$17 \cdot 32$ In an acute triangle $\triangle ABC$, the angle bisector of $\angle A$ intersects side $BC$ at $D$. A circle with center $B$ and radius $BD$ intersects side $AB$ at $M$; a circle with center $C$ and radius $CD$ intersects side $AC$ at $N$. The statement that is always true is
(A) $\angle CND + \angle BMD - \angle DAC = 120^{\circ}$.
(B) $AMDN$ is a trapezoid.
(C) $BC$ is parallel to $MN$.
(D) $AM - AN = \frac{3(DB - DC)}{2}$.
(E) $AB - AC = \frac{3(DB - DC)}{2}$.
(26th American High School Mathematics Examination, 1975)
［Solution］As shown in the figure, by the property of the internal angle bisector of a triangle, we have
$$
\frac{B D}{C D}=\frac{A B}{A C},
$$

Furthermore, since $C N=C D, B M=B D$, it follows that $\frac{B M}{C N}=\frac{A B}{A C}$, hence $M N \parallel C B$.
Options $(A)$, $(B)$, $(D)$, and $(E)$ are not valid. If $\angle A=90^{\circ}-\theta, \angle B=60^{\circ}, \angle C=30^{\circ}+\theta$, where $\theta$ is any sufficiently small positive angle, this serves as an example to show that the above conclusion does not hold.
Therefore, the correct choice is $(C)$.