In the diagram below, $A$ and $B$ mark the centres of two circles and $XY$ a chord common to both circles.
Prove that the segment joining the centres, $AB$, is a perpendicular bisector of the common chord $XY$.
We shall add radii $AX$, $AY$, $BX$, and $BY$ to the diagram and mark the intersection of $AB$ with $XY$ as the point $M$.
Consider triangles $AXB$ and $AYB$: $AX = AY$ (radii), $BX = BY$ (radii), and $AB$ is common to both. As all the lengths are equal, triangles $AXB$ and $AYB$ are congruent, which means that $\angle MAX = \angle MAY$.
Now consider triangles $AMX$ and $AMY$: as $AM$ is common to triangles, $\angle MAX = \angle MAY$, and $AX = AY$, by S(ide) A(ngle) S(ide) we deduce that they are congruent. Therefore $MX = MY$ and $\angle AMX = \angle AMY$.
But as angles $AMX + AMY = 180$ degrees, it follows that angle $AMX = 90$ degrees and we prove that $AB$ is a perpendicular of $XY$.