11.231 The base of the pyramid $S A B C$ is a right isosceles triangle $A B C$, with the length of the hypotenuse $A B=4 \sqrt{2}$. The lateral edge $S C$ of the pyramid is perpendicular to the plane of the base, and its length is 2. Find the measure of the angle and the distance between the lines, one of which passes through point $S$ and the midpoint of edge $A C$, and the other through point $C$ and the midpoint of edge $A B$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-279.jpg?height=289&width=471&top_left_y=1210&top_left_x=190)

Let $K$ and $M$ be the midpoints of edges $AC$ and $AB$, and $N$ be the midpoint of segment $AM$.

$CM \| KN \Rightarrow CM \| SKN \Rightarrow$ the distance $x$ between $CM$ and $SK$ is equal to the distance between $CM$ and the plane $SKN$, or, in other words, it is equal to the height of the pyramid $MSKN$ dropped from $M$ to $SKN$. Additionally, the angle between $CM$ and $SK$ is equal to $\angle SKN$ (or $180^\circ - \angle SKN$).

$CM$ is the median of the right $\triangle ABC \Rightarrow KN = \frac{CM}{2} = \frac{AB}{4} = \sqrt{2}$,

$SK = \sqrt{KC^2 + SC^2} = 2\sqrt{2} \quad \left(AC = \frac{AB}{\sqrt{2}} = 4\right), \quad SM = \sqrt{CM^2 + SC^2} = 2\sqrt{3}$,

$SN = \sqrt{MN^2 + SM^2} = \sqrt{14}$. By the cosine theorem for $\triangle SKN$,
$\cos \angle SKN = \frac{KN^2 + SK^2 - SN^2}{2 \cdot KN \cdot SK} = -\frac{1}{2}, \angle SKN = 120^\circ \Rightarrow$

$\Rightarrow S_{SKN} = \frac{KN \cdot SK \cdot \sin 120^\circ}{2} = \sqrt{3}, V_{MSKN} = \frac{x}{3} \cdot S_{SKN} = \frac{x}{\sqrt{3}}$.

On the other hand, $V_{MSKN} = \frac{1}{3} SC \cdot S_{KMN} = \frac{2}{3} \cdot \frac{S_{ABC}}{8} = \frac{2}{3} \Rightarrow x = \frac{2\sqrt{3}}{3}$.

Answer: $\frac{\pi}{3} ; \frac{2\sqrt{3}}{3}$.