<s>[INST] Area of a square inside a square created by connecting point-opposite midpoint 
Square $ABCD$ has area $1cm^2$ and sides of $1cm$ each.
$H, F, E, G$ are the midpoints of sides $AD, DC, CB, BA$ respectively.
What will the area of the square formed in the middle be?
I know that this problem can be solved by trigonometry by using Area of triangle ($\frac{1}{2}ab\sin{c}$) but, 
is there another method or visual proof?
 [/INST] Here's a solution using analytic geometry which doesn't require any particular insight:
Set up a coordinate system such that $B = (0, 0)$ and $C = (1, 0)$.  Then the equation of line $BH$ is $y - 2x = 0$, and the equation of line $CG$ is $2y + x = 1$.  Therefore, their intersection is $(\frac{1}{5}, \frac{2}{5})$.  Similarly, the equation of line $ED$ is $y - 2x = -1$, so the intersection of lines $ED$ and $CG$ is $(\frac{3}{5}, \frac{1}{5})$.  This gives that the side of the inner square is the distance between these two points, which is
$$\sqrt{\left(\frac{1}{5} - \frac{3}{5}\right)^2 + \left(\frac{2}{5} - \frac{1}{5}\right)^2} = \frac{1}{\sqrt{5}}.$$
Therefore, the area of the square is the square of this length, or $\frac{1}{5}$.
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