2. From an internal point $M$ of triangle $A B C$, lines parallel to the sides of triangle $A B C$ are drawn. They divide the triangle into six parts, three of which are triangles with areas $P_{1}, P_{2}$, and $P_{3}$. Express the area of triangle $A B C$ in terms of the areas $P_{1}, P_{2}$, and $P_{3}$.
Solution. Let $P$ be the area of $\triangle ABC$. Each of the triangles $RMS$, $KLM$, and $MPQ$ is similar to triangle $ABC$, so

$$
\frac{\sqrt{P_{1}}}{\sqrt{P}}=\frac{\overline{R M}}{\overline{A B}}, \frac{\sqrt{P_{2}}}{\sqrt{P}}=\frac{\overline{K L}}{\overline{A B}} \text { and } \frac{\sqrt{P_{3}}}{\sqrt{P}}=\frac{\overline{M P}}{\overline{A B}}.
$$

By summing up the above equations, considering that $\overline{R M}=\overline{A K}$ and $\overline{M P}=\overline{L B}$, we get

$$
\frac{\sqrt{P_{1}}+\sqrt{P_{2}}+\sqrt{P_{3}}}{\sqrt{P}}=\frac{\overline{A K}+\overline{K L}+\overline{L B}}{\overline{A B}}=\frac{\overline{A B}}{\overline{A B}}=1
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_50e976f6299e4a88a6a3g-04.jpg?height=385&width=486&top_left_y=973&top_left_x=978)

from which it follows that $P=\left(\sqrt{P_{1}}+\sqrt{P_{2}}+\sqrt{P_{3}}\right)^{2}$.