centre it shall cut it at right angles; and if it cut it at right angles it shall bisect it.
Let ABC be a circle; and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, at the point F: CD shall cut AB at right angles.
Take E the centre of the circle; and join EA, EB.[III. 1.
Then, because AF is equal to FB, [Hypothesis.
and FE is common to the two triangles AFB, BFE;
the two sides AF, FE are equal to the two sides BF,FE, each to each;
and the base EA is equal to the base EB; [I. Def. 15.
therefore the angle AFE is equal to the angle BFE. [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 each of the angles AFE, BFE is a right angle.
Therefore the straight line CD, drawn through the centre, bisecting another AB which does not pass through the centre, also cuts it at right angles.
But let CD cut AB at right angles: CD shall also bisect AB; that is, AF shall be equal to FB.
The same construction being made, because EA, EB, drawn from the centre, are equal to one another, [I. Def. 15.
the angle EAF is equal to the angle EBF. [I. 5.
And the right angle AFE is equal to the right angle BFE.
Therefore in the two triangles EAF, EBF, there are two angles in the one equal to two angles in the other, each to each;
and the side EF, which is opposite to one of the equal angles in each, is common to both;
therefore their other sides are equal; [I. 26.
therefore AF is equal to FB.
Wherefore, if a straight line &c. q.e.d.
