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the Angle CAD be equal to that at F, and ACD, CDA, each equal to the Angles G or H. Wherefore the Angles ACD, CDA, are cach double to the Angle CAD. This being done, bifect* ACD, CDA, by the Right Lines CE, DB, and join AB, BC, DE, EA.

*

Then because each of the Angles ACD, CDA, is double to CAD, and they are bifected by the Right Lines CE, DB; the five Angles DAC,ACE, ECD, CD B, BDA, are equal to each other. But equal Angles ftand upon equal Circumferences. Therefore the five Circumferences A B, BC, CD, DE, EA, are equal to each other. But equal Circumferences fubtend equal Right Lines. Therefore the five Right Lines AB, BC, CD, DE, EA, are equal to each other. Wherefore ABCDE is an equilateral Pentagon. I fay, it is alfo equiangular; for because the Circumference AB is equal to the Circumference DE, by adding the Circumference BCD, which is common, the whole Circumference ABCD is equal to the whole Circumference EDCB, but the Angle AED ftands on the Circumference A B C D, and BAE on the Circumference EDCB: Therefore the Angle BAE is equal to the Angle A ED. For the fame Reason, each of the Angles ABC, BCD, CDE, is equal to BAE, or AED. Wherefore the Pentagon ABCDE is equiangular; but it has been prov'd to be alfo equilateral. And confequently there is an equilateral and equiangular Pentagon inscribed in a given Circle; which was to be done.

PROPOSITION XII.

PROBLEM.

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

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

Let A, B, C, D, E, be the angular Points of a Pentagon fuppofed to be infcribed in the Circle; fo that the Circumferences A B, BC, CD, DE, E A, be

equal;

equal; and let the Right Lines GH, HK, KL, LM,
MG, be drawn touching † the Circle in the Points † 17.3.
A, B, C, D, E: Let F be the Center of the Circle
ABCDE, and join FB, FK, FC, FL, FD.

18. 3.

Then because the Right Line KL touches the Circle ABCDE in the Point C, and the Right Line FC is drawn from the Center F to C, the Point of Contact; FC will be ‡ perpendicular to KL: And fo both the Angles at Care Right Angles. For the fame Reafon, the Angles at the Points B, D, are Right Angles. And because FCK is a Right Angle, the Square of FK will be * equal to the Squares of FC, 47. 1. CK: And for the fame Reason, the Square of FK is equal to the Squares of FB, BK. Therefore the Squares of FC, 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 fhall be equal to the Square BK; and fo BK is equal to CK. And because F B is equal to FC, and FK is common; the two Sides BF, FK, are equal to the two CF, FK, and the Bafe BK is equal to the Bafe KC; and fo the Angle BF K fhall be † equal to the Angle + 5. 1. KFC, and the Angle BK F to the Angle F K C. Therefore the Angle B F C is double to the Angle KFC, and the Angle B K C double to the Angle FKC: For the fame Reason, the Angle CFD is double to the Angle CF L, and the Angle CLD double to the Angle CLF. And becaufe the Circumference BC is equal to the Circumference CD, the Angle BFC fhall be equal to the Angle CFD. ‡ 27. 3. 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 fo FKC, FLC, are two Triangles having two Angles of the one equal to two Angles of the other, each to each, and one Side of the one equal to one Side of the other, viz. the common Side FC; wherefore they fhall have the other Sides of the one equal † 26. 1. 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 fince KC is equal to CL, KL fhall be double to KC. And by the fame Reason, we prove that

HK

+4. I.

HK is double to BK. Again, because BK has been prov'd equal to KC, and KL the double to KC, as alfo HK the double of BK, HK fhall be equal to KL. So likewife, we prove that GH, GM, and M L, are each equal to H K, or K L. Therefore the Pentagon GHKLM is equilateral. I fay also, it is equiangular; for because the Angle FK C is equal to the Angle FLC; and the Angle HKL has been prov'd to be double to the Angle FKC; and alfo KLM double to FLC: Therefore the Angle HKL fhall be equal to the Angle KLM. By the fame Reafon we demonftrate, that every 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 themfelves. And fo the Pentagon GHKLM is equiangular, and it has been proved likewife to be equilateral, and defcribed about the Circle ABCDE; which was to be done.

PROPOSITION XIII

PROBL E M.

Tonfcribe 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 Lines CF, DF, and from the Point F wherein CF, DF, meet each other, let the Right Lines FB, FA, FE, be drawn. Now because BC is equal to CD, and CF is common, the two Sides B C, C F, are equal to the two Sides DC, CF; and the Angle BCF is equal to the Angle DCF. Therefore the Bafe BF is equal to the Bafe FD; and the Triangle 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 CBF fhall 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

2

Angle

Angle A B C, 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, FKC 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 FC common to each of them. And fo the other Sides of the one will be f 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 defcribed on the Center F, with either of the Distances F G, FH, FK, FL, FM, will pafs thro' the other Points, and fhall touch the Right Lines AB, BC, CD, DÉ, 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. fcribed 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 eqilateral 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.

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PRO

Preced.

PROPOSITION XIV.

PROBLEM.

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

LET ABCDE be an equilateral and equiangular Pentagon. It is requir'd to describe a Circle 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 An*Cor. of gles CBA, BAE, AED, shall be bifected * by the Right Lines B F, FA, FE. And fince the Angle BCD, is equal to the Angle CDE; and the Angle FCD, is half the Angle BCD, as likewife CDF, half CDE; the Angle FCD, will be equal to the Angle FDC; and fo the Side CFt, equal to the Side FD. We demonftrate in like Manner, that FB, FA, or F E, is 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 defcrib'd on the Center F, with any of the Distances FA, FB, FC, FD, FE, will pass thro' the other Points, and will be defcrib'd about the equilateral and equiangular Pentagon ABCDE; which was to be done.

† 6.1.

PROPOSITION XV.

PROBLEM.

To infcribe an equilateral and equiangular Hexagon in a given Circle.

ET ABCDEF be a Circle given. It is requir'd to infcribe 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 defcrib'd; join EG, GC, which produce to the Points B, F: Likewife join AB, BC, CD, DE, EF,

FA.

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