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

Let be the center of the circle. Draw , a chord and let be its midpoint.

Construct perpendicular to latex R as shown above. We show that is on .

**Theorem**

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

**Proof**

Join , and .

Since is the center of the given circle, because both of them are radii of the same circle.

Also, since is the midpoint of . In addition, because a segment is congruent to itself.

So, by the **SSS Congruence Theorem**, .

Also, and both of them are right angles. Therefore, is a part of the two right triangles which means that it is on . 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