4. Point $K$ lies on side $B C$ of parallelogram $A B C D$, and point $M$ lies on its side $A D$. Segments $C M$ and $D K$ intersect at point $L$, and segments $A K$ and $B M$ intersect at point $N$. Find the maximum value of the ratio of the areas of quadrilaterals $K L M N$ and $A B C D$.
4.I.4. Without loss of generality, we can assume that the area of parallelogram $ABCD$ is 1. Let's introduce the following notations: $S_{1}$ - the area of triangle $BKN$, $S_{2}$ - the area of triangle $AMN$, $S_{3}$ - the area of triangle $CKL$, $S_{4}$ - the area of triangle $DLM$, $x$ - the area of triangle $KMN$, and $y$ - the area of triangle $KLM$ (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-076.jpg?height=360&width=814&top_left_y=400&top_left_x=632)

Triangles $KMA$ and $ABM$ have the same area, hence triangles $KMN$ and $ABN$ also have equal areas. We have, $\frac{x}{S_{1}}=\frac{AN}{NK}$ and $\frac{S_{2}}{x}=\frac{AN}{NK}$, so $\frac{x}{S_{1}}=\frac{S_{2}}{x}$, from which $x=\sqrt{S_{1} S_{2}}$. Similarly, $y=\sqrt{S_{3} S_{4}}$. Since $S_{1}+S_{2} \geqslant 2 \sqrt{S_{1} S_{2}}$, we have

$2(x+y)=1-\left(S_{1}+S_{2}+S_{3}+S_{4}\right) \leqslant 1-2\left(\sqrt{S_{1} S_{2}}+\sqrt{S_{3} S_{4}}\right)=1-2(x+y)$, therefore $x+y \leqslant \frac{1}{4}$. Equality is achieved if and only if $S_{1}=S_{2}$ and $S_{3}=S_{4}$, i.e., if and only if $KM \| AB$.

Answer: $\frac{1}{4}$.