## Answers

**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,**

**AB-AP=AC-AP**

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

**BP=BQ**

**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.**

*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.*