CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE. B G A M E Then, because 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, 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 other angles, each to each, 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 CBA, and the angle CDF to CBF, therefore the angle CBA is double of the angle CBF, that is, the angle ABC is bisected by the straight line BF: And in like manner it may be demonstrated, that the angles BAE, AED are bisected by the straight lines AF, EF. H C K D From the point F, draw FG, FH, FK, FL, FM perpendiculars on AB, BC, CD, DE, EA: Then, because the angle FCH is equal to the angle FCK, and the right angle FHC to the right angle FKC, therefore in the two triangles FHC, FKC there are two angles of 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, and therefore FH is equal to FK: And in like 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, 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 touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right angles (3. 16): Wherefore a circle has been inscribed in the given equilateral and equiangular pentagon. Q. E. F. 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. Bisect the angles BCD, CDE by the straight lines CF, DF, meeting in F, and join 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: And, because the angle BCD is equal B E to the angle CDE, and that FCD is the half of BCD, and CDF of CDE, therefore the angle FCD is equal to the angle FDC, and the side FC to the side FD: And 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 any 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. 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, and draw the diameter AGD; and from the centre D, at the distance DG, describe the circle EGCH; join EG, CG, produce them to the points B, F, and join AB, BC, CD, DE, EF, FA: the hexagon ABCDEF is equilateral and equiangular. F G A B D H For, 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; therefore GE is equal to DE, and the triangle EGD is equilateral, and therefore also equiangular (1.5. Cor.): But the three angles of a triangle are together equal to two right E angles; therefore the angle EGD is the third part of two right angles: And, in like manner, it may be shewn 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 together equal to two right angles, therefore the remaining angle CGB is also the third part of two right angles, and therefore the angles EGD, DGC, CGB are equal to one another: And to these are equal the vertical opposite angles BGA, AGF, FGE; therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another: But equal angles stand upon equal 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 six straight lines are equal to one another, and the hexagon ABCDEF is equilateral: It is also equiangular: For, because the circumference AF is equal to the circumference ED, to each of these add the circumference ABCD; therefore the whole circumference FABCD is equal to the whole circumference ABCDE: And the angle FED stands upon the circumference FABCD, and the angle AFE upon the circumference ABCDE; therefore the angle FED is equal to the angle AFE: And, in like 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 equiangular; and it has been shewn to be equilateral: Wherefore, in the given circle, has been inscribed an equilateral and equiangular hexagon. 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 radius 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 it, as 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. Let AC be the side of an equilateral triangle inscribed in the circle (4. 2), and AB the side of an equilateral and B E equiangular pentagon inscribed in the same (4. 11): Then, of such equal parts as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third part of the whole, contains five, and the circumference AB, being the fifth part of the whole, contains three; therefore the circumference BC contains two of the same parts: Bisect BC in E (3. 30); then BE, EC are each of them the fifteenth part of the whole circumference: Therefore, if the straight lines BE, EC be drawn, and straight lines equal to them be placed round in the whole circle (4.1), an equilateral and equiangular quindecagon will be inscribed in it. Q.E.F. 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 will 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. |