Question

ABC is an isosceles triangle, in which AB=AC circumscribe about a circle.Show that BC is bisects at the point of contact.

Answer 1

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.

Answer 2

Answer:

Step-by-step explanation: