7. Let the square be $A B C D$. On the side $\overline{A B}$, there is a point $M$ that divides this side in the ratio $1: 3$ (starting from vertex $A$). On the side $\overline{B C}$, there is a point $N$ such that the area of triangle $M B N$ is in the ratio $1: 4$ to the area of the given square. In what ratio does point $N$ divide the side $\overline{B C}$ (starting from vertex $B$)?
7. Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_4ffb1f624eaee9da51d7g-3.jpg?height=425&width=417&top_left_y=330&top_left_x=405)

Let the side length of the given square be denoted by $a$.

Then the area of the square is $P_{k}=a^{2}$.

From the problem statement, it follows that $|\overline{B M}|=\frac{3}{4} a$.

Triangle $M B N$ is a right triangle with legs $\overline{B M}$ and $\overline{B N}$. 1 POINT

If we denote $b=|\overline{B N}|$, then the area of the right triangle $M B N$ is

$P_{t}=\frac{|B M| \cdot|B N|}{2}=\frac{\frac{3}{4} a \cdot b}{2}=\frac{3}{8} a \cdot b$.

From the problem statement, it follows that $\frac{P_{t}}{P_{k}}=\frac{\frac{3}{8} a \cdot b}{a \cdot a}=\frac{1}{4}$

so it follows that $b=|B N|=\frac{2}{3} a$.

This means that $|C N|=\frac{1}{3} a$

so $\frac{|B N|}{|C N|}=\frac{\frac{2}{3} a}{\frac{1}{3} a}=\frac{2}{1}=2: 1$.

Therefore, point $N$ divides the segment $\overline{B C}$ in the ratio $2: 1$ (starting from vertex $B$).