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

Given: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$, where AD and PM are medians of △ABC and △PQR respectively.

To prove: △ABC ~ △PQR

Proof:

Since AD is a median of △ABC, D is the mid-point of BC.
∴ BC = 2BD

Since PM is a median of △PQR, M is the mid-point of QR.
∴ QR = 2QM

Given: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$

∴ $\dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{AD}{PM}$

i.e., $\dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$

∴ △ABD ~ △PQM (SSS similarity criterion)

∴ ∠ABD = ∠PQM, i.e., ∠B = ∠Q

Now, in △ABC and △PQR:
$$\frac{AB}{PQ} = \frac{BC}{QR} \quad \text{(given)}$$
$$\angle B = \angle Q \quad \text{(proved above)}$$

∴ △ABC ~ △PQR (SAS similarity criterion) $\hspace{2cm}$ Hence proved.

Source: Chapter 6, Section 6.4 (SSS and SAS Similarity Criteria)

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• Key idea: You cannot directly apply SSS similarity to △ABC and △PQR because a median is not a side. So you first work with the half-triangles △ABD and △PQM where BC/QR = BD/QM (halves are in the same ratio), then use SSS to get ∠B = ∠Q, and finally apply SAS similarity to the original triangles.
• Examiners look for: correctly halving BC and QR, applying SSS to △ABD ~ △PQM to extract the angle, then concluding with SAS for △ABC ~ △PQR.
• Write "Hence proved" at the end.