Let complex numbers $a,b,c,d$ lying on unit circle centered at $0$ be our vertices. Concyclicity of $ABEK$ gives $$\overline{k}=\frac{(d-c)k+bc-ad}{(bd-ac)k+ab(c-d)}$$With second concylicity we get $$k=\frac{(c+d)ab-(a+b)cd}{ac+bd-2cd}$$This number exists: if $ac+bd=2cd$ then $\frac{1}{ac}+\frac{1}{bd}=\frac{2}{cd}$ which yields $cd=ab$ and that's a contradiction with parmenides51 wrote: $F$ is the intersection point of the lines $AB$ and $CD$. We have $$\frac{k-m}{k-n}\cdot\frac{f-n}{f-m}=\frac{(a-b)(bd-ac)}{2(ab+cd)(c+d)-ac^2-bd^2-3cd(a+b)}\cdot\frac{2cd(a+b)-(c+d)(ab+cd)}{2ab(c+d)-(ab+cd)(a+b)}$$which is a real number.

Synthetic solution

Attachments:

Let $O$ be the circumcentre. Then, note that $F, O, M, N$ are concyclic. We prove $E$ lies on the radical axis of $(FMN)$ and $(ABCD)$. To do this, note that the intersections of $(FMN)$ and $(ABCD)$ are points from where tangents to $(ABCD)$ pass through $F$. So, the radical axis is the polar of $F$ in $(ABCD)$, so $E$ lies on the radical axis. Also if $N'$ is intersection of $(FMN)$ and $(NE)$, define $M'$ similarly, ($M'$ lies on $(ABCD)$ by the previous claim), $N'E\cdot EN = ME\cdot M'E = AE\cdot CE$, so $N'$ lies on $(ANC)$. Consider the inversion about $E$ preserving $(ABCD)$, then $N$ goes to $N'$, but $C$ goes to $A$ and $D$ goes to $B$, so $N'$ lies on $(ABE)$, so we are done (Since $K = N'$).