Triangle ABC is isosceles in which AB=AC circumscribed about a circle. Prove that base is bisected by the point of contact.

@Studious: You had provided the correct answer. Keep Posting!!

Still the detailed solution is as:

We know that the tangents drawn from an exterior point to a circle are equal in length.

∴ AP = AQ  (Tangents from A)  ..... (1)

 BP = BR  (Tangents from B)  ..... (2)

 CQ = CR  (Tangents from C)  ..... (3)

Now, the given triangle is isosceles, so given AB = AC

Subtract AP from both sides, we get

AB – AP = AC – AP

⇒ AB – AP = AC – AQ  (Using (1))

BP = CQ

⇒ BR = CQ  (Using (2))

⇒ BR = CR  (Using (3))

So BR = CR, shows that BC is bisected at the point of contact.