In the figure,$PQ > PR$ and $QS$ and $RS$ are the bisectors of $\angle Q$ and $\angle R$,respectively. Show that $SQ > SR$.

(N/A) Given: In $\triangle PQR$,$PQ > PR$ and $QS$,$RS$ are bisectors of $\angle Q$ and $\angle R$ respectively.
Since $PQ > PR$,the angle opposite to the longer side is greater. Therefore,$\angle R > \angle Q$.
Since $QS$ and $RS$ are bisectors of $\angle Q$ and $\angle R$ respectively,we have:
$\angle SQR = \frac{1}{2} \angle Q$
$\angle SRQ = \frac{1}{2} \angle R$
Since $\angle R > \angle Q$,it follows that $\frac{1}{2} \angle R > \frac{1}{2} \angle Q$.
Therefore,$\angle SRQ > \angle SQR$.
In $\triangle SQR$,since $\angle SRQ > \angle SQR$,the side opposite to the greater angle is longer. Thus,$SQ > SR$.