Where does this property involving quadrilaterals come from?

$\triangle AFD$ is similar to$\triangle KFB$ so $|AF|/|KF|=|FD|/|FB|$.

$\triangle AFB$ is similar to$\triangle EFD$ so $|EF|/|AF|=|FD|/|FB|$.

So $|EF|/|AF|=|AF|/|KF|$ proving the claim.

You don't need a square. All you need is a parallelogram, the parallel sides enable the angle congruence that make the relevant triangles similar.

Prove that $FC$ is a tangent line to the circumcircle of $\Delta KCE$, for which prove that $$\measuredangle CEK=\measuredangle FAB=\measuredangle FCK.$$ After this use $AF=CF$ and $$FC^2=FK\cdot FC.$$