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Book IV.

PROP. XIII. PROB,

To inscribe a circle in a given equilateral and equian

gular pentagon.

2. 9. I.

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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 Fin which they meet draw the straight lines FB, FA,
FE. therefore since BC is equal to CD, and CF common to the tri-
angles 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 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;
CBA is also double of the angle
CBF; therefore the angle ABF is

A
equal to the angle CBF'; wherefore
the angle ABC is bifected by the

M
Itraight line BF. in the same man-
may

be demonstrated that the B
angles BAE, AED are bifected by
the straight lines AF, FE. from the H
point F draw o FG, FH, FK, FL,
FM perpendiculars to the straight
lines AB, BC, CD, DE, EA. and CK D
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 side FC, which is opposite to one of

the equal angles in each, is common to both; therefore the other d. 26.1. fides shall be 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 circle described from the center F, at the distance of one of these five, shall pass thro’ the extremities of the other four, and touch the straight lines AB, BC, CD,

DE,

C. 12. 1.

DE, EA, because the angles at the points G, H, K, L, M are right Book IV. angles; and that a straight line drawn from the extremity of the di- m

er of a circle at right angles to it, touches the circle. there- c. 16. 3. fie each of the straight lines AB, BC, CD, DE, EA touches the

wherefore it is inscribed in the pentagon ABCDE. Which
hy done.

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PROP. XIV. PROB.
'O describe a circle about a given equilateral and equi.

angular 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, FD, and a. 9. 1.
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 the same manner as

А.
in the preceeding proposition, that the
angles CBA, BAE, AED are bisec-
ted by the straight lines FB, FA, FE. B

F

E
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

D equal to FDC; wherefore the side CF is equal to the side FD. in like manner it may be demonstrated b. 6.1. 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 thro' the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done.

PROP.

To infcribe an equilateral and equiangular hexagon in

Book IV.

PROP. XV. PROB.

O inscribe Sec N.

a 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 diameter 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.

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 isosceles a. 5. 1. triangle are equal". and the three angles of a triangle are equal to b. 32. 1. 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

A
also the third part of two right angles.
and because the straight line GC makes

F

B with EB the adjacent angles EGC, CGB

G
C. 13. 1. equal

equal to two right angles ; the re-
maining angle CGB is the third part of
two right angles; therefore the angles

E
EGD, DGC, CCB are equal to one
d. 15. 1. another. and to these are equal - the

D
vertical opposite angles BGA, AGF,
FGE. therefore the fix angles EGD,
DGC, CGB, BGA, AGF, FGE, are e-
qual to one another. but equal angles

H c. 26. 3. stand upon equal circumferences; therefore the fix circumferences

AB, BC, CD, DE, EF, FA are equal to one another. and equal f. 29. 3. circumferences are subtended by equal f straight lines ; therefore

the fix 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 angle FED stands upon the

circumference

circumference FABCD, and the angle AFE upon EDCBA; there- Book IV. fore 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 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 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.

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PROP. XVI. PROB. "O inscribe an equilateral and equiangular qyindeca- See N.

gon in a giren 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 the a. 2 4. circle, and AB the side of an equilateral and equiangular pentagon inscribed in the same; therefore of such equal parts as the whole b. 11. 4. circumference ABCDF contains fifteen, the circumference ABC, being the third part of the whole, contains five; and the circumference AB,

А which is the fifth part of the whole, contains three ; therefore BC their difference contains two of the fame B

F parts. bisect CBC in E; therefore

E BE, EC are, each of them, the fifteenth part of the whole circumfe.

С

D rence ABCD, therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placed d around in the whole circle, an equi- , d. 1. 4. lateral and equiangular quindecagon shall be inscribed in it. Which was to be done.

And

c. 30. 3.

Book IV.

And in the same manner as was done in the pentagon, if thro the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon shall be described about it, and likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangu. lar quindecagon, and circumscribed about it.

THE

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