Two pairs of parallel lines $A B \| C D$ and $A C \| B D$ are given. Point $E$ lies on line $B D$, point $F$ is the midpoint of segment $B D$, point $G$ is the midpoint of segment $C D$, and the area of triangle $A C E$ is $20 \mathrm{~cm}^{2}$.

Determine the area of triangle $D F G$.

(V. Semeráková)

Hint. Compare the areas of triangles $A C E$ and $A B C$.
The area of a triangle depends on the length of its side and the height to this side. Since lines $A C$ and $B D$ are parallel and point $E$ lies on line $B D$, the area of triangle $A C E$ is always the same for any chosen point $E$. In particular, the area of triangle $A C E$ is the same as the area of triangle $A C D$. For the same reason, the area of triangle $A C D$ is the same as the area of triangle $B C D$. Therefore,

$$
S_{A C E}=S_{A C D}=S_{B C D}=20 \mathrm{~cm}^{2}
$$

Now let's compare the areas of triangles $B C D$ and $D F G$:

![](https://cdn.mathpix.com/cropped/2024_04_17_c15208fdb3e01b0f90f8g-2.jpg?height=409&width=834&top_left_y=281&top_left_x=611)

Triangles $D F G$ and $F B G$ have a common height from vertex $G$ and point $F$ is at the midpoint of side $B D$, so these triangles have the same area. Triangles $D F G$ and $F B G$ together form triangle $D B G$, so $S_{D F G}=\frac{1}{2} S_{D B G}$. For the same reason, $S_{D B G}=\frac{1}{2} S_{D B C}$. Therefore,

$$
S_{D F G}=\frac{1}{4} S_{D B C}=\frac{1}{4} \cdot 20=5\left(\mathrm{~cm}^{2}\right)
$$

Note. The previous expression of the ratio of the areas of triangles $D F G$ and $D B C$ implicitly refers to their similarity, which can also be used in the reasoning ( $F G$ is the midline of triangle $D B C$, so all corresponding sides are proportional in the ratio $1: 2$ ). Without referring to the concept of similarity, it is possible to directly compare, for example, the bases $D F$ and $D B$ and the corresponding heights (both in the ratio $1: 2$ ). This can also be considered for triangles $D F G$ and $A C E$ with any $E \in B D$ (i.e., without the transformations used above).