Example 10. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $B D=10 \mathrm{~cm}, \angle D_{1} B D=30^{\circ}, \angle A O D=60^{\circ}$. Find the distance between $A C$ and $B D_{1}$ (Figure 18).
Let the angle between $B D_{1}$ and $O A$ be $\theta$.
The dihedral angle $D_{1}-B D-A=90^{\circ}$.

By the corollary of the three-line theorem,
$$
\begin{aligned}
\cos \theta & =\cos 30^{\circ} \cos 60^{\circ} \\
& =\frac{\sqrt{3}}{4}, \\
\sin \theta & =\frac{\sqrt{13}}{4} . \\
\text { Therefore, } d & =\frac{5 \cdot \sin 30^{\circ} \sin 60^{\circ} \sin 90^{\circ}}{\sin \theta} \\
& =\frac{5 \sqrt{39}}{13}(\mathrm{~cm}) .
\end{aligned}
$$