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