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

Given: AD and PS are medians of △ABC and △PQR respectively, and △ABD ~ △PQS.

To Prove: △ABC ~ △PQR

Proof:

Since △ABD ~ △PQS,
$$\frac{AB}{PQ} = \frac{BD}{QS} \quad \text{and} \quad \angle B = \angle Q \tag{1}$$

Since AD is median of △ABC, BD = BC/2, so BC = 2BD.
Since PS is median of △PQR, QS = QR/2, so QR = 2QS.

$$\therefore \frac{BC}{QR} = \frac{2BD}{2QS} = \frac{BD}{QS} = \frac{AB}{PQ} \tag{2}$$

From (1) and (2):
$$\frac{AB}{PQ} = \frac{BC}{QR} \quad \text{and} \quad \angle B = \angle Q$$

Therefore, △ABC ~ △PQR (by SAS similarity criterion). [Proved]

Source: Criteria for Similarity of Triangles, Chapter 6

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• The key idea is that medians bisect the opposite side, so BC = 2BD and QR = 2QS. This lets you replace the ratio BD/QS with BC/QR.
• Since △ABD ~ △PQS gives you AB/PQ = BD/QS and ∠B = ∠Q, combining with the above gives two sides in ratio and included angle equal → SAS similarity.
• Examiners expect you to explicitly state why BC = 2BD (because AD is a median), then chain the ratios clearly, and name the criterion at the end.