# Is there a proof that a perpendicular bisector of a chord passes through the center of it's circle?

It seems intuitive because any line perpendicular to the circle passes through the center of the circle, and a chord is just a tangent line translated into the circle. So is this part of the definition of a circle?

Specifically, what brings to mind this is the construction of the center given an arbitrary circle. Two chords and their perp bisectors should work, but how would I prove that's true?

## 2 Answers

Because the perpendicular bisector of a chord $AB$ it's the locus of points $M$, for which $MA=MB$.

Since by definition all locus is a set of all point with a common property and $OA=OB$, then $O$ must be placed on the perpendicular bisector of the chord.

Here $O$ is a center of the circle.

Of course. Vertices of the chord are at equaly distance from the circumcenter and thus the circumcenter is on segment bisector. (Remember that segment bisector is exactly the set of all points which are at equaly distance from the vertices of segment.)