<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] I hope that the following makes sense. I would have been better if you had labelled points on the figure.
The side length of the square is $1$.
The area of one of the large triangles is $A_{\text{large}} = 1\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$.
The hypotenuse of the large triangle is $\sqrt{1^2 + \frac{1}{4}} = \sqrt{5}/2$.
One of the smaller triangles is congruent to the the larger triangle and it has hypotenuse $1/2$. The side lengths of the smaller triangle is therefore
$$
a=  (\frac{1}{2} /\frac{\sqrt{5}}{2}) \cdot 1 = \frac{1}{\sqrt{5}} \\
b = (\frac{1}{2} /\frac{\sqrt{5}}{2}) \cdot \frac{1}{2} = \frac{1}{2\sqrt{5}}.
$$
So the area of a smaller triangle is 
$$
A_{\text{small}} = \frac{1}{20}.
$$
Now you can probably figure out the rest. I think maybe I get the area of the smaller square to be $\frac{4}{20}$.
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