Book IV.

2. 9. 1.

b. 4. 1.

12. I.

d. 26. z.

Ο

PROP. XIII. PRO B.

To infcribe a circle in a given equilateral and equi

angular pentagon.

b

Let ABCDE be the given equilateral and equiangular penta-
gon; it is required to infcribe a circle in the pentagon ABCDE.
Bifect the angles BCD, CDE by the ftraight lines CF, DF,
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 the angle DCF; there→
fore the base BF is equal to the base FD, and the other angles
to the other angles, to which the equal fides are oppofite; 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 alfo double
of the angle CBF; therefore the an-
gle ABF is equal to the angle CBF;
wherefore the angle ABC is bifect-
ed by the ftraight line BF. in the
B
fame manner it may be demonftrat-
ed that the angles BAE, AED are
bifected 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,
EA. and because the angle HCF is

C

A

M

E

CK D

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 op-
pofite to one of the equal angles in each, is common to both;
therefore the other fide fhall be equal, each to each; wherefore
the perpendicular FH is equal to the perpendicular FK. in the
fame manner it may be demonftrated 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-
fcribed from the center F, at the distance of one of these five, fhall
pals thro' the extremities of the other four, and touch the straight

fines 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. 16. 3. CD, DE, EA touches the circle; wherefore it is inscribed in the pentagon ABCDE. Which was to be done.

PROP. XIV. PRO B.

To defcribe a circle about a given equilateral and

T°

equiangular pentagon.

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

Bisect 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 demonftrated, in the fame man-
ner as in the preceding propofition,
that the angles CBA, BAE, AED are
bifected by the straight lines FB, FA, B
FE. and because the angle BCD is e-
qual 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

b

F

A

E

D

may be demon-b. 6. 4.

CF is equal to the fide FD. in like manner it
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 defcribed from the center F, at the
diftance of one of them, fhall pass thro' the extremities of the other
four, and be described about the equilateral and equiangular pen.
tagon ABCDE, Which was to be done.

1

Book IV.

See N.

a. 5. I.

b. 32. I.

C. 13. 1.

d. 15. 1.

c. 26. 3.

f. 29. 3.

T

PROP. XV. PRO B.

O infcribe 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 dia-
meter 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 bafe of an ifofceles
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 demonftrated that the angle DGC is
alfo the third part of two right angles.
and because the straight line GC makes F
with EB the adjacent angles EGC, CGB
equal to two right angles; the re-
maining angle CGB is the third part of.
two right angles; therefore the angles E

d

A

D

H

B

EGD, DGC, CGB are equal to one
another. and to thefe are equal 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
ftand upon equal circumferences; therefore the fix circumferences
AB,BC, CD, DE, EF, FA are equal to one another. and equal
circumferences are fubtended by equal f ftraight lines; therefore
the fix ftraight lines are equal to one another, and the hexagon
ABCDEF is equilateral. It is alfo equiangular; for fince the cir-
cumference AF is equal to ED, to each of these add the circum-
ference ABCD; therefore the whole circumference FABCD shall
be equal to the whole EDCBA. and the angle FED stands upon

the circumference FABCD, and the angle AFE upon EDCBA; Book IV. therefore the angle AFE is equal to FED. in the fame manner it may be demonftrated that the other angles of the hexagon ABCDEFare 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 fide of the hexagon is equal to the ftraight line from the center, that is, to the femidiameter 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 demonftrated from what has been faid of the pentagon; and likewise a circle may be infcribed in a given equilateral and equiangular hexagon, and circumfcribed about it, by a method like to that used for the pentagon.

PROP. XVI. PRO B.

To infcribe an equilateral and equiangular quinde-see N. cagon 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 fide of an equilateral triangle inscribed in the a. 2. 4. circle, and AB the fide of an equilateral and equiangular pentagon inscribed in the fame; therefore of such equal parts as theb. 11. whole circumference ABCDF contains fifteen, the circumference

ABC, being the third part of the
whole, contains five; and the cir-
cumference AB, which is the fifth
part of the whole, contains three;
therefore BC their difference con-B
tains two of the fame 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

A

F

c. 30. 3.

drawn, and straight lines equal to them be placed around in thed. 1. 4. whole circle, an equilateral and equiangular quindecagon fhall be infcribed in it. Which was to be done.

Book IV.

And in the fame manner as was done in the pentagon, if thro' the points of divifion made by infcribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon fhall be described about it. and likewise, as in the pentagon, a circle may be infcribed in a given equilateral and equiangular quindecagon, and circumfcribed about it