Example 7 As shown in Figure 7, in quadrilateral $A B C D$, $A B=$ $B C, \angle A B C=\angle C D A=90^{\circ}, B E \perp A D$ at point $E, S_{\text {quadrilateral } A B C D}=8$. Then the length of $B E$ is ( ).
$\begin{array}{ll}\text { (A) } 2 & \text { (B) } 3\end{array}$
(C) $\sqrt{3}$
(D) $2 \sqrt{2}$
(2003, Wuhan City, Hubei Province Junior High School Mathematics Competition)
Solve As shown in Figure 7, draw a perpendicular line from point $B$ to $CD$, intersecting the extension of $DC$ at point $F$. Then
$$
\begin{array}{l}
\angle C B F=\angle A B E, \\
\angle A E B=\angle C F B=90^{\circ} .
\end{array}
$$

Since $A B=B C$, we have
$$
\triangle A B E \cong \triangle C B F \text {. }
$$

Therefore, $B E=B F$.
Thus, quadrilateral $B E D F$ is a square.
Then $S_{\text {square } B E D F}=S_{\text {quadrilateral } A B C D}=8$.
So, $B E=\sqrt{S_{\text {square } B E D F}}=2 \sqrt{2}$.
Hence, the answer is D.