Example 4 In hexagon $A B C D E F$, $\angle A=\angle B$ $=\angle C=\angle D=\angle E=\angle F$, and $A B+B C=$ $11, F A-C D=3$. Then $B C+D E=$ ?
(1994, Beijing Junior High School Mathematics Competition)
Solution: As shown in Figure 7, each angle of the hexagon is equal, all being $120^{\circ}$, so each exterior angle is $60^{\circ}$. Let the lines $A B, C D$, and $E F$ intersect at $P$, $Q$, and $R$, respectively, forming four equilateral triangles, i.e., $\triangle R A F$, $\triangle R P Q$, $\triangle B P C$, and $\triangle E D Q$ are all equilateral triangles.
$$
\begin{array}{l}
\therefore B C+D E=P Q-C D=P R-C D \\
=F A+A B+B C-C D \\
=(A B+B C)+(F A-C D) \\
=14 .
\end{array}
$$

Here, the key is to use the fact that each exterior angle of the hexagon is $60^{\circ}$, constructing equilateral triangles, which makes the solution very straightforward.