2. In quadrilateral $ABCD$, the diagonals intersect at point $P$, diagonal $AC$ is divided into four equal segments by points $P, N$, and $M$ $(|AP|=|PN|=|NM|=|MC|)$, and diagonal $BD$ is divided into four equal segments by points $L, K$, and $P$ $(|BL|=|LK|=|KP|=|PD|)$. Determine the ratio of the area of quadrilateral KLMN to the area of quadrilateral ABCD.
SOLUTION. Triangles $A P D$ and $N P K$ are symmetrically related with respect to the center $P$ (Fig. 1), so $A D$ and $N K$ are parallel and $|A D|=|N K|$. From the equality of the corresponding segments, it follows that triangles $K N P, L M P$, and $B C P$ are similar, hence $N K\|M L\| B C$ and moreover $|L M|=2|K N|$ and $|B C|=3|K N|$. If we denote by $s$ the area of triangle $A P D$, then the area of triangle $N K P$ is $s$ and the area of triangle $M L P$ is $4 s$ (it has twice the height from vertex $P$ compared to triangle $N K P$ from the same vertex, and its side $M L$ is twice as long as side $N K$). Therefore, the area of trapezoid $K L M N$ is $3 s$.

![](https://cdn.mathpix.com/cropped/2024_04_17_7fa1855ed15af0dc549ag-2.jpg?height=352&width=740&top_left_y=239&top_left_x=665)

Fig. 1

Side $A P$ of triangle $A P D$ is four times smaller than side $A C$ of triangle $A C D$, and the heights from vertex $D$ are the same in both triangles, so the area of triangle $A C D$ is $4 s$. Side $P N$ of triangle $P N K$ is four times smaller than side $A C$ of triangle $A C B$, while the height of triangle $P N K$ from vertex $K$ is three times smaller than the height of triangle $A B C$ from vertex $B$, so the area of triangle $A C B$ is $12 s$. The area of quadrilateral $A B C D$ is equal to the sum of the areas of triangles $A B C$ and $A C D$, i.e., $16 s$.

The ratio of the areas of quadrilaterals $K L M N$ and $A B C D$ is $3: 16$.