Ꭰ

d 8. I.

and FE is equal to FB, wherefore DE, EF are equal to DB, Book III. BF; and the bafe FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal to the angle DBF; but DEF is a right angle, therefore alfo DBF is a right angle: And FB, if produced, is a diameter, and the ftraight line which is drawn at right angles to a diameter, from the B extremity of it, touches the circle: Therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D.

e 16. 3.

F

A

Rectilineal figure is faid to be infcribed in another recti lineal figure, when all the angles of the infcribed figure are upon the fides of the figure in which it is

infcribed, each upon each.

II.

In like manner, a figure is faid to be defcribed
about another figure, when all the fides of
the circumfcribed figure pafs through the an-

gular points of the figure about which it is defcribed, cach
through each.

IV.

A rectilineal figure is faid to be defcribed about a circle, when each fide of the circumfcribed figure

touches the circumference of the circle.

VI.

A circle is faid to be defcribed about a rectilineal figure, when the circumference of the circle paffes through all the angular points of the figure about which it is defcribed.

VII.

A ftraight line is faid to be placed in a circle, when the extremities of it are in the circumference of the circle.

Book IV.

PROP. I. PROB.

IN a given circle to place a straight line, equal to a given straight line not greater than the diameter of the circle.

Let ABC be the given circle, and D the given ftraight line, not greater than the diameter of the circle.

2

Draw BC the diameter of the circle ABC; then, if BC is equal to D, the thing required is done; for in the circle ABC a ftraight line BC is placed equal to D: But, if it is not, BC is greater than D; make CE equal to D, and from the centre C, at the diftance CE, defcribe the circle AEF, and join CA: Therefore, because C is the centre of the circle AEF, CA is equal to CE; but D is D

A

E

a 3. I.

B

F

equal to CE; therefore D is equal to CA: Wherefore, in the circle ABC, a ftraight line is placed equal to the given ftraight line D, which is not greater than the diameter of the circle. Which was to be done.

IN

a given circle to inscribe a triangle equiangular to a given triangle.

G 4

Let

Book IV.

Let ABC be the given circle, and DEF the given triangle ; w it is required to infcribe in the circle ABC a triangle equiangular to the triangle DEF.

a 17.3.

b 23. I.

€ 32.3.

d 32. I.

a 23. I.

b 17. 3.

c 18. 3.

a

Draw the ftraight line GAH touching the circle in the point A, and at the point A, in the ftraight line AH, make the angle HAC equal to the angle DEF; and at the point A, in the ftraight line

AG, make the angle

G

GAB equal to the
angle DFE, and join
BC: Therefore, be-
caufe HAG touches
the circle ABC, and E
AC is drawn from
the point of contact,
the angle HAC is

equal to the angle

B

A

H

d

C

ABC in the alternate fegment of the circle: But HAC is equal to the angle DEF; therefore alfo the angle ABC is equal to DEF: For the fame reason, the angle ACB is equal to the angle DFE; therefore the remaining angle BAC is equal to the remaining angle EDF: Wherefore the triangle ABC is equiangular to the triangle DEF, and it is infcribed in the circle ABC. Which was to be done.

A

PROP. III. PRO B.

BOUT a given circle to describe a triangle equian, gular to a given triangle.

Let ABC be the given circle, and DEF the given triangle; it is required to defcribe a triangle about the circle ABC equiangular to the triangle DEF.

Produce EF both ways to the points G, H, and find the centre K of the circle ABC, and from it draw any straight line KB; at the point K in the ftraight line KB, make the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the ftraight lines LAM, MBN, NCL touching the circle ABC: Therefore, becaufe LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, the angles at the points A, B, C, are right angles: And because the four angles of the quadrilateral figure

AMBK

AMBK are equal to four right angles, for it can be divided in- Book IV. to two triangles; and that two of them KAM, KBM are right n angles, the other

two AKB, AMB

are equal to two right angles But

L

D

the angles DEG,

DEF are likewise

d 13. I.

equal totworight A

K

angles; therefore

the angles AKB,

DEF, of which

AKB is equal to DEG; wherefore the remaining angle AMB is equal to the remaining angle DEF: In like manner, the angle LNM may be demonftrated to be equal to DFE; and therefore the remaining angle MLN is equal to the remaining e 32. I. angle EDF: Wherefore the triangle LMN is equiangular to the triangle DEF: And it is defcribed about the circle ABC. Which was to be done.

To

PROP. IV. PRO B.

O infcribe a circle in a given triangle.

Let the given triangle be ABC; it is required to inscribe a circle in ABC.

A

See N.

Bifect the angles ABC, BCA by the ftraight lines BD, CD a 9. t. meeting one another in the point D, from which drawb DE, DF, b 12. r. DG perpendiculars to AB, BC, CA: And because the angle EBD is equal to the angle FBD, for the angle ABC is bifected by BD, and that the right angle BED, is equal to the right angle BFD, the two triangles EBD, FBD have two angles of the one equal to two angles of the other, and the fide BD, which is oppofite to one of the equal angles in each, is common to B both;

therefore their other ides fhall be equal where

E

G

D

F

c 26. I. fore