If a straight line passing through the center of a circle bisects a straight line not passing through the center, then it also cuts it at right angles; and if it cuts it at right angles, then it also bisects it.

Let a straight line *CD* passing through the center of a circle *ABC* bisect a straight line *AB* not passing through the center at the point *F.*

I say that it also cuts it at right angles.

Take the center *E* of the circle *ABC,* and join *EA* and *EB.*

Then, since *AF* equals *FB,* and *FE* is common, two sides equal two sides, and the base *EA* equals the base *EB,* therefore the angle *AFE* equals the angle *BFE.*

But, when a straight line standing on another straight line makes the adjacent angles equal to one another, each of the equal angles is right, therefore each of the angles *AFE* and *BFE* is right.

Therefore *CD,* which passes through the center and bisects *AB* which does not pass through the center, also cuts it at right angles.

Next, let *CD* cut *AB* at right angles.

I say that it also bisects it, that is, that *AF* equals *FB.*

For, with the same construction, since *EA* equals *EB,* the angle *EAF* also equals the angle *EBF.*

But the right angle *AFE* equals the right angle *BFE,* therefore *EAF* and *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 opposite one of the equal angles. Therefore they also have the remaining sides equal to the remaining sides

Therefore *AF* equals *FB.*

Therefore *if a straight line passing through the center of a circle bisects a straight line not passing through the center, then it also cuts it at right angles; and if it cuts it at right angles, then it also bisects it.*

Q.E.D.

This proposition is used in the next one, a few others in Book III, and XII.16.