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 $120^{\circ}$ and $90^{\circ}$, respectively. Determine in what ratio the segment $B D$ is divided by side $A C$.
Answer: $1: \sqrt{3}$.

Solution. By the inscribed angle theorem, angle $DCA$ is half of arc $AD$, and angle $DBC$ is half of arc $CD$. Therefore, $\angle DCH=60^{\circ}, \angle HBC=45^{\circ}$. Then triangle $BHC$ is a right and isosceles triangle, $BH=HC$. But $HD=CH \operatorname{tg} 60^{\circ}=CH \sqrt{3}$. Therefore, $BH: HD=1: \sqrt{3}$.