Q) In the given figure, if a circle touches the side QR of a △PQR at S and extended sides PQ and PR at M and N respectively, then prove that : PM =1/2 (PQ + QR + PR).

(Q27 B – 30/1/3 – CBSE 2026 Question Paper)

Ans: 

From the diagram, we have a circle which touches QR at S and extended sides PQ and PR at M and N respectively.

Here, we need to prove: PM = (PQ + QR + PR)

Step 1: By circle’s property of tangents, the tangents drawn from an external point to a circle are equal.

∴ PM = PN        (tangents from point P)

and QS = QM    (tangents from point Q)

and RS = RN     (tangents from point R)

Step 2: Let’s express sides of Δ PQR in terms of tangents:

∴ PQ = PM – QM = PM – QS           

(∵ QS = QM)

and PR = PN – RN = PN – RS           

(∵ RS = RN) 

and QR = QS + RS

Step 3: By adding all sides, we get:

∴ PQ + PR + QR = (PM – QS) + (PN – RS) + (QS + RS)

∴ PQ + PR + QR = 2 PM – QS – RS + QS + RS

∴ PQ + PR + QR = 2 PM –

∴ PQ + PR + QR = 2 PM

∴ PM = (PQ + PR + QR)

Hence Proved !

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