18. (12 points) As shown in Figure 4, in the equilateral $\triangle ABC$, points $D$ and $E$ are on sides $AC$ and $AB$ respectively, and $AD = \frac{1}{3} AC$, $AE = \frac{2}{3} AB$, $BD$ intersects $CE$ at point $F$.
(1) Prove that points $A$, $E$, $F$, and $D$ are concyclic;
(2) If the side length of the equilateral $\triangle ABC$ is 2, find the radius of the circle passing through points $A$, $E$, $F$, and $D$.
18. (1) From $A E=\frac{2}{3} A B$, we know $B E=\frac{1}{3} A B$.

In the equilateral $\triangle A B C$, from $A D=\frac{1}{3} A C \Rightarrow A D=B E$.
Also, $A B=B C, \angle B A D=\angle C B E$, then $\triangle B A D \cong \triangle C B E \Rightarrow \angle A D B=\angle B E C$.
Therefore, $\angle A D F+\angle A E F=\pi$.
Thus, points $A, E, F, D$ are concyclic.
(2) As shown in Figure 6, take the midpoint $G$ of side $A E$ and connect $G D$. Then
$$
\begin{array}{l}
A G=G E=\frac{1}{2} A E \\
=\frac{1}{3} A B=\frac{2}{3} . \\
\text { Also, } A D=\frac{1}{3} A C=\frac{2}{3},
\end{array}
$$
$\angle D A E=60^{\circ}$, then $\triangle A G D$ is an equilateral triangle.
Thus, $G D=A G=A D=\frac{2}{3}$, which means
$$
G A=G E=G D=\frac{2}{3} \text {. }
$$

Therefore, $G$ is the center of the circle passing through points $A, E, F, D$, and the radius of $\odot G$ is $\frac{2}{3}$.