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d arches

d 26. 3.

lines CE, DB; and join AB, BC, DE, EA. ABCDE is the Book IV. pentagon required.

Because each of the angles ACD, CDA is double of CAD, and they are bifected by the straight lines CE, DB, the five angles DAC, ACF, ECD,

A CDB, BDA are equal to one another : But equal angles stand upon equal

;

B therefore the five arches AB, BC, CD, DE, EA are equal to one another : And equal arches are subtended by equal e straight lines; therefore the

H

€ 29. 3. five straight lines AB, BC,

с CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also equiangular ; because the arch AB is equal to the arch DE: If to each be added BCD, the whole ABCD is equal to the whole EDCB: And the angle AED stands on the arch ABCD, and the angle BAE on the arch EDCB; therefore the angle BAE is equal to the f 27. 3. angle AED: For the same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: Therefore the pentagon ABCDE is equiangular; and it has been shown, that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done.

PROP. XII. PROB.

To describe an equilateral and equiangular penta- See N.

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

Let the angles of a pentagon, inscribed in the circle by the last propofition, be in the points A, B, C, D, E, so that the arches AB, BC, CD, DE, EA are equal ? ; and through the a 11. 4. points A, B, C, D, E draw GH, HK, KL, LM, MG, touching the circle; take the centre F, and join FB, FK, FC, FL, b 17. 3. FĎ: And because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the centre F, FC is perpendicular C to KL; therefore each of the c 18.3, angles at C is a right angle : For the same reason, the angles at the points B, D are right angles : Join BC; and because the angles CBK, BCK are less than two right angles, CK, BK

do

e

A

B

Book IV. do meet *, if produced : For the same reason, I. V meets KL,

v GM, and GH meets GM, HK. And becaule BF is equal to *12.Ax. 1. FC, and FK common to the triangles GFK, CEK, the illo BF,

FK are equal to the two CF, FK; and the baie BK is equal to dCor.37-3. the base Kc, d because they are drawn from the point k to e 8. 1. touch the circle ; therefore the angle BFK is equal

to the angle KFC, and the angle BKF to 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 arch BC is equal f 27. 3. to CD, the angle BFC is equal to CFD; and BFC is double

of KFC, and CFD double of
CFL; therefore the angle KFC

E
is equal to CFL; and the right
angle FCK is equal to FCL:
Therefore, in the two triangles

HL

N
FKC, FLC, there are two angles
of one equal to two of the other,
each to each, and the fide FC,

D
adjacent to the equal angles, is
common to both; therefore the .
other fides shall be equal 8 to the

K с L
other sides, and the third angle to
the third angle: Therefore KC is equal to CL, and the angle
FKC to FLC : And because KC is equal to CL, KL is double
of KC: In the same manner, it may be shown, that HK is
double of BK : And because BK is equal to KC, as was de-
monstrated, and that KL is double of KC, and HK double of
BK, HK shall be equal to KL: In like manner, it may be
fhown, 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 FLC,
and that HKL is double of FKC, and KLM double of 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 HKL or KLM: Therefore 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.

26. I.

T

PROP. XIII. PROB. To infcribe a circle in a given equilateral and equi. angular pentagon.

Let

C I 2. I

Let ABCDE be the given equilateral and equiangular penta- Book IV. gon; it is required to inscribe a circle in it. Bisect the angles BCD, CDE by the straight lines CF, DF,

a 9. I.
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
DCF therefore the base BF is equal to FD, and the angle b 4. 1.
CBF to CDF: And because the angle CDE is double of CDF,
and that CDE is equal to CBA, and CDF to CBF; CBA is
also double of CBF; therefore the angle ABF is equal to CBF;
wherefore the angle ABC is bi-

A
{ected by BF; In the same man-
ner, it may be demonstrated, that G

M
the angles BAE, AED are bi-
fected by AF, FE: From F draw B

E
FG, FH, FK, FL, FM perpen-
diculars to AB, BC, CD, DE,
EA: And because the angle HCF H
is equal to KCF, and the right
angle FHC equal to FKC; in the
triangles FHC, FKC there are two

ск

D angles of one equal to two of the other, and the side FC, opposite to one of the equal angles in cach, is common to both; therefore the other fides shall be equal d, each to each ; wherefore the perpendicular FH is equal d 26. 1. to FK: In the same manner, it may be demonstrated, that FL, FM, FG are each of them equal to FH or FK: Wherefore the circle described from the centre F, at the distance of one of these five, shall pass through the extremities of the other four : And because the angles at the points G, H, K, L, M are right angles, and that a straight line from the extremity of a diameter at right angles to it, touches e the circle ; therefore each of e 26. 3. 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.

Tocquiangular pentagon.

'O describe a circle about a given equilateral and

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

Biseat the angles BCD, CDE by the straight lines CF, FD, a 9. 1. and from the point F, in which they meet, draw the straight

lines

b 6. I.

b

BOOK IV. lines FB, FA, FE to the points B, A, E. It may be demon

strated, ; in the same manner as in the preceding propofition,
that the angles- CBA, BAE, AED are bisected by the itraight
lines FB, FA, FE: And because the angle BCD is equal to
the angle CDE, and that FCD is the

A
half of the angle BCD, and CDF the
half of CDE ; the angle FCD is equal
to FDC, wherefore the fide CF is B
equal to the fide FD: In like man.
ner, 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

D
to one another; and the circle described
from the centre F, at the distance of one of them, thall pass
through the extremities of the other four, and be described
about the equilateral and equiangular pentagon ABCDE.
Which was to be done.

à I. 3•

PROP. XV. PROB. 10 inscribe an equilateral and equiangular hexa

gon 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, and draw the diameter AGD; and in b 1. 4. the circle place 6 DC, DE, each equal to DG, and join EG,

CG, and produce them to the points B, F; and join AB, BC, EF, FA: The hexagon ABCDEF is equilateral and equiangu. lar.

Because DE, EG are each of them equal to DG, the triangle EGD is equilateral ; and therefore its three angles EGD,

GDE, DEG are equal to one anoc Cor.5.1. ther; and the three angles of a trid 32. 1. angle are equal to two right angles;

B
therefore the angle EGD is the third
part of two right angles : For the
same reason, the angle DGC is also
the third part of two right angles :

T. And because the adjacent angles EGC, € 13. 1. CGB are equal o to two right angles ;

the remaining angle CGB is also the
third part of two right angles; there-
fore the angles EGD, DGC, CGB

H are equal to one another : And to

these

G

these are equal the vertical angles BGA, AGF, FGE : Book IV. Therefore the fix angles at the point G are equal to one another: But equal angles stand upon equal & arches; therefore f 15. 1. the six arches AB, BC, CD, DE, EF, FA are equal to one

g 26. 3. another : And equal arches are subtended by equal 5 straight k 29. 3. lines ; therefore the fix it raight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular; for, since the arch AF is equal to ED, to each of these add the arch ABCD ; therefore the whole arch FABCD Thall be equal to the whole EDCBA : And the angle FED stands upon the arch FABCD, and the angle AFE upon EDCBA ; therefore the angle AFE is equal to FED: In the 1 27. 36 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 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 or radius 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 circuinscribed about it, by a method like to that used for the pentagon.

PROP. XVI. PROB.

T
10 inscribe an equilateral and equiangular quin-

decagon in a given circle.
Let ABCD be the given circle; it is required to inscribe
an equilateral and equiangu-
lar quindecagon in the circle
ABCD.

Let AC be the side of an equilateral triangle infcribed B

T in the circle, and AB the side of an equilateral and equian- E gular pentagon infcribed in the fame; therefore, of such equal parts as the whole circumference ABCDF contains fifteen, the arch ABC, being

the

a 2. 4.

bi1.4.

0 2

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