AHSME 1962 Problem 7

Let the bisectors of the exterior angles at $B$ and $C$ of triangle $ABC$ meet at D$.$ Then, if all measurements are in degrees, angle $BDC$ equals: $\textbf{(A)}\ \frac{1}{2}(90-A) \textbf{(B)}\ 90-A \textbf{(C)}\ \frac{1}{2}(180-A)$ $\textbf{(D)}\ 180-A \textbf{(E)}\ 180-2A$