In figure, $\frac{Q R}{O S}=\frac{Q T}{P R}$ and $\angle 1=\angle 2$.

Solution:

In figure, $\angle 1=\angle 2$ (Given)

$\Rightarrow \mathrm{PQ}=\mathrm{PR}$

(Sides opposite to equal angles of $\Delta \mathrm{PQR}$ )

We are given that

$\frac{Q R}{O S}=\frac{Q T}{P R}$

$\Rightarrow \frac{\mathrm{QR}}{\mathrm{OS}}=\frac{\mathrm{QT}}{\mathrm{PQ}} \quad(\because \mathrm{PQ}=\mathrm{PR}$ proved $)$

$\Rightarrow \frac{Q S}{Q R}=\frac{P Q}{Q T} \quad$ (Taking reciprocals) ...(1)

Now, in $\triangle \mathrm{PQS}$ and $\triangle \mathrm{TQR}$, we have

$\angle \mathrm{PQS}=\angle \mathrm{TQR} \quad($ Each $=\angle 1)$

and $\frac{\mathrm{OS}}{\mathrm{OR}}=\frac{\mathrm{PQ}}{\mathrm{OT}}$(By (1))

Therefore, by SAS similarity criterion, we have

$\Delta \mathrm{PQS} \sim \Delta \mathrm{TQR} .$