Question

in the given figure o is the centre of a circle C (o,r) and PO bisects angle APC. Prove that AB=CD​

Answer 1

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

Answer 2

Step-by-step explanation:

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