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In the given figure, $\Delta PQR$ is a right triangle, right-angled at Q and QS $\perp$ PR. Prove that $\Delta PQS \sim \Delta QRS$.
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OR
(b) In $\Delta PQS$ and $\Delta QRS$
$\angle PSQ = \angle QSR$ (each $90^\circ$) (1 Mark)
$\angle PQS = \angle QRS$ (each ($90^\circ – \angle SQR$)) (1 Mark)
$\Delta PQS \sim \Delta QRS$ (by AA similarity)
(b) In $\Delta PQS$ and $\Delta QRS$
$\angle PSQ = \angle QSR$ (each $90^\circ$) (1 Mark)
$\angle PQS = \angle QRS$ (each ($90^\circ – \angle SQR$)) (1 Mark)
$\Delta PQS \sim \Delta QRS$ (by AA similarity)