In the figure,$\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\Delta PQS \sim \Delta TQR$.

(N/A) In $\Delta PQR$,given $\angle PQR = \angle PRQ$.
Since the angles opposite to equal sides are equal,we have $PQ = PR$ $....(i)$
Given,
$\frac{QR}{QS} = \frac{QT}{PR}$
Using $(i)$,we substitute $PR$ with $PQ$ to obtain:
$\frac{QR}{QS} = \frac{QT}{PQ}$ $....(ii)$
Now,in $\Delta PQS$ and $\Delta TQR$:
$\frac{QR}{QS} = \frac{QT}{PQ}$ $[\text{From } (ii)]$
$\angle Q = \angle Q$ $[\text{Common angle}]$
Therefore,by the $SAS$ similarity criterion,$\Delta PQS \sim \Delta TQR$.