8. 1. † 6. 1. II. 2. And fince DA is equal to AB, and AC is common, the two Sides DA, A C, are equal to the two Sides BA, AC; but the Bafe DC is equal to the Base BC. Therefore the Angle DAC will be equal to the Angle BAC: And confequently the Angle D AB is bifected by the Right Line AC. In the fame Manner we prove, that each of the Angles ABC, BCD, CDA, are bifected by the Right Lines AC, DB. * Then because the Angle DAB is equal to the Angle ABC, and the Angle EAB is half of the Angle DAB, and the Angle EBA half of the Angle ABC; the Angle EAB fhall be equal to the Angle EBA: And fo the Side EA is equal to the Side EB. In like manner we demonftrate, that each of the Right Lines, EC, ED, is equal to each of the Right Lines EA, E B. Therefore the four Right Lines EA, EB, EC, ED, are equal between themselves. Wherefore a Circle being defcribed about the Center E, with either of the Distances EA, E B, EC, ED, will also pass thro' the other Points, and will be described about the Square ABCD; which was to be done. PROPOSITION X. PROBLEM. To make an Ifofceles Triangle, having each of the UT any given Right Line AB in the Point C, fo that the Rectangle contained under AB, BC, be equal to the Square of AC; then about the Center A, with the Distance AB, let the Circle BDE +1 of this. be defcribed; and in the Circle BDE apply the Right Line BD equal to AC; which is not greater than the Diameter. This being done, join DA, DC, 5 of this. and describe a Circle ACD about the Triangle ADC. Then because the Rectangle ABC is equal to the Square of A C, and AC is equal to BD, the Rectangle under AB, BC, fhall be equal to the Square of BD. And because fome Point B is taken without the Circle A CD, and from that Point there fall two Right Lines BCA, BD, to the Circle, one of which cuts cuts the Circle, and the other falls on it. And fince the Rectangle under AB, BC, is equal to the Square of BD, the Right Line BD shall touch the Circle * 37. 3. ACD. And fince BD touches it, and D C is drawn from the Point of Contact D, the Angle BDC is equal to the Angle in the alternate Segment of the Circle, vix. equal to the Angle DAC. And fince † 32. 3. the Angle BDC is equal to the Angle DAC; if CDA, which is common, be added, the whole Angle BDA is equal to the two Angles CDA, DAC. But the outward Angle BCD is equal to CDA, ‡ 32. I. ~ DAC. Therefore BDA is equal to B CD. But the Angle BDA is equal to the Angle CBD, be- * 5. I. cause the Side AD is equal to the Side AB. Wherefore DBA fhall be equal to BCD: And fo the three Angles BDA, DBA, BCD, are equal to each other. And fince the Angle DBC is equal to the Angle BCD, the Side BD is equal to the Side DC. But † 6. 1. BD is put equal to CA. Therefore CA is equal to CD. And fo the Angle CDA is equal to the Angle DAC. Therefore the Angles CDA, DAC, taken together, are double to the Angle DAC. But the Angle B CD is equal to the Angles CDA, DAC. Therefore the Angle BCD is double to the Angle DAC. But BCD is equal to BDA, or DBA. Wherefore BDA, or DBA, is double to DAB. Therefore the Ifofceles Triangle ABD is made, having one of the Angles at the Bafe, double to the other Angle; which was to be done. PROPOSITION XI. PROBLEM. To defcribe an equilateral and equiangular Pentagon in a given Circle. L ET ABCDE be a Circle given. It is required to describe an equilateral and equiangular Pentagon in the fame. Make an Ifofceles Triangle F GH, having * each * 10 of this. of the Angles at the Base GH, double to the other Angle F; and defcribe the Triangle ADC in the Circle ABCDE, equiangular + to the Triangle FGH; fo that † 2 of this. the the Angle CAD be equal to that at F, and A CD, CDA, each equal to the Angles G or H. Wherefore the Angles A CD, CDA, are each double to the Angle CAD. This being done, bifect ACD, CDA, by the Right Lines CE, DB, and join A B, BC, DE, EA. * Then because each of the Angles ACD, CDA, is double to CAD, and they are bifected by the Right Lines CE, DB; the five Angles DAC, ACE, ECD, CDB, BDA, are equal to each other. But equal Angles ftand upon equal Circumferences. There fore the five Circumferences AB, BC, CD, DE, EA, are equal to each other. But equal Circumferences fubtend + equal Right Lines, Therefore the five Right Lines AB, BC, CD, DE, EA, 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 AED ftands on the Circumference ABCD, and BAE on the Circumference EDCB: Therefore the Angle BAE is equal to the Angle AED. For the fame Reason, each of the Angles ABC, BCD, CDE, is equal to BAE, or AED. Wherefore the Pentagon ABCDE is equiangular; but it has been proved to be alfo equilateral. And confequently there is an equilateral and equiangular Pentagon inscribed in a given Circle; which was to be done. PROPOSITION XII. PROBLEM. To defcribe an equilateral and equiangular Pentagon about a Circle given. ET ABCDE be the given Circle. It is required to defcribe an equilateral and equiangular Penta gon about the fame. Let A, B, C, D, E, be the angular Points of a Pentagon supposed to be infcribed in the Circle; fo that the Circumferences AB, BC, CD, DE, EA, be equal; equal; and let the Right Lines GH, HK, KL, LM, * * Then because the Right Line KL touches the Circle ABCDE in the Point C, and the Right Line FC is drawn from the Center F to C, the Point of Con-→ tact; FC will be ‡ perpendicular to KL: And so ‡ 18. 3% both the Angles at C are Right Angles. For the fame Reason, the Angles at the Points B, D, are Right Angles. And becaufe FCK is a Right Angle, the Square of F K will be equal to the Squares of FC, 47. L CK: And for the fame Reason, the Square of F K is equal to the Squares of F B, BK. Therefore the Squares of F C, CK are equal to the Squares of FB, BK. But the Square of FC is equal to the Square of F B. Wherefore the Square of CK fhall be equal to the Square BK; and fo BK is equal to CK. And because FB is equal to F C, and FK is common; the two Sides BF, F K, are equal to the two CF, F K, and the Bafe BK, is equal to the Bafe KC; and fo the Angle BFK fhall be + equal to the Angle † 8. x. KFC, and the Angle BKF to the Angle F KC. Therefore the Angle BFC is double to the Angle KFC, and the Angle BKC double to the Angle FKC: For the fame Reafon, the Angle CFD is double to the Angle CFL, and the Angle CLD double to the Angle CLF. "And because the Circumference BC is equal to the Circumference CD, the Angle BFC fhall be equal to the Angle CFD. † 27. 3* But the Angle BFC is double to the Angle KFC, and the Angle DFC double to LFC. Therefore the Angle KFC is equal to the Angle CFL. And fo FKC, FLC, are two Triangles, having two Angles of the one equal to two Angles of the other, tach to each, and one Side of the one equal to one Side of the other, viz. the common Side FC; wherefore they fhall have + the other Sides of the one equal † 26. to the other Sides of the other; and the other Angle of the one, equal to the other Angle of the other. Therefore the Right Line K C is equal to the Right Line CL, and the Angle FKC to the Angle FLC. And fince KC is equal to CL, KL shall be double to KC. And by the fame Reason, we prove that HK 4 HK is double to BK. Again, because BK has been proved equal to KC, and KL the double to KC, as alfo HK the double of BK, HK fhall be equal to KL. So likewife, we prove that GH, GM, and ML, are each equal to HK, or KL. Therefore the Pentagon GHKLM is equilateral. I fay alfo, it is equiangular; for becaufc the Angle FKC is equal to the Angle FLC; and the Angle HKL has been proved to be double to the Angle FKC; and also KLM double to FLC: Therefore the Angle HKL fhall be equal to the Angle KLM. By the fame Reason we demonftrate, that every one 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 between themfelves. And fo the Pentagon GHKLM is equiangular, and it has been proved likewife to be equilateral, and defcribed about the Circle ABCDE; which was to be done. PROPOSITION XIII. PROBLEM. To defcribe á Circle in an equilatéral and equiangular Pentagon. ET ABCDE be an equilateral and equiangular the fame. a Bifect the Angles BCD, CDE, by the Right Lines CF, DF, and from the Point F wherein CF, DF, meet each other, let the Right Lines F B, FA, FE, be drawn. Now because BC is equal to CD, and CF is common, the two Sides BC, CF, are equal to the two Sides DC, CF; and the Angle BCF is equal to the Angle DCF. Therefore the Bafe BF is equal to the Bafe FD; and the Triangle BFC 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 equal to the Angle CDF. And because the Angle CDE is double to the Angle CDF, and the Angle CDE is equal to the Angle |