Nice solution by Burak Buğra Önder Let $\angle KED=\angle KAD=\beta$. We have $\angle BCD=\angle BAD=\angle KAD=\beta$. Then $L,E,C,D$ is cyclic. Let $\angle LDE=\angle LCE=\alpha$. Let $AD\cap BE=X$. We have $\angle EXD=\angle AXB=180^\circ-\angle BAX-\angle ABX=180^\circ-\beta-\angle BCE=180^\circ-\beta-\alpha$. Also, $\angle ECD=\alpha+\beta$. Then $E,C,D,X$ is cyclic. Then $X,L,D,C,E$ is cyclic. Apply Pascal theorem to $EECLDX$. $EC\cap DX=M$ and $CL\cap XE=B$. Then $EE\cap LD,M,B$ are collinear. Then $EE$ pass through $LD\cap MB=N$. Hence, $EN$ is a tangent. Thus, $\angle NEL=LDE$. This completes the proof.

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Basically same as @above Note that $\angle LED = \angle KED = \angle KAD = \angle BAD = \angle BCD = \angle LCD \implies CDLE$ is cyclic. Let $F = AD\cap BE$. Now note that $\angle ABF = \angle ABE = \angle ECL = \angle EDL$ and $\angle BAF = \angle LED$ so this gives $\triangle ABF \sim \triangle EDL$. And so $\angle ELD = \angle AFB = \angle EFD \implies EFLD$ is cyclic. And so $CEFLD$ is cyclic. Now apply Pascal to $EEFDLC$ to get $EE\cap DL , B , M$ are collinear and so $EE\cap DL = N$, implying that $\angle NEL = \angle LDE = \angle NDE$, as desired. $\square$

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$$T = AD\cap BE$$$\angle LED=\angle KAD=\angle BCD=\angle LCD$ $L,E,C,D$ Concyclic. $\angle BTD=\angle BAD+\angle ABE=\angle BCD+\angle BCE=\angle DCE$ $D,L,E,C,T$ concyclic.$EETDLC$ Pacsal theorem. $EN$ Tangent $(DLE)$. $ \angle NEL= \angle TEL + \angle NET = \angle TDL + \angle TDE = \angle LDE = \angle NDE $

very nice problem.