Question

$ABC$ is an isosceles triangle, in which $AB=AC$, circumscribed about a circle. Show that $BC$ is bisected at the point of contact ?

Answer 1

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

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

$BP=BR$ (tangent from $B$)

$CQ=CR$ (tangent from $C$)

It is given that $ABC$ is an isosceler triangle with sides $AB=AC$

$\Rightarrow AB-AP=AC-AP$

$\Rightarrow AB-AP=AC-AQ$

$\Rightarrow BR=CQ$

$\Rightarrow BR=CR$

So, therefore, $BR=CR$ that imples $BC$ is bisected at the point of contact.