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

(N/A) Let there be 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.
Given that $\angle AOB = \angle CO'D$.
In $\triangle AOB$ and $\triangle CO'D$:
$1$. $OA = O'C$ (Radii of congruent circles are equal).
$2$. $OB = O'D$ (Radii of congruent circles are equal).
$3$. $\angle AOB = \angle CO'D$ (Given).
By $SAS$ (Side-Angle-Side) congruence criterion,$\triangle AOB \cong \triangle CO'D$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = CD$.
Hence,the chords are equal.