In the given figure, `O` is the centre of a circle `C(O,r)` and `PO` bisects `/_APC` . Prove that `AB=CD`.

Explanation

Draw `OE _|_ AB` and `OF_|_ CD`.
Now, in `Delta OEP` and `Delta OFP` we have
`/_OEP = /_ OFP ` [ each equal to `90^(@)`]
`/_ OPE = /_ OPF `[ given ]
` OP =OP` [common]
`:. Delta OEP ~= Delta OFP `[ AAS-congruence ]
`implies OE = OF` [c.p.c.t.]
`implies` chords ` AB` and `CD` are equidistnat from the centre `O`
`implies AB = CD` [ chords equidistant from the centre are equal ].
Hence, `AB=CD`