Area of square created by intersection of segments from a square vertexes to their opposite sides There will be an square created when we draw segments from a square vertexes to their opposite sides' middle.

What is the relation between smaller square's area and the side length of the bigger one?

Let $a$ be the side length of the grey square (and $1$ the side lengthg of the original square).
By similarity, the length of the line segment from $D$ to $DE\cap CH$ is also $a$.
Then the triangle with base $AE$ complete the quadrilateral with top edge $DE$ to a square of area $a^2$. We can do the same with the other triangles and conclude that $1^2= 5a^2$.
