Given $\triangle PQR$ with $\overline{RS}$ bisecting $\angle R$, $PQ$ extended to $D$ and $\angle n$ a right angle, then: 

$\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad  \textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) \qquad \textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad \textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad \textbf{(E)}\ \text{none of these is correct}$
Let $\angle PRS$ be $\theta$.
$p+ q + 2\theta = 180$
$m+\theta+90=180 \implies m+\theta=90 \implies 2m+2\theta=180$
$p+q+2\theta=2m+2\theta \implies \frac{p+q}{2}=m \implies \boxed{\textbf{(B) \ } \angle m = \frac{1}{2}(\angle p + \angle q)}$