4. (10 points) In the triangle $A B C$ shown in the figure, $E B=E D, F C=F D, \angle E D F=72^{\circ}$, then $\angle A E D+\angle A F D$ $=(\quad)$
A. $200^{\circ}$
B. $216^{\circ}$
C. $224^{\circ}$
D. $240^{\circ}$
【Analysis】From the problem, $\angle B=\angle E D B, \angle C=\angle F D C$, then using the sum of interior angles, we can draw the conclusion.
【Solution】Solution: From the problem, $\angle B=\angle E D B, \angle C=\angle F D C$,
$$
\begin{array}{l}
\therefore \angle A E D+\angle A F D=360^{\circ}-(\angle A+\angle E D F)=360^{\circ}-(\angle A+\angle E D F) \\
=360^{\circ}-\left(180^{\circ}-\angle B-\angle C+\angle E D F\right) \\
=180^{\circ}+\angle B+\angle C-\angle E D F \\
=180^{\circ}+\angle E D B+\angle F D C-\angle E D F \\
=180^{\circ}+180^{\circ}-\angle E D F-\angle E D F \\
=360^{\circ}-2 \times 72^{\circ} \\
=216^{\circ}
\end{array}
$$

Therefore, the correct choice is: B.