The angle between the opposite edges $A B$ and $C D$ of the pyramid $A B C D$ is $\alpha, A B=a, C D=b$. Find the area of the section of the pyramid by a plane passing through the midpoint of edge $B C$ and parallel to the lines $A B$ and $C D$.
Let $K$ be the midpoint of $B C$, and the secant plane intersects the edges $A C, A D$, and $B D$ at points $L, M$, and $N$ respectively. The plane $B C D$ passes through the line $C D$, which is parallel to the secant plane, and has a common point $K$ with the secant plane, so the line of intersection of these planes is parallel to the line $C D$. Therefore, $K N$ is the midline of triangle $B C D$. Similarly, $K L, M L$, and $M N$ are the midlines of triangles $A B C, A D C$, and $A B D$ respectively. Thus, the section $K L M N$ is a parallelogram with sides $\frac{1}{2} a$, $\frac{1}{2} b$ and the angle between them equal to the angle between the skew lines $A B$ and $C D$, i.e., $\alpha$. Therefore,

$$
S_{K L M N}=K L \cdot K N \sin \alpha=\frac{1}{2} a \cdot \frac{1}{2} b \cdot \sin \alpha=\frac{1}{4} a b \sin \alpha .
$$

## Answer