4. Given a hexagon $A B C D E F$ inscribed in $\odot O$ with an area of 60, and $A B=B C, C D=D E, E F=A F$. Then the area of $\triangle B D F$ is ( ).
(A) 45
(B) 40
(C) 36
(D) 30
4. D.

Fold $\triangle A B F$ along $B F$ to get $\triangle M B F$, and connect $D M$, as shown in Figure 8.
$$
\begin{array}{l}
\text { Given } A B=B C \Rightarrow \overparen{A B}=\overparen{B C} \\
\Rightarrow \angle A F B=\angle B D C(\text { denoted as } \alpha) .
\end{array}
$$

Similarly, $\angle C B D=\angle D F E($ denoted as $\beta$ ).
Since quadrilateral $B C D F$ is a cyclic quadrilateral of $\odot O$, therefore,
$$
\begin{array}{l}
\angle B F D=180^{\circ}-\angle B C D \\
=180^{\circ}-\left(180^{\circ}-\beta-\alpha\right)=\alpha+\beta .
\end{array}
$$

Then $\angle D F M=\angle B F D-\angle B F M$
$$
=\angle B F D-\angle A F B=(\alpha+\beta)-\alpha=\beta \text {. }
$$

Thus, $\triangle D E F \cong \triangle D M F$.
Similarly, $\triangle B C D \cong \triangle B M D$.
Therefore, the area of $\triangle B D F$ is half the area of hexagon $A B C D E F$.
Hence, the area of $\triangle B D F$ is 30.