wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect the angles ACD, CDA by the straight lines CE, DĚ; (1. 9.) and join AB, BC, DE, EA. F B A E G H Then ABCDE shall be the pentagon required. Because each of the angles ACD, CDA is double of CAD, and that they are bisected by the straight lines CE, DB; therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another: but equal angles stand upon equal circumferences; (III. 26.) therefore the five circumferences AB, BC, CD, DE, EA are equal to one another: and equal circumferences are subtended by equal straight lines; (III. 29.) therefore the five straight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also equiangular: for, because the circumference AB is equal to the circumference DE, if to each be added BCD, the whole ABCD is equal to the whole EDCB: (ax. 2.) but the angle AED stands on the circumference ABCD; and the angle BAE on the circumference EDCB; therefore the angle BAE is equal to the angle AED: (III. 27.) for the same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: therefore the pentagon ABCDE is equiangular; and it has been shewn that it is equilateral: wherefore, in the given circle, an equilateral and equiangular pentagon has been described. Q. E. F. PROPOSITION XII. PROBLEM. To describe an equilateral and equiangular pentagon about a given circle. Let ABCDE be the given circle. It is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angular points of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that the circumferences AB, BC, CD, DE, EA are equal; (IV. 11.) and through the points A, B, C, D, E draw GH,ÉK, KL, LM, MG touching the circle; (III. 17.) the figure GHKLM shall be the pentagon required. And because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the center 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 on FK is equal to the squares on FC, CK: (1. 47.) for the same reason, the square on FK is equal to the squares on FB, BK: therefore the squares on FC, CK are equal to the squares on FB, BK; (ax. 1.) of which the square on FC is equal to the square on FB; therefore the remaining square on CK is equal to the remaining square on BK, (ax. 3.) and the straight line CK equal to BK: and because FB is equal to FC, and FK common to the triangles BFK, CFK, the two BF, FK are equal to the two CF, FK, each to each: wherefore the angle BFC is double of the angle KFC, 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; (III. 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; therefore, in the two triangles FKC, FLC, there are two angles of the 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 com mon to both; therefore the other sides are 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, KL is double of KC. In the same manner it may be shewn 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 shewn that GH, GM, ML are each of them equal to HK, or KL: therefore the pentagon GHKLM is equilateral. for, since the angle FKC is equal to the angle FLC, 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 being equal to one another, the pentagon GHKLM is equiangular : and it is equilateral, as was demonstrated; and it is described about the circle ABCDE. PROPOSITION XIII. PROBLEM. Q.E.F. To inscribe a circle in a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagon. It is required to inscribe a circle in the pentagon ABCDE. 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: therefore since BC is equal to CD, (hyp.) and CF common to the triangles BCF, DCF, 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 the angle 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. From the point F, draw FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA: (1. 12.) and because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; therefore in the 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 are equal, each to each; (I. 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 center F, at the distance of one of these five, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, 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 it is inscribed in the pentagon ABCDE. Q.E.F. PROPOSITION XIV. PROBLEM. To describe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagon. It is required to describe a circle about ABCDE. A E D Bisect the angles BCD, CDE by the straight lines CF, 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. It may be demonstrated, in the same manner as 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 CDF the half of CDE; therefore 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 center F, at the distance of one of them, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Q. E. F. PROPOSITION XV. PROBLEM. To inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle. It is required to inscribe an equilateral and equiangular hexagon in it. Find the center G of the circle ABCDEF, and from D, as a center, at the distance DG, describe the circle EGCH, the hexagon ABCDEF shall be equilateral and equiangular. and because D is the center of the circle EGCH, wherefore GE is equal to ED, (ax. 1.) and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG, are equal to one another: (1. 5. Cor.) but the three angles of a triangle are equal to two right angles; (1. 32.) therefore the angle EGD is the third part of two right angles: in the same manner it may be demonstrated, that the angle DGC is also the third part of two right angles: and because the straight line GC makes with EB the adjacent angles EGC, CGB equal to two right angles; (1. 13.) the remaining angle CGB is the third part of two right angles: therefore the angles EGD, DGC, CGB are equal to one another: and to these are equal the vertical opposite angles BGA, AGF, FGE: (1. 15.) therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another : but equal angles stand upon equal circumferences; (III. 26.) therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another: and equal circumferences are subtended by equal straight lines: (III. 29.) therefore the six straight lines are equal to one another, It is also equiangular: for, since the circumference AF is equal to ED, to each of these equals add the circumference ABCD; therefore the whole circumference FABCD is equal to the whole EDCBA: |