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A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each.

II.

In like manner a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each.

III.

A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle.

IV.

A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle.

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V.

In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure.

VI. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described.

VII. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.

PROP. I.-PROBLEM. In a given circle to place a straight line equal to a gwen straight line which is not greater than the diameter of the circle.

(References - Prop. I. 3.) Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle.

It is required to place in the circle ABC a straight line equal to D.

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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 straight line BC is placed equal to D.
But, if it is not, BC is greater than D; (hyp.)
make CE equal to D, (I. 3)
and from the centre C, at the distance CE, describe the circle AEF

and join CA.
Then CA shall be equal to D.

DEMONSTRATION

Because C is the centre of the circle AEF,

therefore CA is equal to CE; (I. def. 15) but D is equal to CE; (constr.)

therefore D is equal to CA; (ax. 1) Wherefore, in the circle ABC, a straight line CA is placed equal to

the given straight line D, which is not greater than the diameter of the circle.

Q. E. F.

PROP. II.-PROBLEM. In a given circle to inscribe a triangle equiangular to a given triangle.

(References — Prop. I. 23, 32; III. 17, 32.) Let ABC be the given circle, and DEF the given triangle.

It is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.

D

H

E

B

CONSTRUCTION

Draw the straight line GAH touching the circle in the point A ; (III.

17) and at the point A, in the straight line AH, make the angle HAC

equal to the angle DEF; (1. 23) and at the point A, in the straight line AG, make the angle GAB

equal to the angle DFE, and join BC. Then ABC shall be the triangle required.

DEMONSTRATION

Because HAG touches the circle ABC, and AC is drawn from the

point of contact, the angle HAC is equal to the angle ABC in the alternate segment of

the circle ; (111. 32) but HAC is equal to the angle DEF; (constr.)

therefore also the angle ABC is equal to DEF; (ax. 1) for the same reason, the angle ACB is equal to the angle DFE ; therefore the remaining angle BAC is equal to the remaining angle EDF;

(1. 32, and ax. 1) wherefore the triangle ABC is equiangular to the triangle

DEF, and it is inscribed in the circle ABC.

Q. E, F.

PROP. III. - PROBLEM. About a given circle to describe a triangle equiangular to a given triangle.

(References – Prop. I. 13, 23, 32; III. 1, 17, 18.) Let ABC be the given circle, and DEF the given triangle.

It is required to describe a triangle about the circle ABC equiangular to the triangle DEF.

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Produce EF both ways to the points G, H; find the centre K of the circle ABC, (II. 1) and from it draw any

straight line KB; at the point K, in the straight line KB, make the angle BKA equal to

the angle DEG, (1. 23) and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the straight lines LAM, MBN,

NCL, touching the circle ABC. (111. 17.)
Then LMN shall be the triangle required.

DEMONSTRATION

Because 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 ; (III. 18) 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 that two of them, KAM, KBM, are right angles,

the other two AKB, AMB, are equal to two right angles ; (ax. 3) but the angles DEG, DEF, are likewise equal to two right angles;

(1. 13) therefore the angles AKB, AMB, are equal to the angles DEG, DEF,

(ax. 1) of which AKB is equal to DEG ; (constr.) wherefore the remaining angle AMB, is equal to the remaining angle

DEF. (ax. 3.)

In like manner, it may be demonstrated that the angle LNM is equal to

DFE ; and therefore the remaining angle MLN is equal to the remaining angle

EDF; (1. 32, and ax. 3) wherefore the triangle LMN is equiangular to the triangle

DEF; and it is described about the circle ABC. Q. E. F.

PROP. IV.- PROBLEM.
To inscribe a circle in a given triangle.

(References — Prop. I. 9, 12, 26; III. 16.)
Let the given triangle be ABC.
It is required to inscribe a circle in ABC.

A

G

E

B

CONSTRUCTION

Bisect the angles ABC, BCA, by the straight lines BD, CD, meeting

one another in the point D, (1.9) from which draw DE, DF, DG, perpendiculars to AB, BC, CA ;(I. 12) from the centre D, with either of these distances, describe the circle

EFG.
Then the circle EFG shall be inscribed in the triangle

ABC.

DEMONSTRATION

Because the angle EBD is equal to the angle FBD, for the angle ABC

is bisected by. BD, and that the right angle BED is equal to the right angle BFD; (ax.

11) then, in the triangles EBD, FBD, the two angles of the one are equal to two angles of the other, each

to each ; and the side BD, which is opposite to one of the equal angles in each,

is common to both;

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