II. (50 points) As shown in Figure 3, $BC$ is the diameter of the circumcircle $\odot O$ of $\triangle CEF$, $A$ is a point on the arc $\overparen{EF}$, and $D$ is the midpoint of the arc $\overparen{AB}$. A line through point $O$ parallel to $DA$ intersects $AC$ at point $J$. If $J$ is the incenter of $\triangle CEF$, find the measure of $\angle ACF$.

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$$
\begin{array}{l}
AF, FJ, FO. \\
\text {Let } \angle AOD = \angle BOD \\
= \alpha, \angle ACF = \beta.
\end{array}
$$

Since $D$ is the midpoint of $\overparen{AB}$, we know that
$$
\begin{array}{l}
\angle BOD = \angle ACB, \\
OD \parallel AC.
\end{array}
$$

Also, $AD \parallel OJ$, so quadrilateral $ADOJ$ is a parallelogram, and $DO = AJ$.
Since $AC$ is the angle bisector of $\angle ECF$, we have
$$
\overparen{AE} = \overparen{AF}, OA \perp EF.
$$

From $\angle AFC = \frac{1}{2}(180^\circ + 2\alpha) = 90^\circ + \alpha$, we know that
$$
\angle FAJ = 180^\circ - (90^\circ + \alpha) - \beta = 90^\circ - \alpha - \beta.
$$
$$
\text{Also, } \angle ACF = \angle ACE = \angle AFE = \beta, \text{ thus, }
$$
$$
\begin{array}{l}
\angle EFJ = \angle JFC = \frac{1}{2}(\angle AFC - \beta) \\
= \frac{1}{2}(90^\circ + \alpha - \beta), \\
\angle AFJ = \angle EFJ + \beta = \frac{1}{2}(90^\circ + \alpha + \beta), \\
\angle AJF = 180^\circ - \angle FAJ - \angle AFJ \\
= 180^\circ - (90^\circ - \alpha - \beta) - \frac{1}{2}(90^\circ + \alpha + \beta) \\
= \frac{1}{2}(90^\circ + \alpha + \beta).
\end{array}
$$

Therefore, $\angle AFJ = \angle AJF$, so $AF = AJ = DO = OA = OF$, and $\triangle AOF$ is an equilateral triangle.
Thus, $\angle ACF = 30^\circ$.