Prove that if chords of congruent circles subtend equal angles at their centres,then the chords are equal.

(N/A) Given: Two congruent circles with centres $O$ and $O'$. Let $AB$ be a chord of the first circle and $CD$ be a chord of the second circle such that $\angle AOB = \angle CO'D$.
To prove: $AB = CD$.
Proof:
In $\Delta AOB$ and $\Delta CO'D$:
$AO = CO'$ (Radii of congruent circles)
$BO = DO'$ (Radii of congruent circles)
$\angle AOB = \angle CO'D$ (Given)
Therefore,by the $SAS$ (Side-Angle-Side) congruence criterion:
$\Delta AOB \cong \Delta CO'D$
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = CD$.