7. As shown in Figure 3, in 
$\square A B C D$, $B C$ $=2 A B, C E \perp A B$, $E$ is the foot of the perpendicular, $F$ is the midpoint of $A D$. If $\angle A E F=54^{\circ}$, then $\angle B=$ ( ).
$\begin{array}{lll}\text { (A) } 54^{\circ} & \text { (B) } 60^{\circ}\end{array}$
(C) $66^{\circ}$
(D) $72^{\circ}$
7.D.

As shown in Figure 6, take the midpoint $G$ of $BC$, and connect $FG$ to intersect $EC$ at $K$. It is easy to see that quadrilateral $FGCD$ is a rhombus, thus
$$
\angle 2=\angle 3 \text {. }
$$

Since $EK=KC$, and $FK \perp EC$, we have $\angle 1=\angle 2$. Therefore, $\angle A=2 \angle AEF$. Hence, $\angle B=72^{\circ}$.