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As tangents drawn from an external point to a circle are equal in length

So, therefore, we get, AP=AQ (tangents from A) 1)

BP=BR (tangents from B) 2)

CQ=CR(tangents from C) 3)

As it is given that ABC is an isosceles triangle with sides AB=AC

Subtracting AP from both sides, we have,


implies AB-AP=AC-AQ (from 1)


implies BR=CQ (from 2)

implies BR=CR(from 3)

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

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Let the circle touches the side AB at P and side AC at Q and side BC at R
We know that Tangents drawn from external points are equal.
Then we have Tangents from point A  i.e AP = AQ , 
                     Tangents  from point B gives BP = BR ,
                     Tangents  from point C gives RC = CQ.
We have AB=AC
           ⇒ AP+PB=AQ +QC        as AP= AQ
           ⇒  PB = QC
           ⇒  BR = RC
This gives that BC is bisected at point of contact.

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