Question

In the figure
$\angle QPR=\angle UTS={90}^o$ and PR || TS
Prove that $△PQR\sim △TUS$

Answer 1

In $\Delta PQR$ and $\Delta TUS$, it is given that $PR||TS$. Also,

$\angle QPR=\angle UTS={90}^0$ (Given)

$\angle PRQ=\angle TSU$ (Alternate angles)

$\angle PQR=\angle STU$ (Alternate angles)

Therefore, $△PQR\sim △TUS$ (AA criterion)

Hence proved.