T FGH; fo that the Angle CAD be equal to that at F, and A CD, CDA, each equal to the Angles Gor H. Wherefore the Angles ACD, CD A, are each double to the Angle C A B. This being done, bife& the Angles* A CD, CD A, by the Right Lines * CE, DB, and join A B, BC, DE, E A. 9. I. Then, because each of the Angles AC D, CDA, is double to CAD, and they are bifected by the Right Lines CE, DB; the five Angles D'A C, AČE, ECD, CD B, BDA, are equal to each other. But equal Angles ftand + upon equal Cir- † 26. 3. cumferences. Therefore, the five Circumferences AB, RC, CD, DE, EA, are equal to each other. But equal Circumferences fubtend equal Right I 29. 3Lines. Therefore the five Right Lines A B, BC, CD, DE, E A, are equal to each other. Wherefore ABCDE is an equilateral Pentagon. I fay, it is alfo equiangular: For because the Circumference A B is equal to the Circumference DE; by adding the Circumference B CD, which is common, the whole Circumference ABCD is equal to the whole Circumference EDCB: But the Angle' A E D ftands on the Circumference A B C D, and BA E on the Circumference EDC B; therefore the Angle BAE is equal to the Angle A ED. For the fame Reafon, each of the other Angles A B C, B CD, CDE, is equal to B A E, or A ED; wherefore the Pentagon ABCDE is equiangular. But it has been proved to be alfo equilateral: And, confequently, there is an equilateral and equiangular Pentagon infcribed in a given Circle; which was to be done. PROPOSITION XII. PROBLEM. To defcribe an equilateral and equiangular Pentagon about a Circle given. L ET ABCDE be the given Circle. It is required to defcribe an equilateral and equiangular Pentagon about the fame. Let A, B, C, D, E, be the angular Points of a Pentagon fuppofed to be infcribed in the Circle; fo that By 11. of the Circumferences A B, B C, CD, DE, É A, be this. equal i + 17.3. +18.3. 47. I. † 8.1. 127.3. 26. I. equal; and let the Right Lines GH, HK, KL, LM, Then, because the Right Line KL touches the to to BK. Again, because BK has been proved equal to K C, and K L the double of K C, as alío HK the double of B K; HK fhall be equal to K L. So likewife we prove, that GH, GM, and M L, are each equal to HK, or KL: Thelore the Pentagon GHKLM is equilateral. I fay, allo it is equianular. For because the Angle FKC is equal to the Angle F LC, and the Ang e H K L has been proved to be double to the Angle FKC; and alfo KLM double to FLC: Therefore the Angle H K L thall be equal to the Angle KLM. By the fame Realin we demonftrate, that every one of the Angles K HG, HGM, GML, is equal to the Angle H K L, or KLM. Therefore the five Angles GHK, HK L, KLM, LM G, MG H, are equal between themfelves. And fo, the Pentagon GHKLM is equiangular; and it has been proved likewife to be equilateral, and defcribed about the Circle A B C D E; which was to be done. PROPOSITION XIII. To defcribe a Circle in an equilateral and equian. LET ABCDE be an equilateral and equiangu lar Pentagon. It is required to infcribe a Circle in the fame. * Bifect the Angles BCD, CDE, by the Right 9.1. Lines CF, DF; and from the Point F, wherein CF, DF, meet each other, let the Right Lines F B, F A, FE, be drawn. Now, because BC is equal to CD, and C F is common, the two Sides BC, C F, are equal to the two Sides DC, CF; and the Angle BC F is equal to the Angle DCF. Therefore the Bafe B F is equal to the Bafe F D ; and the Triangle BF C equal to the Triangle DCF, and the other Angles of the one equal to the other Angles of the other, which are fubtended by the equal Sides: Therefore the Angle CBF fhall be equai to the Angle CDF. And becaufe the Angle CDE is double to the Angle CDF, and the Angle C D E is equal to the Angle A B C, as alfo C D F equal to CBF; +4.. 12. I. + 26. 1. † 16. 3. the Angle CBA will be double to the Angle C B F; and fo the Angle A B F equal to the Angle CBF: Wherefore the Angle A B C is bifected by the Right Line B F. After the fame manner we prove, that either of the Angles BA E, or A ED, is bifected by the Right Line AF, or F E. From the Point F draw *F G, FH, FK, FL, FM, perpendicular to the Right Lines A B, BC, CD, DE, EA: Then, fince the Angle HCF is equal to the Angle K CF, and the Right Angle FHC equal to the Right Angle FKC; the two Triangles F H C, F K C, fhall have two Angles of the one equal to two Angles of the other, and one Side of the one equal to one Side of the other, viz. the Side F C common to each of them: And fo the other Sides of the one will be + equal to the other Sides of the other, and the Perpendicular FH equal to the Perpendicular F K. In the fame manner we demonftrate, that FL, FM, or FG is equal to FH, or FK. Therefore the five Right Lines F G, FH, FD, FL, F M, are equal to each other, and so a Circle described on the Centre F, with either of the Distances F G, FH, FK, FL, FM, will pafs through the other Points, and fhall touch the Right Lines A B, BC, CD, DE, EA; fince the Angles at G, H, K, L, M, are Right Angles. For, if it does not touch them but cuts them, a Right Line drawn from the Extremity of the Diameter of a Circle, at Right Angles to the Diameter, will fall within the Circle; which is abfurd. Therefore, a Circle defcribed on the Centre F, with the Distance of any one of the Points G, H, K, L, M, will not cut the Right Lines AB, BC, CD, DE, EA, and fo will neceffarily touch them; which was to be done. Coroll. If two of the neareft Angles of an equilateral and equiangular Figure be bifected, and from the Point in which the Lines bifecting the Angles meet, there be drawn Right Lines to the other Angles of the Figure, all the Angles of the Figure will be bifected. PROPOSITION XIV. PROBLEM. To defcribe a Circle about a given equilateral and equiangular Pentagon. LET ABCDE be an equilateral and equiangu lar Pentagon. It is required to defcribe a Circle about the fame. Bifect both the Angles BCD, CDE, by the Right Lines CF, FD; and draw FB, FA, FE, from the Point F, in which they meet. Then each of the other Angles CBA, BAE, AED, fhall be bifected by Cor. of the Right Lines BF, FA, FE. And fince the An- preced. gle BCD is equal to the Angle CDE, and the Angle FCD is half the Angle BCD; as likewife CDF, half C DE; the Angle F CD will be equal to the Angle FDC; and fo the Side CF + equal to the Side + 6. 1. FD. We demonftrate, in like manner, that FB, FA, or FE, is equal to F C, or F D. Therefore the five Right Lines F A, FB, FC, FD, FE, are equal to each other. And fo, a Circle being defcribed on the Centre F, with any of the Distances F A, FB, F C, F D, F E, will pass through the other Points, and will be defcribed about the equilateral and equiangular Pentagon ABCDE; which was to be done. PROPOSITION XV. To infcribe an equilateral and equiangular Hexagon L ETA B C D E F be a Circle given. It is required to infcribe an equilateral and equiangular Hexagon therein. Draw A D, a Diameter of the Circle ABCDEF, and let G be the Centre; and about the Point D. as a Centre, with the Distance D G, let a Cucle, EGCH, be defcribed; join EG, GC, which produce to the Points B, F: Likewife join A B, BC, CD, DE, EF, I 2 FA: |