Example 12 In a convex hexagon $A B C D E F$, $\angle A=\angle B=\angle C=$ $\angle D=\angle E=\angle F$, and $A B+B C=11, F A-C D=3$, find $B C+D E$.
From the problem, we know that $A F // C D, A B // E D$, so we can complete the hexagon to form a parallelogram $M C N F$, as shown in Figure 1.4.13. It is clearly a parallelogram, one of its internal angles is $60^{\circ}$, and $\triangle A M B$ and $\triangle D E N$ are both equilateral triangles. Thus,
$$
\begin{aligned}
M C & =A B+B C=11 \\
F A=M F-A M & =C N-A B=C D+D E-A B
\end{aligned}
$$

Therefore, $F A-C D=E D-A B$, and since $F A-C D=3$, then $E D - A B=3$, hence $B C+D E=14$.
Note: This problem can also be solved by completing it into an equilateral triangle or a trapezoid.