My solution: Let $\angle BAC=d$, $\angle DAC=a$, $\angle ABD=b$, $\angle ACB=c$, $\angle CDK=x$ and $\angle AFE=y$. From Law of Sine on $\triangle CDK$ we have $\frac{sin(d-b-x)}{sin(x)}=\frac{CD}{CL}=\frac{sin(a)}{sin(c)}$. From here we have $ctg(x)=\frac{\frac{sin(a)}{sin(c)}-cos(b-d)}{sin(b-d)}$. From Law of Sine on $\triangle AFE$ and $\triangle DFE$ we have that $\frac{sin(d-b-y)}{sin(y)}=\frac{DE}{AE}=\frac{sin(a)}{sin(c)}$(since $\angle EDF=\angle EAF$. From here we have $ctg(y)=\frac{\frac{sin(a)}{sin(c)}-cos(b-d)}{sin(b-d)}$. So we are done.

My solution: Lemma: If $\theta+\alpha=\beta+\omega$ and $\sin\theta.\sin\omega=\sin\beta.\sin\alpha$ and $0^{\circ}<\alpha,\beta,\theta,\omega<180^{\circ}$ $\Longrightarrow$ $\theta=\beta$ and $\alpha=\omega$ Proof: (It's well-known) In the problem: $\angle BAC=\angle BKC=\angle BDC$(since $ABCD$ is cyclic and $BACK$ is parallelogram) and $\angle CBD=\angle CAD$ and $\angle KBC=\angle BCA=\angle BDA$ $\Longrightarrow$ By Ceva's theorem in $\triangle AFD$ we get: $\sin\angle EAD.\sin\angle EFA=\sin\angle EFD.\sin\angle EDA...(1)$ and By Ceva's theorem in $\triangle DBK$ we get: $\sin\angle KBC.\sin\angle CKD=\sin\angle DBC.\sin\angle CDK...(2)$ $\Longrightarrow$ By $...(1)$ and $...(2)$ we get: $\sin\angle AFE.\sin\angle CKD=\sin\angle EFC.\sin\angle CDK$ and $\angle AFE+\angle EFC=\angle CDK+\angle CKD$ $\Longrightarrow$ By lemma we get: $\angle EFC=\angle CKD$ and $\angle AFE=\angle CDK$...

Let $ M $ be the Miquel point of the complete quadrilateral $ ABDC $. From the converse of Pivot theorem we get $ M $ $ \in $ $ EF $. Since $ M $ is the center of the spiral similarity of $ AB $ $ \mapsto $ $ CD $, so $ \triangle MAB $ $ \sim $ $ \triangle MCD $ $ \Longrightarrow $ $ \frac{CD}{CK} $ $ = $ $ \frac{CD}{AB} $ $ = $ $ \frac{MD}{MB} $, hence combine $ \measuredangle DCK $ $ = $ $ \measuredangle CFA $ $ = $ $ \measuredangle DMB $ we conclude that $ \triangle CDK $ $ \sim $ $ \triangle MDB $ $ \Longrightarrow $ $ \measuredangle AFE $ $ = $ $ \measuredangle BDM $ $ = $ $ \measuredangle KDC $.

This is a subproblem of I hid it to not spoil others

Let $AC$ and $DK$ intersect each other in point $G$. We need to find some similar triangles. Draw point $H$ such that $DH \parallel CK \parallel AB$, because $ABKC$ is parallelogram. Now, triangle $DHC$ is similar with triangle $FDB$ (because $\angle DCH=\angle DBA$-same arc, $\angle DFB=\angle HDC$). We now have this: $\frac{HG}{GC}=\frac{HD}{CK}=\frac{HD}{AB}=\frac{DE}{EB}$ $\implies$ triangle $GDC$ is similar with triangle $FEB$ $\implies$ $\angle CDK=\angle AFE$. That is all