Question

A circle touches the side $BC$ of $△ABC$ at $P$ and touches $AB$ and $AC$ produced at $Q$ and $R$ respectively. Prove that $AQ=\frac{1}{2}$ (perimeter of $△ABC$).

Answer 1

Given:A circle touching the side $BC$ of $△ABC$ at $P$ and $AB$,$AC$ produced at $Q$ and $R$ respectively.

Proof: Lengths of tangents drawn from an external point to a circle are equal.

$\Rightarrow AQ=AR,BQ=BP,CP=CR$.

Perimeter of $△ABC=AB+BC+CA$

$=AB+(BP+PC)+(AR–CR)$

$=(AB+BQ)+\left(PC\right)+(AQ–PC)$ since $[AQ=AR,BQ=BP,CP=CR]$

$=AQ+AQ$

$=2AQ$

$\Rightarrow AQ=\frac{1}{2}$(Perimeter of $△ABC$)

$\therefore AQ$ is the half of the perimeter of $△ABC$