One, (50 points) In $\triangle A B C$, $\angle A=45^{\circ}, \angle B=60^{\circ}, O$ is the circumcenter of the triangle. A line through point $A$ parallel to $O B$ intersects the extension of $C B$ at point $D$. Find the value of $\frac{B C}{D B} \sin D$.
As shown in Figure 8, connect radii $O A, O C$, and let $B O$ intersect $A C$ at point $E$.
Since $O$ is the circumcenter of $\triangle A B C$,

we have
$$
\begin{array}{l}
\angle B O C=2 \angle A=90^{\circ}, \\
\angle A O C=2 \angle B=120^{\circ} . \\
\text { Therefore, } \angle O B C=\angle O C B \\
=45^{\circ}, \\
\angle O A C=\angle O C A=30^{\circ}, \\
\angle A O E=\angle A O C-\angle E O C=120^{\circ}-90^{\circ}=30^{\circ} .
\end{array}
$$

Thus, $\triangle A O E$ is an isosceles triangle, and $\triangle C O E$ is a right triangle.
Hence, $C E=2 O E=2 E A$.
Since $A D / / O B$, we have
$$
\left\{\begin{array}{l}
\frac{B C}{D B}=\frac{C E}{E A}=2, \\
\angle A D C=\angle O B C=45^{\circ} .
\end{array}\right.
$$

Therefore, $\frac{B C}{D B} \sin D=2 \sin 45^{\circ}=\sqrt{2}$.