18.109 In trapezoid $A B C D$, $A D / / B C$. If $\angle B=40^{\circ}, \angle C=50^{\circ}$, and $E, M, F, N$ are the midpoints of $A B, B C, C D, D A$ respectively, and $E F=a, M N=b$, then $B C$ is
(A) $a+b$.
(B) $a+2 b$.
(C) $2 a+b$.
(D) $2 a-b$.
(2nd "Jinyun Cup" Junior High School Mathematics Invitational, 1985)
[Solution] Consider the extreme case: when $A D=0$, at this point, points $A, D, N$ coincide.
$$
\begin{array}{l}
\because \angle B=40^{\circ}, \angle C=50^{\circ}, \\
\therefore \angle B A C=90^{\circ} .
\end{array}
$$

And $N M$ is the median of the hypotenuse of the triangle. Also,
$E F$ is the midline of the triangle, then
$$
B C=2 M N=2 b,
$$

and $B C=2 E F=2 a$, so $a=b$.
At this point, $(B),(C),(D)$ are $3 a, 3 a, a$ respectively; and substituting into $(A)$ gives $2 a$. Therefore, $(A)$ is correct.
Hence, the answer is $(A)$.