PROPOSITION 2. 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 the straight

line GAH touching the circle at the point

A;

[III. 17.

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;

G

H

and join BC. ABC shall be the triangle required. Because GAH touches the circle ABC, and AC is drawn from the point of contact A,

[Construction.

therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle.

[III. 32. But the angle HAC is equal to the angle DEF. [Constr. Therefore the angle ABC is equal to the angle 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. [I. 32, Axioms 11 and 3. 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.

from K draw any radius KB;

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

DEG, and the angle BKC equal to the angle DFH ; [I. 23. and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC. [III. 17. LMN shall be the triangle required.

Because LM, MN, NL touch the circle ABC at the points A, B, C,

[Construction. to which from the centre are drawn KA, KB, KC, therefore the angles atthepointsA,B,Carerightangles.[III.18. And because the four angles of the quadrilateral figure AMBK are together 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 AKB, AMB are together equal to two right angles.

[Axiom 3. But the angles DEG, DEF are together equal to two right angles. Therefore the angles AKB, AMB are equal to the angles DEG, DEF;

[I. 13.

of which the angle AKB is equal to the angle DEG ; [Constr. therefore the remaining angle AMB is equal to the remaining angle DEF. [Axiom 3. In the same manner the angle LNM may be shewn to be equal to the angle DFE.

Therefore the remaining angle MLN is equal to the remaining angle EDF [I. 32, Axioms 11 and 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 ABC be the given triangle: it is required to inscribe a circle in the triangle ABC.

Bisect the angles ABC, ACB, by the straight lines BD, CD, meeting one another at the point Ꭰ ; [I. 9. and from D draw DE, DF, DG perpendiculars to AB, BC, CA. [I. 12.

Then, because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, [Construction.

and that the right angle BED is equal to the right angle BFD;

[Axiom 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;

therefore DE is equal to DF.

For the same reason DG is equal to DF. Therefore DE is equal to DG.

[I. 26.

[Axiom 1.

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 one of them, will pass through the extremities of the other two;

and it will 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. Cor. Therefore the straight lines AB, BC, CA do each of them touch the circle, and therefore the circle is inscribed in the triangle ABC.

Wherefore a circle has been inscribed in the given triangle. Q.E.F.

PROPOSITION 5. PROBLEM.

To describe a circle about a given triangle.

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

Bisect AB, AC at the points D, E; from these points draw DF, EF, at right AB, AC;

DF, EF, produced, will meet one another;

for if they do not meet they are parallel,

therefore AB, AC, which are at right angles to them are parallel; which is absurd:

let them meet at F, and join FA; also if the point F be not in BC, join BF, CF.

Then, because AD is equal to BD,

and DF is common, and at right angles to AB, therefore the base FA is equal to the base FB.

[Construction.

[I. 4.

In the same manner it may be shewn that FC is equal to FA. Therefore FB is equal to FC;

and FA, FB, FC are equal to one another.

[Axiom 1.

Therefore the circle described from the centre F, at the distance of any one of them, will pass through the extremities of the other two, and will be described about the triangle ABC.

Wherefore a circle has been described about the given triangle. Q.E.F.

COROLLARY. 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, each of them being in a segment greater than a semicircle; and when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle; and when the centre

falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle.

[III. 31.

Therefore, conversely, if the given triangle be acuteangled, 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. PRÓBLEM.

To inscribe a square in a given circle.

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

Draw two diameters AC, BD

of the circle ABCD, at right angles to one another; [III. 1, I. 11. and join AB, BC, CD, DA. The figure ABCD shall be the square required.

Because BE is equal to DE,

for E is the centre;

and that EA is common, and at

right angles to BD;

E

therefore the base BA is equal to the base DA.

[I. 4.

And for the same reason BC, DC are each of them equal to BA, or DA.

Therefore the quadrilateral figure ABCD is equilateral. It is also rectangular.

For the straight line BD being a diameter of the circle ABCD, BAD is a semicircle;

[Construction. [III. 31.

therefore the angle BAD is a right angle. For the same reason each of the angles ABC, BCD, CDA is a right angle;

therefore the quadrilateral figure ABCD is rectangular.

And it has been shewn to be equilateral; therefore it is a square.

Wherefore a square has been inscribed in the given circle. Q.E.F.