$ABCD$ is a square. $|AF|=6$, $|FK|=2$, and $DE \parallel AB$. What is $|EK|=?$
My geometry book has a property for this:
Can you show me where does this property come from in simple terms?
$\triangle AFD$ 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 $$FC^2=FK\cdot FC.$$