71. A segment $AB$ of unit length, which is a chord of a sphere with radius 1, is positioned at an angle of $\pi / 3$ to the diameter $CD$ of this sphere. The distance from the end $C$ of the diameter to the nearest end $A$ of the chord $AB$ is $\sqrt{2}$. Determine the length of the segment $BD$.
71. Draw a line through $C$ parallel to $A B$, and take a point $E$ on it such that $|C E|=|A B|$, making $A B E C$ a parallelogram. If $O$ is the center of the sphere, then since $\widehat{O C E}=\pi / 3$ and $|C E|=1$ (as follows from the condition), $\triangle O C E$ is equilateral. Therefore, point $O$ is equidistant from all vertices of the parallelogram $A B E C$. This implies that $A B E C$ is a rectangle, the projection of $O$ onto the plane $A B E C$ is the point $K$ - the center of $A B E C$, and $|B D|=2|O K|=$ $=2 \sqrt{|O C|^{2}-\frac{1}{4}|B C|^{2}}=1$.