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. B 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. 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. Find the centre G of the circle ABCDEF, [III. 1. and draw the diameter AGD; join EG, CG, and produce them For, because C 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; 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; [TTT. 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. 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. 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. Let AC be the side of an and let AB be the side of an [IV. 11. Then, of such equal parts as the whole circumference E ABCDF contains fifteen, the arc ABC, which is the third part of the whole, contains five, and the arc AB, which is the fifth part of the whole, contains three; therefore their difference, the arc BC, contains two of the same parts. Bisect the arc BC at E; [III. 30. therefore each of the arcs BE, EC is the fifteenth part of the whole circumference ABCDF. Therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placed round in the whole circle, [IV. 1. an equilateral and equiangular quindecagon will be inscribed in it. Q.E.F. And, in the same manner as was done for the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon will be described about it; and also, as for the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it. 1. A LESS magnitude is said to be a part of a greater magnitude, when the less measures the greater; that is, when the less is contained a certain number of times exactly in the greater. 2. A greater magnitude is said to be a multiple of a less, when the greater is measured by the less; that is, when the greater contains the less a certain number of times exactly. 3. Ratio is a mutual relation of two magnitudes of the same kind to one another in respect of quantity. 4. Magnitudes are said to have a ratio to one another, when the less can be multiplied so as to exceed the other. 5. The first of four magnitudes is said to have the same ratio to the second, that the third has to the fourth, when any equimultiples whatever of the first and the third being taken, and any equimultiples whatever of the second and the fourth, if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth, and if the multiple of the first be equal to that of the second, the multiple of the third is also equal to that of the fourth, and if the multiple of the first be greater than that of the second, the multiple of the third is also greater than that of the fourth. 6. Magnitudes which have the same ratio are called proportionals. |