Let the area of triangle $A D F$ be denoted by $t$. In triangle $A E F$, $F D$ is a median, and in triangle $A E C$, $E F$ is a median. Therefore, $t_{A E F}=2 t_{A D F}=2 t$ and $t_{A E C}=2 t_{A E F}=4 t$. The heights of triangles $A E C$ and $E B C$ from $C$ are the same, so the ratio of their areas is the same as the ratio of $A E$ to $E B$, which is 2. Thus, $t_{E B C}=\frac{1}{2} t_{A E C}=2 t$. Therefore, the area of triangle $A B C$ is $6 t$, or $t=\frac{1}{6}$.

In triangle $A C H$, $H F$ is a median, so $t_{A F H}=t_{F C H}$. In triangle $A B H$, $D$ divides $A B$ into a ratio of 1:2, so

$$
2 \cdot t_{A D H}=t_{D B H}, \quad \text { or } \quad 2 \cdot\left(\frac{1}{6}+t_{F C H}\right)=5 \cdot \frac{1}{6}+t_{F C H}
$$

from which we get $t_{F C H}=\frac{1}{2}$. Since $E$ also divides $A B$ into a ratio of 1:2, $2 \cdot t_{E G B}=t_{A G E}$, and since $G F$ is a median in triangle $A G C$,

$$
t_{A G F}=t_{F G C}, \quad \text { or } \quad 2 \cdot \frac{1}{6}+2 t_{E G B}=4 \cdot \frac{1}{6}+t_{E G B}
$$

from which we get $t_{E G B}=\frac{1}{3}$.

Therefore, the desired area $t_{F G H}=t_{E G B}+t_{E B C F}+t_{F C H}=\frac{1}{3}+\frac{4}{6}+\frac{1}{2}=\frac{3}{2}$.

Erika Sors (Budapest, Szent István Gymnasium, 12th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_6a410fc359d419197b84g-1.jpg?height=488&width=1087&top_left_y=1038&top_left_x=519)
In a unit area triangle $ABC$, let the points that trisect side $AB$ be $D$ and $E$, and the midpoint of side $AC$ be $F$. The line $CB$ is intersected by line $FE$ at point $G$, and by line $FD$ at point $H$. Determine the area of triangle $FGH$.

---

Translate the above text into English, preserving the original text's line breaks and formatting. Directly output the translation result.