In the given figure, $ABCD$ is a square whose side measures $1 \, \text{cm}$, $E$ is the midpoint of the diagonal $BD$, and $F$ is the midpoint of the segment $BE$. What is the area of the triangle $\triangle CBF$?

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-015.jpg?height=223&width=240&top_left_y=1305&top_left_x=1588)
The diagonals of the square $A B C D$ divide the square into 4 equal triangles, so the area of triangle $\triangle B C E$ is one fourth of the area of the square, that is,

$$
1 \div 4=0.25 \mathrm{~cm}^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-065.jpg?height=260&width=286&top_left_y=2006&top_left_x=1590)

Since the length of $B F$ is half of $B E$ and $C E$ is the common height to the bases $B F$ and $B E$, we conclude that the area of triangle $\triangle C B F$ is necessarily half the area of triangle $\triangle C B E$. Thus, the area of triangle $\triangle C B F$ is $0.125 \mathrm{~cm}^{2}$.