to B K. Again, becaufe BK has been proved equal to K C, and KL the double of KC, as also H K the double of BK; HK fhall be equal to K L. So likewife we prove, that GH, GM and M L, are each equal to HK, or KL: Therefore the Pentagon GHKLM is equilateral. I fay, alfo it is equiangular. For because the Angle FKC is equal to the Angle FLC, and the Angle H K L has been proved to be double to the Angle F K C; and alfo K L M double to FLC: Therefore the Angle HKL fhall be equal to the Angle KLM. By the fame Reason we demonftrate, that every one of the Angles K HG, HGM, GML, is equal to the Angle HK L, or KLM. Therefore the five Angles, GHK, HKL, KLM, LMG, MGH, are equal between themfelves. And fo, the Pentagon GHKLM is equiangu lar; and it has been proved likewife to be equilateral, and defcribed about the Circle ABCDE; which was to be done.

PROPOSITION XIII.

PROBLEM.

To defcribe a Circle in an equilateral and equiangular Pentagon.

LET ABCDE be an equilateral and equiangular Pentagon. It is required to infcribe a Circle in the fame.

Bifect the Angles BCD, CDE, by the Right 9. t. 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, becaufe BC is equal to CD, and C F 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 Bafe B F is equal to the Bafe FD; and the Triangle † 4, 1. BFC equal to the Triangle DCF, and the other Angles of the one equal to the other Angles of the other, which are fubtended by the equal Sides: Therefore the Angle C B F fhall be equal to the Angle CDF. And because the Angle CDE is double to the Angle CDF, and the Angle C D E is equal to the Angle ABC, as alfo CDF equal to CBF;

12. I.

† 26. 1.

16. 3°

the Angle C B A 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 B F. After the fame manner we prove, that either of the Angles B A E, or A ED, is bifected by the Right Line AF, or FE. From the Point F draw *FG, FH, FK, FL, FM, perpendicular to the Right Lines A B, 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 F H C, F K C, fhall 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 F C common to each of them : And fo the other Sides of the one will be † equal to, the other Sides of the other, and the Perpendicular. FH equal to the Perpendicular F K. In the fame manner we demonstrate, that F L, F M, or F G, is equal to FH, or F K. Therefore the five Right Lines FG, FH, FK, FL, F M, are equal to each other, and fo a Circle described on the Centre 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 deferibed on the Centre 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

PROPOSITION XIV.

PROBLE M.

To defcribe a Circle about a given equilateral and equiangular Pentagon.

LET ABCDE be an equilateral and equiangular Pentagon. It is required to defcribe a Circle about the fame.

*

Bifect both the Angles B CD, CDE, by the Right Lines CF, FD; and draw FB, FA, FE, from the Point F, in which they meet. Then each of the other Angles CBA, BAE, AED, fhall be bifected by Cor. of the Right Lines B F, FA, FE. And fince the An-preted. gle B C D is equal to the Angle CDE, and the Angle FCD is half the Angle BCD; as likewife CDF, half CDE; the Angle F CD will be equal to the Angle FDC; and fo the Side CF + equal to the Sidet 6. 1. FD. We demonftrate, in like manner, that F B, FA, or F E, is equal to F C, or FD. Therefore the five Right Lines F A, FB, FC, FD, FE, are equal to each other. And fo, a Circle being defcribed on the Centre F, with any of the Distances F A, FB, FC, FD, FE, will pass thro' the other Points, and will be defcribed 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.

L

ET ABCDEF be a Circle given. It is required to infcribe an equilateral and equiangular Hexagon therein.

Draw A D, a Diameter of the Circle ABCDEF, and let G be the Centre; and about the Point D, as a Centre, with the Distance DG,let a Circle, E GCH, be described; join EG, GC, which produce to the Points B, F: Likewife join A B, BC, CD, DE, EF,

13. 1.

15. 1.

FA: I fay, ABCDEF is an equilateral and equiangular Hexagon.

*

:

For, fince the Point G is the Centre of the Circle ABCDEF, G E will be equal to GD. Again, because the Point D is the Centre of the Circle EGCH, DE fhall be equal to DG: But GE has been proved equal to GD; therefore G E is equal to E D. And fo EGD is an equilateral Triangle; and confequently the three Angles thereof, E G D, GDE, DEG, are Cor. 5. 1. equal between themselves. But the three Angles of † 32. I. a Triangle are equal to two Right 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 EGC, CGB; therefore the other Angle, C G B, 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 the Angles EGD, DGC, CGB. Wherefore the fix Angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another. But equal Angles ftand t on equal Circumferences: Therefore the fix Circumferences A B, BC, CD, DE, EF, FA, are equal to each other. But equal Right Lines fubtend equal Circumferences: Therefore the fix Right Lines are equal between themselves; and accordingly the Hexagon ABCDEF is equilateral. I fay, it is also equiangular. For, becauie the Circumference AF is equal to the Circumference E D, add the common Circumference A BCD, and the whole Circumference FABCD is equal to the whole Circumference EDCBA. But the Angle FED ftands on the Circumference FABCD; and the Angle A F E, on the Circumference ED CBA: Therefore the Angle AFE is equal to the Angle DEF. In the fame manner we prove, that the other Angles of the Hexagon ABCDEF are feverally equal to A FE, 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.

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$29.3.

27.3.

Coroll.