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a 3. I.

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

V.

In like manner, a circle is faid to be in-
fcribed in a rectilineal figure, when
the circumference of the circle touches
each fide of the figure.

VI.

A circle is faid to be defcribed about a
rectilineal figure, when the circumfe-
rence 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.

IN

PROP. I. PROB.

Na given circle to place a ftraight line, equal to a given ftraight line not greater than the diameter of the circle.

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

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 a to D, and from the centre C, at the di

F

stance CE, defcribe the circle D

E

B

AEF, and join CA: Therefore, because C is the centre of

the

circle AEF, CA is equal to CE; but D is equal to CE; Book IV. 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

PROP. II. PROB.

a given circle to infcribe a triangle equiangular to a given triangle.

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

Draw a the ftraight line GAH touching the circle in the a 17. 3. point A, and at the point A, in the ftraight line AH, make b b 23. I. the angle HAC equal to the angle DEF; and at the point A,

in the ftraight line AG, make the angle GAB equal to the angle DFE,and join BC: Therefore, becaufe HAG touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal to the angle

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ABC in the alternate fegment of the circle: But HAC is equal to the angle DEF; therefore also 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 d to the remaining angle EDF: Wherefore the triangle d 32. 1. ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Which was to be done.

PROP.

Book IV.

a 23. I.

PROP. III. PROB.

ABOUT a given circle to describe a triangle e

quiangular 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 straight line KB, make a the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH; and through the points A, B, C, b 17. 3. draw the ftraight lines LAM, MBN, NCL touching b 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 c 18. 3. right angles. And because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it can be divided into two triangles, and because two of them, KAM, KBM are right

L

angles, the other

two AKB,AMB

D

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are equal to the angles DEG, 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 demonstrated to be equal to DFE; and therefore the e 32. 1. remaining angle MLN is equale to the remaining 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.

PROP.

Book IV.

PROP. IV. PROB..

To infcribe a circle in a given triangle.

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

Bifect a the angles ABC, BCA by the ftraight lines BD, a 9. 1. CD meeting one another in the point D, from which draw b b 12. 1. DE, DF, DG perpendiculars

Co AB, BC, CA: And be-
cause the angle EBD is equal
to the angle FBD, the angle
ABC being bifected by BD;
and because 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

B

E

A

G

angles in each, is common to both; therefore their other fides are equal; wherefore DE is equal to DF. For the fame c 26. . reafon, DG is equal to DF; therefore the three ftraight lines DE, DF, DG are equal to one another, and the circle decribed from the centre D, at the distance of any of them, will pass through the extremities of the other two, and will touch the traight lines AB, BC, CA, because the angles at the points E, F, G are right angles, and the ftraight line which is drawn rom the extremity of a diameter at right angles to it, touches the circle: Therefore the straight lines AB, BC, CA, do d 16.3. ach of them touch the circle, and the circle EFG is infcribed n the triangle ABC. Whigh was to be done.

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Book IV.

a IO. I.

b II. I.

PRO P. V. PRO B.

To defcribe a circle about a given triangle.

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

Bifecta AB, AC in the points D, E, and from these points draw DF, EF at right angles b to AB, AC; DF, EF produced

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C 4. I.

meet one another; for, if they do not meet, they are parallel,
wherefore AB, AC, which are at right angles to them, are
parallel, which is abfurd: Let them meet in F, and join FA;
also, if the point F be not in BC, join BF, CF: then, because
AD is equal to DB, and DF common, and at right angles to
AB, the base AF is equal to the base FB.
In like manner,

it

may be shown that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, FC are equal to one another; wherefore the circle described from the centre F, at the distance of one of them, fhall pass through the extremities of the other two, and be defcribed about the triangle ABC, which was to be done.

COR. And it is manifeft, that when the centre of the circle falls within the triangle, each of its angles is lefs than a right angle, each of them being in a fegment greater than a femicircle; but, when the centre is in one of the fides of the triangle, the angle oppofite to this fide, being in a femicircle, is a right angle; and, if the centre falls without the triangle, the angle oppofite to the fide beyond which it is, being in a fegment lefs than a femicircle, is greater than a right angle: Wherefore,

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