From a point `P` outside a circle with centre `O`, tangents `PA` and `PB` are drawn to the circle. Prove that `OP` is the right bisector of the line segment `AB`.

Explanation

GIVEN `PA` and `PB` are tangents to a circle with centre `O` from an external point `P`.
TO PROVE `OP` is the right bisector of `AB` .
CONSTRUCTION Join `AB`. Let `AB` intersect `OP` at `M`.
PROOF In `DeltaMAP` and `DeltaMBP`, we have
`PA=PB` [`:'` tangents to a circle from an external point are equal]
`MP=MP` [common]
`/_MPA=/_MPB` [`:'` tangents from an external point are equally inclined to the line segment joining the centre to that point i.e., `/_OPA=/_OPB`]
`:.MAP ~= DeltaMBP` [by SAS-congruence]
And so, `MA=MB`[cpct]
and `/_AMP=/_BMP` [cpct]
But `/_AMP+/_BMP=180^(@)` [linear pair]
`:./_AMP=/_BMP=90^(@)`
Hence, `OP` is the right bisector of `AB`.