In the given figure, $\frac{QR}{QS}=\frac{QT}{PR}$ and $\angle 1 = \angle 2$. show that $∆PQR \sim ∆TQR$.
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Given:

$\frac{QR}{QS}=\frac{QT}{PR}$ and $\angle 1 = \angle 2$.

To do:

We have to show that $∆PQR \sim ∆TQR$.

Solution:

$\angle 1=\angle 2$

This implies,

$PQ=PR$            (Sides opposite to equal angles are equal)

In $\triangle PQS$ and $\triangle TQR$,

$\frac{QR}{QS}=\frac{QT}{PR}$

$\frac{QR}{QS}=\frac{QT}{PR}$       (Since $PQ=PR$)

$\angle PQS=\angle TQR=\angle 1$

Therefore, by SAS criterion,

$\Delta PQS \sim \Delta TQR$

Hence proved.

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Updated on: 10-Oct-2022

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