Question

Consider the figure, $PQRS$ is a rectangle. $PT$ and $RU$ are perpendicular from $P$ and $R$ on $SQ$. Prove that $\Delta PTS\cong \Delta QRU$

Answer 1

REF.Image.

In $△PTS$ and $△QUR$

$\angle PTS=\angle QUR={90}^{\circ }$ (given)

$\angle TSP=\angle UQR$ [alternate angles ]

$\Rightarrow \angle TPS=\angle QRU$

$SP=QR$ (given)

$\therefore$ By Angle side Angle (A.S.A) axiom of congruence.

$△PTS\cong △RUQ$