In Fig.-2, 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 $PM = \frac{1}{2}(PQ + QR + PR)$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:28 · grounding rag
Model Answer
To Prove: $PM = \dfrac{1}{2}(PQ + QR + PR)$
Proof:
Since tangents drawn from an external point to a circle are equal (Theorem 10.2):
- From external point P: $PM = PN$ … (i)
- From external point Q: $QM = QS$ … (ii)
- From external point R: $RS = RN$ … (iii)
Now, the perimeter of △PQR:
$$PQ + QR + PR = PQ + (QS + SR) + PR$$
$$= PQ + QM + RN + PR \quad \text{[using (ii) and (iii)]}$$
$$= (PQ + QM) + (PR + RN)$$
$$= PM + PN \quad \text{[since M is on PQ extended, } PQ + QM = PM\text{]}$$
$$= PM + PM \quad \text{[using (i)]}$$
$$= 2\,PM$$
Therefore,
$$\boxed{PM = \dfrac{1}{2}(PQ + QR + PR)}$$
Hence proved.
Source: Chapter 10, Section 10.3 (Theorem 10.2)
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Explanation
- The key theorem used is "tangents from an external point are equal" — apply it to all three vertices P, Q, R separately.
- Note: M lies on the extension of PQ beyond Q, so $PM = PQ + QM$. Similarly $PN = PR + RN$. Students often miss this step and lose marks.
- The proof is essentially showing perimeter = 2PM, then halving — examiners expect each step (i), (ii), (iii) stated clearly with the reason "tangents from external point are equal."