9. As shown in Figure 4, in $\triangle A B C$, $A B=3, A C=4$, $B C=5$, $\triangle A B D$, $\triangle A C E$, and $\triangle B C F$ are all equilateral triangles. Then the area of quadrilateral $A E F D$ is $\qquad$
II, 9.6.
Since $B D=A B$,
$$
\begin{array}{l}
\angle D B F=\angle D B A-\angle F B A \\
=\angle F B C-\angle F B A=\angle A B C, \\
B F=B C,
\end{array}
$$

Therefore, $\triangle D B F \cong \triangle A B C$.
Similarly, $\triangle E F C \cong \triangle A B C$.
Thus, $A D=A B=E F, D F=A C=A E$.
Hence, quadrilateral $A E F D$ is a parallelogram.
In $\triangle A B C$, $A B=3, A C=4, B C=5$, so $A B^{2}+A C^{2}=3^{2}+4^{2}=5^{2}=B C^{2}$.
Therefore, $\angle B A C=90^{\circ}$,
$$
\angle D A E=360^{\circ}-90^{\circ}-60^{\circ}-60^{\circ}=150^{\circ} \text {. }
$$

Thus, $\angle A D F=\angle A E F=30^{\circ}$.
Also, $D F=A C=4$, so the distance from point $F$ to $A D$ is 2.
Since $A D=A B=3$, the area $S_{\square \triangle A S P D}=3 \times 2=6$.