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the remaining square of BK, and the straight line CK equal Book IV.
to 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 e8. 1.
the angle BKF to FKC; wherefore the angle BFC is double
the angle KFC, and BKC double FKC: For the same reason,
the angle CFD is double the angle CFL, and CLD double
CLF: And because the circumference BC is equal to the
circumference CD, the angle
BFC is equal to the angle
CFD: and BFC is double the

angle KFC, and CFD double
CFL; therefore the angle KFC
is equal to the angle CFL; and H
the right angle FCK is equal
to the rightangle FCL: There-
fore, in the two triangles FKC, B
FLC, there are two angles of
one equal to two angles of the
other, each to each, and the

side FC, which is adjacent to

G

f 27. 3.

A

E

M

F

D

K

C

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the equal angles in each, is common to both; therefore the other sides are equal to the other sides, and the third angle g 26. 1. 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 KC: In the same manner, it may be shown that HK is double BK: And because BK is equal to KC, as was demonstrated, and KL is double KC, and HK double BK, HK is equal to KL: In like manner, it may be shown that GH, GM, ML are each of them equal to HK or KL: Therefore the pentagon GHKLM is equilateral. It is also equiangular: for, since the angle FKC is equal to the angle FLC, and the angle HKL is double the angle FKC, and KLM double FLC, as was before demonstrated, the angle HKL is equal to KLM: And in like manner, it may be shown 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 demonstrated; and it is described about the circle ABCDE. Which was to be done.

Book IV.

a 9. 1.

b 4. 1.

c 12 1.

PROP. XIII. PROB.

To inscribe a circle in a given equilateral and equiangular pentagon.

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 line, 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 sides 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 sides are opposite: therefore the angle CBF is equal to the angle CDF: And because the angle CDE is double CDF, and that CDE is

equal to CBA, and CDF to
CBF; CBA is also double the
angle CBF; therefore the angle
ABF is equal to the angle CBF;
wherefore the angle ABC is bi-B
sected by the straight line BF:
In the same manner, it may be
demonstrated, that the angles H
BAE, AED are bisected by the
straight lines AF, FE: From
the point F draw FG, FH,
FK, FL, FM perpendiculars
to the straight lines AB, BC,
CD, DE, EA: And because

A

G

M

CK D

L

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 the one equal to two angles of the other, and the side FC, which is opposite to one of the equal angles in each, is common to both: therefore the other sides are d 26. 1. equal, equal, each to each; wherefore the perpendicular FH is equal to the perpendicular FK: In the same 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 centre F, at the distance of one of these Book IV. five, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, because the 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: There- e Cor. 16.3. fore each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore the circle is inscribed in the pentagon ABCDE. Which was to be done.

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PROP. XIV. PROB.

To describe a circle about a given equilateral and equiangular pentagon.

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, a 9. 1. FD, and from the point F, in which they meet, draw the

A

F

E

straight lines FB, FA, FE, to
the points B, A, E. It may be
demonstrated, in the same man-
ner as in the preceding proposi-
tion, that the angles CBA, BAE,
AED are bisected by the straight B
lines FB, FA, FE: And because
the angle BCD is equal to the an-
gle CDE, and because FCD is
the half of the angle BCD, and
CDF the half of CDE; the an-
gle FCD is equal to FDC; where-
fore the side CF is equal to the side FD: In like manner,
it may be demonstrated, 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 centre F, at the distance of one of them,
will pass through the extremities of the other four, and be
described about the equilateral and equiangular pentagon
ABCDE. Which was to be done.

b

b 6. 1.

Book IV.

See N.

a 5. 1.

b 32. 1.

PROP. XV. PROB.

To 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 centre G of the circle ABCDEF, and draw the diameter AGD; and from D as a centre, 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.

с

b

F

a

A

B

G

Because G is the centre of the circle ABCDEF, GE is equal to GD: And because D is the centre 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 EGĎ, GDE, DEG are equal to one another, because the angles at the base of an isosceles 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 same manner, it may be demonstrated that the angle DGC is also the third part of two right angles: And because the straight line GC makes with EB the adjacent c 13. 1. 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 the vertical opposite angles BGA, AGF, FGE: Therefore the six angles EGD, DGC, CGB, BGA, AĞF, FGE, are equal to one another: But equal angles at the centre stand upon equal circumferences; therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another: And equal circumferences are subtended by equal straight lines; therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular; for since the circumference AF is equal to ED, to each of these add the circumference ABCD; therefore the whole circumference FABCD shall be equal to the whole EDCBA: And the

d 15. 1.

e 26. 3.

f 29. 3.

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D

H

angle FED stands upon the circumference FABCD, and the Book IV. angle AFE upon EDCBA; therefore the angle AFE is equal to FED: In the same 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; it is also equilateral, as was shown; and it is inscribed 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 centre, that is, to the semidiameter of the circle.

And if through the points A, B, C, D, E, F, there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonstrated from what has been said of the pentagon; and likewise 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 inscribe an equilateral and equiangular quin- See N. decagon in a given circle.

Let ABCD be the given circle; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle inscribed in a 2. 4. the circle, and AB the side of an equilateral and equiangular pentagon inscribed in the same; therefore, of such 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 difference contains two of the same

B

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if the straight lines BE, EC be

drawn, and straight lines equal to

A

F

c 30. 3.

D

them be placed around in the whole circle, an equilateral and d 1. 4.

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