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

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#### 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.

Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles

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