Since $CL=BA,$ then $\omega \equiv \odot(EBC)$ and $\odot(AEL)$ meet again at the center $M$ of the rotation that swaps $\overline{CL}$ and $\overline{BA}.$ Thus, it follows that $MB=MC$ ($M$ is the midpoint of the arc $BEDC$) and $MA=ML.$ Similarly $MK=MA$ $\Longrightarrow$ $MA=MK=ML$ $\Longrightarrow$ $M$ is the circumcenter of $\triangle AKL.$

My solution: Let $O$ be circumcentre of $(AKL)$. We have: $\Delta BAK=\Delta CLA$, so: $AK=AL$ and: $\widehat{BAK}=\widehat{CLA}$, $\widehat{AKB}=\widehat{ALC}$ Since: $\widehat{AEL}=180^\circ-\widehat{BAC}-\widehat{ACE}=\widehat{LAK}$ (easy by angle-chasing) $\widehat{AOL}=2\widehat{AKL}=180^\circ-\widehat{LAK}$ Hence, $A,O,L,E$ are concyclics, likewise, $A,O,D,K$ are concyclics. We have that: $\widehat{EOD}=360^\circ-\widehat{AOE}-\widehat{AOD}=360^\circ-\widehat{ALE}-\widehat{AOD}=\widehat{ALC}+\widehat{AKB}=\widehat{ALC}+\widehat{LAC}=180^\circ-\widehat{DCE}$. This implies $O$ lies on $\omega$, as desired.

Attachments:

Luis González wrote: Since $CL=BA,$ then $\omega \equiv \odot(EBC)$ and $\odot(AEL)$ meet again at the center $M$ of the rotation that swaps $\overline{CL}$ and $\overline{BA}. How do we know this?

My solution Let $O=w\cap (\triangle ADK)$ ,then $\triangle OAK \sim \triangle OCB$ $\Rightarrow$ $OA/OC=OK/OB$ $\Rightarrow$ $OAC \sim OKB$,since $AC=KB$ then $OA=OK$ and $OB=OC$. Similarly define $O^{i} = (\triangle AEL) \cap w$ $\Rightarrow$ $O=O^{i}$ ,That means $OA=OK=OL$

After guessing the center of the $\odot (AKL ) $ problem becomes fairly easy. My solution, Assume that $M $is the midpoint of arc $BEDC $. Then $\widehat{DBM} =\widehat{DCM}$, $BL=AC $ and $BM=MC $. $\implies \triangle AMC \equiv \triangle LMB \implies LM=AM $ By combining them we get $ML=MA=MK $. Hence $M $ is the circumcenter of the $\triangle AKL $. Similarly since $\widehat{MBE}=\widehat {MCE} $, $AB=KC $ and $MB=MC$ $\implies \triangle AMB \equiv \triangle KMC \implies MA=MK $ By combining them we get that $MA=MK=ML $ hence $M $ is the circumcenter of the $\triangle AKL $. Hence circumcenter of the $\triangle AKL $ lis on $\odot (O) $

Konigsberg wrote: Luis González wrote: Since $CL=BA,$ then $\omega \equiv \odot(EBC)$ and $\odot(AEL)$ meet again at the center $M$ of the rotation that swaps $\overline{CL}$ and $\overline{BA}$. How do we know this? I will explain it in different notation. Let $\sigma$ be a rotation about center $O$ with rotational angle $\varphi$. Let $\sigma$ map a line segment $g=XY$ onto $g'=X' Y'$. Then it is a well known property of a rotation $\sigma$ that $\varphi=\angle XOX' =\angle YOY'=\angle gg'$ because each line $g$ and its image $g'=g\sigma$ will enclose the rotational angle $\varphi$. Thus $XX'OS$ as well as $YY'OS$ are concyclic with $S:=g\cap g'$. Luis González used this property to state that $\odot(XX'S)$ and $\odot(YY'S)$ intersect in the center of rotation $O$.