In the given figure, $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ and ∠1 = ∠2. Show that △ PQS ~ △ TQR.

In Δ PQR,

∠1 = ∠2 (Given)

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

In Δ PQS and Δ TQR

⇒ ∠PQS = ∠TQR (Both = ∠1)

Given,

$\Rightarrow \dfrac{QR}{QS} = \dfrac{QT}{PR}$

Since, PR = PQ.

$\therefore \dfrac{QR}{QS} = \dfrac{QT}{PQ}$

∴ Δ PQS ~ Δ TQR (By SAS criterion)

Hence, proved that △ PQS ~ △ TQR.