$17 \cdot 27$ As shown in the figure, in $\triangle A B C$, $\angle A=42^{\circ}$. If the trisectors of $\angle B$ and $\angle C$ intersect at $D, E$. Then the degree measure of $\angle B D C$ is
(A) $67^{\circ}$.
(B) $84^{\circ}$.
(C) $88^{\circ}$.
(D) $110^{\circ}$.
(China Beijing Junior High School Mathematics Competition, 1990);
[Solution] Extend $A D$. By the exterior angle property of a triangle,
$$
\angle 1=\angle 3+\angle 5, \angle 2=\angle 4+\angle 6,
$$

Adding these two equations, we get
$$
\begin{aligned}
\angle B D C & =\angle A+\frac{1}{3} \angle B+\frac{1}{3} \angle C \\
& =\angle A+\frac{1}{3}(\angle B+\angle C) \\
& =\angle A+\frac{1}{3}\left(180^{\circ}-\angle A\right)=60^{\circ}+\frac{2}{3} \angle A=88^{\circ} .
\end{aligned}
$$

Therefore, the answer is $(C)$.