Book IV. Wherefore because each of the angles ACD, CDA is double of the angle CAD; and they have been cut in halves by the ftraight lines CE, DB; therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another; but (by 26. 3.) equal F B angles ftand upon equal circumfe- G H C rences AB, BC, CD, DE, EA are equal to one another but (by To circumfcribe an equilateral and equiangular pentagon about a given circle. Let ABCDE be the given circle; it is required to circumfcribe an equilateral and equiangular pentagon about the circle ABCDE. Let A, B, C, D, E be understood to be the points of the angles of a pentagon inscribed (by 11.4.) fo that the circumferences AB, BC, CD, DE, EA may be equal: And through the points A, B, C, D, E let GH, HK, KL, LM, MG be drawn touching the circle: and let F the center of the circle ABCDE be taken; and let FB, FK, FC, FL, FD be joined. And G Book IV. E A H M B C L K C D And because the straight line KL touches the circle ABCDE in the point C ; and from the center F to the contact at C the line FC hath been drawn ; therefore (by 18. 3.) FC is perpendicular to KL; therefore each of the angles at the point C is a right angle. For the fame reason alfo the angles at the points B, D are right angles and because the angle FCK is a right angle; therefore (by 47. 1.) the Square of FK is equal to the Squares of FC, CK: Certainly for the fame reason alfo the Square of FK is equal to the Squares of FB, BK ; therefore (by com. not. 1.) the fquares of FC, CK are equal to the Squares of FB, BK; of which the fquare of FC is equal to the Square of FB; therefore the remaining Square of CK is equal to the remaining Square of BK; therefore BK is equal to CK: And becaufe FB is equal to FC and FK common; certainly the two BF, FK are equal to the two CF, FK; and the base BK is equal to the base CK; therefore (by 8.1.) the angle BFK is equal to the angle CFK; and the angle BKF; equal to FKC; therefore the angle BFC is double of KFC and BKC is double of FKC. Certainly for the fame reason alfo CFD is double of CFL; and the angle CLD is double of CLF. And because the circumference BC is equal to the circumference CD (by the conft.); the angle BFC is also equal (by 27. 3.) to the angle CFD; and BFC is the double of KFC; and the angle DFC is the double of LFC; therefore the angle KFC is equal to CFL; certainly FKC, FLC are two triangles having the two angles equal to the two angles, each to each; and one fide equal to one fide; FC common to them both; therefore (by 26. 1.) they will allo have the remaining fides equal to the remaining fides; and the remaining angle equal to the remaining angle; therefore the straight line KC is equal to CL; and the angle FKC to FLC and fince KC is equal to CL; therefore KL is the double of KC. Certainly in the fame manner HK will be demonftrated to be double of BK: and because BK has been demonftrated to be equal to KC; and KL is the double of KC; and HK the double of BK; therefore HK is equal to KL; certainly in the fame manner each of the lines P 2 GH, Book IV. GH, GM, ML will be demonftrated to be equal to either of the lines HK, KL; therefore the pentagon GHKLM is equilateral. I say that it is also equiangular; for because the angle FKC is equal to FLC; and HKL has been demonftrated to be double of FKC; and KLM to be double of FLC; therefore HKL is alfo equal to KLM certainly in the fame manner each of the angles KHG, HGM, GML will be demonftrated to be equal to either of the angles HKL, KLM; therefore the five angles GHK, HKL, KLM, LMG, MGH are equal to one another: therefore the pentagon GHKLM is equiangular; but it has been alfo demonftrated to be equilateral; and it has been circumfcribed about the given circle ABCDE. Which was to be done. To infcribe a circle, in a given pentagon which is equilateral and equiangular. Let ABCDE be the given pentagon, which is equilateral and equiangular it is required to infcribe a circle in the pentagon ABCDE. Let each of the angles BCD, CDE be cut in halves, by each. of the straight lines CF, DF; and from the point F, in which the ftraight lines CF, DF meet one another, let the straight lines FB, FA, FE be drawn. And because BC, CD are equal (by supp.); and CF common; certainly the two BC, CF are equal to the two DC, CF; and the angle BCF is equal to the angle DCF (by conft.); therefore (by 4. 1.) the base BF is equal to the base DF; and the triangle BFC is equal to the triangle DCF; and the remaining angles are equal to the remaining angles, under which the equal fides are extended; therefore the angle CBF is equal to the angle CDF and because the angle CDE is double of CDF; but CDE is equal (by fupp.) to ABC; and the angle CDF to CBF; therefore CBA is the double of CBF; therefore the angle ABF is equal to FBC therefore the angle ABC is cut in halves by the straight line BF. Certainly in the fame manner it will be demonftrated, that each of the angles BAE, AED hath been cut in halves by cach of the ftraight lines FA, FE. Let Let the perpendiculars FG, FH, FK, FL, FM be drawn from the point F to the straight lines AB, BC, CD, DE, EA; A Book IV. G M B E H C K D and because the angle HCF is equal to Wherefore a circle has been infcribed in a given pentagon, which is equilateral and equiangular. Which was to be done. PRO P. XIV. To circumfcribe a circle about a given pentagon, which is equilateral and equiangular. Let ABCDE be the given pentagon, which is equilateral and equiangular; it is required to circumfcribe a circle about the tagon ABCDE. pen Let: Book IV. Let each of the angles BCD, CDE be cut in halves by each of the straight lines CF, DF; and from the point F in which the ftraight lines meet, let the ftraight lines FB, FA, FE be drawn to the points B, A, E; certainly in the fame manner as in the propofition before this it will be demonftrated, that each of the angles CBA, BAE, AED hath been cut in halves by each of the straight lines BF, FA, FE. And because the angle BCD is equal to the angle CDE; and the angle FCD is the half of BCD; and the angle CDF is the half of CDE; therefore also the angle FCD is equal to FDC: so that (by 6. 1.) the fide FC is equal to the fide FD. Certainly in the fame manner it will be demonftrated that each of the lines FB, FA B A E F FE is equal to either of the lines FC, FD: therefore the five straight lines FA, FB, FC, FD, FE are equal to one another ; wherefore the circle defcribed with the center F and at ahe distance of any one of the lines FA, FB, FC, FD, FE will also pass through the remaining points; and will be circumfcribed about the pentagon ABCDE, which is equilateral and equiangular; let it be circumfcribed and let it be ABCDE. Wherefore a circle has been circumfcribed about a given pentagon, which is equilateral and equiangular. Which was to be done. To infcribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle; it is required to infcribe an equilateral and equiangular hexagon in the circle ABCDEF. Let AD the diameter of the circle ABCDEF be drawn: and let G the center of the circle be taken; and with the center D and at the distance DG let the circle EGCH be defcribed; and EG, CG being joined, let them be produced to the points B, F; and let AB, BC, CD, DE, EF, FA be drawn; I fay that the hexagon ABCDEF is equilateral and equiangular. For |