Triangle $A B C$ is divided by segments as shown in the figure. Segments $D E$ and $A B$ are parallel. Triangles $C D H, C H I, C I E, F I H$ have the same area, which is $8 \mathrm{dm}^{2}$.

Determine the area of quadrilateral $A F H D$.

(E. Semerádová)

![](https://cdn.mathpix.com/cropped/2024_04_17_5ae3f20c38a35ad29464g-7.jpg?height=442&width=1008&top_left_y=453&top_left_x=524)
Triangles $C D H$ and $C H I$ have a common side $C H$, so they have the same height from the common vertex $C$. These triangles have the same area, so segments $D H$ and $H I$ are congruent. Triangles $C H I$ and $F I H$ have a common side $H I$, so they have the same height from the common vertex $I$. These triangles have the same area, so segments $C H$ and $H F$ are congruent.

The above conclusions mean that $H$ is the midpoint of segments $D I$ and $C F$, which are the diagonals of quadrilateral $C D F I$. This quadrilateral is therefore a parallelogram, which is divided into four triangles of equal area by its diagonals. The area of parallelogram $C D F I$ is thus equal to four times the area of triangle $C D H$.

In particular, lines $A C$ and $F I$ are parallel, so quadrilateral $A F I D$ is also a parallelogram. This parallelogram shares triangle $D F I$ with parallelogram $C D F I$, which forms half of both the first and the second parallelogram. The area of parallelogram $A F I D$ is therefore the same as the area of parallelogram $C D F I$.

The area of quadrilateral $A F H D$ can be expressed as follows:

$$
\begin{aligned}
S_{A F H D} & =S_{A F I D}-S_{F I H}= \\
& =S_{C D F I}-S_{C D H}= \\
& =4 \cdot S_{C D H}-S_{C D H}=3 \cdot 8=24\left(\mathrm{dm}^{2}\right) .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_5ae3f20c38a35ad29464g-7.jpg?height=448&width=1008&top_left_y=1989&top_left_x=524)

Notes. Quadrilateral $A F H D$ is actually a trapezoid. Points $B$ and $E$ play no role in solving the problem. From the initial observations, several additional facts can be derived, which can be used in solving the problem:

Segments $D I, I F$, and $F D$ are the midlines of triangle $A G C$, and they divide this triangle into four congruent triangles. The area of each of these triangles is equal to twice the area of the reference triangle $C D H$ (marked as $S$ in the figure). Thus, $S_{A F H D}=3 \cdot S_{C D H}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5ae3f20c38a35ad29464g-8.jpg?height=437&width=721&top_left_y=587&top_left_x=676)

Triangles $C D H$ and $C A F$ are similar with a ratio of 2. The area of triangle $C A F$ is therefore four times the area of triangle $C D H$. Thus, $S_{A F H D}=S_{C A F}-S_{C D H}=3 \cdot S_{C D H}$.