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the remaining square of BK, and the straight 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 base BK is equal to the base KC: therefore the angle BFK is equal to the angle KFC, and the angle BKF to c 8. 1. FKC; wherefore the angle BFC is double of the angle KFC, and BKC double of FKC: For the same 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 f to the

G

angle CFD; and BFC is dou

ble of the angle KFC, and

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CFD double of CFL; therefore the angle KFC is equal to

the angle CFL; and the right H

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angle FCK is equal to the right

angle FCL: Therefore, in the two triangles FKC, FLC, there are two angles of one equal to

B

D

two angles of the other, each

K

to each, and the side FC, which

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£ 27.3.

is adjacent to the equal angles in each, is common to both; therefore the other fides shall be equal & to the other fides, and the third g 26. 1. angle to the third angle: Therefore the straight 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 same manner, it may be shewn that HK is double of BK: And because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK shall be equal to KL: In like manner, it may be shewn 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 demonstrated, the angle HKL is equal to KLM: And in like manner it may be shewn, 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 described about the cirele ABCDE. Which was to be done.

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Book IV.

PROP. XIII. PRO B.

O inscribe a circle in a given equilateral and equi

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2 9. 1.

b 4. I.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to inscribe a circle in the pentagon ABCDE. Bisect the angles BCD, CDE by the straight lines CF, DF, and from the point F in which they meet draw the straight lines FB, FA, FE: Therefore, since 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 base BF is equal to the base FD, and the other angles to the other angles, to which the equal fides are opposite; 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; CВА

C 12. 1.

fected by the straight lines AF, H

d 26. 1.

is also double of the angle CBF;
therefore the angle ABF is equal
to the angle CBF; wherefore the
angle ABC is bisected. by the
straight line BF: In the same man-

ner, it may be demonftrated, that B

the angles BAE, AED are bi

FE: From the point F draw FG, FH, FK, FL, FM perpen-
diculars to the straight lines AB,
BC, CD, DE, EA: And be-
cause 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 shall be equald, 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 straight lines, FG,
FH, FK, FL, FM are equal to one another: Wherefore the cir-
cle described from the center F, at the distance of one of these

five, shall pass through the extremities of the other four, and
touch

A

G

M

F

E

L

CKD

touch the straight lines AB, BC, CD, DE, EA, because the Book IV. angles at the points G, H, K, L, M are right angles; and that a straight 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 inscribed in the pentagon ABCDE. Which

was to be done.

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PROP. XIV. PROB.

O describe a circle about a given equilateral and e-
quiangular pentagon

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

Bisect the angles BCD, CDE by the straight lines CF, FD, 29. 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 demonstrated, in

A

the fame manner as in the preceeding

proposition, that the angles CВА, BAE, AED are bisected by the B straight lines FB, FA, FE: And

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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

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CF is equal to the fide FD: In like manner it may be demon- b 6. 1. ftrated that FB, FA, FE are each of them equal to FC or FD: Therefore the five straight lines FA, FB, FC, FD, FE are equal to one another; and the circle described from the center F, at the distance of one of them, shall pass through the extremities of the other four, and be described about the equilateral

and equiangular pentagon ABCDE. Which was to be done.

1

H2

PROP.

7

Book IV.

See N.

T

PROP. XV. PROB.

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

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

Find the center G of the circle ABCDEF, and draw the di ameter AGD; and from D as a center, at the distance DG, describe 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.

a 5. 1.

b 32. Ie

• 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 ifosceles 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 demonstrated that

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d 15. 1.

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f 9.3.

the angle DGC is also the third part
of two right angles: And because the
straight line GC 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 these are equal d 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 stand upon equal circumfe-
rences; therefore the fix circumfe-
rences AB, BC, CD, DE, EF, FA are equal to one another:
And equal circumferences are fubtended by equal f straight
lines; therefore the fix straight lines are equal to one another,
and the hexagon ABCDEF is equilateral. It is alfo equiangu-
lar; for, fince the circumference AF is equal to ED, to each of
these add the circumference ABCD; therefore the whole cir-
cumference FABCD shall be equal to the whole EDCBA:
And

A

B

G

C

D

H

And the angle FED stands 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 shewn; and it is infcribed in the given circle ABCDEF. Which was to be done.

COR. From this it is manifest, that the side of the hexagon is equal to the straight line from the center, that is, to the semidiameter of the circle.

And if thro' the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangu. Jar hexagon shall be described about it, which may be demonstrated from what has been faid of the pentagon; and likewife a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used 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 infcribe an equilateral and equiangular quindecagon in the circle ABCD.

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 inscribed in the same; therefore, of fuch equal parts b 11. 4. as the whole circumference ABCDF contains fifteen, the cir

cumference ABC, being the third

part of the whole, contains five;
and the circumference AB, which
is the fifth part of the whole, con-
tains three; therefore BC their dif B

A

F

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straight lines BE, EC be drawn, and

straight lines equal to them be placed d around in the whole cir-d 1.4

cle, an equilateral and equiangular quindecagon shall be in

scribed in it. Which was to be done.

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