A,B and C are any three points on a circle.The bisector of <BAC cuts BC at D and the tangent at A meets BC at O when extended,then prove that AO=OD.
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Join BC. Let O be the centre of the circle. Join OA and OB.

In △BCT, △ACT,
∠BTC = ∠ATC = 36˚
∠ACT = ∠ABC = 48˚
∠BAC = ∠ACT + ∠ATC
= 48˚ + 36˚ = 84˚
∴ ∠BCA = 180˚ - (∠ABC + ∠BAC)
= 180˚ – (48˚ + 84˚) = 48˚
∴ ∠BOA = 2∠BCA
= 2 × 48˚
= 96˚
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