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MATHEMATIC& PLANE GEOMETRY. [BOOK vi. FBOP. xvi. xx 



PROPOSITION XVI. THEOREM. 

 If four straight lines (AB, CD, E, Y) an proportionals, the 

 reetangU contained by the extreme* (AB, F) i* equal to 

 that contained by On mean* (CD, E): ami if the rtctwjtt 

 contained by the extreme* be equal to that contained by 

 the meant, the four stniiyht line* art proportional*. 

 From A, C draw A G, C H, perpendiculars to A B, 

 ill. CD;* make A G - F, and C H - E, and 

 4 ii i. complete the parallelograms B G, D H.t It 

 is to be proved that these rectangles are equal. 



Because A B : C D : : C H : A G, /.the sides of 

 the parallelograms about the equal angles A, C, are 



reciprocally proportional ; they 



are therefore equal to one 

 14 vi. another ;* and B G is 

 tho rectangle contained by A B, 

 F, and DH the rectangle con- 

 tained by CD, E; .'. AB'F = 

 CD-E. And if ABF=CDE, 

 then shall A B : C 1) : : E : F. 

 ' The same construction being 

 made, the rectangle B G = D H ; and as they are equi- 

 + 14 vi. angular also, .-. AB : CD : : CH : A G ;t 

 that isAB:CD::E:F; .'. if four straight lines, 

 &c. Q. E. D. 



NOTE. The next proposition is merely a corollary to this : it ii that 

 particular cue of it in which the meant are equal. 



PROPOSITION XVII. THEOREM. 



If three itraiqht lines be proportionals, the rectangle con- 

 tained by the extremes is equal to tlie square of the mean, 

 and if 0>e rectangle contained by tlie extreme* i* equal to 

 the square of the mean, the three straight lines are 

 proportionals. 

 COR. The first of the three proportionals (A, B, C) is 



to the third, as the square of the first to the square of 



the second. 

 . i vi For* A : C : : rectangle A'A : rectangle A'C ; 



but A-C = B 2 , /. A : C : : A a : B. 



PROPOSITION XVIIL PROBLEM. 



Upon a given straight line (A B) to describe a rectilineal 

 figure similar, and similarly situated, to a given recti- 

 lineal figure. 

 First, let the given rectilineal figure, C, D, E, F have 



four sides : it is required to describe on A B a figure 



similar, and similarly situated to C, D, E, F. 



Draw D F; and at the points A, B, in A B, make the 



angle A = C, and ABG = CDF; then AGB = CFD;t 

 f 35 1 . ' . the triangles F C D, GAB, are equiangular. 



At the points G,B in GB, make the angle BGH = DFE, 



MI. andGBH = FDE; thenGHB = FED;*.-, 

 the triangles F D E, G B H are equiangular. 



Because the angle 

 AGB = CFD, and 

 BGH = DFE;.-. 

 AGH=CFE. For 

 like reasons A B H 

 = CDE, .'. .since 

 A = C, and G H B 

 = ED, the figures 

 A B H G, C D F, F, are equiangular. Moreover these figures 

 have their sides about the equal angles proportionals ; 

 GAB, FCD being equiangular triangles, .-.BA : AG 

 * 4 vi. : : D C : C F ; and A G : G B : : C F : F D :t 

 also by reason of the equiangular triangles B G H, D F E, 

 GB:GH::FD:FE; .-.AG:GH::CF:FE. 

 In like manner it may be proved that A B : B H : : C D : 

 DE, andGH : HB :: FE : ED: .'. the equiangular 

 figures AB H G, CD EF hate their sides about the equal 



M. l VI. angle* proportionals, . . they are similar.* 

 Next, let the given rectilinear figure CDKEF have 



five sides. Draw D E ; and upon A B describe the figure 

 A BH O, similar and similarly situated, to the quadri- 

 lateral C D E F, by the former case : and at the points 

 B, II, in B II, make the angle H BL - EDK, and 

 MI. BHL-DEK, thcnL- K.' 

 Becatwe the figures A B H G, C D E F arc similar, tho 

 ngU GHB-FED;andBHLwas made = D K K ; 

 .'.GHL-FEK. Forlike reason* A BL- CDK; 



. . the figitre* ABLHG, CDKEF, are equiari'in'ar. 

 And because the figures A BHG, CD K F ;n, similar, 

 .-. GH :HB ::FE :ED;LutHB :HL ::ED : EK, 

 .-. GH: HL : : FK : K K. For like reasons AB : 

 BL::CD : D K, andBL:LH ::DK :KE; .-. tht 

 t'/uiangular figures ABLHG, CDKEF have their 

 suits about the equal angles proportionali, .'. tltey are 

 similar. And in like manner may a rectilineal figure of 

 six sides, similar to a given one, be described upon A B ; 

 and so on. Which was to be done. 



PROPOSITION XIX. THEOREM. 



Similar triangles (A B C, D E F) are to one another, as 

 the squares of their homologous sides. 



The triangles being similar, having the angle B = E, 

 Def. 9 v. and the side B C homologous to E F,* that is 

 such that A B : B C : : D K : E K, it is to be proved that 

 tri-A B C : tri'D K F : : B C 2 : E F 2 - 

 + n vi. Take B G, a third proportional to B C, E F ;f 

 : : E F : B G ; and draw G A, 



GA. 



. 1)E :EF, alternately,* 

 K F, but B C : K K : : K I 

 AB:DE::EF:BG;t 

 A 



so that B C : E F 



Then because A B : B C 



n v. A B : 1) E : : B C 



t2V - :BG (Const),.-, 

 but triangles which have the 

 sides about two equal angles 

 reciprocally proportional .ire 



is Vi. equal,* .'. ABG 

 = D E F. And because 



B C : E F : : K F : B G, . . ^ 



B C : B G : : B C 1 : E F 2 ;t B 



t IT vi. Cor. but BC: BG :: triangle ABC : tri;in,'l'' 



* l VI - ABG;*.'. triangle ABC: triangle Am: 

 BC 2: EF 2 , but ABG = DEF, .-. triangle ABC 



similar triangles, &c. 



triangle D E F : : B C* : E F ; .- 

 Q. E. D. 



COR. If three straight lines be proportionals, then as 

 the first is to the third, so is any triangle upon the first, 

 to a similar and similarly situated triangle upon the 

 second. 



PROPOSITION XX. THEOREM. 



Similar polygons may be divided into the same number of 

 similar triangles, which are to one another as the poly- 

 gons themselves : and the polygons are to one another as 

 the squares of their homologous sides. 

 Let ABODE, FGHKLbe similar polygons, and 

 let A B be the side homologous to F G. 



From E, L draw the diagonals. Because the polygons 

 are similar, the angle B A E = G F L, and B A : A K : : 

 G F : F L ; . . the triangles ABE, F G L have an angle 

 in one equal to an angle in the other, and the sides about 

 those equal angles proportionals, .-. the triangles are 

 6 & 4 VI. equiangular and similar,* . -. the angle A B 1C 

 = F G L : but since the polygons are similar, the whole 

 angle A B C = the whole angle F GH, .-. the remaining 

 angle EBC = LGH. 

 Also E B : B A : : 

 L G : G F, and from 

 the similar polygons, 

 AB: BC :: FG : 

 GH; .-. E B : BO 

 ::LG :GH,t that 

 i: v. Cor. is, the sides 

 about tho equal angles E B C, L G H are proportionals ; 

 .6 VI. .'.the triangles EEC, LGH are similar.* 

 Forlike reasons the triangle BCD, LH K are similar, 

 . . thr. polygons are divided into the same number of sim i/n r 

 triiimjles. It is now to be proved that those triangles are, 

 each to each, as the polygons themselves, and that these 

 are to each other as tho squares of their like sides. 

 Because the triangles A B K, F G L are similar, as 

 .19 vi. also BEC, OLH, .-. 



triangle ABE: triangle F GL : : B E : O Ir*l . , ,, K 

 triangle BEC: triangle G L H : : B E 2 : G L 2 / ' 

 uv. :FGL ::BEC :GLH.t 



In like manner, because the similar triangles I! K C, 

 G L H, as also the similar triangles E C D, L H K, are an 

 EC 2 to LH, .'. It F,C :(i 1. H :: KC1) :LII K, . . 

 A B E : F G L : : B E C : G L H : : K C 1) : L Ii K. Con- 



