Draw within it any straight line AB and bisect AB at D; [I. 10. from the point D draw DC at right angles to AB; [I. 11. produce CD to meet the circumference at E, and bisect CE at F. [I. 10.
The point F shall be the centre of the circle ABC.
For if F be not the centre, if possible, let G be the centre; and join GA, GD, GB. Then, because DA is equal to DB, [Construction. and DG is common to the two triangles ADG, BDG; the two sides AD, DG are equal to the two sides BD, DG, each to each;
and the base GA is equal to the base GB, because they are drawn from the centre G; [I. Definition 15.
therefore the angle ADG is equal to the angle BDG. [I. 8. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; [I. Definition 10.
therefore the angle BDG is a right angle.
But the angle BDF is also a right angle. [Construction.
Therefore the angle BDG is equal to the angle BDF, [Ax.ll
the less to the greater; which is impossible.
Therefore G is not the centre of the circle ABC.
In the same manner it may be shewn that no other point out of the line CE is the centre;
and since CE is bisected at F, any other point in CE
divides it into unequal parts, and cannot be the centre.
Therefore no point but F is the centre;
that is, F is the centre of the circle ABC:
which was to he found.
Corollary. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the straight line which bisects the other.
