Book IV. triangle FGH, so that the angle CAD be equal to the ☑ angle at F, and each of the angles ACD, CDA F angles ACD, CDA is double of the angle CAD. B с 9. 1. Bisect the angles ACD, CDA by the straight lines CE, DB; and join AB, Because angles the ACD, CDA are each of them double of CAD, and are bisected by the straight lines CE, DB, the five angles DAC, ACE, ECD, CDB, BDA are equal to one anod 26 3. ther; but equal angles stand upon equal d circumferences; therefore the five circumferences AB, BC, CD, DE, EA are equal to one another: and equal circumferences are fubtended by e 29. 3. equal e straight lines; 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; 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: and the angle AED stands on the circumference ABCD, and the angle BAE on the circumference f 27. 3. EDCB; therefore the angle BAE is equal f to the angle AED: for the fame reafon, 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 shown that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done. T PROP. XII. PROB. O describe an equilateral and equiangular pentagon about a given circle. Let に i Let ABCDE be the given circle, it is required to describe Book IV. an equilateral and equiangular pentagon about the circle ABCDE. C 18. 3. Let the angles of a pentagon, inscribed in the circle, by the laft proposition, be in the points A, B, C, D, E, so that the circumferences AB, BC, CD, DE, EA are equala; and a 11,4. through the points A, B, C, D, E draw GH, HK, KL, LM, MG, touching b the circle: take the centre F, and join FB, b 17. 3. FK, FC, FL, FD: and 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 c to KL; therefore each of the angles at C is a right angle: for the fame reason, the angles at the points B, D are right angles; and because FCK is a right angle, the square of FK is equal d to the d 47. 1. squares of FC, CK. 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, of which the square of FC is equal to the square of FB; the remaining square of CK is therefore equal to the remaining square of BK, 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; and the base BK is equal to the base KC; therefore the angle BFK is equale to the e 8. 1. angle KFC, and the angle BKF to FKC; 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 f to the angle CFD; and BFC is double of f 27.3 the angle KFC, and CFD G double of CFL; therefore angles of the other, each to each, and the fide FC, which is adjacent to the equal angles Book IV. g 26. 1. in each, is common to both; therefore the other fides shall be equal g to the other fides, and the third angle to the third angle: 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 shown tha 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, HK shall be equal to KL: 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, fince 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: 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 being equal to one another, the pentagon GHKLM is equiangular: and it is equilateral, as was demonftrated; and it is described about the circle ABCDE. Which was to be done. a 9. 1. b 4. 1. PROP. XIII. O inscribe a PROB. circle in a given equilateral and Tangular pentagon Let ABCDE be the given equilateral and equiangular pentagon; it is required to infcribe a circle in the pentagon ABCDE. Bisect a the angles BCD, CDE by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE: therefore, fince BC is equal to CD, and CF common to the triangles BCF, DCF, the two fides BC, CF are equal to the two DC, CF; and the angle ECF is equal to the angle DCF; therefore the base BF is equal b to the base FD, and the other angles to the other angles, to which the equal fides are opposite; therefore the angle CBF is equal to the angle CDF: and because the angle angle CDE is double of CDF, and that CDE is equal to Book IV. CBA, and CDF to CBF; CBA is also double of the A M angle CBF; wherefore the FK, FL, FM perpendiculars to the straight lines AB, BC, C K D CD, DE, EA: and because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; in the triangles FHC, FKC there are two angles of one equal to two angles of the other, and the fide FC, which is oppofite to one of the equal angles in each, is common to both; there fore the others fides shall be equal d, each to each; wherefore d 26. 1. the perpendicular FH is equal to the perpendicular FK: in the fame 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 touch the straight lines AB, BC, CD, DE, EA, because that 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 e the circle: therefore each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore it is infcribed in the pentagon ABCDE. Which was to be done. 1 PROP. Book IV. a 9. 1. PROP. XIV. PROB. 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 it. b6. г. Bisect a the angles BCD, CDE by the straight lines CF, FD, 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 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 the half of CDE; the angle FCD is equal to FDC; wherefore the fide CF is equal b to the fide FD: 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, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done. A E B F D C PROP. XV. PROB. To infcribe an equilateral and quiangular hexagon in a given circle. Let |