A geometry problem by Wildan Bagus Wicaksono

Let $x$ be the side length of the square.

By the pythagorean theorem , $AC=\sqrt{2}x$

$FC=AC-x=\sqrt{2}x-x=x(\sqrt{2}-1)$

Since $AE=FC$ ,

$EF=AC-2(FC)$

$8\left (2-\sqrt{2}\right)=\sqrt{2}x-2x(\sqrt{2}-1)$

$8\left (2-\sqrt{2}\right)=\sqrt{2}x-2x\sqrt{2}+2x$

$8\left (2-\sqrt{2}\right)=2x-\sqrt{2}x$

$8\left (2-\sqrt{2}\right)=x\left (2-\sqrt{2}\right)$

$x=8$

Finally the area of the square is $x^2=8^2=$ $\color{#D61F06}\large \boxed{64}$

Length of $AE=FC=x\sqrt { 2 } -x$ .

$AC = AE + EF + FC$

$x\sqrt { 2 } =\left( x\sqrt { 2 } -x \right) +\left( 16-8\sqrt { 2 } \right) +\left( x\sqrt { 2 } -x \right) \\ 2x-x\sqrt { 2 } =16-8\sqrt { 2 } \\ x\left( 2-\sqrt { 2 } \right) =8\left( 2-\sqrt { 2 } \right) \\ x=8$

Thus, the area of the square $ABCD$ is $64$ .