102
A circle touches the side $BC$ of a $\triangle ABC$ at a point $P$ and touches $AB$ and $AC$ when produced at $Q$ and $R$ respectively. Show that $AQ = \frac{1}{2}$ (Perimeter of $\triangle ABC$).
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$AQ = AR$
$2AQ = AQ + AR$
$= AB + BQ + AC + CR$
$= AB + AC + (BP + CP)$
$= AB + AC + BC$
$AQ = \frac{1}{2} (AB + AC + BC) = \frac{1}{2} (\text{Perimeter of } \triangle ABC)$
$2AQ = AQ + AR$
$= AB + BQ + AC + CR$
$= AB + AC + (BP + CP)$
$= AB + AC + BC$
$AQ = \frac{1}{2} (AB + AC + BC) = \frac{1}{2} (\text{Perimeter of } \triangle ABC)$