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.

Question:

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.

 

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 ΔOEP and ΔOFPwe have:
∠OEP = ∠OFP         (90° each)
OP = OP                   (Common)
OPE = ∠OPF         (∵ OP bisects ∠BPD )
Thus, ΔOEP ≅ ΔOFP      (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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