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