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))



⇒ 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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let base be bisected at x ,then,ap=aq(tangents).


That is PB=CQ----1

But BP=BX and QC=CX (tangets) -----2

Fom 1 and 2,we prove BX=CX

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