Question

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

Answer 1

$\to$ As lengths of tangents

drawn from a fixes point

to a circle is equal,

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

$Now,2AQ=AQ+AQ$

$=AQ+AR$

$=AB+BQ+AC+RC$

$=AB+BP+AC+PC$

2AQ = perimeter of $△ABC$

$AQ=\frac{1}{2}$ (perimeter)