Angle ABC, as alfo CDF equal to CBF; the Angle CBA will be double to the Angle CBF; and fo the Angle ABF equal to the Angle CBF. Wherefore the Angle ABC is bifected by the Right Line BF. After the fame Manner we prove, that either of the Angles BAE or AED is bifected by the Right Lines AF, FE. From the Point F draw *FG, FH, FK, FL, FM, perpendicular to the* 12. 1. Right Lines AB, BC, CD, DE, EA. Then fince the Angle HCF is equal to the Angle KCF; and the Right Angle FHC equal to the Right Angle FKC; the two Triangles FHC, FK C fhall have two Angles of the one equal to two Angles of the ether, and one Side of the one equal to one Side of the other, viz. the Side FC common to each of them. And fo the other Sides of the one will be † equal to † 26, 1. the other Sides of the other: And the Perpendicular FH equal to the Perpendicular FK. In the fame Manner we demonftrate, that FL, FM, or F G, is equal to FH, or FK. Therefore the five Right Lines FG, FH, FK, FL, FM, are equal to each other. And fo a Circle described on the Center F, with either of the Distances FG, FH, FK, FL, FM, will pass thro' the other Points, and fhall touch the Right Lines AB, BC, CD, DE, EA; fince the Angles at G, H, K, L, M, are Right Angles: For if it does not touch them, but cuts them, a Right Line drawn from the Extremity of the Diameter of a Circle at Right Angles to the Diameter, will fall within the Circle; which is abfurd. Therefore a Circle de-‡ 16. 3. scribed on the Center F with the Distance of any one of the Points G, H, K, L, M, will not cut the Right Lines AB, BC, CD, DE, EA; and fo will neceffarily touch them; which was to be done. Coroll. If two of the nearest Angles of an equilateral and equiangular Figure be bifected, and from the Point in which the Lines bifecting the Angles meet, there be drawn Right Lines to the other Angles of the Figure, all the Angles of the Figure will be bifected. PRO Cor. of Preced + 6.1. PROPOSITION XIV. PROBLEM. To defcribe a Circle about a given equilateral and equiangular Pentagon. ET ABCDE be an equilateral and equiangular about the fame. Bifect both the Angles BCD, CDE, by the Right Lines CF; FD, and draw FB, FA, FE, from the Point F; in which they meet. Then each of the Angles CBA, BAE, AED, fhall be bifected by the Right Lines BF, FA, FE. And fince the Angle BCD, is equal to the Angle CDE; and the Angle FCD is half the Angle BCD, as likewise CDF, half CDE; the Angle FCD; will be equal to the Angle FDC; and fo the Side CF +, equal to the Side FD. We demonftrate in like Manner, that FB, FA, or FE, equal to FC, or FD. Therefore the five Right Lines, FA, FB, FC, FD, FE, are equal to each other. And fo a Circle being described on the Center F, with any of the Distances FA, FB, FC, FD, FE, will pafs thro' the other Points, and will be described about the equilateral and equiangular Pentagon ABCDE; which was to be done. PROPOSITION XV. PROBLEM. To infcribe an equilateral and equiangular Hexagon in a given Circle. ET ABCDEF be a Circle given. It is required to inscribe an equilateral and equiangular Hexagon therein. Draw AD a Diameter of the Circle ABCDEF, and let G be the Center; and about the Point D; as a Center, with the Distance DG, let a Circle EGCH, be described; join EG, GC, which produce to the Points B, F: Likewise join AB, BC, CD, DE, EF, FA FA. I fay ABCDEF is an equilateral and equiangular Hexagon. * * For fince the Point G is the Center of the Circle ABCDEF, GE will be equal to GD. Again, because the Point D is the Center of the Circle EGCH, DE fhall be equal to DG: But GE has been proved equal to GD. Therefore GE is equal to ED. And fo EGD is an equilateral Triangle; and confequently the thrée Angles thereof, EGD, GDE, DEG, are equal between themselves: But the Cor. 5 1 three Angles of a Triangle are † equal to two Right † 32. 1. Angles. Therefore the Angle EGD, is a third Part of two Right Angles. In the fame Manner we demonftrate, that DGC is one third Part of two Right Angles: And fince the Right Line CG, ftanding upon the Right Line EB, makes the adjacent Angles † 13. 1. EGC, CGB; the other Angle CGB, is alfo one third Part of two Right Angles. Therefore the Angles EGD, DGC, CGB, are equal between themfelves: And the Angles that are vertical to them, viz. the Angles BGA, AGF, FGE, are equal to thè * 15. 1. Angles EGD, DGC, CGB. Wherefore the fix Angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another. But equal Angles ftand † on † 26. 3. equal Circumferences. Therefore the fix Circumferences, AB, BC, CD, DE, EF, FA, are equal to each other. But equal Right Lines fubtend equal Circumferen- 29. 3. ces. Therefore the fix Right Lines are equal between themselves; and accordingly the Hexagon ABCDEF is equilateral: I fay it is alfo équiangular. For, because the Circumference AF is equal to the Circumference ED, add the common Circumference ABCD, and the whole Circumference F ABCD, is equal to the whole Circumference EDCBA. But the Angle FED, ftands on the Circumference FABCD; and the Angle AFE, on the Circumference EDCBA. Therefore the Angle AFE is equal to the Angle * 27. 3. DEF. In the fame Manner we prove, that the o ther Angles of the Hexagon ABCDEF, are feverally equal to AFE, or FED. Therefore the Hexagon ABCDEF is equiangular. But it has been proved to be alfo equilateral, and is infcribed in the Circle ABCDEF; which was to be done |