Question

In the given figure $PQRS$ is a cyclic quadrilateral $PQ$ and $SR$ produced meet at $T$. Find $SP$ if $TP=18cm,RQ=4cm$ and $TR=6cm$.

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 $AA$)

In similar triangles $△TPS$ and $TRQ$
$\frac{TP}{TR}=\frac{PS}{PQ}$
$\frac{18}{6}=\frac{PS}{4}$
$PS=\frac{72}{6}$
$=12cm$.