalf's Steel. A Manual for Steel-users i2mo, 2 00 Metcalfe's Cost of Manufactures — And the Administration of Workshops, Public and Private 8vo, 5 00 Meyer's Modern Locomotive Construction 4to, 10 00 Morse's Calculations used in Cane-sugar Factories i6mo, morocco, 1 50 * Reisig's Guide to Piece-dyeing 8vo, 25 00 Smith's Press-working of Metals 8vo, 3 00 Spalding's Hydraulic Cement i2mo, 2 00 Spencer's Handbook for Chemists of Beet-sugar Houses i6mo, morocco, 3 00 HandbooK tor sugar Manutacturers and their Chemists.. ,i6mo, morocco, 2 00 Thurston's Manual of Steam-boilers, their Designs, Construction and Opera- tion 8vo, 5 00 * Walke's Lectures on Explosives 8vo, 4 00 West's American Foundry Practice i2mo, 2 50 Moulder's Text-book i2mo, 2 50 Wiechmann's Sugar Analysis Small 8vo, 2 50 Wolff's Windmill as a Prime Mover 8vo, 3 00 Woodbury's Fire Protection of Mills 8vo, 2 50 Wood's Rustless Coatings: Corrosion and Electrolysis of Iron and Steel. . .8vo, 4 00 10 MATHEMATICS. Baker's Elliptic Functions 8vo, i 50 * Bass's Elements of Differential Calculus i2mo, 4 00 Briggs's Elements of Plane Analytic Geometry * 2mo v x °° Compton's Manual of Logarithmic Computations i2mo, 1 50 Davis's Introduction to the Logic of Algebra 8vo, 1 50 * Dickson's College Algebra Large i2mo, 1 50 * Answers to Dickson's College Algebra 8vo, paper, 25 * Introduction to the Theory of Algebraic Equations Large i2mo, 1 25 Halsted's Elements of Geometry 8vo, 1 75 Elementary Synthetic Geometry 8vo, 1 50 Rational Geometry nmo, x 75 •Johnson's Three-place Logarithmic Tables: Vest-pocket size paper, 15 100 copies for 5 00 * Mounted on heavy cardboard, 8 X 10 inches, 25 10 copies for 2 00 Elementary Treatise on the Integral Calculus Small 8vo, 1 50 Curve Tracing in Cartesian Co-ordinates i2mo, 1 00 Treatise on Ordinary and Partial Differential Equations Small 8vo, 3 50 Theory -of Errors and the Method of Least Squares i2mo, 1 50 * Theoretical Mechanics i2mo, 3 00 Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.) i2mo, 200 * Ludlow and Bass. Elements of Trigonometry and Logarithmic and Other Tables 8vo, 3 00 Trigonometry and Tables published separately Each, 2 00 * Ludlow's Logarithmic and Trigonometric Tables 8vo, 1 00 Maurer's Technical Mechanics 8vo, 4 00 Merriman and Woodward's Higher Mathematics 8vo, 5 00 Merriman's Method of Least Squares $vo, 2 00 Rice and Johnson's Elementary Treatise on the Differential Calculus . Sm., 8vo, 3 00 Differential and Integral Calculus. 2 vols, in one Gmall 8vo, 2 50 Sabin's Industrial and Artistic Technology of Paints and Varnish. (7n press.) Wood's Elements of Co-ordinate Geometry 8vo, 2 00 Trigonometry: Analytical, Plane, and Spherical nmo, x 00 MECHANICAL ENGINEERING. MATERIALS OF ENGINEERING, STEAM-ENGINES AND BOILERS. Baldwin's Steam Heating for Buildings nmo, 2 50 Barr's Kinematics of Machinery 8vo, 2 50 * Bartlett's Mechanical Drawing 8vo, 3 00 * " " " Abridged Ed 8vo. 1 50 Benjamin's Wrinkles and Recipes i2mo, 2 00 Carpenter's Experimental Engineering 8vo, 6 00 Heating and Ventilating Buildings 8vo, 4 00 Cary's Smoke Suppression in Plants using Bituminous CoaL {In prep- aration.) Clerk's Gas and Oil Engine Small 8vo, 4 00 Coolidge's Manual of Drawing 8vo, paper, 1 00 Coolidge and Freeman's Elements of General Drafting for Mechanical En- gineers. (In press.) Cromwell's Treatise on Toothed Gearing i2mo, 1 50 Treatise on Belts and Pulleys i2mo, 1 50 Durley's Kinematics of Machines 8vo, 4 00 Flather's Dynamometers and the Measurement of Power nmo, 3 00 Rope Driving l2mo, 2 00 U Gill's Gas and Fuel Analysis for Engineers « . i2mo, i Hall's Car Lubrication i2mo, i Hering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 2 Hutton's The Gas Engine 8vo, 5 Jones's Machine Design: Part I. — Kinematics of Machinery 8vo, 1 Part IL-^-Form, Strength, and Proportions of Parts 8vo, 3 Kent's Mechanical Engineer's Pocket-book i6mo, morocco, 5 Kerr's Power and Power Transmission 8vo, 2 MacCord's Kinematics; or, Practical Mechanism 8vo, 5 Mechanical Drawing 4to, 4 Velocity Diagrams 8vo, 1 Mahan's Industrial Drawing. (Thompson.) .8vo, 3 Poole's Calorific Power of Fuels 8vo, 3 Reid's Course in Mechanical Drawing „ 8vo. 2 Text-book of Mechanical Drawing and Elementary Machine Design. .8vo, 3 Richards's Compressed Air nrao, 1 Robinson's Principles of Mechanism 8vo, 3 Smith's Press-working of Metals 8vo t 3 Thurston's Treatise on Friction and Lost Work in Machinery and Mill Work * •• 8vo , 3 Animal as a Machine and Prime Motor, and the Laws of Energetics . nmo, 1 Warren's Elements of Machine Construction and Drawing 8