Question

In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

Answer 1

∆ABC is an isosceles triangle where AB = AC and circumscribed a circle.

The circle touches its sides BC, CA and AB at P, Q and R respectively.

To Prove : P bisects the base BC, i.e. BP = PC

Now form figure,

BR and BP are tangents to the circle.

So, BR = BP …..(1)

AR and AQ are tangents to the circle.

AR = AQ But AB = AC

AB – AR = AC – AQ

BR = CQ …..(2)

Similarly, CP and CQ are tangents to the circle.

CP = CQ ……(3)

From (1), (2) and (3)

BP = PC

Hence Proved.