Question

In the given figure, $SP$ is bisector of $\angle RPT$ and $PQRS$ is a cyclic quadrilateral. Prove that: $SQ=SR$.

Answer 1

$PQRS$ is a cyclic quadrilateral

$\therefore \angle QRS+\angle QPS={180}^{\circ }$ ......$\left(1\right)$

(pair of opposite angles in a cyclic quadrilateral are supplementary)

Also,$\angle QPS+\angle SPT={180}^{\circ }$ ......$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\angle QRS=\angle TPS$ ......$\left(3\right)$

Also,$\angle RQS=\angle RPS$ .....$\left(4\right)$

Angle subtended by the same chord on the circle are equal.

$\therefore PS$ bisects $\angle RPT$ .......$\left(5\right)$

From $\left(3\right),\left(4\right)$ and $\left(5\right)$

$\angle QRS=\angle RQS$

$\Rightarrow SQ=SR$