As $\overline{BD}$ and $\overline{CE}$ are diameters of the circumference, we have $\measuredangle EAC = 90^o$ such as $\measuredangle BAD=90^o \Rightarrow \measuredangle DAL=90^o$. Besides that $\overline{AE} = \overline{AK}$ and $\overline{AD} = \overline{AL}$, so triangles $AEK$ and $DAL$ are both rectangle and isosceles $\Rightarrow$ There’s a spiral similarity that takes $AEK$ to $DAL$ and then the angle between the lines $EK$ and $DL$ is the angle of the spiral similarity. Now if $F=EK \cap DL$, then $\measuredangle EFD= \measuredangle EAD\Rightarrow F\in (ABC) _\blacksquare$

baby_shark wrote: As $\overline{BD}$ and $\overline{CE}$ are diameters of the circumference, we have $\measuredangle EAC = 90^o$ such as $\measuredangle BAD=90^o \Rightarrow \measuredangle DAL=90^o$. Besides that $\overline{AE} = \overline{AK}$ and $\overline{AD} = \overline{AL}$, so triangles $AEK$ and $DAL$ are both rectangle and isosceles $\Rightarrow$ There’s a spiral similarity that takes $AEK$ to $DAL$ and then the angle between the lines $EK$ and $DL$ is the angle of the spiral similarity. Now if $F=EK \cap DL$, then $\measuredangle EFD= \measuredangle EAD\Rightarrow F\in (ABC) _\blacksquare$ After using spiral similarity you can find that $DFE$ + $EAD$ is 180

Without spiral similarity?

Pick a phantom point $F$ on $c$, such that $OF \perp AO$. Now show that $F,K,E$ and $L,F,D$ are collinear, which comes from easy angle chase.

My solution... (really easy to observe ) Let LD and EK intersect at T. See that the angles ALD, ADL, AEK, AKE are all equal, which gives us that AKTL is circumscribed. Finally, we will show that EBDT is also circumscribed. But this is obvious because of the fact that OBC=OCB=EAB=DAC.