BC be not equal to D, it is greater than D. Make CF equala Book IV. to D, and from the centre C, at the distance CF, describe the circle AEF, and join CA. Because C is the centre of the a 3. 1. circle AEF, CA is equal to CF. But D is equal to CF; therefore D is equal to CA. Wherefore in the circle ABC a straight line is placed equal to the given straight line D, which is not greater than the diameter of the circle. Which was to be done.

PROP. II. PROB.

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 in the a 17. 3. point A, and at the point A, in the straight line AH, makeb b 23.1. the angle HAC equal to the angle DEF; and at the point A,

in the straight line

HAC is equal to

A

H

c 32. 3.

the angle ABC in the alternate segment of the circle. But HAC is equal to the angle DEF; therefore also the angle ABC is equal to DEF. For the same reason the angle AČB is equal to the angle DFE. Therefore the remaining angle BAC is equal to the remaining angle EDF. Wherefore d 32.1. the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Which was to be done.

PROP. III. PROB.

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, 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 the angle BKA equal to the angle DEG, and the angle BKC

b 17. 3.

c 18. 3.

equal to the angle DFH; and through the points A, B, C draw the straight lines LAM, MBN, NCL touching the circle ABC. 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 righte angles. Because the four angles of the quadrilateral figure AMBK are equal to four right angles*, and because two of them, KAM, KBM, are right angles, the other two, AKB, AMB, are equal to two right angles. But the angles DEG,

* Euclid assumes this proposition without proof. It is evident that any quadrilateral figure may be divided into two triangles, and the three angles of each triangle are equal to two right angles, therefore all the angles of both triangles are equal to four right angles.

DEF are equal to two right angles; therefore the angles Book IV. AKB, AMB are equal to the angles DEG, DEF. But the angle AKB is equal to the angle DEG; wherefore the re- d 13. 1. maining angle AMB is equal to the remaining angle DEF. In like manner the angle LNM may be demonstrated to be equal to DFE. Therefore the remaining angle MLN is equale to the remaining angle EDF. Wherefore the trian- e 32. 1. gle LMN is equiangular to the triangle DEF, and it is described about the circle ABC. Which was to be done.

PROP. IV. PROB.

TO inscribe a circle in a given triangle.

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

A

Bisecta the angles ABC, BCA by the straight lines BD, a 9. 1. CD meeting each other in the point D, from which drawb b12. 1. DE, DF, DG perpendicular to AB, BC, CA. Because the angle EBD is equal to the angle FBD, and the right angle BED to the right angle BFD, the two triangles EBD, FBD have two angles of one equal to two angles of the other; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore DE is equal

E

B

F

to DF. For the same reason DG is equal to DF. Therefore c 26. 1. the three straight lines DE, DF, DG are equal to one another; therefore the circle described from the centre D, at the distance of any 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 dangles. Therefore the straight lines AB, BC, CA touch the d Cor. 16.3. circle, and the circle EFG is inscribed in the triangle ABC. Which was to be done.

Book IV.

a 10. 1.

b 11. 1.

PROB. V. PROB.

TO describe a circle about a given triangle.

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

Bisecta AB, AC in the points D, E, and from these points draw DF, EF at right angles to AB, AC; then DF, EF

c 4.1.

produced will meet each other; 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; and, if the point F be not in BC, join BF, CF. 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. 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, will pass through the extremities of the other two, and will be described about the triangle ABC. Which was to be done.

COR. When the centre of the circle falls within the triangle, each of its angles is less than a right angle, because each of them is in a segment greater than a semicircle. 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

* This demonstration is imperfect, for Euclid assumes a property of parallel lines which, though true, has not been proved, and is not self-evident.

angle. When the centre falls without the triangle, the angle Book IV. opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle. Wherefore if the given triangle be acute angled, the centre of the circle falls within it; if it be right angled, the centre is in the side opposite to the right angle; and, if it be obtuse angled, the centre falls without the triangle, beyond the side opposite to the obtuse angle.

PROP. VI. PROB.

TO inscribe a square in a given circle.

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

A

a 4. 1.

E

Draw the diameters AC, BD at right angles to each other, and join AB, BC, CD, DA. Because BE is equal to ED, E being the centre, and because EA is at right angles to BD, and common to the triangles ABE, ADE, the base BA is equala to the base AD. For the same reason BC, CD are each of them equal to BA or AD. Therefore the quadrilateral figure ABCD is equilateral. It is also rectangular; for the straight line BD being a diameter of the circle ABCD,

B

C

BAD is a semicircle; wherefore the angle BAD is a right b 31. 3. 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 shown to be equilateral; therefore it is a square; and it is inscribed in the circle ABCD. Which was to be done.