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Question
In Fig. 4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.
Solution
Given: An isosceles ΔABC with AB = AC, circumscribing a circle.
To prove: P bisects BC
Proof: AR and AQ are the tangents drawn from an external point A to the circle.
∴ AR = AQ (Tangents drawn from an external point to the circle are equal)
Similarly, BR = BP and CP = CQ.
It is given that in ΔABC, AB = AC.
⇒ AR + RB = AQ + QC
⇒ BR = QC (As AR = AQ)
⇒ BP = CP (As BR = BP and CP = CQ)
⇒ P bisects BC
Hence, the result is proved.
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