Let: $\angle BCE = \angle BAF = \alpha$ and $\angle FAD = \angle ECD = \beta$ $\Longrightarrow \angle CED = \angle AFD = \alpha$ Let $DL=a$, $LF=b$ and $FC=c$ $\Longrightarrow a+b=b+c$ then $a=c, DL=FC$, analogously $AE=KD$ Let $AE=KD=a$, $EK=r$, $DL=FC=b$ and $LF=k$ note that $\triangle AFD \sim \triangle CED$ $\Longrightarrow \frac{2a+r}{b+k} = \frac{2b+k}{a+r} ...(1)$ but by (1), we obtain: $\triangle ABK \sim \triangle CBL$ $\Longrightarrow \angle ABK = \angle LBC$ thus $\angle BKD = \angle BLD$

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$\angle BAF = \angle BCE \implies \angle EAF = \angle ECF \implies AEFC$ is cyclic. $\angle DAF = \angle ECD , \angle ADF = \angle CDE \implies ADF$ and $CDE$ are similar. $\angle BAK = \angle BCL$ and $\frac{BA}{BC} = \frac{CD}{AD} = \frac{ED}{DF} = \frac{AK}{CL}$ so $BKA$ and $BLC$ are similar so $\angle BKD = \angle BLD$.