141. In space, points $A, B, C$ and $D$ are given, such that $|A B|=|B C|=|C D|, \quad \widehat{A B} C=\widehat{B C D}=\widehat{C D A}=\alpha$. Find the angle between the lines $A C$ and $B D$.
141. Let's fix $\triangle A B C$, then in $\triangle A D C$ two of its sides $|A C|$ and $|D C|$ and the angle $\widehat{A D C}=\alpha$ are known. Construct in the plane of $\triangle A D C$ a circle of radius $|A C|$ with center at $C$ (Fig. 25, a). If

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-083.jpg?height=344&width=675&top_left_y=486&top_left_x=290)

Fig. 25.

$\alpha \geqslant 60^{\circ}$, then there is only one triangle with the given sides and angle (the second point $A_{1}$ will be on the other side of point $D$), which is a triangle congruent to triangle $A B C$. In this case, $A C$ and $B D$ are perpendicular.

If, however, $\alpha<60^{\circ}$, then there is a second possibility (in Fig. 25, a this is $\triangle A_{1} D C$), for which $\widehat{C A_{1} D}=90^{\circ}+\frac{\alpha}{2}$, $\widehat{A_{1} C D}=90^{\circ}-\frac{3 \alpha}{2}$. But in this case, at vertex $C$ (Fig. 25, b) the angles $\widehat{B C A}_{1}=90^{\circ}-\frac{\alpha}{2}$, $\widehat{B C D}=\alpha$, $\widehat{1_{1} C D}=90^{\circ}-\frac{3 \alpha}{2}$ meet, and since $90^{\circ}-\frac{\alpha}{2}=\left(90^{\circ}-\frac{3 \alpha}{2}\right)+\alpha$, the points $A_{i}, B, C$ and $D$ lie in the same plane and the angle between $A_{1} C$ and $B D$ will be $\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-083.jpg?height=34&width=920&top_left_y=1376&top_left_x=194)
if $\alpha<60^{\circ}$, then the angle between $A C$ and $B D$ can be either $90^{\circ}$ or $\alpha$.