Solution. It is clear that $h_{1}+h_{2}+h_{3}=h$ (see the diagram). Triangles $A B C$ and $Q P M$ are similar, so we have $P_{1}: P=h_{1}^{2}: h^{2}$.

Similarly, we have $P_{2}: P=h_{2}^{2}: h^{2}$ and $P_{3}: P=h_{3}^{2}: h^{2}$, from which we get:

$$
\begin{aligned}
P & =\frac{h^{2}\left(P_{1}+P_{2}+P_{3}\right)}{h_{1}^{2}+h_{2}^{2}+h_{3}^{2}} \\
& =\frac{\left(h_{1}+h_{2}+h_{3}\right)^{2}}{h_{1}^{2}+h_{2}^{2}+h_{3}^{2}}\left(P_{1}+P_{2}+P_{3}\right) \\
& =\left(1+\frac{h_{2}}{h_{1}}+\frac{h_{3}}{h_{1}}\right)^{2} \frac{P_{1}+P_{2}+P_{3}}{1+\left(\frac{h_{2}}{h_{1}}\right)^{2}+\left(\frac{h_{3}}{h_{1}}\right)^{2}} \\
& =\left(1+\sqrt{\frac{P_{2}}{P_{1}}}+\sqrt{\frac{P_{3}}{P_{1}}}\right)^{2} \frac{P_{1}+P_{2}+P_{3}}{1+\frac{P_{2}}{P_{1}}+\frac{P_{3}}{P_{1}}} \\
& =\left(\sqrt{P_{1}}+\sqrt{P_{2}}+\sqrt{P_{3}}\right)^{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_d61831f2f852c6e23b39g-5.jpg?height=385&width=560&top_left_y=954&top_left_x=898)
2. In triangle $ABC$, a point $M$ is given. Through $M$, three lines $p, q$, and $r$ are drawn parallel to $AB, BC$, and $CA$ respectively. Let $R=p \cap BC, L=p \cap AC$, $P=q \cap AB, T=q \cap AC, Q=r \cap AB, N=r \cap BC$. If the areas $P_{1}=P_{P Q M}, P_{2}=P_{M R N}$, and $P_{3}=P_{L M T}$ are known, find the area $P$ of triangle $ABC$.