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

Given: In △ABC and △PQR, AD and PM are medians such that
$$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$$

To prove: △ABC ~ △PQR

Construction: Produce AD to E such that DE = AD, and produce PM to N such that MN = PM. Join BE and CE, QN and RN.

Proof:

Since AD is a median, D is the mid-point of BC, so BD = DC.
Since DE = AD (construction), ABEC is a parallelogram (diagonals AE and BC bisect each other).
∴ BE = AC ... (1)

Similarly, PQNR is a parallelogram, so QN = PR ... (2)

Now, AE = 2AD and PN = 2PM.

Given: $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$

So, $\dfrac{AB}{PQ} = \dfrac{BE}{QN} = \dfrac{AE}{PN}$ [using (1), (2) and AD/PM = AE/PN]

∴ △ABE ~ △PQN (SSS similarity)
∴ ∠BAE = ∠QPN, i.e., ∠BAC = ∠QPR

Now in △ABC and △PQR:
$$\frac{AB}{PQ} = \frac{AC}{PR} \quad \text{and} \quad \angle BAC = \angle QPR$$

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

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

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• The key trick is the construction — producing the median to double its length to form a parallelogram — which lets you express AC as BE and PR as QN.
• This converts the three given ratios (AB/PQ, AC/PR, AD/PM) into three sides of △ABE and △PQN, giving SSS similarity, which yields ∠BAC = ∠QPR.
• Then use SAS similarity on △ABC and △PQR (two sides proportional + included angle equal).
• Examiners award marks for: correct construction (1 mark), parallelogram argument (1 mark), SSS step (1 mark), angle equality (1 mark), final SAS conclusion (1 mark).