8.27. In space, points $A, B, C$ and $D$ are given, such that $A B=B C=C D$ and $\angle A B C=\angle B C D=\angle C D A=$ $=\alpha$. Find the angle between the lines $A C$ and $B D$.
8.27. In triangles $A B C$ and $C D A$, sides $A B$ and $C D$ are equal, as are angles $B$ and $D$, and side $A C$ is common to both. If $\triangle A B C = \triangle C D A$, then $A C \perp B D$. Now consider the case when these triangles are not equal. Take a point $P$ on ray $B A$ such that $\triangle C B P = \triangle C D A$, i.e., $C P = C A$ (Fig. 62). It can only not coincide with point $A$ if $\angle A B C < \angle A P C = \angle B A C$, i.e., $\alpha < 60^{\circ}$. In this case, $\angle A C D = \angle P C B = (90^{\circ} - \alpha / 2) - \alpha = 90^{\circ} - 3 \alpha / 2$. Therefore, $\angle A C D + \angle D C B = (90^{\circ} - 3 \alpha / 2) + \alpha = 90^{\circ} - \alpha / 2 = \angle A C B$. Thus, points $A, B, C$, and $D$ lie in the same plane, and point $D$ lies inside angle $A C B$. Since $\triangle A B C = \triangle D C B$ and these triangles are isosceles, the angle between lines $A C$ and $B D$ is $\alpha$.

Therefore, if $\alpha \geqslant 60^{\circ}$, then $A C \perp B D$, and if $\alpha < 60^{\circ}$, then either $A C \perp B D$ or the angle between lines $A C$ and $B D$ is $\alpha$.