If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Let the two circles have centres O and Q and they intersect at two points A and B. Then, AB is the common chord of the two circles. OQ is the line joining the centres of the two circles. Let OQ cuts AB at P. We need to prove OQ is the perpendicular bisector of AB. Draw line segment OA, OB, QA and QB. In △OAQ and △OBQ, OA=OB (Radius of the circle) QA=QB (Radius of the circle) and OQ=OQ (common side) △OAQ ≅ △OBQ SSS test of congruence ⇒∠AOQ=∠BOQ c.a.c.t. (1) [Since ∠AOQ=∠AOP and ∠BOP=∠BOQ] In triangles AOP and BOP, OA=OB (Radius of the circle) ∠AOP=∠BOP. From (1) OP=OP. Common side △AOP ≅ △BOP SAS test of congruence, ⇒AP=BP and ∠APO=∠BPO But, ∠APO+∠BPO=180° 2∠APO=180° ⇒∠APO=90° AP=BP and ∠APO=∠BPO=90° Hence, OQ is the perpendicular bisector of AB.