# Where does this property involving quadrilaterals come from?

$$ABCD$$ 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?

$$\triangle AFD$$ 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.
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 $$FC^2=FK\cdot FC.$$
• $CF$ is tangent, maybe? – Oscar Lanzi Jan 11 at 12:01