4. The area of a convex quadrilateral $A B C D$ is 40. Points $K, L, M, N$ are chosen on sides $A B, B C, C D, A D$ respectively, such that $A K: A B=B L: B C=C M: C D=D N: A D$. Find the ratio $A K: K B$ if the area of quadrilateral $K L M N$ is 25.
Solution: $A K: A B=x, D N: A D=x$, $A N: A D=(A D-N D): A D=1-x$, $S_{A K N}=x(1-x) S_{A B D}$. Similarly, we get $S_{C L M}=x(1-x) S_{\text {BCD }}$. Then $S_{A K N}+S_{C L M}=x(1-x) S_{A B C D}=100 x(1-x)$.

Similarly $\quad S_{B L K}+S_{D N M}=x(1-x) S_{A B C D}=40 x(1-x) . \quad$ From this, $S_{\text {KLMN }}=40-80 x(1-x)$. We have the equation $25=40-80 x(1-x)$, $16 x^{2}-16 x+3=0, x=1 / 4$ or $x=3 / 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_03a96cebf8326f3655efg-4.jpg?height=534&width=511&top_left_y=1875&top_left_x=1435)

Answer: $1 / 4$ or $3 / 4$.