Answer: $\sqrt{3}: 1$.

Solution. By the inscribed angle theorem, angle $D C A$ is half of arc $A D$, and angle $D B C$ is half of arc $C D$. Therefore, $\angle D C H=30^{\circ}, \angle H B C=45^{\circ}$. Then triangle $B H C$ is a right and isosceles triangle, $B H=H C$. But $H D=C H \operatorname{tg} 30^{\circ}=\frac{C H}{\sqrt{3}}$. Therefore, $B H: H D=\sqrt{3}: 1$.
4. The continuation of the height $B H$ of triangle $A B C$ intersects the circumscribed circle around it at point $D$ (points $B$ and $D$ lie on opposite sides of line $A C$). The degree measures of arcs $A D$ and $C D$, not containing point $B$, are $60^{\circ}$ and $90^{\circ}$, respectively. Determine in what ratio the segment $B D$ is divided by side $A C$.