6. Origami. Mitya is going to fold a square sheet of paper $A B C D$. Mitya calls a fold beautiful if side $A B$ intersects side $C D$ and the four resulting right triangles are equal.

Before that, Vanya chooses a random point $F$ on the sheet. Find the probability that Mitya can make a beautiful fold passing through point $F$.
Solution. Let's unfold a beautiful fold (see fig.). The diagonal $B D$ and the fold line $S T$ intersect at point $O$. Triangles $B S O$ and $D T O$ are equal (by side and two angles). Therefore, $B O=O D$, and thus $O$ is the center of the square. This means that the fold line $S T$ passes through the center of the square. It is also clear that the converse is true - if the fold line passes through the center of the square, then the fold will be beautiful.
![](https://cdn.mathpix.com/cropped/2024_05_06_c76ac3d7d759b54fc6bag-08.jpg?height=734&width=1448&top_left_y=1872&top_left_x=290)

Point $S$ can take any position between $B$ and $C$, and point $T$ is located between $D$ and $A$. Therefore, to make a beautiful fold through point $F$, point $F$ must belong to triangle $B O C$ or triangle $A O D$.

The area of the figure bounded by these triangles is half the area of the square.

Answer: $\frac{1}{2}$.