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.

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