BOOK iv. PEOP. i. vi.] MATHEMATICS. PLANE GEOMETRY. 



573 



PROPOSITION L PROBLEM. 



In a giren circle (ABC), to 

 place a straight line equal to a 

 giren straight line (D), vhich 

 it not greater than the di- 

 ameter of the circle. 



Draw BC, any diameter 

 of the circle. Then if C B 

 = D, the thing required is 

 done ; but if B C be not = D, 



Hyp. it is greater ;* make C E = D ; and with centre 

 C and radius C E, describe the circle A E F ; join C A; 

 C A shall be = D. 

 Because C is the centre of the circle A E F, C A = 



cont. C E ; but D = C E,* .-. D = C A ; therefore in 

 the circle A B C, o straiyht line, equal to D, w placed as 

 required. 



NOTI. The enunciation of this proposition might hare been~as 

 follows : From a given point, in the circumference of a given 

 circle, to draw a chord equal to a given straight line, which ia not 

 greater than the diameter. 



PROPOSITION H. PROBLEM. 



In a yiren circle (A B C) to intcribe a triangle equiangvlar to a 

 giren triangle (D E F). 



Take any point A in the circumference, and draw 

 1 17 in. G A H to touch the circle in that point, t Make 



M i. the angle H A C = E, and G A B = F * and 

 draw B C : ABC 



shall be the tri- 



Becanse HAG 

 touches the circle, 

 and A C is drawn 

 from the point of 

 contact, the angle 

 H A C = B, in 

 the alternate seg- 



i 32 in. ment:t but H A C = E, . . B = E. Fora 

 like reason, C = F, . . the remaining angle B A C = the 



32 i. remaining angle D ; * . . the triangle ABC, 

 inscribed in the circU A B C, is equiangular to D E F. 

 Which was to be done. 



Von. If the angle H A C has been made equal to F, 

 insti-ad of ., the renulting triangle would (till have been 

 equiangular to DEP : but the con.truction above furnishes 

 a triangle not onljr equiangular, but similar In petition to 

 DEP. Like remarks appljr to the nejt proposition. 



PROPOSITION III. PROBLKK. 

 About a given circle (A B C) to deicribe a triangle 



equiangular to a giren triangle (D E F). 

 Prolong E F both ways to G and H : find the 



i in. centre K of the circle,* and draw any 

 radius K B. Make the angle B K A = D E G, 



+ 53 1. and the angle BKC D F H : f and 

 through A, B, C, draw LM, 

 M N, N L to touch the 



17 1. circle : * these 

 lines will meet and form a 

 triangle L M N equiangu- 

 lar to D E F. 



Because LM.MN.NL 

 touch the circle at A, B, C, 

 the lines K A, KB, K C 

 from the centre make the 

 angles at A, B, C ruiht 



tin in. angles, t .. AM, 

 B M would make with a M. 

 line joining A, B, angles 

 which are together lest than two right angles, .-. AM, 



AI. IS. BM, must meet.* In like manner A L, C L, 

 m ;et ; as also B N, C N : and because the four angles of 

 the quadrilateral A M B K, are = four right angles, for 

 it can be divided into two triangles, and that two of the 

 angles K A M, K B M, are right angles, .- . the other two, 

 A K 11, A M B make two right angles. But 1) E G -f 



+ ni. D K F = two right angles, t . . A K B + A M B 

 - D E G -f D E F ; but A K B = I) E G, .-. A M B = 

 D E F. In like manner it may be proved that L N M = 



D F E, .. the remaining angle L = the remaining angle 



32 i. D ; * . . the triangle L M N is equiangular to 

 the triangle D E F ; and it is described about the circle. 

 Which was to be done. 



PROPOSITION IV. PROBLEM. 

 To inscride a circle in a given triangle (A B C). 



91. Bisect the angles ABC, B C A* by the 

 straight lines B D, CD, meeting one another in D, from 

 which draw D E perpendicular to one of the sides.f 



+ 121. The circle described with centre D, and radius 

 D E, will be inscribed in the 

 triangle. 



For draw D F, D G, perpen- 

 diculars to the other sides :* 



12 1 then, because the angle 

 EBD = FBD, for ABC is 



t Const, bisected by B D,t and 



Ai.n. thatBED=BFD,* 

 .'. the triangles EBD, FBD 

 have two angles of the one = two 

 of the other, each to each ; and 

 the side B D, opposite to one of the equal angles in each, 



+ 26 1. common to both, .'. the other sides are equal ;t 

 that is, D E = D F. For a like reason, D G = D F, .-. 

 DE = DG, .-. DE, DF, DG are all equal; and the 

 circle described with centre D, and either of these for a 

 radius, will pass through the extremities of all ; and 

 will touch A B, B C, C A, because the angles at E, F, G 



18 III. are right angles;* .-. the circle E F G is in- 

 scribed in the triangle ABC. Which was to be done. 



PROPOSITION V. PROBLEM. 

 To deicribe a circle about a given triangle (A B C). 



101. Bisect AB, AC in D, E ;* and from these 

 points draw D F, E F, at right angles to A B, A C.f 



+ n I. Then D F, E K must meet one another, because 

 they make with a line joining D, E, angles on the same 



Ax. 12. aide, together less than two right angles. * Let 

 them meet in F, and draw F A, then a circle described 

 with centre F, and radius FA, will circumscribe the 

 triangle ABC. 



For if the point F be not in B C, draw B F, C F. 

 Then, because A 1) = I) B, and D F is common, and at 



+ 41. right angles to A B, A F = F B.f In like man- 

 ner, it may be shown that CF = AF, .-. FB = FC, .-. 

 FA, F B, F C are all equal ; .. the circle described with 

 centre F, and either of these lines for radius, will pass 

 through the extremities of all, and be circumscribed about 

 the triangle ABC. Which was to bo done. 



PROPOSITION VI. PROBLEM. 

 To inicribe a tqvare in a girrn circle (A B C D). 



Draw any two diameters AC, B D at right angles to 

 one another : draw A B, B C, 

 CD, DA: the figure ABCD 

 shall be thr> square required. 



Because E is the centre, E B 

 = E D ; also E A is common, 

 and at right angles to B D, 



.41. A B =. A D.* For 

 a like reason C B, C D are re- 

 spectively = A B, A D, . . the 

 figure A B C D is equilateral. It 

 is also reitangular ; for B D being a diameter of the 

 circle, B A D is a semicircle, . . B A D is a right angle, f 

 + 31 in. For the same reason each of the angles 

 ABC, B C D, C D A, like D A B, is a right ang.'e, .-. 



