$[$ [extension of a tetrahedron to a parallelepiped]

Segment $A B(A B=1)$, being a chord of a sphere with radius 1, is positioned at an angle of $60^{\circ}$ to the diameter $C D$ of this sphere. The distance from the end $C$ of the diameter to the nearest end $A$ of the chord $A B$ is $\sqrt{2}$. Find $B D$.

#