In the given figure, O is the centre of a circle in which chords AB an

1.	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» 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.

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»

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.

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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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