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 given straight 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; it is required to place in the circle ABC a straight line equal to D.

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 (Hyp.) than D; make CE equal (I. 3.) to D, and from the centre C, at the distance CE, describe the circle AEF, and join CA.

D

Because C is the centre of the circle AEF, (I. Def. 15.) 1. CA is equal to CE;

but D is equal (Constr.) to CE; therefore (Ax. 1.) 2. D is equal to CA.

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. 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.

Draw (III. 17.) the straight line GAH touching the circle in the point 4; and at the point 4, in the straight line AH, make (I. 23.) the angle HAC equal to the angle DEF; and at the point 4, in the straight line AG, make the angle GAB equal to the angle DFE, and join BC.

for the same reason,

E

H

Because HAG touches the circle ABC, and AC is drawn from the point of contact, (III. 32.)

1. The angle HAC is equal to the angle ABC

in the alternate segment of the circle: but HAC is equal (Constr.) to the angle DEF; therefore also (Ax. 1.)

2.

The angle ABC is equal to DEF;

3. The angle ACB is equal to the angle DFE:

therefore (I. 32. and Ax. 1.)

4. The remaining angle BAC is equal to the remaining angle EDF.

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. 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.

Produce EF both ways to the points G, H; find (III. 1.) 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 (I. 23.) 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 straight lines LAM, MBN, NCL, touching (III, 17.) the circle ABC. D

A K

GE FH

M B

Because LM, MN, NL, touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, (III. 18.)

1. The angles at the points A, B, C, are 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 that two of them, KAM, KBM, are right angles, the other two (Ax. 3.)

2. AKB, AMB, are equal to two right angles:

but the angles DEG, DEF, are likewise equal (I. 13.) to two right angles i therefore (Ax. 1.)

3. The angles AKB, AMB, are equal to the angles DEG,
DEF,

of which AKB is equal (Constr.) to DEG; wherefore (Ax. 3.)
4. The remaining angle AMB is equal to the remaining
angle DEF.

In like manner, it may be demonstrated that

5. The angle LNM is equal to DFE;

and therefore (I. 32. and Ax. 3.)

6. The remaining angle MLN is equal to the remaining angle EDF.

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.

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

in ABC.

Bisect (I. 9.) the angles ABC, BCA, by the straight lines BD, CD, meeting one another in the point D, from which draw (I. 12.) DE, DF, DG, perpendiculars to AB, BC, CA.

E

And 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 (Ax. 11.) to the right angle BFD; therefore the two triangles EBD, FBD, have two angles of the one 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; therefore their other sides are equal (I. 26.); wherefore 1. DE is equal to DF:

for the same reason,

2. DG is equal to DF:

therefore (Ax. 1.)

3. DE is equal to DG:

therefore the three straight lines DE, DF, DG, are equal to one another; and the circle described from the centre D, at the distance of any of them, will pass through the extremities of the other two, and touch the straight lines AB, BČ, CA, because the angles at the points E, F, G, are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches the circle (III. 16.); therefore 4. The straight lines AB, BC, CA, do each of them touch the circle,

and therefore the circle EFG is inscribed in the triangle ABC. Q.E.F.

To describe a circle about a given triangle.

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

Bisect (I. 10.) AB, AC, in the points D, E, and from these points draw DF, EF, at right angles (I. 11.) to AB, AC. [DF, EF, produced, 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 absurd.] Let them meet in F.

B

PROP. V.-PROBLEM.

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, (I. 4.)

1. The base AF is equal to the base FB.

In like manner it may be shown that 2. CF is equal to FA ;

and therefore (Ax. 1.)

3. BF is equal to FC;

F