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.
(N/A) Consider two congruent circles with centres $O$ and $O'$. Let $AB$ and $CD$ be equal chords of these circles respectively,such that $AB = CD$.
We need to prove that $\angle AOB = \angle CO'D$.
In $\Delta AOB$ and $\Delta CO'D$:
$AO = CO'$ (Radii of congruent circles are equal)
$BO = DO'$ (Radii of congruent circles are equal)
$AB = CD$ (Given)
By the $SSS$ (Side-Side-Side) congruence criterion,$\Delta AOB \cong \Delta CO'D$.
Since congruent triangles have equal corresponding parts,we have:
$\angle AOB = \angle CO'D$.
Hence,equal chords of congruent circles subtend equal angles at their centres.
We need to prove that $\angle AOB = \angle CO'D$.
In $\Delta AOB$ and $\Delta CO'D$:
$AO = CO'$ (Radii of congruent circles are equal)
$BO = DO'$ (Radii of congruent circles are equal)
$AB = CD$ (Given)
By the $SSS$ (Side-Side-Side) congruence criterion,$\Delta AOB \cong \Delta CO'D$.
Since congruent triangles have equal corresponding parts,we have:
$\angle AOB = \angle CO'D$.
Hence,equal chords of congruent circles subtend equal angles at their centres.
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