Question

In Fig, $PR>PQ$ and $PS$bisects $QPR.$Prove that $PSR>PSQ.$

Answer 1

Given : $PR>PQ$

$PQR>PRQ....\left(i\right)$ (Since $PR>PQ$ as angle opposite to the larger side is always larger)

$QPS=RPS....\left(ii\right)$ (As $PS$bisects $\angle QPR$)

$PSR=PQR+QPS....\left(iii\right)$ (Since the exterior angle of a triangle equals to the sum of opposite interior angles)

$PSQ=PRQ+RPS....\left(iv\right)$ (Since the exterior angle of a triangle equals to the sum of opposite interior angles)

On adding $\left(i\right)$and $\left(ii\right)$

$PQR+QPS>PRQ+RPS$

$\Rightarrow PSR>PSQ$ [From $\left(iii\right)and\left(iv\right)$]

Hence proved.