In the figure, an isosceles triangle ABC, with AB=AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.
Hint: We have to use the tangents concept to solve the given problem. Tangent segments drawn from an external point to a circle are congruent.
Given an isosceles triangle ABC with AB=AC, circumscribing a circle. We need to prove P bisects BC. Proof: AR and AQ are the tangents drawn from an external point A to the circle. Therefore, AR=AQ (Tangents drawn from an external point to the circle are equal) Similarly, BR=BP and CP=CQ. It is given that in triangle ABC, AB=AC. $$ \Rightarrow $$ AR+RB=AQ+QC $$ \Rightarrow $$ BR=QC (As AR=AQ) $$ \Rightarrow $$ BP=CP (as BR=BP and CP=CQ) $$ \Rightarrow $$ P bisects BC. Hence proved.
Note: Here we used the property that the tangents drawn from an external point to the circle are congruent. It is because that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
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