$17 \cdot 44$ In the figure, $B O$ bisects $\angle C B A$, $C O$ bisects $\angle A C B$, and $M N$ is parallel to $B C$. If $A B=12, B C=24, A C=18$, then the perimeter of $\triangle A M N$ is
(A) 30.
(B) 33.
(C) 36.
(D) 39.
(E) 42.
(33rd American High School Mathematics Examination, 1982)
[Solution] In the figure, since $M N / / B C, \angle M O B=\angle C B O=\angle O B M$,
so
$$
\angle C O N=\angle O C B=\angle \mathrm{NCO} \text {, }
$$

and $A M+M O+O N+A N$
$$
\begin{array}{l}
=(A M+M B)+(A N+N C) \\
=A B+A C=12+18=30 .
\end{array}
$$

Therefore, the answer is $(A)$.