Example 5 In 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 $B C+D E$.
(1994, Beijing Junior High School Mathematics Competition)
Solution: As shown in Figure 5, let $F A$ and $C B$ intersect at $P$, and $F E$ and $C D$ intersect at $Q$. It is easy to see that quadrilateral $P C Q F$ is a parallelogram, and $\triangle A P B$ and $\triangle Q E D$ are both equilateral triangles. Therefore,
$$
\begin{aligned}
& F A + A B \\
= & F P = Q C \\
= & D E + C D .
\end{aligned}
$$

It is known that
$$
\begin{aligned}
& D E - A B \\
= & F A - C D \\
= & 3, \\
& A B + B C \\
= & 11 .
\end{aligned}
$$

Adding the two equations, we get $D E + B C = 14$.