| Construct the section of the triangular prism $A B C A 1 B 1 C 1$ by a plane passing through points $A 1$ and $C$ and parallel to the line $B C 1$. In what ratio does this plane divide the edge $A B$?
The plane $B B 1 C 1$ passes through the line $B C 1$ and intersects the secant plane along the line $a$, passing through point $C$. Since the line $B C 1$ is parallel to the secant plane, $a \| B C 1$. From this, the following construction follows. Through point $C$, draw a line $a$ parallel to $B C 1$. Let $P$ and $Q$ be the points of intersection of this line with the extensions of the edges $B 1 C 1$ and $B 1 B$, respectively, and let $M$ be the point of intersection of $A 1 Q$ and $A B$. Then the desired section is the triangle $A 1 M C$. Since $B C$ is the midline of triangle $Q B 1 P$, the segment $B M$ is the midline of triangle $A 1 Q B 1$, therefore

$$
B M=\frac{1}{2} A 1 B 1=\frac{1}{2} A B
$$

Thus, $A M: M B=1: 1$.

## Answer

$1: 1$.