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.

Answer 1

Consider △ OEP and △ OFP

We know that

∠OEP = ∠OFP = 90^o

OP is common i.e. OP = OP

From the figure we know that OP bisects ∠BPD

It can be written as

∠OPE = ∠OPF

By ASA congruence criterion

△ OEP ≅ △ OFP

OE = OF (c. p. c. t)

We know that AB and CD are equidistant from the centre

So we get

AB = CD

Therefore, it is proved that AB = CD.