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

G

H

до

E

Then ABC shall be the triangle required.

Proof. Because HAG touches the circle ABC, and AC is drawn from the point of contact, therefore

1. The angle HAC is equal to the angle ABC,

in the alternate segment of the circle (III. 32), but HAC is equal to the angle DEF (constr.); therefore also

2. The angle ABC is equal to DEF (Ax. 1);

for the same reason,

therefore

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

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

PROPOSITION 3.-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.

H;

find

Construction. Produce EF both ways to the points G, the centre K of the circle ABC (III. 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 (I. 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 (III. 17). Then LMN shall be the triangle required,

Proof. Because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, therefore 1. 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; therefore the other two

2. AKB, AMB are equal to two right angles (Ax. 3) ; but the angles DEG, DEF are likewise equal to two right angles (I. 13); therefore

3. The angles AKB, AMB are equal to the angles DEG, DEF (Ax. 1),

of which AKB is equal to DEG (constr.); wherefore

4. The remaining angle AMB is equal to the remaining angle DEF (Ax. 3).

In like manner, it may be demonstrated that

5. The angle LNM is equal to DFE;

and therefore

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

PROPOSITION 4.-Problem.

To inscribe a circle in a given triangle.

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

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

Proof. 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 to the right angle BFĎ (Ax. 11), 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

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, BC, 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.

PROPOSITION 5.-Problem.

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.

Construction. Bisect AB, AC in the points D, E (I. 10), and from these points draw DF, EF at right angles to AB, AC (I. 11); 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, and join FA; also, if the point F be not in BC, join BF, CF.

Proof. Then, because AD is equal to DB, and DF common, and at right angles to AB, therefore

1. The base AF is equal to the base FB (I. 4). In like manner, it may be shown that

2. CF is equal to FA;

and therefore

3. BF is equal to FC (Ax. 1);

and FA, FB, FC are equal to one another; wherefore the circle described from the centre F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC. Q.E.F.

Cor. And it is manifest, that when the centre of the circle falls within the triangle, each of its angles is less than a right angle (III. 31), each of them being in a segment greater than a semicircle; but, when the centre is in one of the sides of the triangle, the angle opposite to this side is a right angle, being in a semicircle (III. 31);

and, if the centre falls without the triangle, the angle opposite to the side beyond which it is, is greater than a right angle, being in a segment less than a semicircle (III. 31) ;

therefore, conversely, if the given triangle be acute-angled, the centre of the circle falls within it; if it be a right-angled triangle, the centre is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.

PROPOSITION 6.-Problem.

To inscribe a square in a given circle.

Let ABCD be the given circle. It is required to inscribe a square in ABCD.

Construction. Draw

C

diameters AC, BD at right angles to one

another (III. 1 and I. 11), and join AB, BC, CD, DA.

Then the figure ABCD shall be the square required.

Proof. Because BE is equal to ED, for E is the centre, and that EA is common, and at right angles to BD;

1. The base BA is equal to the base AD (I. 4);

and, for the same reason,