tadt ; equal; and let the Right Lines GH, HK, KL, LM, Then because the Right Line K L touches the Circle ABCDE in the Point C, and the Right Line FC is drawn from the Center F to C, the Point of Con FC will be I perpendicular to KL: And so I 18. 36 both the Angles at Care Right Angles. For the same Reason, the Angles at the Points Bs D, are Right Angles. And because FCK is a Right Angle, the Square of FK will be * equal to the Squares of FC, * 47. to CK: And for the same Reason, the Square of FK is equal to the Squares of FB, BK. Therefore the Squares of F C, CK are equal to the Squares of FB, BK. But the Square of FC is equal to the Square of FB. Wherefore the Square of CK shall be equal to the Square BK; and fo BK is equal to CK. And because FB is equal to FC, and FK is common ; the two Sides BF, F K, are equal to the two CF, FK, and the Base BK, is equal to the Base KC; and fo the Angle BFK shall be t equal to the Angle † 8. I. KFC, and the Angle BKF to the Angle FKC. Therefore the Angle BFC is double to the Angle KFC, and the Angle BKC double to the Angle FKC: For the same Reason, the Angle CFD is double to the Angle CFL, and the Angle CLD double to the Angle CLF: "And because the Cir. cumference BC is equal to the Circumference CD, the Angle BFC shall be f equal to the Angle CFD. 197. 38 But the Angle BFC is double to the Angle KFC, and the Angle DFC double to LFC. Therefore the Angle KFC is equal to the Angle CFL. And so F KC, FLC, are two Triangles, having two Angles of the one equal to two Angles of the other, tach to each, and one side of the one equal to one Side of the other, viz. the common Side FC; wherefore they fhall have t the other Sides of the one equal † 26. t. to the other sides of the other; and the other Angle of the one, equal to the other Angle of the other. Therefore the Right Line KC is equal to the Right Line CL, and the Angle FKC to the Angle FLC: And since KC is equal to CL, KL shall be double to KC. And by the same Reason, we prove that НК. HK is double to BK. Again, because BK has been proved equal to KC, and KL the double to K C, as also HK the double of BK, HK shall be equal to KL. So likewise, we prove that GH, GM, and ML, are each equal to HK, or KL. Therefore the Pentagon GHKLM is equilateral. I say also, it is equiangular; for becausc the Angle FKC is equal to the Angle FLC; and the Angle HKL has been proved to be double to the Angle FKC; and also KLM double to FLC: Therefore the Angle HKL fhall be equal to the Angle KLM. By the same Reason we demonstrate, that every one of the Angles KHG, HGM, GML, is equal to the Angle HKL, or KLM. Therefore the five Angles, GHK, HKL, KLM, LMG, MGH, are equal between themselves. And so the Pentagon GHKLM is equiangular, and it has been proved likewise to be equilateral, and described about the Circle ABCDE; which was to be done. PROPOSITION XIII. PROBLEM. gular Pentagon. ET ABCDE be an equilateral and equiangular the same. Bisect * the Angles BCD, CDE, by the Right Lines CF, DF, and from the Point F wherein CF, DF, meet each other, let the Right Lines F B, FA, FE, be drawn. Now because BC is equal to CD, and CF is common, the two Sides BC, CF, are equal to the two Sides DC, CF; and the Angle BCF is equal to the Angle DCF. Therefore the Base BF is equal to the Base FD; and the Triangle BF C equal to the Triangle DCF, and the other Angles of the one equal to the other Angles of the other, which are subtended by the equal Sides: Therefore the Angle CBF shall be equal to the Angle CDF. And because the Angle CDE is double to the Angle CDF, and the Angle CDE is equal to the Angle Angle ABC, as also CDF equal to CBF; the Angle A B C is bisected 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. I. Right Lines, AB, BC, CD, DE, EA. : Then since the Angle HCF is equal to the Angle KCF; and the Right Angle_FHC equal to the Right Angle FKC; the two Triangles F HC, FKC Thall have two Angles of the one equal to two Angles of the other, and one side of the one equal to one side of the other, viz. the Side FC common to each of them. And so the other Sides of the one will be t 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 FG, 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 shall 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 s absurd. Therefore a Circle de- 1 16.30 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 so will neceffarily touch them; which was to be done. Coroll. If two of the nearest Angles of an equilateral and equiangular Figure be bisected, and from the PRO PROPOSITION XIV: PROBLEM. equiangular Pentagon. about the same. Bifect both the Angles BCD, CDE, by the Right Lines F; FD, and draw FB, FA, FÉ, from the Point F; in which they meet. Then each of the An*Cor. of gles CBA, BAE, AED, shall be bifected * by the Preced. Right Lines BF, FA, FE. And since the Angle half CDE; the Angle FCD; will be equal to the + 6.1. Angle FDC; and so the Side CFt, equal to the Side FD. We demonstrate 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 pass thro' the other Points, and will be described about the equilateral and equiangular Pentagon ABCDE; which was to be done, PROPOSITION XV. PROBLEM. in a given Circle. 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 jóin AB, BC, CD, DE, EF, FA: FA. I fay ABCDEF is an equilateral and equiángular 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 shall be equal to DG: But GE has been proved equal to GD. Therefore G E is equal to ED. And so EGD is an equilateral Triangle; ånd consequently the thrée Angles thereof, EGD, GDE, DEG, are * equal between themselves : But the * Cor. 5. t. three Angles of a Triangle are + equal to two Right # 32. I. Angles. Therefore the Angle EGD, is a third Part of two Right Angles. In the fame Manner we demonstrate, that DGC is one third Part of two Right Angles: And fince the Right Line CG, standing upon the Right Line EB, makes † the adjacent Angles † iz. , EGC, CGB; the other Angle CGB, is also one third Part of two Right Angles. Therefore the Anglés ËĞD, DGC, CGB, are equal between themfelves: And the Angles that are vertical to them, viz. the Angles BGA, AGF, FGE, are * equal to the * 15. 1. Angles E GD, DGC, CGB. Wherefore the six Angles E GD, DGC, CGB BGA, AGF, FGE, are equal to one another. But équal Angles stand + 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 I equal Circumferen- † 29. 3. ces. Therefore the fix Right Lines are equal between themselves; and accordingly the Hexagon ABCDEF is equilateral : I say it is also equiangular. For, because the circumference AF is equal to the Circumference ED, add the common Circumference ABCD, and the whole Circumference FABCD, is equal to the whole Circumference EDCBA. But the Angle FED, stands 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 severally equal to AF E, or FED. Therefore the Hexagon ABCDEF is equiangular. But it has been proved to be also equilateral, and is inscribed in the Circle ABCDEF; which was to be done. |