f 8. I. Book IV. equal to CF, FK, and the bafe BK equal to KC, the angle BFK is equal to CFK f. For the fame reafon, CFL is equal to LFD; but the whole angle BFC is equal to the whole angle CFD ; therefore the angle KFC is equal to the angle CFL"; and the bafe KC to CL; but KC is equal to BK; therefore HK is equal to KL. For the fame reafon, KL is equal to LM therefore the figure is equilateral. h Ax. 7. I, i 4. I. k Cor. 7. It is likewife equiangular; for the angles B and C are each. right ones; and the two angles BFC, BKC, equal to two right angles. For the fame reason, CFD, CLD, are equal to two 32, I. right angles; but the angles CFD, CFB, are equal; therefore CLD, BKC, are likewife equal. For the fame reafon, the whole angle at L is equal to the angle at M; therefore the figure is equiangular; and likewife proved equilateral. Wherefore, &c. a 10, I. b 11. I. PROP. XIII. PRO B. To infcribe a circle in a given equilateral and equiangular pentagon. It is required to infcribe a circle in the equilateral and equis angular pentagon ABCDE. Bifed the fides BC, CD, DE, in the points H, K, L'; from which draw the right lines HF, KF, LF, at right angles to BC, CD, DE ; from the point F, where the right lines HF, KF, intersect each other, draw FC, FD, FE; then, because the angles FHC, FKC, are right angles, the fquare of FC is equal to the fquares of FH, HC, and likewife to the fquares of FK, KC; therefore the fquares of FH, HC, are ed Ax. 1. 1. qual to the fquares of FK, KCd. Take the equal fquare of HC, C 47. I. € 8. I. CK, from both, there remains the fquare of HF equal to the fquare of FK; that is, HF equal FK; therefore HF, FC, are equal to FC, FK, and the base HC to CK; therefore the angle HFC is equal to KFC; but the angles FHC, FKC, are right ones; therefore the remaining angle HCF is equal to KCF; therefore the angle BCD is bifected by the right line FC. For the fame reason, FK is equal to FL, and the angle CDE bifected by FD; therefore, becaufe FH, FK, FL, are equal, if, with the center F, and either of these distances, a circle is described, it will touch the fides of the pentagon in the points G, H, K, L, M; wherefore the circle GHKLM is infcribed in the equilateral and equiangular pentagon ABCDE: Which was res quired. COR COR. If two of the nearest fides of an equilateral and equian- Book IV, gular figure be bisected, and from the point where these lines cut each other, there be lines drawn to all the angles of the figure, thefe lines will bifect all the angles of the figure. PRO P. XIV. PRO B. To defcribe a circle about a given equilateral and equiangular pentagon. It is required to defcribe a circle about the equilateral and equiangular pentagon ABCDE. Bifect the right lines AB, BC, in the points Hand G; a 1o. i. from which draw the right lines HF, GF, interfecting each other in the point F, and at right angles, to AB, BC; from the b 1. 1. point F draw the right lines BF, FĂ, FE, FD, FC, they will bifect the angles at A, B, C, D, E. Then, because AB is e- c cor. 1 3. equal to AE, and AF common, and the angle BAF equal to EAF, the bafe BF will be equal to EFd. For the fame reason,d 4. 1. EF is equal to FC; therefore, if, with the center F, and distance B, E, or C, a circle is defcribed, it will pafs through the points A, B, C, D, E, of the equilateral and equiangular pentagon : Which was required. PRO P. XV. PRO B. To infcrive an equilateral and equiangular hexagon in a given circle. It is required to infcribe the equilateral and equiangular hexagon in the given circle ABCDEF. Draw AĎ a diameter to the circle ABCDEF, whose center is G; with the point D as a center, and distance DG, defcribe a circle EGCH; join EG, GC; which produce to the points B, F; join AB, BC, CD, DE, EF, FA; then ABCDEF is an equilateral and equiangular hexagon. a For, fince G is the center of the circle ABCDEF, GC is equal to GD; and fince D is the center of the circle CGEH, a defis. ĜD is equal to DC; therefore CGD is an equilateral triangle ;ь 1. 1. but it is likewife equiangular. For the fame reafon, GDE is c coг. 5. an equilateral and equiangular triangle, and equal to the triangle A d 15. 1. e 13. I. Book IV. triangle CGD; and, because BG, GA, are equal to DG, GE, and the angle BGA to DGE, the bafe AB is equal to DE. For the fame reafon, AF is equal to CD; but CD is equal to DE; therefore AB is equal to AF. Again, because CG falls upon BE, the angles CGB, CGE, are equal to two right angles; but CGD, DGE, are each one third of two right angles f; f cor. 32. 1. therefore CGB is likewife one third of two right angles; therefore BG, GC, are equal to CG, GF, and the angle BGC equal to the angle CGD; therefore the bafe BC is equal to CD; but likewife BG, GC, are equal to FG, GE, and the angle BGC to FGE 4; therefore the bafe BC is equal to FE; but BC is proved equal to CD, and CD to DE; therefore the fix fides BC, CD, DE, EF, FA, AB, are equal; therefore the figure is equilateral; it is likewife equiangular; for the two angles GDC, GDE, are equal to the two angles GCD, GCB; for each is one third of two right angles f; therefore the whole angle BCD, is equal to the whole CDE. For the fame reason, all the other angles are equal to one another; therefore the figure is likewife equiangular. Wherefore, &c. d. COR. Hence the fide of a hexagon is equal to the femi-diameter of the circle. And, if through the points A, B, C, D, E, F, tangents to the circle, be drawn, an equilateral and equiangular hexagon will be described about the circle, as may be proved in the same manner as the pentagon: And fo likewife a circle may be inscribed and described about a given hexagon. a 15. b II. PRO P. XVI. PRO B. то It is required to infcribe an equilateral and equiangular quindecagon in the given circle ABCD. Let AC be the fide of an equilateral and equiangular triangle infcribed in the circle, and AB the fide of an equilateral and equiangular pentagon, drawn from the point A; then, if the circle is divided into fifteen parts, the fide of the triangle AC will fubtend five of them, and the fide of the pentagon AB will fubtend three; therefore BC will be two of faid parts; therefore bifect BC in E; BE or CE will be one fifteenth part of the cir-. cum |