O is the centre of the circle and PO bisects the angle APD. Prove that AB = CD.

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Solution

Given : A circle with centre O. Chords AB
and CD meet at point P. PO bisects the angle APD.
To prove : AB = CD
Construction : Draw $O M ⊥ A B$ and $O N ⊥ C D$ .
Proof : In $Δ O M P a n d Δ O N P$ ,
$∠ O M P = ∠ O N P . . . (E a c h 90^{\circ})$
$O P = O P$ ...(Common)
$∠ O P M = ∠ O P N$ ...(Given)
$\Rightarrow Δ O M P ≅ Δ O N P$ (AAS congruency)
$\Rightarrow O M = O N (C P C T)$
$\Rightarrow$ Chords AB and CD are equidistant from centre.
We know that chords, which are equidistant from the centre of a circle, are also equal.
$∴ A B = C D$

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