An isosceles triangle ABC, with AB = AC, circumscribes a circle, touching Bc at P, AC at Q and AB at R. Prove that the contact point P bisects BC.

Question

Philoid · Accepted Answer

Consider a triangle ABC that circumscribes a circle touching AB, BC and AC at R, P and Q respectively.

To prove: P bisects BC i.e. BP = CP

We know that, tangents drawn from an external point to a circle are equal

⇒ AR = AQ [1]

⇒ BR = BP [2]

⇒ CP = CQ [3]

Also, ABC is an isosceles triangle with AB = AC [4]

Subtracting [1] From [4] gives,

AB – AR = AC – AQ

⇒ BR = CQ

From [2] and [3]

BP = CP

Hence, proved!