Book IV. CBA is also double of the с FK, FL, 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, in the triangles FHC, FKC two angles of the one are equal to two angles of the other, and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore, the other sides are d 26. 1. equal, each to each; that is, 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 that 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 • Cor. 16. 3. right angles to it touches the circle: therefore each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore the circle is inscribed in the pentagon ABCDE. Which was to be done. PROP. XIV. 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. a Book IV. Bisect the angles BCD, CDE by the straight lines a 9. 1. CF, FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE to the points B, A, E. 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 that the angle BCD is equal to the angle CDE, and that FCD is the half of the angle BCD, and B E F CDF the half of CDE; the angle FCD is equal to FDC; wherefore the side CF is equal to the side FD: In like b 6. 1. manner it may be demonstrated, that FB, FA, FE are each of them equal to FC or FD: therefore the five straight lines FÀ, 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. Book IV. Find the centre G of the circle ABCDEF, 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 is 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 EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one a Cor. 5. 1. another; and the three angles of a triangle are equal b 32. 1. to two right 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 is also the third part of two right angles: and because the straight line GC makes with F EB the adjacent angles EGC, c 13. 1. CGB equal to two right angles: the remaining angle CGB is the third part of two right angles; D therefore the angles EGD, DGC, CGB, are equal to one another: and also the angles vertical to d 15. 1. them, BGA, AGF, FGE; DGC, CGB, BGA, AGF, FGE H But e 26. 3. equal angles at the centre stand upon equal e arches; therefore the six arches AB, BC, CD, DE, EF, FA are equal to one another: and equal arches are subtended f 29. 3. by equal f straight lines; therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular; for, since the arch AF is equal to ED, to each of these add the arch ABCD; therefore the whole arch FABCD shall be equal to the whole EDCBA: and the angle FED stands upon the arch 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 equi angular; it is also equilateral, as was shown: and it is in- Book IV. scribed 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 radius 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 shall 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. PROP. XVI. PROB. 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. in A a 2. 4. b 11. 4. F Let AC be the side of an equilateral triangle inscribed a in the circle, and AB the side of an equilateral and equiangular pentagon inscribed the same; therefore, of such equal parts as the whole cir- B cumference ABCDF contains fifteen, the arch ABC, being E the third part of the whole, contains five; and the arch AB, which is the fifth part of the whole, contains three; there D с fore BC their difference contains two of the same parts, c 30. 3. 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 around in the whole d 1.4. Book IV. 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 quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon may be described about it: And likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it. |