Let's divide a square into three equal area parts using lines parallel to one of its diagonals.
It is sufficient to determine the line $MN$ parallel to the diagonal $AC$ such that the area of $\triangle BMN$ is $\frac{2}{3}$ of the area of $\triangle ABC$. Indeed, in this case, $MN$ and the line symmetric to it with respect to $AC$ will meet the requirements of the problem.

![](https://cdn.mathpix.com/cropped/2024_05_02_9f4bd85571fead488fcdg-1.jpg?height=244&width=290&top_left_y=325&top_left_x=904)

Clearly, the following holds:

$\triangle BMN : \triangle BAC = \overline{BM}^2 : \overline{BA}^2$, so it is necessary that $\overline{BM}^2 : \overline{BA}^2 = 2 : 3$.

Therefore, let us take an arbitrary distance $PR$ and divide it into two parts in the ratio $2:3$: $PQ : QR = 2 : 3$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9f4bd85571fead488fcdg-1.jpg?height=269&width=425&top_left_y=836&top_left_x=839)

The semicircle corresponding to the diameter $PR$ intersects the perpendicular to $PR$ at point $Q$ at point $S$. Then: $\overline{PS}^2 : \overline{RS}^2 = 2 : 3$. On the line $SR$, measure the distance $ST = AB$ and draw a line parallel to $PR$ from point $T$; this line intersects $SP$ at point $U$ such that $SU = BM$.

Note. If $BM = x, AB = a$, then $x^2 = \frac{2a^2}{3}$. Based on this relationship, $x$ can be constructed as the geometric mean of $a$ and $\frac{2a}{3}$. Alternatively, $x = \frac{a\sqrt{2}}{\sqrt{3}}$ represents the smaller leg of a right triangle whose larger leg is $a\sqrt{2}$, the diagonal of the given square, and whose acute angles are $30^\circ$ and $60^\circ$.