BOOK vi. PROP, ii. iv.] MATHEMATICS. PLANE GEOMETRY. 



585 



SI. them, each equal to B C ;* andKL, etc., any 

 number of them, each equal to D K ; and draw A G, 

 A H, drc., and A L, (fee. ; then the triangles A B C, A G B, 



+ 38 i. A H G, (fee., are all equal ;t and if the base of 

 the triangle which is the sum of all these be m times 

 B C, the triangle itself will be m times the triangle 



38 I. ABC.* In like manner if D L be n times D K, 

 the triangle A D L will be n times AD K. 



base B C : base D K : : triangle ABC: triangle A D K, 



m. BC: n. DK:: m. ABC : . ADK 

 Hence if m . B C = n. D K, then m. A B C = n. A D K ; 

 if m. B C > n. D K, then m. A B C > . A D K ; and if 

 m. B C = n. D K, then m. A B C = . A D K, and m, n. 

 are any whole numbers whatever, . . (Def. A., page 573). 



base B C : base D K : : triangle ABC: triangle ADK. 

 And because the parallelogram E C is double of the tri- 

 angle ABC, and the parallelogram F D double of the 

 triangle ADK, . ' . triangle ABC: triangle ADK:: 



+ s v. parallelogram E C : parallelogram F D ;t and, 

 comparing this proportion with the last, base B C : base 



2 v. D K : : parallelogram E C : parallelogram F.D,* 

 . . triangles and parallelograms, (fee. Q. E. D. 



COB. If triangles or parallelograms have equal alti- 

 tudes, they are to one another as their bases. 



For if the figures be go placed as to have their bases 

 in the same straight line, and perpendiculars be drawn 

 from the vertices to the bases, the straight line joining 

 the vertices will be parallel to that in which are the 



2J I. bases,* since the perpendiculars are both equal 



+ 28 1. and parallel ;f and . . if the above construc- 

 tion be made, the demonstration will be the same. 



PROPOSITION H. THF.OREM. 



If a strait/hi line (D E) If drawn parallel to one of the 

 fides (B C) of a triangle, it shall cut the other sides, or 

 these produced, proportionally ; and if the tides, or the 

 tides produced, be cut proportionally, the straight line 

 which joins the points of section shall be parallel to the 

 remaining side of the triangle. 



For draw B E, C D. Then the triangle B D E = C D E, 



because they are on the same base D E, and are between 



37 i. the same parallels D E, B C ;* and A D E is 

 another triangle, . . BDE:ADE::CDE:ADE. 



ti vi. But BDE :ADE::BD:DA.f For a 

 similar reason CDE : ADE :: CE :EA, 



2V. /.BD :DA :.CE :EA.* 



Again, let the sides A B, AC, or these produced, be 

 cut proportionally in the points D, E ; that is, so that 

 BD:DA::CE:EA;and draw D E ; DE shall be 

 parallel to BC. 



The same construction being made, because 



BD :DA : : CE : EA, and B D : 



+ ivi. DA :: BDE : ADE.f 



and CDE : ADE :: CE: 



i vi. E A.* 



ADE ::CDE : 

 ADE. 



But as the consequents are equal the antecedents are 



*ov. Cor. i. equal, t triangle BDE = triangle 



CDE; and they are on the same base D E, and are 



between the same parallels, .'. D E is parallel to BC ; 



.'. if a straight line, &o. Q. E. D. 



PROPOSITION III. THEOREM. 



If an angle (BAG) of a triangle, be divided into two 



equal angles, by a straight Itne which cuts the opposite 



bate, the segments of the bane (B D, D C; shall be to 



each other as the remaining sides of the triangle ; and if 



VOL. j. 



the segment* of the base are to each other as the remain- 

 ing sides of the triangle, the straight line drawn from 

 the vertex to the point of section, shall divide the vertical 

 angle into two equal angles. 



311. 





Draw C E parallel to D A ;* and let B A 

 produced meet C E in E. 

 Because A C meets the 

 parallels AD, EC, the 

 angle ACE = C A D ;t 

 + 29 I. but by hypothesis, 

 CAD= BAD; .-.BAD 

 = A C E. Again, because 

 B E meets the parallels 

 A D, E C, the angle BAD 

 A E C ;* but it was proved that A C E = 

 A C E = A E C, and . -. A E = A C ; and 







29 I. 



BAD, 



since A D is parallel to E C, a side of the triangle B C E ; 

 + 2 vi. .'. B D : D C : : B A : A E ;f but A E = A C, 



/. B D : D C : : B A : A C. 



Next let B D : D C : : B A : A C ; and draw AD; the 

 angle BAG shall be divided into two equal angles 

 by A D. 



The same construction being made, A D IB parallel to 



Const. E C,* and because 



B D : D C : : B A : A C, and \ ,'. (2 V.) B A : A C 

 + avi. BD : DC :: BA : ::BA : AE, .'.AC 

 9V. Cor. AE,t ) =AE,* 



and consequently the angle A E C = A C E ; but AEG 

 + 291. = BAD.tand ACE=CAI),* .'. B A 1) = 



29 I. C A 1), that is, the angle A C is divided into 

 two equal angles by A D ; /. if an angle, &c. Q. E. D. 



PROPOSITION A. THEOREM. 



If the outward angle (C A E) of a triangle (A B CT) made 

 by producing one of its sides, be divided in two equal 

 angles, by a straight line (A D) which cuts, the base pro- 

 duced, the segments (BD, DC) between the dividing 

 line and the extremities of the base, are to each other as 

 the remaining sides of the triangle and if the segments 

 of the base produced are to each otjicx. as the remaining 

 tides of the triangle, the straight, U/ne drawn from the 

 vertex to the point of section, divides tha- outward angle 

 of the triangle into two equal angles. 



31 L Through C draw C F parallel to A ID.* Then 



because A C meeta the paral- 

 lels AD, FC, the angle ACF 

 + 291. =CAD;f but (hyp.) 

 CAD = DAE, /. DAE = 

 ACF. Again, because F E 

 meets the parallels AD, F C, 

 the angle DAE = CFA; 

 29l. but as just proved, 

 DAE = ACF .'. CFA = 



+ 61. ACF; .'. AC=AF.f And because A D is 

 parallel to F C, a side of the triangle B C F, .'. B D : 

 D C : : B A : A F ; but A C - A F, .-. 



BD:DC::BA:AC. 



Next, let B D : DC : : B A : A C ; and draw A D ; 

 the angle CAD shall be equal to D A E. 



The same construction being made, because 

 B D : D C : : B A : A C, and ) . . (2 V.) BA : AC : : BA 

 2 vi. BD:DC::BA: AF,* I : AF, .-. AC = AF, 

 .-. the angle AFC = A C F ;t but AFC = 

 BAD, and AFC CAD,' .-. EAD = 

 CAD; .'.if the outward angle, &c. Q. E. D. 



PROPOSITION IV. -THEOREM. 



The sides about the equal angles of equiangular triangles 

 (A B C, D C E) are proportiotials ; and thofe which are 

 opposite to the equal angles are homologous sides ; that 

 ts, are the antecedents, or the consequents of the ratio. 



Let the angle ABC=DCE;ACB = DEC; and 

 consequently H A C = C D E ; and let the triangle 1) C E 

 be placed so that its side C E may bo contiguous to B C, 



4 r 



