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

Given: △ABC ~ △PQR; AD and PM are medians of △ABC and △PQR respectively.

To prove: $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$

Proof:

Since △ABC ~ △PQR,

$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \tag{1}$$

and ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(2)

Since AD is a median of △ABC, D is the mid-point of BC, so BD = $\dfrac{BC}{2}$.

Since PM is a median of △PQR, M is the mid-point of QR, so QM = $\dfrac{QR}{2}$.

From (1): $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$

$$\Rightarrow \frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM} \tag{3}$$

Also, ∠B = ∠Q …from (2)

In △ABD and △PQM:

$$\frac{AB}{PQ} = \frac{BD}{QM} \text{ and } \angle B = \angle Q$$

∴ △ABD ~ △PQM (SAS similarity criterion)

$$\therefore \frac{AB}{PQ} = \frac{AD}{PM}$$

Hence proved.

Source: Chapter 6, Section 6.4 (Criteria for Similarity of Triangles)

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• The key idea is to use the given similarity to establish that AB/PQ = BD/QM (by halving BC and QR), and then combine this with the equal included angle ∠B = ∠Q to apply SAS similarity on △ABD and △PQM.
• Examiners award marks for: correctly stating the ratio from similarity (1 mark), using medians to write BD = BC/2 and QM = QR/2 (1 mark), setting up the SAS condition (1–2 marks), concluding the result (1 mark).
• Do not skip the step showing BD/QM = AB/PQ — this is the critical link.