# The Perpendicular Bisector of A Chord Passes The Center of a Circle

The perpendicular bisector of a chord of a circle passes through its center. In this post, we prove this claim.

Let $O$ be the center of the circle. Draw $\overline{PQ}$, a chord  and let $R$ be its midpoint.

Construct $\overline{ST}$ perpendicular to $\overline{PQ} at$latex R as shown above. We show that $O$ is on $\overline{ST}$.

Theorem

The perpendicular bisector of a chord passes through the center of a circle.

Proof

Join $\overline {OP}$,  $\overline{OR}$ and $\overline{OQ}$.

Since $O$ is  the center of the given circle, $\overline{OP} \cong \overline{OQ}$ because both of them are radii of the same circle.

Also, $\overline{PR} \cong \overline{RQ}$ since $R$ is the midpoint of $PQ$. In addition, $OR \cong OR$ because a segment is congruent to itself.

So, by the SSS Congruence Theorem, $\triangle ORP \cong \triangle ORQ$.

Also, $\angle ORP \cong \angle ORQ$ and both of them are right angles. Therefore, $O$ is a part of the two right triangles which means that it is on $\overline{ST}$. This is what we want to prove.

Since the perpendicular bisectors of a chord of a circle passes through its center, two perpendicular bisectors of a chord can be used to find the center of the given circle