$Q$ is a point on the side $SR$ of a $\triangle PSR$ such that $PQ = PR$. Prove that $PS > PQ$.

(N/A) Given: $PQ = PR$.
To prove: $PS > PQ$.
Proof: In $\triangle PQR$,we have:
$PR = PQ$ (Given).
Therefore,$\angle PQR = \angle PRQ$ (Angles opposite to equal sides of a triangle are equal).
In $\triangle PQS$,$\angle PQR$ is an exterior angle.
We know that the exterior angle of a triangle is greater than each of the interior opposite angles.
Therefore,$\angle PQR > \angle S$.
Since $\angle PQR = \angle PRQ$,we have $\angle PRQ > \angle S$.
In $\triangle PSR$,since $\angle PRQ > \angle S$,the side opposite to $\angle PRQ$ must be greater than the side opposite to $\angle S$.
Therefore,$PS > PR$.
Since $PR = PQ$,we can substitute $PQ$ for $PR$.
Thus,$PS > PQ$.
Hence proved.