Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

(N/A) Let there be two circles with centres $O$ and $O^{\prime}$ respectively,which intersect each other at points $P$ and $Q$. We need to prove that $\angle OPO^{\prime} = \angle OQO^{\prime}$.
Join $OP, O^{\prime}P, OQ, O^{\prime}Q$ and $OO^{\prime}$.
In $\Delta OPO^{\prime}$ and $\Delta OQO^{\prime}$:
$OP = OQ$ (Radii of the same circle with centre $O$)
$O^{\prime}P = O^{\prime}Q$ (Radii of the same circle with centre $O^{\prime}$)
$OO^{\prime} = OO^{\prime}$ (Common side)
Therefore,by the $SSS$ (Side-Side-Side) congruence criterion:
$\Delta OPO^{\prime} \cong \Delta OQO^{\prime}$
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$\angle OPO^{\prime} = \angle OQO^{\prime}$.