Mathematics — Standard (041) — CBSE Class 10 board question

Given: Circle touches QR at S, and extensions of PQ and PR at M and N respectively.

To prove: $PM = \dfrac{1}{2}(PQ + QR + PR)$

Proof:

Since tangents drawn from an external point to a circle are equal in length:

• From external point P: PM = PN ...(i)
• From external point Q: QM = QS ...(ii)
• From external point R: RN = RS ...(iii)

Now,

$$PM = PN$$
$$PM = PQ + QM = PQ + QS \quad \text{[from (ii)]}$$
$$PN = PR + RN = PR + RS \quad \text{[from (iii)]}$$

Adding both expressions:

$$2PM = PQ + QS + PR + RS = PQ + PR + (QS + RS)$$

$$2PM = PQ + PR + QR \quad \text{[since } QS + RS = QR\text{]}$$

$$\boxed{PM = \dfrac{1}{2}(PQ + QR + PR)}$$

Hence proved.

Source: Chapter 10, Section 10.3 (Theorem 10.2)

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• The key theorem used is: tangents from an external point are equal (Theorem 10.2).
• Note that M and N are on extensions beyond Q and R, so PM = PQ + QM and PN = PR + RN — students often make the mistake of writing PM = PQ − QM.
• Since QS + RS = QR (S lies on QR), the substitution completes the proof neatly.
• Examiners award marks for: stating the theorem, writing all three pairs of equal tangents, and the algebraic steps clearly leading to the result.