Proof. Therefore, since BC is equal to CD (hyp.), and CF common to the triangles BCF, DCF, the two sides BC, CF are equal to the two DC, CF, each to each; and the angle BCF is equal to the angle DCF (constr.); therefore therefore B wherefore H C therefore M 1. The base BF is equal to the base FD (I. 4) ;' and the other angles to the other angles, to which the equal sides are opposite; therefore E 2. 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; also 3. CBA is double of the angle CBF ; 4. The angle ABF is equal to the angle CBF; 5. The angle ABC is bisected by the straight line BF; in the same manner it may be demonstrated, that 6. The angles BAE, AED, are bisected by the straight lines AF, FE. 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 (I. 26); wherefore 7. The perpendicular FH is equal to the perpendicular FK; in the same manner it may be demonstrated, that 8. FL, FM, FG are each of them equal to FH, or FK; 9. The five straight lines FG, FH, FK, FL, FM are equal to one another; B H C 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 the circle (III. 16); therefore 10. Each of the straight lines AB, BC, CD, DE, EA touches the circle: Q.E.F. wherefore it is inscribed in the pentagon ABCDE. B E 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 ABCDE. F D E Construction. Bisect the angles BCD, CDE by the straight lines CF, FD (I. 9), and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E. Proof. It may be demonstrated, in the same manner as the preceding proposition, that wherefore 1. 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 2. The angle FCD is equal to FDC (Ax. 7) ; therefore 3. The side CF is equal to the side FD (I. 6); in like manner it may be demonstrated, that 4. FB, FA, FE, are each of them equal to FC or FD; 5. 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. Q.E.F. PROPOSITION 15.-Problem. 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. C Construction. Find the centre G of the circle ABCDEF, and draw the diameter AGD (III. 1); 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, FÀ. wherefore and therefore E Then the hexagon ABCDEF shall be equilateral and equiangular. Proof. Because G is the centre of the circle ABCDEF, D 1. GE is equal to GD; and because D is the centre of the circle EGCH, 2. DE is equal to DG; H therefore B and 3. GE is equal to ED (Ax. 1), and the triangle EGD is equilateral; but the three angles of a triangle are equal to two right angles (I. 32) ; therefore 4. The three angles EGD, GDE, DEG are equal to one another (I. 5. Cor.) ; 5. The angle EGD is the third part of two right angles; in the same manner it may be demonstrated, that also 6. The angle DGC is 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 (I. 13), 7. The remaining angle CGB is the third part of two right angles; 8. The angles EGD, DGC, CGB are equal to one another; therefore 9. To EGD, DGC, CGB are equal the vertical opposite angles BGA, AGF, FGE (I. 15); 10. The six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another; but equal angles stand upon equal circumferences (III. 26); therefore 11. The six circumferences AB, BC, CD, DE, EF, FA are equal to one another; and equal circumferences are subtended by equal straight lines (III. 29); therefore the six straight lines are equal to one another, and 12. The hexagon ABCDEF is equilateral. It is also equiangular; for, since the circumference AF is equal to ED, to each of these equals add the circumference ABCD; therefore 1. The whole circumference FABCD is equal to the whole EDCBA; and the angle FED stands upon the circumference FABCD, and the angle AFE upon EDCBA; therefore 2. The angle AFE is equal to FED (III. 27) ; 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 3. The hexagon ABCDEF is equiangular; and it is equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Q.E.F. 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 semi-diameter 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 16.-Problem. 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. |