Prove that equal chords of congruent circles subtend equal angles at their centres.
(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 such that $AB = CD$.
We need to prove 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).
$3. AB = CD$ (Given).
By $SSS$ (Side-Side-Side) congruence criterion,$\triangle AOB \cong \triangle CO'D$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$\angle AOB = \angle CO'D$.
Let $AB$ be a chord of the first circle and $CD$ be a chord of the second circle such that $AB = CD$.
We need to prove 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).
$3. AB = CD$ (Given).
By $SSS$ (Side-Side-Side) congruence criterion,$\triangle AOB \cong \triangle CO'D$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$\angle AOB = \angle CO'D$.
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