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.

perpendicular bisector

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

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