the remaining fquare of BK, and the ftraight line CK equal to Book IV. BK; And because FB is equal to FC, and FK common to the triangles BFK, CFK, the two BF, FK are equal to the two CF, FK; and the bafe BK is equal to the bafe KC: therefore the angle BFK is equal to the angle KFC, and the angle BKF to e 6. 1. FKC; wherefore the angle BFC is double of the angle KFC, and BKC double of FKC: For the fame reason, the angle CFD is double of the angle CFL, and CLD double of CLF: And because the circumference BC is equal to the circumference CD, the angle BFC is equal to the

£27.3.

angle CFD; and BFC is dou

G

E

CFD double of CFL; there

two angles of the other, each to each, and the fide FC, which

K

is adjacent to the equal angles in each, is common to both; therefore the other fides (hall be equal to the other fides, and the third g 26. 1. angle to the third angle: Therefore the ftraight line KC is equal to CL, and the angle FKC to the angle FLC: And because KC is equal to CL, KL is double of KC: In the fame manner, it may be fhewn that HK is double of BK: And becaufe BK is equal to KC, as was demonftrated, and that KL is double of KC, and HK double of BK, HK fhall be equal to KL: In like manner, it may be fhewn that GH, GM, ML are each of them equal to HK or KL: Therefore the pentagon GHKLM is equilateral. It is also equiangular; for, fince the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and KLM double of FLC, as was before demonftrated, the angle HKL is equal to KLM: And in like manner it may be fhewn, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM: Therefore the five angles GHK, HKL, KLM, LMG, MGH being equal to one another, the pentagon GHKLM is equiangular : And it is equilateral, as was demonftrated; and it is defcribed about the circle ABCDE. Which was to be done.

Book IV.

2 9. 1.

b 4. I.

€ 12. 1.

8 26. 1.

PROP. XIII. PROB.

O infcribe a circle in a given equilateral and equi

Tangular pentagon.

B

G

A

M

Let ABCDE be the given equilateral and equiangular pentagon; it is required to infcribe a circle in the pentagon ABCDE. Bifect the angles BCD, CDE by the ftraight lines CF, DF, and from the point F in which they meet draw the straight lines FB, FA, FE: Therefore, fince BC'is equal to CD, and CF common to the triangles BCF, DCF, the two fides BC, CF are equal to the two DC, CF; and the angle BCF is equal to the angle DCF; therefore the bafe BF is equal to the base FD, and the other angles to the other angles, to which the equal fides are op pofite; therefore the angle CBF is equal to the angle CDF: And because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF; CBA is alfo double of the angle CBF; therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bifected. by the ftraight line BF: In the fame manner, it may be demonftrated, that the angles BAE, AED are bifected by the ftraight lines AF, FE: From the point F draw c FG, FH, FK, FL, FM perpendiculars to the ftraight lines AB, BC, CD, DE, EA: And because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; in the triangles FHC, FKC there are two angles of one equal to two angles of the other, and the fide FC, which is oppofite to one of the equal angles in each, is common to both; therefore the other fides fhall be equal, each to each; wherefore the perpendicular FH is equal to the perpendicular FK: In the fame manner it may be demonstrated that FL, FM, FG are each of them equal to FH or FK; therefore the five ftraight lines, FG, FH, FK, FL, FM are equal to one another: Wherefore the circle defcribed from the center F, at the distance of one of these five, fhall pass through the extremities of the other four, and

H

CK D

Ꮮ

touch

touch the ftraight lines AB, BC, CD, DE, EA, because the Book IV. angles at the points G, H, K, L, M are right angles; and that in a ftraight line drawn from the extremity of the diameter of a circle at right angles to it, touches the circle: Therefore each e 16, 3. of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore it is infcribed in the pentagon ABCDE. Which was to be done.

PROP. XIV. PRO B.

O defcribe a circle about a given equilateral and e

Tquiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to defcribe a circle about it.

Bifect the angles BCD, CDE by the ftraight lines CF, FD, a 9. 1.

and from the point F in which they meet draw the straight

lines FB, FA, FE to the points B,
A, E. It may be demonftrated, in
the fame manner as in the preceeding
propofition, that the angles CBA,
BAE, AED are bifected by the B
ftraight lines FB, FA, FE: And
because the angle BCD is equal to
the angle CDE, and that FCD is
the half of the angle BCD, and CDF
the half of CDE; the angle FCD is
equal to FDC; wherefore the fide

A

F

E

CF is equal to the fide FD: In like manner it may be demon- b 6. 8. ftrated that FB, FA, FE are each of them equal to FC or FD: Therefore the five ftraight lines FA, FB, FC, FD, FE are equal to one another; and the circle defcribed from the center F, at the distance of one of them, fhail pafs through the extre mities of the other four, and be defcribed about the equilateral and equiangular pentagon ABCDE. Which was to be done.

7

Book IV.

See N.

a 5. I. b 32. I.

C13. I.

d IS. I.

€ 26.3.

f 9.3.

P.RO P. XV. PROB.

O infcribe an equilateral and equiangular hexagon in a given circle.

Let ABCDEF be the given circle; it is required to infcribe an equilateral and equiangular hexagon in it.

Find the center G of the circle ABCDEF, and draw the diameter AGD; and from D as a center, at the distance DG, defcribe the circle EGCH, join EG, CG, and produce them to the points B, F; and join AB, BC, CD, DE, EF, FA: The hexagon ABCDEF is equilateral and equiangular.

F

A

B

G

Because G is the center of the circle ABCDEF, GE is equal to GD: And because D is the center of the circle EGCH, DE is equal to DG; wherefore GE is equal to ED, and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, because the angles at the base of an ifofceles triangle are equal; and the three angles of a triangle are equal to two right angles; therefore the angle EGD is the third part of two right angles: In the fame manner it may be demonftrated that the angle DGC is alfo the third part of two right angles : And because the ftraight line GČ makes with EB the adjacent angles EGC, CGB equal to two right angles; the remaining angle CGB is the third part of two right angles; therefore the angles EGD, DGC, CGB are equal to one another: And to thefe are equal the vertical oppofite angles BGA, AGF, FGE: Therefore the fix angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another: But equal angles ftand upon equal circumferences; therefore the fix circumferences AB, BC, CD, DE, EF, FA are equal to one another : And equal circumferences are fubtended by equal f straight lines; therefore the fix ftraight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is alfo equiangular; for, fince the circumference AF is equal to ED, to each of thefe add the circumference ABCD; therefore the whole circumference FABCD fhall be equal to the whole EDCBA:

e

D

H

And

And the angle FED ftands upon the circumference FABCD, Book IV. and the angle AFE upon EDCBA; therefore the angle AFE is equal to FED: In the fame 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 is equilateral, as was fhewn; and it is infcribed in the given circle ABCDEF. Which was to be done. COR. From this it is manifeft, that the fide of the hexagon is equal to the ftraight line from the center, that is, to the femidiameter of the circle.

And if thro' the points A, B, C, D, E, F there be drawn ftraight lines touching the circle, an equilateral and equiangu Jar hexagon fhall be defcribed about it, which may be demonstrated from what has been faid of the pentagon; and likewise a circle may be infcribed in a given equilateral and equiangular hexagon, and circumfcribed about it, by a method like to that ufed for the pentagon.

PROP. XVI. PROB.

TO infcribe an equilateral and equiangular quindeca- See N.

gon in a given circle.

Let ABCD be the given circle; it is required to inferibe an equilateral and equiangular quindecagon in the circle ABCD.

a

Let AC be the fide of an equilateral triangle infcribed in a 2, 4. the circle, and AR the fide of an equilateral and equiangular pentagon infcribed in the fame; therefore, of fuch equal parts b 11. 4. as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three; therefore BC their dif B ference contains two of the fame parts: Bifect BC in E; therefore E BE, EC are, each of them, the fifteenth part of the whole circumfe- C rence ABCD: Therefore, if the

ftraight lines BE, EC be drawn, and

A

F

C 39. 3.

D

ftraight lines equal to them be placed around in the whole cir- d 1.4. cle, an equilateral and equiangular quindecagon fhall be infçribed in it. Which was to be done.