In the figure,the side $QR$ of $\Delta PQR$ is produced to a point $S$. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point $T$,then prove that $\angle QTR = \frac{1}{2} \angle QPR$.

(N/A) In $\Delta PQR$,the side $QR$ is produced to $S$.
$\therefore$ Exterior $\angle PRS = \text{Sum of the interior opposite angles}$
$\Rightarrow \angle PRS = \angle P + \angle Q$
Since $QT$ and $RT$ are bisectors of $\angle PQR$ and $\angle PRS$ respectively,
$\therefore \angle TQR = \frac{1}{2} \angle PQR$ and $\angle TRS = \frac{1}{2} \angle PRS$.
Now,in $\Delta QRT$,the exterior angle $\angle TRS = \angle TQR + \angle QTR$.
Substituting the values:
$\frac{1}{2} \angle PRS = \frac{1}{2} \angle PQR + \angle QTR$
$\angle QTR = \frac{1}{2} \angle PRS - \frac{1}{2} \angle PQR$
$\angle QTR = \frac{1}{2} (\angle PRS - \angle PQR)$
Since $\angle PRS = \angle P + \angle PQR$ (exterior angle property),
$\angle QTR = \frac{1}{2} (\angle P + \angle PQR - \angle PQR)$
$\angle QTR = \frac{1}{2} \angle P$
i.e.,$\angle QTR = \frac{1}{2} \angle QPR$.