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] [1]

QC = CP [ Tangents from point C] [2]

AR = AQ [ Tangents from point A] [3]

As ABC is an isosceles triangle,

AB = BC [Given] [4]

Now substract [3] from [4]

AB - AR = BC - AQ

BR = QC

BP = CP [ From 1 and 2]

∴ P bisects BC

Hence Proved.

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