[Solution] As shown in the figure, in $\triangle A B C$, extend $B N$ to intersect $A C$ at $E$.
Since $\angle B A N=\angle E A N, A N=A N$, and $\angle A N B=\angle A N E=90^{\circ}$,
we have $\triangle A B N \cong \triangle A E N$.
Therefore, $N$ is the midpoint of $B E$, and $A B=A E=14$.
Thus, $E C=A C-A E=19-14=5$.

Since $M$ is the midpoint of $B C$, $N M$ is the midline of $\triangle E B C$. Therefore, $M N=\frac{1}{2} E C=\frac{5}{2}$.
Hence, the answer is $(B)$.
17.52 As shown in the figure, in $\triangle A B C$, $M$ is the midpoint of side $B C$, $A N$ bisects $\angle B A C$, $B N \perp A N$, and the measure of $\angle B A C$ is $\theta$. If the lengths of $A B$ and $A C$ are 14 and 19, respectively, then the length of $M N$ is
(A) 2.
(B) $\frac{5}{2}$.
(C) $\frac{5}{2}-\sin \theta$.
(D) $\frac{5}{2}-\frac{1}{2} \sin \theta$.
(E) $\frac{5}{2}-\frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$.
(32nd American High School Mathematics Examination, 1981)