1962 AHSME 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:

\begin{array}{l} \textbf{(A)}\ \frac {1}{2} (90 - A) \qquad \textbf{(B)}\ 90 - A \qquad \textbf{(C)}\ \frac {1}{2} (180 - A) \qquad \\ \textbf{(D)}\ 180-A\qquad \textbf{(E)}\ 180-2A \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.