In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and ∠ 1 =∠ 2. Show that ∆ PQS ~ ∆ TQR.

In ΔPQR,

∠1 = ∠2

⇒ PR = PQ (In a triangle sides opposite to equal angles are equal)

In ΔPQS and ΔTQR

∠PQS = ∠TQR = ∠1 (same angle)

 $\frac{QR}{QS} = \frac{QT}{PR}$(Since PR = PQ)

⇒ ΔPQS ~ ΔTQR (SAS criterion)