Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of another triangle PQR. Show that $\triangle ABC \sim \triangle PQR$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:24 · 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, BD = BC/2.
Since PM is a median of △PQR, QM = QR/2.
Given: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$
Since $\dfrac{BC}{QR} = \dfrac{AB}{PQ}$, we have $\dfrac{2BD}{2QM} = \dfrac{AB}{PQ}$
$$\therefore \dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$$
So, △ABD ~ △PQM (SSS similarity criterion)
$$\therefore \angle ABD = \angle PQM$$
i.e., $\angle B = \angle Q$
Now in △ABC and △PQR:
$$\dfrac{AB}{PQ} = \dfrac{BC}{QR} \quad \text{(given)}$$
$$\angle B = \angle Q \quad \text{(proved above)}$$
Therefore, △ABC ~ △PQR (SAS similarity criterion) $\blacksquare$
Source: Chapter 6, Section 6.4 (Criteria for Similarity of Triangles)
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Explanation
- Key idea: You cannot directly apply SSS to △ABC and △PQR because you only have two sides and a median (not all three sides). So the trick is to first prove △ABD ~ △PQM using SSS (three sides in proportion), extract ∠B = ∠Q, then use SAS on the original triangles.
- Common mistake: Students forget to halve BC and QR (since D and M are midpoints) before applying SSS to the smaller triangles.
- Examiner focus: Clear logical steps — establishing BD = BC/2, showing all three ratios equal for △ABD and △PQM, then correctly citing SAS for the final result. Naming the similarity criterion at each step earns method marks.