Prove that equal chords of a circle subtend equal angles at the centre.
(N/A) Given : $AB$ and $CD$ are equal chords of a circle with centre $O,$ i.e.,$AB = CD$.
To prove : $\angle AOB = \angle COD$.
Proof: In $\Delta OAB$ and $\Delta OCD$:
$OA = OC$ (Radii of the same circle)
$OB = OD$ (Radii of the same circle)
$AB = CD$ (Given)
Therefore,$\Delta OAB \cong \Delta OCD$ ($SSS$ congruence rule).
Therefore,$\angle AOB = \angle COD$ (by $CPCT$).
To prove : $\angle AOB = \angle COD$.
Proof: In $\Delta OAB$ and $\Delta OCD$:
$OA = OC$ (Radii of the same circle)
$OB = OD$ (Radii of the same circle)
$AB = CD$ (Given)
Therefore,$\Delta OAB \cong \Delta OCD$ ($SSS$ congruence rule).
Therefore,$\angle AOB = \angle COD$ (by $CPCT$).
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