In Fig. 2, if a circle touches the side $QR$ of $ riangle 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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Question

In Fig. 2, if a circle touches the side $QR$ of $	riangle 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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Accepted Answer

Step 1: We know that tangents drawn from the external point to the circle are equal
$	herefore QS = QM$
$RS = RN$
$PM = PN$
Step 2: Now $2PM = PM + PN$
$= (PQ + QM) + (PR + RN)$
$= PQ + QS + PR + RS$
$= PQ + (QS + RS) + PR$
$= PQ + QR + PR$
Step 3: $	herefore PM = \frac{1}{2}(PQ + QR + PR)$

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