Question

In figure, $PR>PO$ and $PS$ bisect $OPR$. Prove that $\angle PSR>\angle PSO$

Answer 1

Given:

$PR>PO$ and $PS$ bisects $\angle OPR$

To prove:

$\angle PSR>\angle PSO$

Proof:

$\angle POR>\angle PRO$ $..........\left(1\right)$ $(PR>PO$ as angle opposite to larger side is larger.)

$\angle OPS=\angle RPS$ $............\left(2\right)$ $(PS$ bisects $\angle OPR$)

$\angle PSR=\angle POR+\angle OPS$ $...........\left(3\right)$

(exterior angle of a triangle equals to the sum of opposite interior angles)

$\angle PSO=\angle PRO+\angle RPS$ $........\left(4\right)$

(exterior angle of a triangle equals to the sum of opposite interior angles)

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

$\angle POR+\angle OPS>\angle PRO+\angle RPS$

$\Rightarrow \angle PSR>\angle PSO$ $[$ from $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)]$

Hence, this is the answer.