Really liked this problem First some new points: Let $AD$ and $AE$ intersect the circumcircle again at $M$ and $N$. Let $S$ be the center of $\omega$ which lies on $\Gamma$ Let $T$ be the other intersection of $\omega$ with $\Gamma$. $EA’$ intersects $AT$ at $R$. Now because $K$ and $L$ are interchangeable, we standardize their names (and by this removing the $or$ part of the problem) start from $A$ in $\omega$ and turn clockwise call the first intersection you reach with $EA’$, $L$ and the second $K$. Now we solve the problem: First, $M$ and $S$ are diametrically opposite in $\Gamma$ Now notice that $$\widehat{REC} = \widehat{BEA’} = \widehat{BEA} = \widehat{ANM} = \widehat{ASM}= \widehat{TSM}= \widehat{TAM}= \widehat{TAD}$$So $ARED$ is cyclic. Thus $\widehat{ARD}= \widehat{AED}$ also having $\widehat{AED}= \widehat{ASM} = \widehat{ATM}$ we get $RD$ and $TM$ are parallel so $\frac{AD}{DM}=\frac{AR}{RT}$ (*). Also see that $\widehat{ADB} = \widehat{ARK}$ (**) Now using the angle equalities we have reach it is easy to see that the spiral similarity with center $A$ that takes $\omega$ to $\Gamma$ also takes $T$ to $M$ and now using (*) and (**) we get that said spiral similarity takes $R,K,L$ to $D,B,C$ respectively so the triangles $ABC$ and $AKL$ are similar. We now claim that $T,K,B$ and $C,T,L$ are collinear. Which is easy because $\widehat{ATK}= \widehat{ALK}= \widehat{ACB}$ so $KT$ passes through $B$. Similarly for $C$. So $BK$ and $CL$ concur on $T$ which is on $\Gamma$. This concludes the proof.

Now using the angle equalities we have reach it is easy to see that the spiral similarity with center $A$ that takes $\omega$ to $\Gamma$ also takes $T$ to $M$ and now using (*) and (**) we get that said spiral similarity takes $R,K,L$ to $D,B,C$ respectively so the triangles $ABC$ and $AKL$ are similar. why?Could you please tell me more detail?

Of course $\widehat{TAM}= \widehat{AED}= \widehat{ANM}$ so the arcs $AM$ in $\Gamma$ and $AT$ in $\omega$ are equal so the spiral similarity takes $T$ to $M$. Now using (*) we have that said spiral similarity takes $R$ to $D$. Now let the spiral take $K$ to $B’$ and $L$ to $C’$, we have $\widehat{B’DA}= \widehat{KRA}= \widehat{BDA}$ (this is due to (**)) and because how we named them at the start, $B$ and $B’$ are on the same side of $AM$ (Note that they are both on $\Gamma$) so they are the same point. Similarly for $C$ So now we get that the spiral similarity takes $R,K,L$ to $D,B,C$. Knowing that $A$ is the center of the spiral, we conclude that the spiral takes $AKL$ to $ABC$ so they are similar.

Another solution: First of all, assume without loss of generality that $E$ lies to the same semiplane with respect to $AA'$ as $C$ (to avoid configuration issues. This way, $BL$ and $CK$ meet on $\Gamma$. The other case is analogous.) Let $O'$ be the circumcenter of $\omega$ and $L'$ be the intersection of the circumcircle of $\triangle{AEB}$ and $A'E$. We will prove that $L\equiv{L'}$. It suffices to show that $O'A=O'L'$. Note that $\angle{AL'B}=\angle{AEB}$ and $\angle{BAL'}=\angle{BEA'}=\angle{AEB}$, so $\angle{AL'B}=\angle{BAL'} \Longleftrightarrow BA=BL'$ (*). Plus, $\angle{DAL'}=\angle{BAL'}-\angle{BAD}=\angle{AEB}-\angle{CAE}=\angle{ACB}$. Since $O'A\perp{DA}$, we have $\angle{OAL'}=90^\circ-\angle{ACB}$. $AO'CB$ is cyclic, so $\angle{AO'B}=\angle{ACB}$. Hence, $BO'\perp{AL'}$ and because of (*) we have that $BO'$ is the perpendicular bisector of $AL'$ and finally, $O'A=O'L'$ as desired. Similarly, one can prove that the intersection of the circumcircle of $\triangle{AEC}$ and $A'E$ is $K$. Now, let $BL$ intersect $\Gamma$ at $M$. We have: $\angle{ACM}=\angle{ABM}=\angle{ABL}=\angle{AEL}=\angle{AEK}=\angle{ACK}$ which means that $C$, $M$, $K$ are collinear.

Attachments:

Let $H$ be the foot of the perpendicular from $A$ to $BC$.Let $L$ be the reflection of $A$ over $BO$ and let $AL$ cut $BO$ at $M$.Let $N$ be the midpoint of $AE$.Assume $D$ lies between $B$ and $E$.Then $$\angle ABO= 90-\angle C-\angle BAD =90-\angle C-\angle CAE =\angle EAH =\angle AHN$$Also , since $(ABHM)$ is cyclic, we have that $$\angle AHM =\angle ABO =\angle AHN \implies H-M-N$$Then , the homothety at $A$ with ratio $2$ gives that $A'-L-E$ , and also we have that $OL=OA$ , such $L$ is the mentioned point in the problem.Thus , if $P$ denotes the $A-$ antipode of $\omega$ , then $\angle ALP=90 \implies BO\parallel LP$. Then , if $F=\Gamma\cap \omega, $ , then converse of Reim's gives that $B=-L-F$.Similarly , $C-K-F$.Thus we are done

Let $L'$ be reflection of $A$ across $BO$. Let $AL'$ meet $BO$ at $S$. Let $AA'$ meet $BC$ at $H$. $\angle ASB = \angle 90 = \angle AHB \implies ABHS$ is cyclic. $\angle AA'E = \angle A'AE = \angle 90 - \angle C - \angle CAE = \angle OCA = \angle OBA = \angle SHA$ and we have $AS = SL'$ and $AH = HA'$ so $A',E,L'$ are collinear so $L'$ is $L$. Let $\Gamma$ meet $\omega$ at $P$. $\angle AEL = 2\angle A'AL = 2\angle ABO = \angle ABL \implies ABEL$ is cyclic so $\angle B = \angle ALK = \angle APK \implies K,P,C$ are collinear. with same approach we used on $L$ for $K$ we have $K$ is reflection of $A$ across $CO$ so $\angle PBA = \angle PCA = \angle KCA = 2\angle KA'A = \angle LEA = \angle LBA \implies P,L,B$ are collinear so Now we have $BL$ and $CK$ meet at $P$ on $\Gamma$.

Let $(ABE) \cap A'E=L'$ and $(ACE) \cap A'E=K'$ now let $BL' \cap (ABC)=G$ and now by angle chase $\angle AGL=\angle ACE=\angle AKL$ which means that $AKGL$ cyclic, similarily u can do the same with $CK' \cap (ABC)=G'$ and get that $G'=G$ so $BL',CK'$ meet at the intersection of $(AL'K')$ and $(ABC)$, now let $AC \cap (AK'GL')=J$ and by angle chase $\angle AK'G=\angle AEB=\angle K'EC=\angle K'AJ$ which means that arcs $AG,JK'$ in $(AK'GL')$ are equal but that means that arcs $AJ, GK'$ are equal in $(AK'GL')$ so by more angle chase $\angle DAJ=\angle BAE=\angle K'L'G=\angle AK'J$ so $(AK'GL'J)$ is tangent to $AD$ and now since $AK'GJ$ is an ISLsusceles Trapezoid we have that $\triangle ACK'$ is isosceles so let $O_1$ the center of $(AK'GL'J)$ and by angle chasing $\angle AO_1G+\angle ACG=2\angle AK'G+180-2\angle AK'G=180$ which says that $O_1$ lies in $(ABC)$ so $K'=K$ and $L'=L$ thus we are done

Really fun problem, not too easy not too hard. This is the diagram we shall we referring to. [asy][asy] import geometry; size(10cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; path rect; rect= (2,-1.5)--(2,1)--(-2,1)--(-2,-1.5)--cycle; pair A,B,C,D,E,A_1,B1,L,K,R; C=dir(60); A=dir(210); B=dir(330); D = midpoint(B--C); line AD= line(A,D); line l =sector(2,line(A,B),line(A,C)); transform isog = reflect(l); transform reflectacrossBC = reflect(line(B,C)); line AE = isog*l; E = intersectionpoint(AE,line(B,C)); A_1=reflectacrossBC*A; line perpA = perpendicular(A,AD); B1= intersectionpoints(circle(A,B,C),perpA)[1]; L = intersectionpoints(circle(A,E,B),line(A_1,E))[0]; K = intersectionpoints(circle(A,E,C),line(A_1,E))[0]; R = intersectionpoints(circle(A,B,C),line(C,K))[0]; filldraw((path)(A--B--C--cycle), white+0.1*pri, pri); filldraw(circumcircle(A,B,C), tfil, tri); draw(A--D); draw(A--E); filldraw(circumcircle(A,E,C), tfil, dashed+ pri); filldraw(circumcircle(E,B,A), tfil, dashed+ pri); filldraw(circumcircle(K,L,A), tfil, sec); draw(K--C); draw(B--R); draw(K--A_1,dashed+sec); dot("$A$",A,dir(230)); dot("$B$",B,dir(300)); dot("$C$",C,dir(90)); dot("$D$",D,dir(330)); dot("$E$",E,dir(160)); dot("$A_1$",A_1,dir(90)); dot("$B'$",B1,dir(90)); dot("$L$",L,dir(90)); dot("$K$",K,dir(90)); dot("$R$",R,dir(90)); [/asy][/asy] Let $B'$ be the center of $\omega$. Let $R$ be the second intersection of circles $\omega$ and $\Gamma$. We solve the case where $B'$ lies on the same side of $AB$ as $C$. Here, lines $CK$ and $BL$ intersect at $R$. The other case is entirely similar where $CL$ and $BK$ intersect at $R$. To prove this, we do some phantom pointing. Then, let $L' = \overline{BR} \cap \omega$ and $K'=\overline{CR} \cap \omega$. We start off with a couple of observations. Claim : Quadrilaterals $ABEL'$ and $AECK'$ are cyclic. Proof : This is easy to see since, \[\measuredangle ECA = \measuredangle BCA = \measuredangle BRA = \measuredangle L'RA = \measuredangle L'K'A = \measuredangle EK'A\]from which it is clear that $AECK'$ is cyclic. The other is entirely similar. From this it then follows that points $K'$, $L$ and $E$ are clearly collinear since, \[\measuredangle ELA = \measuredangle EBA = \measuredangle CBA = \measuredangle K'RA = \measuredangle K'LA\]Now we are almost there. Note that, \[\measuredangle AEK' = \measuredangle ACK = \measuredangle ACR = \measuredangle ABR = \measuredangle AB'R = 2\measuredangle ARB'\]Further, \[\measuredangle ARB' = \measuredangle BRB' + \measuredangle ARB = \measuredangle BAB' + \measuredangle ACB\]Thus, \[\measuredangle AEK' = 2\measuredangle ARB' = 2(\measuredangle BAB' + \measuredangle ACB) = 2(\measuredangle BAD + \measuredangle ACB) \]We now finish off by noting that \[\measuredangle BAD + \measuredangle ACB = \measuredangle BAC + \measuredangle CAD + \measuredangle ACB = \measuredangle EAB + \measuredangle ABC\]Thus, we have that $\measuredangle AEK' = 2(\measuredangle EAB + \measuredangle ABC)$. But it is also quite clear that, \[\measuredangle EAA_1 = \measuredangle EAB + \measuredangle BAA_1 = \measuredangle EAB + 90 + \measuredangle ABC\]which implies that \[2\measuredangle EAA = 2(\measuredangle EAB + 90 + \measuredangle ABC) = 2(\measuredangle EAB + \measuredangle ABC) = \measuredangle AEK' \]which implies that $A_1$ also lies on $\overline{EK'}$ which implies that in fact $K'$ and $L'$ are the intersections of line $A_1E$ with $\omega$. Thus, $K'=K$ and $L'=L$ which finishes the problem.