2. Let $K$ and $L$ be points on sides $BC$ and $AC$ of triangle $ABC$, respectively, such that $|BK|=\frac{1}{3}|BC|,|AL|=\frac{1}{3}|AC|$. Let $M$ be the intersection of segments $AK$ and $BL$. Calculate the ratio of the areas of triangles $ABM$ and $ABC$.
2. Let $v$ be the height of triangle $ABC$ on side $AB$, $v_{1}$ the height of triangle $ABM$ on side $AB$, and $v_{2}$ the height of triangle $KLM$ on side $KL$ (Fig. 1). From the similarity of triangles $LKC$ and $ABC$ (guaranteed by the AA similarity criterion), it follows that $|KL| = \frac{2}{3}|AB|$. By comparing their heights from the common vertex $C$, we see that the height $v$ of triangle $ABC$ is three times the distance of the line segment $KL$ from side $AB$, i.e., $v = 3(v_{1} + v_{2})$. Since $AK$ and $BL$ are transversals of the parallel lines $KL$ and $AB$, the similarity of triangles $ABM$ and $KLM$ follows from the congruence of the corresponding alternate interior angles.

![](https://cdn.mathpix.com/cropped/2024_04_17_ccc8d169f6ddeb4279cbg-2.jpg?height=551&width=876&top_left_y=1849&top_left_x=590)

Fig. 1

Since $|KL| = \frac{2}{3}|AB|$, it also follows that $v_{2} = \frac{2}{3} v_{1}$, and thus $v_{1} + v_{2} = \frac{5}{3} v_{1}$, therefore

$$
v = 3(v_{1} + v_{2}) = 5 v_{1}
$$

Triangles $ABM$ and $ABC$ share the side $AB$, so their areas are in the ratio of the heights to this side, meaning the area of triangle $ABC$ is five times the area of triangle $ABM$.

For a complete solution, award 6 points. For determining the similarity ratio of triangles $ABM$ and $KLM$, award 2 points. For determining the similarity ratio of triangles $ABC$ and $LKC$, award 2 points. For the final derivation of the area ratio of triangles $ABM$ and $ABC$, award 2 points.