In the unit square $ABCD$, the midpoint of side $AB$ is $E$, and the midpoint of side $DC$ is $F$. The line $DE$ intersects the diagonal $AC$ at point $G$, and the line $BF$ intersects the diagonal $AC$ at point $H$. What is the length of segment $GH$?
The triangle $E G A$ is similar to the triangle $C G D$ due to the equality of their angles. We also know that the ratio of their corresponding sides is $E A: D C=1: 2$. This implies that $A G=\frac{1}{3} A C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3c0f94a4516212e35751g-1.jpg?height=534&width=586&top_left_y=286&top_left_x=756)

Similarly, it follows that the triangle $C H F$ is similar to the triangle $B H A$, and $C H=\frac{1}{3} A C$. From these two equalities, it follows that $G H=\frac{1}{3} A C$. The diagonal of a unit square is $\sqrt{2}$, so $G H=\frac{\sqrt{2}}{3}$.

Note. The problem was very easy. This is proven by the fact that many people obtained the correct result. However, most of the proofs relied on the "clearly visible" properties of the diagram. The most common mistake in the proofs that used the property $E D \| F B$ was that they either did not prove it at all, or if they realized it needed to be proven, they only wrote as justification that $D F=E B$. However, the equality of two segments does not imply the parallelism of the lines connecting their endpoints!