# In the figure, an isosceles triangle ABC, with AB=AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

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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.

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.

Last updated date: 30th May 2023

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