PROPOSITION 3.
If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it. Let ABC be a circle, and in it let a straight line CDthrough the centre bisect a straight line AB not through the centre at the point F; I say that it also cuts it at right angles. For let the centre of the circle ABC
be taken, and let it be E; let EA, EB be joined. Then, since AF is equal to FB, and FE is common,
I say that it also bisects it. that is, that AF is equal to FB. For, with the same construction,
therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles;