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Book IV.

PRO P. XIII. PRO B.

To infcribe a circle in a given equilateral and equi

.

1

Let ABCDE be the given equilateral and equiangular penta

gon; it is required to inscribe a circle in the pentagon ABCDE. a. y. I. 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 sides BC, CF are equal to the

two DC,CF; and the angle BCFis equal to the angle DCF; thereb. 4. Si fore the base BF is equal o to the base FD, and the other angles

to the other angles, to which the equal fides are opposite; there
fore the angle CBF is equal to the angle CDF. and because the an-
gle 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 an-

А.
gle ABF is equal to the angle CBF;
wherefore the angle ABC is bisect-

M
ed by the straight line BF. in the

the B
same manner it may be demonstrat-
ed that the angles BAE, AED are
bisected by the straight lines AF,FE. H
from the point F draw · FG, FH,
FK, FL, FM perpendiculars to the
straight lines AB, BC, CD, DE, C K D
EA. and because the angle HCF is
equal toʻKCF, 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 op-

posite to one of the equal angles in each, is common to both; d. 26.1. therefore the other side shall be equal d, 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 de-
fcribe! from the center F, at the distance of one of these five, shall
pals thro' the extremities of the other four, and touch the straight

f. 12. I.

lines AB, BC, CD, DE, EA, because the angles at the points G, Book IV. 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 of the straight lines AB, BC,c. 16. 3. CD, DE, EA touches the circle ; wherefore it is inscribed in the pentagon ABCDE. Which was to be done.

PRO P. 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.

Bifect 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 FB, FA, FE to the points B, A, E. It may be demonstrated, in the fame man

A ner as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA,B

F

E 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

D is equal to FDC; wherefore the side CF is equal to the side FD. in like manner it may be demon-b. 6. t. strated 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 pen. tagon ABCDE. Which was to be done.

See N.

A. $. I.

Book IV.

PRO P. XV. PRO B. "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 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 ifofceles

triangle are equal ?. and the three angles of a triangle are equal bto b. 32. 1.

two right angles; therefore the angle EGD is the third part of two
right angles. in the same manner it may

A
be demonstrated that the angle DGC is
also the third part of two right angles.
and because the straight line GC makes F

with EB the adjacent angles LGC,CGB C. 13. 1. equal to two right angles; the re

maining angle CGB is the third part of,
two right angles ; therefore the angles E

EGD, DGC, CGB 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
equal to one another. but equal angles

H 6. 26. 2

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 fubtended 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

3.

the circumference FABCD, and the angle AFE upon EDCBA ; Book IV.
therefore the angle AFE is equal to FED. in the same manner it may
be demonstrated that the other angles of the hexagon ABCDEFare
each of them equal to the angle AFE or FED. therefore the hexa-
gon is equiangular. and it is equilateral, as was shewn; 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 center, that is, to the semi-
diameter 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 circumfcribed about it, by a method like to that used for the pentagon.

!

PROP. XVI. PROB.

Tinscribe an equilateral and equiangular quinde- see N.

cagon 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 inscribed in the a. 2.6
circle, and AB the side of an equilateral and equiangular penta-
gon inscribed 6 in the same; therefore of such equal parts as theb. 11. 4
whole circumference ABCDF contains fifteen, the circumference
ABC, being the third part of the

А
whole, contains five; and the cir-
cumference AB, which is the fifth
part of the whole, contains three;
therefore BC their difference con-

'B

F
tains two of the same

parts.
bifect

E
BC in E; therefore BE, EC are, each
of them, the fifteenth part of the
whole circumference ABCD. there-
fore if the straight lines BE, EC be
drawn, and straight lines equal to them be placed around in the d. S. A
whole circle, an equilateral and equiangular quindecagon shall be
inscribed in it. Which was to be done.

c. 30. 3.

Book IV. And in the same manner as was done in the pentagon, if thro'

the points of division made by infcribing the quindecagon, straighe
lines be drawn touching the circle, an equilateral and equiangu-
lar quindecagon shall be described about it. and likewise, as in the
pentagon, a circle may be inscribed in a given equilateral and
cquiangular quindecagon, and circumscribed about it

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