BOOK iv. PROP, xm. xv.] MATHEMATICS. PLANE GEOMETRY. 



575 



17 ill. the circle :* the figure G H K L M shall be 

 the pentagon required. 



From the centre F draw FB, FK, F 0, FL, FD. 

 Then because K L touches the circle in 0, F C is per- 

 + 18 III. pendicular to KL,f .' the angles at C are 

 right angles : for a like reason the angles B, D are right 

 angles. And because F C K is a right angle, F K 2 = 

 471. FC ! +CK 2 :* for a like reason F K 2 = F B 2 

 + BK 1 , .-. FC' + CK 2 , = FB 2 + BK 2 ; but FB 2 = 

 FC 2 , .-.BK 2 = CK 2 , .-.BK=CK. Again: because 

 FB = FC, the two sides FB, FK = the two FO, FK; 

 and B K = C K, as just proved, . . the angle B F K = 

 + 8 i. CFK,f and BKF = CKF, .'. ~BFC = twice 

 C F K, and B K C = twice C K F. For a similar reason 

 CFD = twice CFL, and CLD= twice CLF : and 

 because the arc B C = arc C D, the angle B F C = 

 27 in. CFD :* and BFC=twice CFK, and CFD 

 = twice CFL, .-. CFK = CFL; .-. in the two 

 triangles F K C, FLO, there are two angles of the one 

 equal to two of the other, each to each, and the side 

 F C, adjacent to equal angles in each, is common to both, 

 281. .-. KC=LC,* and the angle FKC=FLC, 

 and K L is twice K C. In a similar manner it may be 

 shown that HK is twice BK, 

 and consequently, since it 

 was proved that B K = K C, 

 and that K L = twice K C, 

 and H K = twice BK, H K 

 = K L. In like manner it 

 may be shown that GH, GM, 

 M L are each = H K or K L: 

 . . the pentagon G H K L M 

 is equilateral. It is also equi- 

 angular ; for, since the angle 

 F K C = F L C, and II K L = twice F K C, and K L M 

 = twice F L C, as already proved, . . H K L = K L M. 

 And in like manner it may be shown that each of the 

 angles KHG, HGM, GMLis-HKLorKLM; 

 .-. the five angles G H K, HKL, KLM, LMG, 

 M G H, being equal to one another, the pentagon is 

 equiangular, and it is described about the circle A B C D. 

 Which was to be done. 



PROPOSITION XIII. PROBLEM. 



To inscribe a circle in a given equilateral and equiangular 

 pentagon (A B C D E). 



Bisect the angles BCD, CDE by the straight lines, 



91. C F, D F ;* and from the point F, in which 

 they meet, draw F K perpendicular to one of the sides ; 

 then if with F as centre and F K as radius a circle be 

 described, it will touch every side of the pentagon. 



t HJ-P. Draw F B, F A, F E ; then since B C = CD,t 



and C F is common to the triangles B C F, D C F, the 



two sides BC, CF = DC, CF each to each ; and the 



41. angle BCF = DCF, .'. BF = FD, and 



the angle C B i = C D F. And because C D E = twice 



t cont. C D F,f and that C D E = C B A, and CDF 

 = C B F, as just proved, .'. C B A = twice C B F, 

 .-. A B F = C B F, .'. A B C is bitected by B F. In a 

 similar manner it may be demonstrated that the angles 

 B A E, A E 1) are bisected by 

 F A, F E. From F draw F G, 

 F H, F L, F M perpendicular 

 to A B, B C, D E, E A ; then 

 because the angle H C F = 

 KCF, and FHC= F K C, 

 being right angles, . . in the 

 triangles FHC, F K C, two 

 angles of the one are equal to 

 two of the other, each to 

 each ; and the side F C oppo- 

 site to one of the equal angles in each is common to both, 



20 I. . . F H = F K. * In like manner it may be 

 proved that F L, F M, F G are each = F K, or F H, 

 . . the five F G, F H, F K, F L, F M, are equal to one 

 another, . . the circle described from centre F with either 

 of them for radius, will pass through the extremities of 

 all ; jt will moreover touch A B, B C, C D, D E, E A ; 



K 



since the angles at G, H, K, L, M, are right angles ;t 

 + 16 III. . ' . the circle is inscribed in the pentagon 

 A B C D E. Which was to be done. 



PROPOSITION XIV. PROBLEM. 



To describe a circle about a given equilateral and equi- 

 angular pentagon (A B C D E). 



Bisect the angles BCD, C D E by the straight lines 



9 I- CF, DF ;* and with the point F, in which they 

 meet, as centre, and either of them as radius, if a circle 

 be described, it will be that required. 



Draw F B, F A, F E. It may be demonstrated, as in 

 the preceding proposition, that the angles C B A, B A E, 

 A E D are bisected by F B, F A, FE ; and because the 

 angle B C D = C D E, and that F C D is half BCD, and 

 CDF half CDE, .-. F C D = 

 + 61. FDC, . . CF = FD.f 

 In like manner it may be demon- 

 strated that F B, F A, F E are 

 each = F C or F D, . . the five 

 F A, F B, F C, FD, F E are equal 

 to one another ; and . . the circle 

 described from centre F, with 

 either of them for radius, will pass 

 through the extremities of all, 

 and will . . be described about the 

 pentagon ABODE. Which was to be done. 



PROPOSITION XV. PROBLEM. 



To imcribe an equilateral and equiangular hexagon in a 

 given circle (A C D F). 



1 ill. Find the centre G of the circle,* and draw the 

 diameter A G D ; from D as centre with radius D G, de- 

 scribe the circle EGCH; draw EG, CG, and prolong 

 them to B,F; and draw A B, BC, CD, DE, EF, FA; AB 

 C D E F shall be the equilateral and equiangular hex agon 

 required. Because G is the centre of the circle A C D F, 

 G b = G I) ; and because D is the centre of the circle 

 E G C H, D E = D G, . . G E = E D, and the triangle 

 K G D is equilateral, . . the angles E G D, G D E, 1) E G, 



+ S I. C. are equal" to one another ;t and as they are 

 32 1. together equal to two right angles,* . . K G D 

 is the third part of two right angles. In a similar man- 

 ner it may be demonstrated that D G C is the third part 

 of two right angles; and because GO makes with EB 

 the adjacent angles E G C, C G B, equal to two right 

 + is I. angles, t the remaining angle C G B is also the 

 third part of two right angles, . . E G D, D G C, C G B, 

 are equal to one another ; and to these are equal the ver- 



is I. tical or opposite angles B G A, A G F, F G E ;* 



. . the six angles at G are equal 

 to one another ; consequently the 

 six arcs which subtend them are 

 + 26 in. equal, f and .. the six 

 29 in. chords of these arcs,* 

 . ' . the hexagon A B C D E F is 

 equilateral. It is also equiangular; 

 for since the arc A F = R D, add 

 A BCD to each; .-.the whole 

 arc F A B C D = E D C B A ; and 

 the angle FED stands upon 

 F A B C D, and A F E upon 

 EDCBA, .-. AFE =FED.t 

 + 27 III. In a similar manner it may be demonstrated 

 that the other angles of the hexagon are each of them = 

 AFE, or F E 1 ) ; . . the hexagon is eqiiiangular ; and it 

 is inscribed in the given circle A C D I 1 . Which was to 

 be done. 



COR. From this it is manifest that a side of the hex- 

 agon is equal to the radius of the circle. And if through 

 A, B, C, D, E, F lines be drawn touching the circle, an 

 equilateral and equiangular hexagon will be described 

 about it, as may be demonstrated from what has been 

 said of the pentagon : and likewise a circle may be in- 

 scribed in a given equilateral and equiangular hexagon, 

 and circumscribed about ifc, by a method like to that 

 used for the pentagon. 



