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Question
In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that point of contact P bisects the base BC.
Solution
We know that tangent segments to a circle from the same external point are congruent
Now, we have
AR = AO, BR = BP and CP = CQ
Now, AB = AC
⇒ AR+ RB= AQ+ QC
⇒ AR + RB = AR + OC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.
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