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$ABCD$ is a square. $|AF|=6$, $|FK|=2$, and $DE \parallel AB$. What is $|EK|=?$

My geometry book has a property for this:

$$|AF|^2=|FK|\cdot|FE|$$

Can you show me where does this property come from in simple terms?

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$\triangle AFD$ 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.

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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 $$FC^2=FK\cdot FC.$$

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    $\begingroup$ $CF$ is tangent, maybe? $\endgroup$ – Oscar Lanzi Jan 11 at 12:01
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    $\begingroup$ @Oscar Lanz It was typo. Thank you! $\endgroup$ – Michael Rozenberg Jan 11 at 13:13

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