8. Two isosceles right triangles $\triangle A B C_{1}$ and $\triangle A B C_{2}$, both with leg lengths of 1, are located in half-planes that form a $60^{\circ}$ dihedral angle. Then the set of possible lengths of the line segment $C_{1} C_{2}$ is $\qquad$ (Supplied by An Zhenping)

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8. $\left\{1, \frac{\sqrt{2}}{2}, \sqrt{2}\right\}$.

As shown in Figure 3, there are three ways to connect two triangular planes.
When the plane figure (1) is folded along $A B$ to form a $60^{\circ}$ dihedral angle, as shown in Figure 4, $\triangle A C_{1} C_{2}$ is an equilateral triangle with a side length of 1, so $C_{1} C_{2}=1$.

When the plane figure (2) is folded along $A B$ to form a $60^{\circ}$ dihedral angle, as shown in Figure 5, take the midpoint $O$ of $A B$, the side length of the equilateral triangle $O C_{1} C_{2}$ is half the diagonal of a unit square, so $C_{1} C_{2}=\frac{\sqrt{2}}{2}$.

When the plane figure (3) is folded along $A B$ to form a $60^{\circ}$ dihedral angle, as shown in Figure 6, take a point $D$ in the plane $A B C_{2}$ such that $\overrightarrow{A D}=\overrightarrow{B C_{2}}$, connect $C_{1} D, C_{2} D$, the side length of the equilateral triangle $A C_{1} D$ is 1, so $C_{1} D=1$. It is easy to know that $C_{2} D=1$, in $\triangle C_{1} C_{2} D$, $\angle C_{1} D C_{2}=90^{\circ}$, so $C_{1} C_{2}=\sqrt{2}$.