Let's inscribe a rectangle in a given triangle such that the area of the rectangle is one-third of the area of the given triangle.
Let $ABC$ be the given triangle and $MNPQ$ the sought right-angled quadrilateral, whose side $PQ$ lies on $BC$ and vertex $M$ lies on $AB$.

Draw the altitude $AD$, then

$$
\frac{MP}{AD}=\frac{MB}{AB}
$$

and

$$
\frac{MN}{BC}=\frac{AM}{AB}
$$

Multiplying these two equations, we get

$$
\frac{MP \cdot MN}{AD \cdot BC}=\frac{MB \cdot AM}{\overline{AB}^{2}}
$$

But since

$$
MP \cdot MN=\frac{AD \cdot BC}{6}
$$

it follows that

$$
AM \cdot MB=\frac{\overline{AB}^{2}}{6}
$$

Let

$$
AM \cdot BM=x^{2}
$$

then

$$
x=\frac{AB}{\sqrt{6}}
$$

Thus, we can construct $x$. If we know $x$, we also know the position of point $M$ on side $AB$ and consequently the position of the right-angled quadrilateral.

Generally, there are two points $M$, and thus two solutions.

(Krisztián György.)

The problem was also solved by: Freibauer E., Lukhaub Gy., Oltay K., Prohászka, Weisz Á., Weisz J.