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Cyclic Quadrilateral

In the given ...

Question

In the given figure, SP is bisector of $∠$ RPT and PQRS is a cyclic quadrilateral. Prove that :

SQ = SR.

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Solution

$PQRS$ is a cyclic quadrilateral

$\angle SPR = \angle SPT = x$

$\angle QPR = 180^o-2x$

$\angle SQR + \angle QPS = 180$ (cyclic quadrilateral)

$\angle SRQ = 180^o-(180^o-x)= x$

Also $\angle RQS = \angle RPS = x$ (angles in the same segment of the circle subtended by the same chord)

$\therefore \angle RQS = \angle QRS$

$\therefore RS=QS$

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