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.

Answer 1

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.