17・19 Let the external angle bisectors of $\angle B$ and $\angle C$ of $\triangle ABC$ intersect at $D$, then the degree measure of $\angle BDC$ is
(A) $\frac{1}{2}\left(90^{\circ}-\angle A\right)$.
(B) $90^{\circ}-\angle A$.
(C) $\frac{1}{2}\left(180^{\circ}-\angle A\right)$.
(D) $180^{\circ}-\angle A$.
(E) $180^{\circ}-2 \angle A$.
(13th American High School Mathematics Examination, 1962)
［Solution］As shown in the figure, by the exterior angle property of a triangle, we have
$$
2 \beta=180^{\circ}-\angle B, \quad 2 \gamma=180^{\circ}-\angle C,
$$

Thus, $\beta+\gamma=180^{\circ}-\frac{1}{2}(\angle B+\angle C)$.
$$
\text { Also, } \angle B+\angle C=180^{\circ}-\angle A \text {, }
$$

Therefore, $\angle B D C=180^{\circ}-(\beta+\gamma)$
$$
\begin{array}{l}
=180^{\circ}-\left[180^{\circ}-\frac{1}{2}\left(180^{\circ}-\angle A\right)\right] \\
=\frac{1}{2}\left(180^{\circ}-\angle A\right) .
\end{array}
$$

Hence, the answer is $(C)$.