In an equilateral triangle $A B C$ with a side length of $8 \mathrm{~cm}$, point $D$ is the midpoint of side $B C$ and point $E$ is the midpoint of side $A C$. Point $F$ lies on segment $B C$ such that the area of triangle $A B F$ is equal to the area of quadrilateral $A B D E$.

Calculate the length of segment $B F$.

(L. Růžičková)
Both quadrilateral $A B D E$ and triangle $A B F$ can be divided into two triangles, of which $A B D$ is common to both. The remaining parts, i.e., triangles $A D E$ and $A D F$, must therefore have the same area. These two triangles have a common side $A D$, so they must also have the same height to this side. This means that points $E$ and $F$ lie on a line parallel to line $A D$.

Since point $E$ is the midpoint of side $A C$, point $F$ is also the midpoint of segment $C D$. Given that point $D$ is the midpoint of segment $B C$, point $F$ is thus at three-quarters of segment $B C$. The length of segment $B F$ is $4+2=6 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ac1663341c936f6e1c9eg-2.jpg?height=529&width=596&top_left_y=1940&top_left_x=730)

Another solution. The equilateral triangle $A B C$ is divided by its medians into four congruent triangles, three of which form quadrilateral $A B D E$. Therefore, the area of triangle $A B F$ is equal to three-quarters of the area of triangle $A B C$. These two triangles have the same height from the common vertex $A$, so the length of side $B F$ is equal to three-quarters of the length of side $B C$. The length of segment $B F$ is $6 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ac1663341c936f6e1c9eg-3.jpg?height=526&width=600&top_left_y=465&top_left_x=728)

Another solution. Segment $E D$ is the median of triangle $A B C$, so it is parallel to $A B$ and has a length of $8: 2=4(\mathrm{~cm})$. Quadrilateral $A B D E$ is a trapezoid with bases of lengths $8 \mathrm{~cm}$ and $4 \mathrm{~cm}$ and a height equal to half the height of triangle $A B C$.

If we denote the height of triangle $A B C$ by $v$, then the area of trapezoid $A B D E$, or the area of triangle $A B F$, is

$$
\frac{(8+4) \cdot \frac{1}{2} v}{2}=3 v, \quad \text { or } \quad \frac{|B F| \cdot v}{2}
$$

According to the problem, these areas are equal, so $|B F|=6 \mathrm{~cm}$.

Evaluation. 2 points for any insight leading to a unique determination of point $F$; 2 points for solving the problem; 2 points based on the quality of the commentary.