G For, since the angle FKC is equal to the angle FLC, therefore the angle HKL is equal [Axiom 6. In the same manner it may be shewn that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM ; H M B Κ therefore the pentagon GHKLM is equiangular. And it has been shewn to be equilateral. Wherefore an equilateral and equiangular pentagon has been described about the given circle. Q.E.F. PROPOSITION 13. PROBLEM. 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. and the angle BCF is equal to the angle DCF; [Constr. therefore the base BF is equal to the base DF, and the [I. 4. other angles to 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 the angle CDF, and that the angle CDE is equal to the angle CBA, and the angle CDF is equal to the angle CBF, therefore the angle CBA is double of the angle CBF; therefore the angle ABF is equal to the angle CBF; therefore the angle ABC is bisected by the straight line BF. In the same manner it may be shewn that the angles BAE, AED are bisected by the straight lines AF, EF. From the point F draw FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA. [Î. 12. Then, because the angle FCH is equal to the angle FCK, and the right angle FHC equal to the right angle FKC; therefore in the two 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 their other sides are equal, each to each, and therefore the perpendicular FH is equal to the perpendicular FK. [I. 26. In the same manner it may be shewn 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, and the circle described from the centre F, at the distance of any one of them will pass through the extremities of the other four; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right angles, [Construction. and the straight line drawn from the extremity of a diameter, at right angles to it, touches the circle; [III. 16. Therefore each of the straight lines AB, BC, CD, DE, EA touches the circle. Wherefore a circle has been inscribed in the given equilateral and equiangular pentagon. Q.E.F. PROPOSITION 14. PROBLEM. 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, DF; [1.9. and from the point F, at which they meet, draw the straight lines FB, FA, FE. Then it may be shewn, as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines BF, AF, EF BA E And, because the angle BCD is equal to the angle CDE, and that the angle FCD is half of the angle BCD, and the angle FDC is half of the angle CDE, therefore the angle FCD is equal to the angle FDC; [Ax. 7. therefore the side FC is equal to the side FD. [I. 6. In the same manner it may be shewn 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 any one of them, will pass through the extremities of the other four, and will be described about the equilateral and equiangular pentagon ABCDE. Wherefore a circle has been described about the given equilateral and equiangular pentagon. Q.E.F. 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, [III. 1. and draw the diameter AGD; join EG, CG, and produce them and because D is the centre therefore GE is equal to DE, and the triangle EGD is equilateral; H [Axiom 1. therefore the three angles EGD, GDE, DEG are equal to one another. [I. 5. Corollary. But the three angles of a triangle are together equal to two right angles; [I. 32. therefore the angle EGD is the third part of two right angles. In the same manner it may be shewn, that the angle DGC is the third part of two right angles. And because the straight line GC makes with the straight line EB the adjacent angles EGC, CGB together equal to two right angles, [I. 13. therefore the remaining 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. [I. 15. Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another. But equal angles stand on equal arcs; [III. 26. therefore the six arcs AB, BC, CD, DE, EF, FA are equal to one another. And equal arcs are subtended by equal straight lines ; [III.29. therefore the six straight lines are equal to one another, and the hexagon is equilateral. It is also equiangular. For, the arc AF is equal to the arc ED; to each of these add the arc ABCD; therefore the whole arc FABCD is equal to the whole arc ABCDE; and the angle FED stands on the arc FABCD, and the angle AFE stands on the arc ABCDE; therefore the angle FED is equal to the angle AFE. H [III. 27. In the same manner it may be shewn 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 has been shewn to be equilateral; and it is inscribed in the circle ABCDEF. Wherefore an equilateral and equiangular hexagon has been inscribed in the given circle. Q.E.F. COROLLARY. 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.. Also, 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 the circle, as may be shewn from what was said of the pentagon; and a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like that used for the pentagon. |