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: 
$\textbf{(A)}\ \frac{1}{2}(90-A)\qquad\textbf{(B)}\ 90-A\qquad\textbf{(C)}\ \frac{1}{2}(180-A)\qquad$
$\textbf{(D)}\ 180-A\qquad\textbf{(E)}\ 180-2A$
Calculating for angles $\angle DBC$ and $\angle DCB$, we get
$\angle DBC$ = $90 - \frac{B}{2}$ and
$\angle DCB$ = $90 - \frac{C}{2}$.
In triangle $BCD$, we have
$\angle BDC$ = $180 - (90 - \frac{B}{2}) - (90 - \frac{C}{2})$ = $\frac{B+C}{2}$ = $\frac{1}{2}\cdot(180 - A)$, so the answer is $\fbox{C}$.