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

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

To prove: △ABC ~ △PQR

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

Proof:

Since AD is a median, D is the mid-point of BC. Also AD = DE, so ABEC is a parallelogram.
∴ BE = AC ... (1)

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

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

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

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

So, △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) $\blacksquare$

Source: Triangles, Section 6.4 (Exercise 6.3, Q.14)

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• The key trick is producing the medians to double their length, forming parallelograms (ABEC and PQNR). This converts the median ratio into a full side ratio.
• This lets you apply SSS to △ABE and △PQN, getting the included angle equal.
• Then SAS finishes the main result on △ABC ~ △PQR.
• Examiners award marks for: correct construction (1 mark), parallelogram argument (1 mark), SSS on helper triangles (1 mark), getting the equal angle (1 mark), final SAS conclusion (1 mark).