Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Given:
Two congruent circles.
To do:
We have to prove that equal chords of congruent circles subtend equal angles at their centres.
Solution:
Consider two circles in which \( \mathrm{AB} \) is the chord of \( \mathrm{C}_{1} \) and ( \mathrm{PQ} \) is the chord of \( \mathrm{C}_{2} \).
\( A B=P Q \)
We have to prove that \( \angle \mathrm{AOB}=\angle \mathrm{PXQ} \).
In $\triangle AOB$ and $\triangle PXQ$
\( \mathrm{AO}=\mathrm{PX} \quad \) (Radius of congruent circles are equal)
\( B O=Q X \) (Radius of congruent circles are equal)
\( A B=P Q \) (Given)
Therefore, by SSS congruence,
\( \Delta \mathrm{AOB} \cong \Delta \mathrm{PXQ} \)
This implies,
\( \angle \mathrm{AOB}=\angle \mathrm{PXQ} \quad(\mathrm{CPCT}) \)
Hence proved.
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