For five mutually distinct points as shown in the figure, it is given that points $C, D, E$ lie on a line parallel to line $A B, F$ is the midpoint of segment $A D, G$ is the midpoint of segment $A C$, and $H$ is the intersection of lines $A C$ and $B E$. The area of triangle $B C G$ is $12 \mathrm{~cm}^{2}$ and the area of quadrilateral $D F H G$ is $8 \mathrm{~cm}^{2}$.

Determine the areas of triangles $A F E, A H F, A B G$ and $B G H$.

![](https://cdn.mathpix.com/cropped/2024_04_17_6aef7ef60e9f55f815aeg-5.jpg?height=414&width=1148&top_left_y=1204&top_left_x=457)

Hint. Can you compare the areas of some figures directly, i.e., without specific values?
Triangles $A B C, A B D$ and $A B E$ have a common side $A B$ and the same height to this side, so they have the same area. Points $F$ and $G$ are the midpoints of segments $A D$ and $A C$, so line $F G$ is parallel to $A B$. Triangles $A B F$ and $A B G$ have a common side $A B$ and the same height to this side, which is half the height of the previous three triangles. Therefore, triangles $A B F$ and $A B G$ have the same area, which is half the area of triangles $A B C, A B D$ and $A B E$. In total, we get that triangles $A F E$, $A B F, B D F, A D G, A B G$ and $B C G$ have the same area, which is $12 \mathrm{~cm}^{2}$.

Triangle $A D G$ is composed of triangle $A H F$ and quadrilateral $D F H G$, which has an area of $8 \mathrm{~cm}^{2}$. The area of triangle $A H F$ is therefore $12-8=4\left(\mathrm{~cm}^{2}\right)$. Similarly, the area of triangle $B G H$ is the same.

The areas of triangles $A F E, A H F, A B G$ and $B G H$ are, in order, $12,4,12$ and $4\left(\mathrm{~cm}^{2}\right)$.

Note. Among the given objects, there are relationships that we have not used, but which allow for alternative methods of solving the problem. For example, it can be directly shown that quadrilaterals $A B C D$ and $A B D E$ are parallelograms, that triangles $A H F$ and $B G H$ have the same area, that their sum is equal to the area of quadrilateral $D F H G$, etc.

For the curious: the area of triangle $A B H$ in the problem is $8 \mathrm{~cm}^{2}$.