A space diagonal of a rectangular prism is $A A^{\prime}$. The plane through the endpoints $B, C, D$ of the edges starting from vertex $A$ divides the diagonal $A A^{\prime}$ in what ratio?
Let the vertices $A, B, C, D$ be denoted by their centrally symmetric counterparts $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_0cf599dc8ea450462bf3g-1.jpg?height=563&width=856&top_left_y=228&top_left_x=602)

The rectangle $A D A^{\prime} D^{\prime}$ contains the space diagonal $A A^{\prime}$, the face diagonal $A D$, which is bisected by the face diagonal $B C$ at point $F$, and the segment $F D$, which intersects the space diagonal $A A^{\prime}$ at point $T$, where the space diagonal $A A^{\prime}$ intersects the plane $B C D$. Due to the equality of angles,

$$
A F T_{\triangle} \sim A^{\prime} D T_{\triangle}
$$

and thus

$$
A T: T A^{\prime}=A F: A^{\prime} D=A F: 2 A F=1: 2
$$

Therefore, the plane in question intersects the space diagonal at one-third of its length.

Gyöngyi Pétertfia (Bp. XIV., Teleki B. lg. III. o. t.)