Problem 3. The base of the pyramid $DABC$ is an equilateral triangle $ABC$, with the length of each side being $4\sqrt{2}$. The lateral edge $DC$ is perpendicular to the plane of the base and has a length of 2. Find the measure of the angle and the distance between the skew lines, one of which passes through point $D$ and the midpoint $M$ of edge $BC$, and the other passes through point $C$ and the midpoint $K$ of edge $AB$ (Fig. 3).
Solution. The required angle is $P M D$. Therefore, the problem has been reduced to solving the right trihedral angle $M C P D$ (with $M$ as its vertex, $C M$ as the edge of the right dihedral angle).

The length of the segment $C H$ is equal to the distance between the skew lines $D M$ and $C K$, because the line $C K$ is parallel to the plane $D P M$. From $\triangle C D M$ we find:

$$
M D=\sqrt{C D^{2}+C M^{2}}=2 \sqrt{3}
$$

Therefore,

$$
\begin{gathered}
\cos \angle C M D=\sqrt{2: 3} \text { and } \cos \angle P M D= \\
=\cos \angle P M C \cos \angle C M D=\cos 30^{\circ} \cdot \sqrt{\frac{2}{3}}=0.5 \sqrt{2} \\
\text { i.e., } \angle P M D=45^{\circ}
\end{gathered}
$$

From the trihedral angle $M C P D$ we get:

$$
\begin{gathered}
\cos \angle D P C=\operatorname{tg} \angle C M P: \operatorname{tg} \angle P M D=\sqrt{3}: 3 \\
\sin \angle D P C=\sqrt{1-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\sqrt{\frac{2}{3}} \\
C H=C P \sin \angle D P C=\sqrt{2} \cdot \sqrt{\frac{2}{3}}=\frac{2}{\sqrt{3}}
\end{gathered}
$$