2. Inside the parallelogram $A B C D$, there is a point $K$ and in the strip between the parallels $B C$ and $A D$ in the half-plane opposite to $\mathrm{k} C D A$, there is a point $L$. The areas of triangles $A B K, B C K, D A K$ and $D C L$ are $S_{A B K}=18 \mathrm{~cm}^{2}, S_{B C K}=8 \mathrm{~cm}^{2}, S_{D A K}=16 \mathrm{~cm}^{2}, S_{D C L}=36 \mathrm{~cm}^{2}$. Calculate the areas of triangles $C D K$ and $A B L$.
2. Triangles $ABK$ and $CDK$ have equal sides $AB$ and $CD$, and the sum of their heights $v_{1}$ and $v_{2}$ (the distances from point $K$ to the lines $AB$ and $CD$, respectively) is equal to the height $v$ of the parallelogram $ABCD$ (the distance between the parallel lines $AB$ and $CD$, see Fig. 1). Therefore, the sum of their areas gives half the sum of the area of the given parallelogram:

$$
S_{ABK} + S_{CDK} = \frac{1}{2}|AB| v_{1} + \frac{1}{2}|CD| v_{2} = \frac{1}{2}|AB| \cdot (v_{1} + v_{2}) = \frac{1}{2}|AB| \cdot v = \frac{1}{2} S_{ABCD}.
$$

Similarly, $S_{BCK} + S_{DAK} = \frac{1}{2} S_{ABCD}$, thus

$$
S_{CDK} = S_{BCK} + S_{DAK} - S_{ABK} = 6 \text{ cm}^2
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_e4196ab5bfde4ebdfd75g-2.jpg?height=520&width=825&top_left_y=1413&top_left_x=618)

Fig. 1

Triangles $ABL$ and $DCL$ have equal sides $AB$ and $CD$. Denote $v_{3}$ as the height of the second one, then the height of the first one is $v + v_{3}$, so for the difference in their areas we have

$$
\begin{aligned}
S_{ABL} - S_{DCL} & = \frac{1}{2}|AB| \cdot (v + v_{3}) - \frac{1}{2}|CD| \cdot v_{3} = \frac{1}{2}|AB| \cdot (v + v_{3} - v_{3}) = \\
& = \frac{1}{2}|AB| \cdot v = \frac{1}{2} S_{ABCD} = S_{BCK} + S_{DAK}.
\end{aligned}
$$

From this it follows that

$$
S_{ABL} = S_{BCK} + S_{DAK} + S_{DCL} = 60 \text{ cm}^2.
$$

For a complete solution, award 6 points. For determining each of the areas, award 3 points. For merely discovering (and proving) the fact that the sum of the areas of "opposite" triangles $ABK$ and $CDK$ or $BCK$ and $AKD$ gives half the area of the entire parallelogram, award 2 points.