4B. On the side $AB$ of triangle $ABC$, a point $F$ is chosen arbitrarily, and points $D$ and $E$ are on sides $BC$ and $AC$ such that $FD \parallel AC$ and $FE \parallel BC$. Express the area $P$ of triangle $CDE$ in terms of the areas $P_{1}$ and $P_{2}$ of triangles $AFE$ and $BDF$.
Solution. From the condition of the problem, it follows that $F D C E$ is a parallelogram, so $\overline{F D}=\overline{E C}$. Triangles $A F E$ and $F D B$ are similar, and due to $\overline{F D}=\overline{E C}$, we get

$$
P_{1}: P_{2}=\overline{A E}^{2}: \overline{F D}^{2}=\overline{A E}^{2}: \overline{E C}^{2}
$$

Furthermore, triangles $A F E$ and $E D C$ have equal heights ($\overline{F H}=\overline{D G}$), so we have

![](https://cdn.mathpix.com/cropped/2024_06_05_84ab8bb1a5b8361dd462g-03.jpg?height=382&width=471&top_left_y=266&top_left_x=997)

$$
P_{1}: P=\overline{A E}: \overline{E C}
$$

From (1) and (2), we get $P_{1}: P_{2}=P_{1}^{2}: P^{2}$, i.e., $P=\sqrt{P_{1} P_{2}}$.

## II year