Question

O is the center of a circle in which chords AB and CD intersect at P such that PO bisects angle BPD. Prove that AB=CD.

Answer 1

Answer:

Given: AB and CD are the chords of a circle whose centre is O.

They intersect at P. PO is bisector of ∠APD

Required to prove: AB = CD

Construction: Draw OR ⊥ AB and OQ ⊥ CD

Proof: In ΔOPR and ΔOPQ

∠OPR = ∠OPQ (given)

OP = OP (Common)

∠ORP = ∠OQP (Construction)

ΔOPR ≅ ΔOPQ. (AAS axiom)

∴ OR = OQ (C.P.C.T)

AB = CD (chords of a circle which are at equidistant from the centre are equal)

Hence proved.

Answer 2

We know that

angle 0EP = angle OFP = 90°

OP is common OP=OP

OP bisects angle BPD

angle OPE = angle OPF

BY ASA congruence criterion

∆OEP is congruent to ∆DFP

OE= OF (C.P.C.T)

AB= CD

therefore it is proved that AB = CD