: * *26. 1. 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 shown that HK is double of BK and because BK is equal to KC, as was demonstrated, and that KL is double of KC, +6 Ax. and HK double of BK, therefore HK is 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, since the angle FKCis 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 +6 Ax. 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 demonstrated; and it is described about the circle ABCDE. Which was to be done. *9. 1. : To inscribe a circle in a given equilateral and equiangular 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 F, in which they meet, draw the straight lines FB, FA, Hyp. FE: therefore since BC is equal † to CD, and CF common to the triangles BCF, DCF, the two sides BC, CF are equal to the two DC, CF, +Const. each to each; and the angle BCF is equal † to the angle DCF; therefore the base BF is equal to the base FD, and the other angles to the 4. 1. * A other angles, to which the equal sides are oppo- M E C KD AED, are bisected by the straight lines AF, FE. From the point F, draw* FG, FH, FK, FL, *12. 1. FM perpendiculars to the straight lines AB, BC, CD, DE, EA: and because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; therefore in the triangles FHC, FKC there are two angles of the one equal to two angles of the other, each to each; and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal,* each to each; * 26. 1. wherefore the perpendicular FH is equal to the perpendicular FK: in the same 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, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, because 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* *16. & the circle therefore each of the straight lines AB, BC, CD, DE, EA touches the circle: wherefore it is inscribed in the pentagon ABCDE. Which was to be done. M *9.1. *6.1. To describe a circle about a given equilateral ana 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 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 B E is the half of the angle BCD, and CDF the half +7 Ax. 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 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. PROP. XV. PROB. To inscribe an equilateral and equiangular hexagon 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, †1.8. and draw the 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 ED, and the triangle †1 Ax. EGD is equilateral; and therefore its three angles EGD, GDE, DEG, are equal to one an- +Cor. 5. other: but the three angles of a triangle are i. equal to two right angles; therefore the angle *32. 1. EGD is the third part of two right angles: in the same manner it may be demonstrated, that the angle DGC 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 angle CGB is the third part of two right angles: therefore the angles EGD, DGC, CGB are E equal to one another and to these are equal* the vertical opposite angles BGA, AGF, FGE: therefore the six an F gles EGD, DGC, CGB, BGA, AGF, FGE, are equal : H 13. 1. 15. 1. 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 the whole cir cumference FABCD is equal to the whole EDCBA: and the angle FED stands upon the circumference FABCD, and the angle AFE upon EDCBA; therefore the angle AFE is +27.3. equal to FED: in the same manner it may be demonstrated that the other angles of the hexgon ABCDEF are each of them equal to the angle AFE or FED: therefore the hexagon is equiangular; and it is equilateral, as was shown; 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. To inscribe an equilateral and equiangular quindecagon 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 *2.4 inscribed in the circle, and ÁB the side of an equilateral and equiangular pen *11.4. tagon inscribed in the same: therefore, of such equal parts B as the whole circumference E ABCDF contains fifteen, the' circumference ABC, being the third part of the whole, contains AF |