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 $$\triangle OAQ $$  and $$\triangle OBQ,$$ 

$$OA = OB$$   ....(Radius of the circle)
$$QA = QB$$   ...(Radius of the circle)
and $$OQ = OQ$$   ....(common side)     
$$\triangle OAQ \cong \triangle OBQ $$    ....SSS test of  congruence

$$\Rightarrow \angle AOQ = \angle BOQ$$         ....c.a.c.t.     ....(1)

[Since $$\angle AOQ= \angle AOP$$ and $$\angle BOP= \angle BOQ$$]

In triangles $$AOP$$ and $$BOP$$, 

$$OA = OB $$    ....(Radius of the circle)
$$\angle AOP= \angle BOP$$        .....From (1)

$$OP = OP$$       .....Common side
$$\triangle AOP \cong \triangle BOP $$    ....SAS test of congruence, 

$$\Rightarrow AP=BP$$ and $$\angle APO = \angle BPO$$

But,  $$\angle APO +\angle BPO= 180^o$$

$$2 \angle APO=180^o \Rightarrow \angle APO=90^o$$

$$AP = BP$$  and $$\angle APO =\angle BPO=90 ^o$$

Hence, $$OQ$$ is the perpendicular  bisector of $$AB$$.