Let's construct a line with a given direction that divides the area of a given triangle in half.

 editar nota: The translation provided is a direct translation of the given Hungarian text into English, maintaining the original format and structure.
Solution. Draw lines parallel to the given direction through the vertices of the triangle (Figure 1). We can assume that one of the lines passes through point $C$. This line intersects the side $A B$ at point $R$ (it is possible that this coincides with one of the vertices). If $R$ is the midpoint of side $A B$, then we are done; otherwise, we can assume that $A R > B R$.

![](https://cdn.mathpix.com/cropped/2024_05_02_dbcc2d2f517925987c46g-1.jpg?height=344&width=594&top_left_y=343&top_left_x=749)

Figure 1

Let $P Q$ be the desired line, with $A P = x$, $A Q = y$, and $y < A R$.

According to the condition:

$$
\frac{b c \cdot \sin \alpha}{2} = x y \cdot \sin \alpha, \quad \text { from which } \quad \frac{b c}{2} = x y
$$

By the theorem of parallel chords: $\frac{x}{y} = \frac{b}{A R}$, multiplying by $y^2$: $x y = \frac{b y^2}{A R}$. Comparing with the previous result:

$$
\frac{b c}{2} = \frac{b y^2}{A R}, \quad \text { from which } \quad \frac{c}{2} = \frac{y^2}{A R}
$$

Thus, $y$ is the geometric mean of the segments $A R$ and $\frac{c}{2}$: $y = \sqrt{A R \cdot \frac{c}{2}}$, which can be constructed using the altitude theorem.

According to Figure 2, we construct $y$, then measure from point $A$ to the obtained point $Q$, and draw a parallel line to the given line through $Q$.

The construction can be carried out uniquely, and there is exactly one solution.

![](https://cdn.mathpix.com/cropped/2024_05_02_dbcc2d2f517925987c46g-1.jpg?height=288&width=480&top_left_y=1447&top_left_x=801)

Figure 2