Q. 104.6( 19 Votes )

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.


Answer :

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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