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In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.
As we know that tangents drawn from an external point to a circle are equal,
BR = BP [ Tangents from point B] 
QC = CP [ Tangents from point C] 
AR = AQ [ Tangents from point A] 
As ABC is an isosceles triangle,
AB = BC [Given] 
Now substract  from 
AB - AR = BC - AQ
BR = QC
BP = CP [ From 1 and 2]
∴ P bisects BC