and through the points A, B, C, D, E, draw GH, HK, KL, LM, MG, touching the circle. (III. 17.) Then the figure GHKLM shall be the equilateral and equiangular pentagon required. Take the centre F, and join FB, FK, FC, FL, FD. DEMONSTRATION Because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the centre F, FC is perpendicular to KL; (III. 18) therefore each of the angles at C is a right angle; for the same reason, the angles at the points B, D, are right angles; and because FCK is a right angle, the square of FK is equal to the squares of FC, CK; (1. 47) for the same reason, the square of FK is equal to the squares of FB, BK; therefore the squares of FC, CK, are equal to the squares of FB, BK; (ax. 1) of which the square of FC is equal to the square of FB; therefore the remaining square of CK is equal to the remaining square of BK; (ax. 3) and the straight line CK equal to BK. Hence in the two triangles BFK, CFK, because FB is equal to FC, and FK common to both, the two BF, FK, are equal to the two CF, FK, each to each; and the base BK was proved equal to the base KC; therefore the angle BFK is equal to the angle KFC, (1.8) and the angle BKF to FKC; (1.4) wherefore the angle BFC is double of the angle KFC, and BKC double of FKC; for the same reason, the angle CFD is double of the angle CFL, and CLD double of CLF; and because the circumference BC is equal to the circumference CD, the angle BFC is equal to the angle CFD; (111. 27) and BFC is double of the angle KFC, and CFD double of CFL; therefore the angle KFC is equal to the angle CFL, (ax. 7) and the right angle FCK is equal to the right angle FCL. Hence in the two triangles FKC, FLC, there are two angles of one equal to two angles of the other, each to each, and the side FC, which is adjacent to the equal angles in each, is common to both; therefore the other sides shall be equal to the other sides, and the third angle to the third angle; (1. 26) therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC; and because KC is equal to CL, therefore KL is double of KC; in the same manner it may be shown that HK is double of BK. And because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, therefore HK is equal to KL; (ax. 6) in like manner, it may be shown that GH, GM, ML, are each of them equal to HK or KL; therefore the pentagon GHKLM is equilateral. It is also equiangular; for, since the angle FKC is equal to the angle FLC, and the angle HKL double of the angle FKC, and KLM double of FLC, as was before demonstrated, the angle HKL is equal to KLM; (ax. 6) and in like manner, it may be shown that each of the angles KHG, HGM, GML, is equal to the angle HKL or KLM; therefore the five angles GHK, HKL, KLM, LMG, MGH, are equal to one another, therefore the pentagon GHKLM is equiangular. and it is equilateral, as was demonstrated; and it is described about the circle ABCDE. Q. E. F. PROP. XIII.-PROBLEM. To inscribe a circle in a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagon. H Bisect the angles BCD, CDE, by the straight lines CF, DF; (1.9) and from the point F, in which they meet, draw the straight lines FB, FA, FE; draw also the lines FG, FH, FK, FL, and FM perpendicular to the sides of the pentagon; (1. 12) and from the centre F, at the distance FG, FH, FK, FL, or FM, describe the circle GHKLM. Then GHKLM shall be inscribed as required in the pentagon ABCDE. DEMONSTRATION Because in the two triangles BCF, DCF, BC is equal to CD (hyp.), and CF common, the two sides BC, CF, are equal to the two DC, CF, each to each; and the angle BCF is equal to the angle DCF; (constr.) therefore the base BF is equal to the base FD, (1. 4) and the other angles to the other angles, to which the equal sides are opposite; therefore the angle CBF is equal to the angle CDF; and because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF; CBA is also double of CBF; therefore the angle ABF is equal to the angle CBF wherefore the angle ABC is bisected by the straight line BF. In the same manner it may be demonstrated that the angles BAE, AED, are bisected by the straight lines AF, FE. And because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; then in the two triangles FHC, FKC, there are two angles of the one equal to two angles of the other, each to each; and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore the other sides shall be equal, each to each; (1. 26) wherefore the perpendicular FH is equal to the perpendicular FK. In the same manner it may be demonstrated that FL, FM, FG, are each of them equal to FH or FK; therefore the five straight lines FG, FH, FK, FL, FM, are equal to one another; wherefore the circle described from the centre F, at the distance of one of these five, shall pass through the extremities of the other four. And because the angles at the points G, H, K, L, M, are right angles, and that a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches the circle; (III. 16) therefore each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore the circle GHKLM is inscribed in the pentagon ABCDE. Q. E. F. PROP. XIV.-PROBLEM. To describe a circle about a given equilateral and equiangular pentagon. (References-Prop. 1. 6, 9.) Let ABCDE be the given equilateral and equiangular pentagon. Bisect the angles BCD, CDE, by the straight lines FC, FD, (1. 9) and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E ; from the centre F, at the distance FA, FB, FC, FD, or FE, describe the circle ABCDE. Then ABCDE shall be described as required about the given pentagon. DEMONSTRATION It may be demonstrated, as in the same manner as in the preceding proposition, that the angles CBA, BAE, AED, are bisected by the straight lines FB, FA, FE; and because the angle BCD is equal to the angle CDE, and that FCD is the half of the angle BCD, and CDF the half of CDE; the angle FCD is equal to FDC; (ax. 7) wherefore the side CF is equal to the side FD. (1. 6.) In like manner it may be demonstrated that FB, FA, FE, are each of them equal to FC or FD; therefore the five straight lines FA, FB, FC, FD, FE, are equal to one another; and the circle described from the centre F, at the distance of one of them, shall pass through the extremities of the other four, wherefore the circle ABCDE is described about the equilateral and equiangular pentagon ABCDE. Q. E. F. PROP. XV.-PROBLEM. To inscribe an equilateral and equiangular hexagon in a given circle. (References-Prop. L. 5 cor., 13, 15, 32; IL 1, 26, 27, 29.) Let ABCDEF be the given circle. It is required to inscribe an equilateral and equiangular hexagon in it. D H B CONSTRUCTION Find the centre G of the circle ABCDEF, (III. 1) and draw the diameter AGD; |