In the given figure, O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects $∠$ BPD. Prove that AB = CD.

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Solution

Given: O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects ∠ BPD.
To prove: AB = CD
Construction: Draw OE ⊥ AB and OF ⊥ CD
Proof: In Δ OEP and Δ OFP, we have:
∠ OEP = ∠ OFP
(90° each)
OP = OP
(Common)
∠ OPE = ∠ OPF
(∵ OP bisects ∠BPD )
Thus, Δ OEP ≅ Δ OFP (AAS criterion)
⇒ OE = OF
Thus, chords AB and CD are equidistant from the centre O.
⇒ AB = CD
(∵ Chords equidistant from the centre are equal)
∴ AB = CD

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