4. As shown in Figure 1, in equilateral $\triangle A B C$, $E$ is the midpoint of $A B$, and $D$ is on $A C$ such that $\frac{A D}{A C}=\frac{1}{3}$. Then ( ).
(A) $\triangle A E D \backsim \triangle B E D$ (B) $\triangle A E D \backsim \triangle C B D$ (C) $\triangle A E D \backsim \triangle A B D$ (D) $\triangle B A D \backsim \triangle B C D$
4.B.

Consider option (A): In $\triangle A E D$, $\angle A=60^{\circ}$, while $\triangle B E D$ does not have a $60^{\circ}$ angle, so we exclude it.

Consider option (C): Since $\angle E A D=\angle B A D$, and $\angle A E D>\angle E B D$, if option $(\mathrm{C})$ holds, then it must be that $\angle A E D=\angle A D B$.
Thus, $\frac{A D}{A B}=\frac{A E}{A D}$.
$$
\text { But } \frac{A D}{A B}=\frac{A D}{A C}=\frac{1}{3}, \frac{A E}{A D}=\frac{\frac{1}{2} A B}{\frac{1}{3} A C}=\frac{1}{6} \text {, which is a contradiction. }
$$

Hence, we exclude it.
Consider option (D): $\angle B A D=\angle B C D$, and $\angle A D B>\angle D B C$, and $\angle A D B \neq \angle B D C$, so it does not hold.
Now, let's prove that option (B) holds:
$$
\begin{array}{l}
\angle B A D=\angle B C D, \text { and } \\
\frac{A D}{C D}=\frac{1}{2}, \frac{A E}{B C}=\frac{A E}{2 A E}=\frac{1}{2},
\end{array}
$$

Thus, $\frac{A D}{C D}=\frac{A E}{B C}$, so $\triangle A E D \backsim \triangle C B D$.