In the 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 $\triangle BCF$?

![](https://cdn.mathpix.com/cropped/2024_05_01_df4451ddcede6035446dg-1.jpg?height=323&width=328&top_left_y=535&top_left_x=1315)
The diagonals of the square $A B C D$ divide the square into 4 equal triangles, so the area of the triangle $\triangle B C E$ is

![](https://cdn.mathpix.com/cropped/2024_05_01_df4451ddcede6035446dg-4.jpg?height=317&width=328&top_left_y=544&top_left_x=1412)

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

Since the length of $B F$ is half of $B E$ and the height relative to the sides $B F$ and $B E$ is $C E$, the area of the triangle $\triangle C B F$ is half the area of the triangle $\triangle C B E$. Therefore, the area of the said triangle is $0.25 \div 2=0.125 \mathrm{~cm}^{2}$.