Question

In the given figure $PQRS$ is a cyclic quadrilateral $PQ$ and $SR$ produced meet at $T$.
Prove $△TPS\sim △TRQ$

Answer 1

In $△TPS$ and $△TRQ$
$\angle PST=\angle RQT$
$\angle SPQ=\angle QRT$
$[\because$ exterior angle of cyclic quadrilateral is equal to the interior opposite angle]
$\angle T=\angle T$ (common)
$\therefore △TPS-△TRQ$ (By $AA$)