4.19. Chord $A B$ of a sphere with radius 1 has a length of 1 and is positioned at an angle of $60^{\circ}$ to the diameter $C D$ of this sphere. It is known that $A C=\sqrt{2}$ and $A C<B C$. Find the length of the segment $B D$.
4.19. Let $O$ be the center of the sphere. Take a point $E$ such that $\overrightarrow{C E}=\overrightarrow{A B}$. Since $\angle O C E=60^{\circ}$ and $C E=1=O C$, then $O E=1$. Point $O$ is equidistant from all vertices of the parallelogram $A B E C$, so $A B E C$ is a rectangle and the projection $O_{1}$ of point $O$ onto the plane of this rectangle coincides with its center, i.e., with the midpoint of segment $B C$. Segment $O O_{1}$ is the midline of triangle $C B D$, therefore $B D=20 O_{1}=2 \sqrt{O C^{2}-B C^{2} / 4} \Rightarrow$ $=2 \sqrt{1-\left(A B^{2}+A C^{2}\right) / 4}=1$.