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

(N/A) Let there be two circles with centres $O$ and $O^{\prime}$ that intersect at points $A$ and $B$.
$AB$ is the common chord of the two circles,and $OO^{\prime}$ is the line segment joining the centres. Let $OO^{\prime}$ and $AB$ intersect at point $M$.
To prove that $OO^{\prime}$ is the perpendicular bisector of $AB$,we join $OA, OB, O^{\prime}A$,and $O^{\prime}B$.
In $\Delta OAO^{\prime}$ and $\Delta OBO^{\prime}$:
$OA = OB$ (radii of the same circle)
$O^{\prime}A = O^{\prime}B$ (radii of the same circle)
$OO^{\prime} = OO^{\prime}$ (common side)
By $SSS$ congruence criterion,$\Delta OAO^{\prime} \cong \Delta OBO^{\prime}$.
Therefore,$\angle 1 = \angle 2$ (by $CPCT$).
Now,in $\Delta AOM$ and $\Delta BOM$:
$OA = OB$ (radii of the same circle)
$OM = OM$ (common side)
$\angle 1 = \angle 2$ (proved above)
By $SAS$ congruence criterion,$\Delta AOM \cong \Delta BOM$.
Therefore,$AM = BM$ (by $CPCT$) and $\angle 3 = \angle 4$ (by $CPCT$).
Since $AB$ is a line segment,$\angle 3 + \angle 4 = 180^{\circ}$ (linear pair).
Thus,$2 \angle 3 = 180^{\circ} \Rightarrow \angle 3 = 90^{\circ}$.
Since $\angle 3 = \angle 4 = 90^{\circ}$ and $AM = BM$,$OO^{\prime}$ is the perpendicular bisector of $AB$.