Let the circumcenter of $\triangle ALE$ and $\triangle ADK$ intersect at $A,O$ $\because \angle ACL=\angle ABK;CL=AB;BK=AC$ $\therefore \triangle ALC\cong \triangle KAB \Rightarrow\angle AOE=\pi -\angle ALE=\pi-\angle KAB$ $\therefore \angle EOD=2\pi -\angle AOE-\angle AOD=2\pi-(\pi-\angle KAB)-(\pi-\angle AKD)=\pi=\angle ABK=\angle EOD$ It means that point $O$ lies on $\omega$. And $\angle AOL=\angle AEL=\angle BEC=\angle BDC=\angle ADK=\angle AOK$ So the radium of $\odot ALO$ ia equal to the radium of $\odot AOK$ From the symmetry of the two circles, we know that$LO=OK$ And $\angle ALO=\angle AEO=\angle OCB=\pi-\angle OEB=\angle AED=\angle ALO$ $\therefore OL=OA$ So we are done There are another two conclusions : 1.$O$ is the midpoint of arc $BEDC$ 2.A fixed point is on the circumcircle of $\triangle ALK$

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