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In the given figure, $\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}}$ and $\angle 1 = \angle 2$, show that $\triangle$ PQS $\sim \triangle$ TQR.
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In $\triangle$ PQR, $\angle 1 = \angle 2 \Rightarrow \text{PR} = \text{PQ}$ (1/2)
$\therefore \frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}} \Rightarrow \frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PQ}}$ (1/2)
Also, $\angle 1 = \angle 1$ (1/2)
$\therefore \triangle$ PQS $\sim \triangle$ TQR (1/2)
$\therefore \frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}} \Rightarrow \frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PQ}}$ (1/2)
Also, $\angle 1 = \angle 1$ (1/2)
$\therefore \triangle$ PQS $\sim \triangle$ TQR (1/2)