In today’s lesson, we will show that a line connecting the centers of two intersecting circles is a perpendicular bisector of the common chord of the two circles, connecting the intersection points.

When two circles intersect, we can connect the two intersection points and create a common chord.

If we connect the centers of these two circles, the connecting line will be a perpendicular bisector of the common chord, as we will now show.

## Problem

Circles O and Q intersect at points A and B. Show that the line OQ connecting the centers of the two circles bisects AB and is perpendicular to AB.

## Strategy

Many problems involving circles and chords make use of the fact that all of a circle’s radii are equal, and use this fact to show congruent triangles, so let draw those radii:

This immediately suggests we could solve this geometry problem by first showing that triangles △AOQ and △BOQ are congruent (using the Side-Side-side postulate – We’ve actually already done that as an exercise here – Congruent Triangles in a Circle ).

And after doing that, we could easily show that triangles △AOC and △BOC are congruent using the Side-Angle-Side postulate, having just shown that ∠AOC is congruent to ∠BOC as corresponding angles in the congruent triangles △AOQ and △BOQ).

But there’s an even easier way. The radii of each circle are equal, OA=OB and QA=QB, which means that OAQB is a kite (just like it looks in the diagram!). And we know that in a kite, the diagonal connecting the corners between the equal sides bisects the other diagonal. We also know that the diagonals are perpendicular to each other. So we are done!

## Proof

(1) OA=OB //radii of a circle are equal

(2) QA=QB //radii of a circle are equal

(3) OAQB is a kite //(1) , (2) , definition of a kite

(4) AC=CB // (3), diagonal connecting the angles formed by the equal sides bisects the other diagonal

(5) AB⊥OQ // (3), diagonals of a kite are perpendicular