Draw the lines parallel to two sides of a triangle that bisect the area of the triangle. In what ratio does the line parallel to the third side of the triangle, passing through their intersection point, divide the area of the triangle?
Solution. In triangle $ABC$, the area bisector line parallel to side $AC$ intersects sides $AB$ and $BC$ at points $E$ and $F$, respectively. Similarly, the area bisector line parallel to side $BC$ intersects side $AB$ at $G$ and side $AC$ at $H$, and the intersection point of the two lines is $M$. Finally, the line through $M$ parallel to $AB$ intersects $BC$ and $AC$ at points $K$ and $L$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_3b47af3d774dbb25dcdeg-1.jpg?height=334&width=504&top_left_y=348&top_left_x=800)

Due to the parallelism, triangle $AGH$ is similar to triangle $ABC$. The corresponding sides $AG: AB = \sqrt{\frac{1}{2}}$ are in the ratio of the square roots of the areas. Therefore, $AG = \frac{\sqrt{2}}{2} AB$, and thus $GB = \frac{2-\sqrt{2}}{2} AB$. Similarly, $AE = \frac{2-\sqrt{2}}{2} AB$. Since $LM \parallel AE$ and $MK \parallel GB$, we have $LK = LM + MK = AE + GB = (2-\sqrt{2}) AB$. This implies that the ratio of the areas of triangle $LKC$ and triangle $ABC$ is $(2-\sqrt{2})^2 = 6 - 4\sqrt{2}$. The line $LK$ divides triangle $ABC$ into triangle $LKC$ and trapezoid $ABKL$. We need to determine the area $T$ of trapezoid $ABKL$:

$$
T = T_{\triangle ABC} - (6 - 4\sqrt{2}) \cdot T_{\triangle ABC} = (4\sqrt{2} - 5) \cdot T_{\triangle ABC}
$$

The desired ratio is:

$$
\frac{T_{\triangle LKC}}{T} = \frac{6 - 4\sqrt{2}}{4\sqrt{2} - 5} = \frac{4\sqrt{2} - 2}{7} \approx 0.5224
$$

Zoltán Török (Gödöllő, Török I. Gymnasium, 11th grade)