Question

In the figure, seg QS $\perp$ side PR, seg RT $\perp$ side PQ, P-S-R, and P-T-Q. Show that (i) $\Delta PQS\sim \Delta PRT$.

Answer 1

$\angle PTR=\angle PSQ={90}^{\circ }\angle QPS=\angle RPT$

In $\Delta PSQ\angle PQS=180-90-\angle P\Delta RTP\angle PRT=180-90-\angle P\angle PQS=\angle PRT$

Since all angles are equal the triangle is similar