Sonkin $M$.

In trapezoid $A B C D$ with area 1, the bases $B C$ and $A D$ are in the ratio $1: 2$. Let $K$ be the midpoint of diagonal $A C$. Line $D K$ intersects side $A B$ at point $L$. Find the area of quadrilateral $B C K L$.
Consider the parallelogram $A D C E$. (see fig.). Then $K$ is the point of intersection of its diagonals. $E C=A D=2 B C$, that is, $B$ is the midpoint of the segment $E C$. $A B$ and $E K$ are medians of triangle $A C E$. By a known property of medians, $A L=2 / 3$ $A B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-13.jpg?height=443&width=740&top_left_y=482&top_left_x=662)

$S_{A C D}=2 S_{\text {BAC }}$ (the heights of these triangles are equal, and the base of the first is twice the base of the second).

Thus, $S_{B A C}=1 / 3 S_{A B C D}=1 / 2$,

$S_{L A K}=2 / 3 \cdot 1 / 2 S_{B A C}=1 / 9, S_{B C K L}=S_{B A C}-S_{L A K}=2 / 9$.

## Answer

$2 / 9$