23. On the image of a cube, construct a segment that defines the distance between two intersecting diagonals of two adjacent faces of the cube. Calculate this distance if the length of the edge of the cube is $a$.
23. Consider $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ (Fig. 80) and draw the diagonals $\left[A D^{\prime}\right]$ and $\left[D C^{\prime}\right]$. To find the distance between the lines ( $A D^{\prime}$ ) and ( $D C^{\prime}$ ), draw two planes through these lines that are parallel to each other. Such planes are ( $A B^{\prime} D^{\prime}$ ) and $\left(B D C^{\prime}\right)$, since $\left(A B^{\prime}\right) \|\left(D C^{\prime}\right)$ and $\left(B C^{\prime}\right) \|\left(A D^{\prime}\right)$. These planes intersect the diagonal of the cube $A^{\prime} C$ at points $M$ and $M^{\prime}$ - the centroids of triangles $B C^{\prime} D$ and $A B^{\prime} D^{\prime}$. Each plane cuts off $\frac{1}{3}$ of the diagonal (see problem 5), so we have:

$$
\left|M M^{\prime}\right|=\frac{1}{3}\left|A^{\prime} C\right|=\frac{1}{3} a \sqrt{3}
$$

By a parallel translation, map $\left[M M^{\prime}\right]$ to the segment $\left[N N^{\prime}\right]$, where $N \in\left[C^{\prime} D\right], N^{\prime} \in\left[A D^{\prime}\right]$. The corresponding vector is $\overrightarrow{M N}=\overrightarrow{M^{\prime} N^{\prime}}=$ $=\frac{1}{3} \overrightarrow{B D}=\frac{1}{3} B^{\prime} D^{\prime}$, since the centroid is at a distance from the base of the triangle equal to $\frac{1}{3}$ of the length of the median. Thus, the desired distance is $\frac{1}{3} a \sqrt{3}$.