In the given figure, $\angle 1 = \angle 2$, show that $\triangle PQS \sim \triangle TQR$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer
In △PQS and △TQR:
$$\frac{QS}{QR} = \frac{QP}{QT} \quad \text{(Given: } \frac{QR}{QS} = \frac{QT}{QP} \text{)}$$
Also, ∠1 = ∠2 (Given), i.e., ∠PQS = ∠TQR (same angle ∠Q is common / ∠1 = ∠2 means ∠PQS = ∠TQR)
Since the sides including the equal angles are proportional:
$$\frac{QP}{QT} = \frac{QS}{QR}$$
and ∠PQS = ∠TQR (∠1 = ∠2, given)
Therefore, by SAS similarity criterion,
$$\triangle PQS \sim \triangle TQR \qquad \textbf{(Proved)}$$
Source: Chapter 6, Section 6.4 — Criteria for Similarity of Triangles (SAS similarity criterion, Exercise 6.3, Q.4)
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Explanation
- This is Exercise 6.3, Q.4. The given condition is $\dfrac{QR}{QS} = \dfrac{QT}{QP}$, which can be rewritten as $\dfrac{QP}{QT} = \dfrac{QS}{QR}$ — sides including ∠Q are proportional.
- ∠1 = ∠2 means ∠PQS = ∠TQR (these are the included angles between the proportional sides).
- Two conditions met: one pair of equal included angles + sides around them proportional → SAS similarity. Always state the criterion by name for full marks.
- Write the similarity in correct vertex order: △PQS ~ △TQR.