Area of a square inside a square created by connecting point-opposite midpoint

By moving small triangles we can make $5$ equal small squares.

By drawing lines along and parallel to the segments within the square, a grid overlapping the original square is produced. If $A$ represents the area of the shaded square, then $9A$ represents the area of the circumscribing square with sides parallel to the shaded square

It is easy to see a right triangle on each side of the original square with its hypotenuse on the side of the original square, each with an area of $\frac{2\cdot A}{2} = A$. Thus the total area of the original square is $5A$, or the ratio of the area of the shaded square to the original square is $1/5$.

Let $GK$ be a perpendicular to $AF$, $GK=x$ be the side-length of the little square

and $BH\cap AF=\{M\}$.

Thus, $AM=GK$ and since $\Delta AMH\sim\Delta ADF$, we obtain: $$\frac{x}{1}=\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{4}}},$$ which gives $x=\frac{1}{\sqrt5}$ and the answer: $\frac{1}{5}$.