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In the given figure, if a circle touches the side $QR$ of $\triangle PQR$ at $S$ and extended sides $PQ$ and $PR$ at $M$ and $N$ respectively, then prove that : $PM = \frac{1}{2} (PQ + QR + PR)$
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$PM = PN$
$QS = QM$
$RS = RN$ (I) (1/2 Mark)
$PM + PN = PQ + QM + PR + RN$ (II) (1/2 Mark)
$2 PM = PQ + QS + PR + RS$
$= PQ + QS + RS + PR$
$= PQ + QR + PR$ (III) (1/2 Mark)
$\therefore PM = \frac{1}{2} (PQ + QR + PR)$ (IV) (1/2 Mark)
$QS = QM$
$RS = RN$ (I) (1/2 Mark)
$PM + PN = PQ + QM + PR + RN$ (II) (1/2 Mark)
$2 PM = PQ + QS + PR + RS$
$= PQ + QS + RS + PR$
$= PQ + QR + PR$ (III) (1/2 Mark)
$\therefore PM = \frac{1}{2} (PQ + QR + PR)$ (IV) (1/2 Mark)