. ll. 4. • 17. 3. 18. 3. Let the angles 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 ; and through the points A, B, C, D, E, draw GH, HK, KL, LM, MG, touching * the circle; the figure GHKLM shall be the pentagon required. Take the centre F, and join FB, FK, FC, FL, FD: and because the straight line KL touches the circle ABCDE in the point C, to which FG is drawn from the centre F, FC is perpendicular * to KL, G therefore each of the angles at C is a right А. angle ; for the same reason, the angles at H Н M the points B, D are right angles; and be Bi cause FCK is a right angle, the of square FK is equal * to the squares of FC, CK ; к с. for the same reason, the square of FK, is equal to the squares of FB, BK; therefore the squares of FC, CK are equal t to the squares of FB, BK; of which the square of FC is equal to the square of FB; therefore the remaining square of CK is 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, each to each ; and the base BK was proved equal to the base KC; therefore the angle BFK is equal * to the 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 * to the angle CFD ; and BFC is double of the angle KFC, and CFD double of CFL; therefore the angle KFC is equal t to the angle CFL: and the right angle FCK is equal to the right angle FCL; therefore in the two triangles FRC, 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 common to both ; therefore the other sides are equal * to the other sides, 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 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: in like +6 Ax. manner it may be shewn, that GH, GM, ML are each of them equal to HK, or KL; therefore the pentagon GHKLM is equilateral. It is also equiangular ; for, since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and KLM double of FLC, as was before demonstrated, therefore the angle HKL is equal + to 7 6 Ax. KLM: and in like manner it may be shewn, 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. Which was to be done. 47. I. tl Ax. 73 Ax. * 8. 1. • 27. 3. 77 Ax. 2 . 26. 1. PROPOSITION XIII. * 9. 1. L PROB.—To inscribe a circle in a given equilateral and equi angular 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, and from the point M F, in which they meet, draw the straight B E lines FB, FA, FE: therefore, since BC is II equal † to CD, and CF common to the + Нур. . triangles BCF, DCF, the two sides BC, CF KD are equal to the two DC, CF, each to each ; and the angle BCF is equal to the angle DCF; therefore the + Constr. base BF is equal * to the base FD, and the other angles to • 4. 1. 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 perpendi * 12. 1. culars to the straight lines AB, BC, CD, DE, EA: and because • 26. 1. the angle HCF is equal to KCF, and the right angle FHC equal • 16. 3. * the PROPOSITION XIV. * 9.1. 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. Bisect * 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 BA 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 half of CDE, therefore the angle FCD is equal + to FDC; wherefore the side CF is equal * to the side 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 + 7 Ax. * 6. 1. them, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done. PROPOSITION XV. PROB.--To inscribe an equilateral and equiangular hexagon See N. in a given circle. Let ABCDEF be the given circle ; it is required to inscribe an equilateral and equiangular hexagon in it. Find + the centre G of the circle ABCDEF, and draw the + 1. 3. diameter AGD; and from D as a centre, at the distance DG, describe the circle EGCH, join EG, CG, and produce them to the points B, F; and join AB, BC, CD, DE, EF, FA: the hexagon ABCDEF shall be equilateral and equiangular. Because G is the centre of the circle ABCDEF, GE is equal to GD ; and because D is the centre of the circle EGCH, DE is equal to DG; wherefore GE is equal + to f 1 Ax. ED, and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG, F are equal t to one another : but the three + Cor. 5. 1. angles of a triangle are equal * to two right E . 32. 1. angles, therefore the angle EGD is the third part of two right angles ; in the same manner it may be demonstrated, that the angle DGC HI 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, the remaining * 13. 1. 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; . 15. 1. therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another : but equal angles stand upon equal * * 26. 3. circumferences, therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another : and equal circumferences are subtended by equal * straight lines, therefore the * 29. 3. six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular ; for, since the circumference AF is equal to ED, to each of these add the circumference ABCD; therefore whole circumference FABCD is equal to the whole EDCBA : and the angle FED * 27. 3. stands upon the circumference FABCD, and the angle AFE upon EDCBA ; therefore the angle AFE is equal + to FED: in the same manner it may be demonstrated, that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED; therefore the hexagon is equiangular; and it is equilateral, as was shewn ; and it is inscribed in the given circle ABCDEF. Which was to be done. CoR.-From this it is manifest, that the side of the hexagon is equal to the straight line from the centre, that is, to the semidiameter of the circle. And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon will be described about it, which may be demonstrated from what has been said of the pentagon : and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon. PROPOSITION XVI. See N. 2. 4. 11. 4. Prob.—To inscribe an equilateral and equiangular quinde cagon in a given circle. Let ABCD be the given circle; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle inscribed * in the circle, and AB the side of an equilateral and equiangular pentagon inscribed * in the same: therefore, of such equal parts, as the whole circumference ABCDF contains fifteen, the circumfer B. F ence ABC, being the third part of the whole, EU contains five; and the circumference AB, which is the fifth part of the whole, contains three; therefore BC, their difference, contains two of the same parts : bisect * BC in E: therefore BE, EC are each of them the fifteenth part of the whole circumference ABCD: therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placed round * in the whole circle, an equilateral and equiangular quindecagon will be inscribed in it. Which was to be done. And in the same manner as was done in the pentagon, if through the points of division made by inscribing the quin 30.3. .). 4. |