5. In quadrilateral ABCD, it is known that $\angle A B C=60^{\circ}, B E$ bisects $\angle A B C$, intersecting $A D$ and $C D$ at points $E$ and $F$ respectively; $A E=3, D E=2, O$ is the circumcenter of $\triangle D E F$. The following conclusions are given:

(1) Quadrilateral $O E D F$ is a rhombus;
(2) $\triangle O A C$ is an equilateral triangle;
(3) $\triangle O A E \cong \triangle O C D$.
Among these, the number of correct conclusions is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
5. D.

As shown in Figure 3, it is easy to see that
$$
\begin{array}{l}
D E=D F, \\
A E=A B=C D, \\
\angle A D C=60^{\circ} . \\
\text { Since } O E=O F,
\end{array}
$$

we know that $O D$ perpendicularly bisects $E F$, and $\triangle O D E$ and $\triangle O D F$ are both equilateral triangles.
Thus, quadrilateral $O E D F$ is a rhombus.
And $\angle O E A=180^{\circ}-\angle O E D$
$$
=180^{\circ}-\angle O D F=\angle O D C \text {, }
$$

Therefore, $\triangle O A E \cong \triangle O C D$.
By rotating $\triangle O C D$ 60° clockwise around point $O$, it coincides with $\triangle O A C$. Thus, $\angle A O C=60^{\circ}$.
Hence, $\triangle O A C$ is an equilateral triangle.