4. As shown in Figure 3, in $\triangle ABC$, it is given that $D$ is a point on side $BC$ such that $AD = AC$, and $E$ is the midpoint of side $AD$ such that $\angle BAD = \angle ACE$. If $S_{\triangle BDE} = 1$, then $S_{\triangle ABC}$ is $\qquad$.
4.4.

Notice that
$$
\begin{array}{l}
\angle E C D=\angle A C D-\angle A C E \\
=\angle A D C-\angle B A D=\angle A B C .
\end{array}
$$

Therefore, $\triangle E C D \backsim \triangle A B C$.
$$
\text { Then } \frac{C D}{B C}=\frac{E D}{A C}=\frac{E D}{A D}=\frac{1}{2} \text {. }
$$

Thus, $D$ is the midpoint of side $B C$.
Hence $S_{\triangle A B C}=2 S_{\triangle A C D}=4 S_{\triangle E C D}=4 S_{\triangle D B E}=4$.