II. (25 points) Given that $O$ is the circumcenter of $\triangle A B C$, $\angle B A C=45^{\circ}$, extend $B C$ to $D$ such that $C D=\frac{1}{2} B C$, and $A D / / O C$. Find the measure of $\angle A B C$.
$$
\begin{array}{l}
\angle B A C=45^{\circ} \\
\Rightarrow \angle B O C=90^{\circ} \\
\Rightarrow \angle O B C=\angle O C B=45^{\circ} .
\end{array}
$$

Given $A D / / O C$
$$
\Rightarrow O E \perp A D, \frac{B O}{O E}=\frac{B C}{C D}=2
$$
$$
\begin{array}{l}
\Rightarrow \angle A B E=\frac{1}{2} \angle A O E=30^{\circ} \\
\Rightarrow \angle A B C=30^{\circ}+45^{\circ}=75^{\circ} .
\end{array}
$$

Let $A D$ intersect $\odot O$ at another point $A_{1}$, and connect $A_{1} B 、 A_{1} C$. Then $\angle B A_{1} C=\angle B A C=45^{\circ}$. Therefore, $\triangle A_{1} B C$ can be another case of the known $\triangle A B C$.
$$
\begin{array}{l}
\text { Also } \angle A C B=180^{\circ}-\angle B A C-\angle A B C \\
=180^{\circ}-45^{\circ}-75^{\circ}=60^{\circ}, \\
\angle D=\angle O C B=45^{\circ},
\end{array}
$$

Thus, $\angle A_{1} B C=\angle C A D=\angle A C B-\angle D=15^{\circ}$.
Therefore, the measure of $\angle A B C$ is $75^{\circ}$ or $15^{\circ}$.