In quadrilateral $PQRS$,the bisectors of $\angle P$ and $\angle Q$ intersect at $M$. Prove that $\angle S + \angle R = 2 \angle PMQ$.

(N/A) In quadrilateral $PQRS$,the sum of all interior angles is $360^{\circ}$.
Therefore,$\angle P + \angle Q + \angle R + \angle S = 360^{\circ}$.
In $\triangle PMQ$,the sum of angles is $180^{\circ}$.
Thus,$\angle PMQ + \angle MPQ + \angle MQP = 180^{\circ}$.
Since $PM$ and $QM$ are bisectors,$\angle MPQ = \frac{1}{2} \angle P$ and $\angle MQP = \frac{1}{2} \angle Q$.
Substituting these into the triangle equation: $\angle PMQ + \frac{1}{2} \angle P + \frac{1}{2} \angle Q = 180^{\circ}$.
Multiplying by $2$,we get $2 \angle PMQ + \angle P + \angle Q = 360^{\circ}$.
From the quadrilateral sum,$\angle P + \angle Q = 360^{\circ} - (\angle R + \angle S)$.
Substituting this into the equation: $2 \angle PMQ + 360^{\circ} - (\angle R + \angle S) = 360^{\circ}$.
Simplifying,we get $2 \angle PMQ = \angle R + \angle S$.