3. As shown in the figure, $A B / / D E, \angle A=120^{\circ}, \angle B=50^{\circ}, \angle D=60^{\circ}, \angle E=140^{\circ}$, then $\angle B C D-\angle A F E$ $=(\quad)$.
(A) $0^{\circ}$.
(B) $10^{\circ}$.
(C) $20^{\circ}$.
(D) $30^{\circ}$.
Then
$$
\angle A+\angle 3=180^{\circ}, \angle E+\angle 4=180^{\circ} \text {, }
$$

So
$$
\angle B C D=\angle B+\angle D=110^{\circ} \text {, }
$$
$$
\begin{array}{c}
\angle A F E=\angle 3+\angle 4=360^{\circ}-\angle A-\angle E=100^{\circ}, \\
\angle B C D-\angle A F E=110^{\circ}-100^{\circ}=10^{\circ} .
\end{array}
$$

Therefore, the answer is B.
Answer: $B$.
Solution: Since $A B / / E D$, draw $C G / / A B, F H / / A B$ through points $C, F$ respectively (as shown in the figure),