Question

ΔABC is an isosceles triangle in which AB = AC and sides of the triangle touch the circle at P, Q and R. Prove that Q is the mid point of the base BC.

Answer 1

We know that the lengths of tangents to a circle drawn from an external point are equal.

Here, AB = AC, BP = BQ, AP = AR and CR = CQ

Using this information, we get:

BQ = BP = AB – AP = AC – AP (AB = AC)

⇒BQ = AC – AR (AP = AR)

⇒BQ = RC

⇒BQ = CQ (CR = CQ)

Hence, Q is the midpoint of BC.