2. As shown in Figure 2, in parallelogram $A B C D$, $\angle A B C=72^{\circ}, A F \perp B C$ at point $F, A F$ intersects $B D$ at point $E$. If $D E=2 A B$, then $\angle A E D=$ $\qquad$
2. $66^{\circ}$.

As shown in Figure 5, take the midpoint $M$ of $DE$, and connect $AM$.
In the right triangle $\triangle ADE$,
$$
AM=EM=\frac{1}{2} DE=AB \text{. }
$$

Let $\angle AED=\alpha$. Then
$$
\begin{array}{l}
\angle AME=180^{\circ}-2 \alpha, \angle ABM=\alpha-18^{\circ} . \\
\text{ Also } \angle ABM=\angle AMB \\
\Rightarrow 180^{\circ}-2 \alpha=\alpha-18^{\circ} \\
\Rightarrow \alpha=66^{\circ} .
\end{array}
$$