Question.
Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:
Let us consider two congruent circles (circles of same radius) with centres as O and O'.
In $\triangle \mathrm{AOB}$ and $\triangle C O^{\prime} D$,
$\angle \mathrm{AOB}=\angle \mathrm{CO}^{\prime} \mathrm{D}($ Given $)$
$O A=O^{\prime} C$ (Radii of congruent circles)
$O B=O^{\prime} D$ (Radii of congruent circles)
$\therefore \triangle \mathrm{AOB} \cong \triangle \mathrm{CO}^{\prime} \mathrm{D}(\mathrm{SAS}$ congruence rule $)$
$\Rightarrow A B=C D(B y C P C T)$
Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Let us consider two congruent circles (circles of same radius) with centres as O and O'.
In $\triangle \mathrm{AOB}$ and $\triangle C O^{\prime} D$,
$\angle \mathrm{AOB}=\angle \mathrm{CO}^{\prime} \mathrm{D}($ Given $)$
$O A=O^{\prime} C$ (Radii of congruent circles)
$O B=O^{\prime} D$ (Radii of congruent circles)
$\therefore \triangle \mathrm{AOB} \cong \triangle \mathrm{CO}^{\prime} \mathrm{D}(\mathrm{SAS}$ congruence rule $)$
$\Rightarrow A B=C D(B y C P C T)$
Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
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