Question

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.

Answer 1

Given, the circle touches the sides AB at R and side AC at Q and side BC at P.

Since, the tangents drawn from external points are equal

Then, we have tangents from points A,

i.e., AR = AQ   ...(i)

Tangents from point B,

i.e., BR = BP   ...(ii)

And tangents from point C,

i.e., CP = CQ   ...(iii)

Given, AB = AC

⇒ AR + BR = AQ + CQ

⇒ AR + BP = AQ + CP   ...[From equations (ii) and (iii)]

⇒ AQ + BP = AQ + CP   ...[From equation (i)]

⇒ BP = CP

Hence, the point of contact P bisects BC.