take away

3. The rectangle BE, EF, together with the square on EG, is equal to the squares on HE, EG;

the square on EG, which is common to both; therefore

4. The rectangle BE, EF is equal to the square on HE. But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore

5. BD is equal to the square on EH;

but BD is equal to the rectilineal figure A (constr.); therefore

6. The square on EH is equal to the rectilineal figure A. Wherefore a square has been made equal to the given rectilineal figure A, namely, the square described upon EH.

Q.E.F.

BOOK III.

DEFINITIONS.

1.

EQUAL circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumferences are equal.

2.

A straight line is said to touch a circle when it meets the circle, and being produced does not cut it.

3.

Circles are said to touch one another, which meet, but do not cut one another.

4.

Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal.

5.

And the straight line on which the greater perpendicular falls, is said to be farther from the centre.

6.

A segment of a circle is the figure contained by a straight line, and the circumference which it cuts off.

7.

The angle of a segment is that which is contained by the straight line and the circumference.

8.

An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment.

9.

An angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle.

10.

A sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them.

11.

Similar segments of circles are those in which the angles are equal, or which contain equal angles.

PROPOSITION 1.-Problem.

To find the centre of a given circle.

Let ABC be the given circle. It is required to find its centre.

Construction. Draw within it any straight line AB, and bisect AB in D (I. 10); from the point D draw DC at right angles to AB (I. 11); produce it to E, and bisect CE in F.

Then the point F shall be the centre of the circle ABC.

Proof. For, if it be not, let, if possible, G be the centre, and join GA, GD, GB. Then, because DA is equal to DB (constr.), and DG common to the two triangles ADG, BDG, the two sides AD, DG, are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, (I. Def. 15), because they are drawn from the centre G; therefore

1. The angle ADG is equal to the angle GDB (I. 8),

but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle (I. Def. 10); therefore

2. The angle GDB is a right angle;

but FDB is likewise a right angle (constr.); wherefore

3. The angle FDB is equal to the angle GDB (Ax. I), the greater to the less, which is impossible; therefore

4. G is not the centre of the circle ABC.

In the same manner it can be shown, that no other point but F is the centre; that is,

5. F is the centre of the circle ABC.

Which was to be found.

Cor. From this it is manifest, that if in a circle a straight line bisects another at right angles, the centre of the circle is in the line which bisects the other.

PROPOSITION 2.-Theorem.

If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.

Let ABC be a circle, and A, B, any two points in the circumference. Then the straight line drawn from A to B shall fall within the circle.

Construction. For if AB do not fall within the circle, let it fall, if possible, without, as AEB; find D, the centre of the circle ABC (III. 1), and join DA, DB; in the circumference AB, take any point F, join DF, and produce it to meet AB in E.

Demonstration. Then because DA is equal to DB (I. Def. 15),

therefore

1. The angle DAB is equal to the angle DBA (I. 5) ; and because AE, a side of the triangle DAE, is produced to B, 2. The exterior angle DEB is greater than the interior and opposite angle DAE (I. 16);

but DAE was proved equal to the angle DBE; therefore

3. The angle DEB is greater than the angle DBE ;

but to the greater angle, the greater side is opposite (I. 19); therefore