Conditions for Diameter to be Perpendicular Bisector

Theorem
If in a circle a diameter bisects a chord (which is itself not a diameter), then it cuts it at right angles, and if it cuts it at right angles then it bisects it.

Geometric Proof

 * Euclid-III-3.png

Let $$ABC$$ be a circle, in which $$AB$$ is a chord which is not a diameter (i.e. it does not pass through the center).

First part
Let $$CD$$ be a diameter which bisects $$AB$$ at the point $$F$$.

Find the center $$E$$ of circle $$ABC$$, and join $$EA$$ and $$EB$$.

Because $$E$$ is the center, $$EA = EB$$.

Because $$F$$ bisects $$AB$$, $$FA = FB$$.

Also, $$FE$$ is common, so from Triangle Side-Side-Side Equality $$\triangle AFE = \triangle BFE$$.

From book 1 definition 10:
 * "When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands."

So $$\angle AFE$$ and $$\angle BFE$$ are both right angles.

Therefore the diameter $$CD$$ cuts $$AB$$ at right angles.

Second part
Let $$CD$$ be a diameter which cuts $$AB$$ at right angles at point $$F$$.

We using the same construction as above.

Because $$E$$ is the center, $$EA = EB$$.

From Isosceles Triangles have Two Equal Angles, $$\angle EAF = \angle EBF$$.

But $$\angle AFE = \angle BFE$$ because both are right angles.

As $$EF$$ is common, it follows from Triangle Side-Angle-Angle Equality that $$\triangle AFE = \triangle BFE$$.

Therefore $$AF = FB$$.

Hence the result.