Sides $AB$, $BC$ and the median $AD$ of $\triangle ABC$ are respectively proportional to sides $PQ$, $QR$ and the median $PM$ of another $\triangle PQR$. Prove that $\triangle ABC \sim \triangle PQR$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
Given: In △ABC and △PQR, $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$, where AD and PM are medians.
To Prove: △ABC ~ △PQR
Proof:
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 $QM = \dfrac{QR}{2}$.
Given: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$
$$\Rightarrow \dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{AD}{PM}$$
$$\Rightarrow \dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$$
Therefore, △ABD ~ △PQM (SSS similarity criterion)
Hence, $\angle ABD = \angle PQM$, i.e., $\angle ABC = \angle PQR$.
Now, in △ABC and △PQR:
$$\dfrac{AB}{PQ} = \dfrac{BC}{QR} \quad \text{(given)}$$
$$\angle ABC = \angle PQR \quad \text{(proved above)}$$
Therefore, △ABC ~ △PQR (SAS similarity criterion). $\hspace{1cm}\blacksquare$
Source: Triangles, Section 6.4 (Criteria for Similarity of Triangles)
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Explanation
- Key idea: You cannot apply SSS directly to △ABC and △PQR because the median is not a side. So you first use the given proportion (after halving BC and QR) to prove △ABD ~ △PQM by SSS, extract the included angle ∠B = ∠Q, and then use SAS on the original triangles.
- Common mistake: Students try to apply SSS directly to △ABC and △PQR with only two sides and a median — that is wrong. The two-step approach (SSS → get angle → SAS) is what examiners expect.
- Mark split (5 marks): ~1 mark for setting up BD = BC/2, QM = QR/2; ~2 marks for proving △ABD ~ △PQM by SSS; ~1 mark for extracting ∠ABC = ∠PQR; ~1 mark for concluding △ABC ~ △PQR by SAS.