Mathematics — CBSE Class 10 board question

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)

---

• 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.