In the figure,$PR > PQ$ and $PS$ bisects $\angle QPR$. Prove that $\angle PSR > \angle PSQ$.

(N/A) In $\Delta PQR$,$PS$ bisects $\angle QPR$ [Given].
Therefore,$\angle QPS = \angle RPS$.
Since $PR > PQ$ [Given],
Therefore,the angle opposite to $PR$ is greater than the angle opposite to $PQ$.
Thus,$\angle PQS > \angle PRS$.
Adding $\angle QPS$ to the left side and $\angle RPS$ to the right side (since $\angle QPS = \angle RPS$):
$\angle PQS + \angle QPS > \angle PRS + \angle RPS$ ... $(1)$.
We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
For $\Delta PQS$,the exterior angle $\angle PSR = \angle PQS + \angle QPS$.
For $\Delta PRS$,the exterior angle $\angle PSQ = \angle PRS + \angle RPS$.
Substituting these into $(1)$,we get $\angle PSR > \angle PSQ$.