Question

A circle is touching the side $BC$ of $\Delta ABC$ at $P$ and is touching $AB$ and $AC$ when produced at $Q$ and $R$ respectively. Prove that $AQ=1/2$
(Perimeter of $\Delta ABC$ ) .

Answer 1

Consider $△ABC$

$BP,CP,CR,AQ,AR$ and $BQ$ becomes tangents to circle

$⟹AQ=AR,BQ=BP,CP=CR$

Consider perimeter of $△ABC$

$AB+BC+AC=AB+BP+PC+AC=AQ+AR=2AQ$

$⟹AQ=\frac{1}{2}($perimeter of $△ABC)$