Let $X \equiv MN \cap CD$ and $P \equiv AC \cap BD.$ By Menelaus' theorem for $\triangle ACD$ and $\triangle BCD$ cut by $MN,$ we obtain $\tfrac{KA}{KC}=\tfrac{LD}{LB}=-\tfrac{XD}{XC}$ $\Longrightarrow$ $\tfrac{KA \cdot KP}{KP \cdot KC}=\tfrac{LD \cdot LP}{LP \cdot LB}$ $\Longrightarrow$ powers of $K$ and $L$ WRT $\odot(PAD)$ and $\odot(PBC)$ are in the same ratio $\Longrightarrow$ $\odot(PAD),$ $\odot(PBC)$ and $\odot(PKL)$ are coaxal, i.e. if $Q$ is the second intersection of $\odot(PAD)$ and $\odot(PBC),$ then $PKQL$ is cyclic. Therefore $Q$ is the Miquel point of $MN$ WRT $\triangle PAD$ $\Longrightarrow$ $Q \in \odot(AMK)$ and likewise $Q \in \odot(BNL).$ Now by Miquel theorem in $\triangle PAB,$ it follows that the 2nd intersection of $\odot(AMKQ)$ and $\odot(BNLQ)$ lies on $AB.$ P.S. The property still holds for all points $M,N$ on $AD,BC$ verifying $\tfrac{MA}{MD}=\tfrac{NC}{NB}.$ The proof is exactly the same.

Let me add another quick solution I think which also works on the general case posted by Luis! Note also that the problem is similar to: http://artofproblemsolving.com/community/c6h84559p490691 Let the $AC\cap BD\equiv O$ and $BC\cap AD\equiv S$ Then the center $T$ of spiral similarity sending $A\mapsto C$ and $D\mapsto B$ is the second point of intersection of $\odot (AOD), \odot (BOC)$ But this is the same sending $A\mapsto C$ and $M\mapsto N$ and the same sending $M\mapsto N$ and $D\mapsto B$ so $T,K,M,A$ and $T,L,N,B$ are cocyclic and this means that $T$ is the Miquel point of the complete quadrilateral $SCANM,$ so $SNTM$ is cyclic, so now by Miquel's pivot theorem in triangle $SAB$ we have the desired. [asy][asy] size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.906622983754532, xmax = 25.19435318221192, ymin = -9.961874542045134, ymax = 9.315467412733959; /* image dimensions */ pen ccqqqq = rgb(0.8,0.,0.); /* draw figures */ draw((-2.1878044470715396,6.550740570496484)--(8.151516483487773,2.3089678810362515)); draw((8.151516483487773,2.3089678810362515)--(12.759664485425345,-4.150158239143364)); draw((12.759664485425345,-4.150158239143364)--(-4.032054729327774,-3.850717430101127)); draw((-4.032054729327774,-3.850717430101127)--(-2.1878044470715396,6.550740570496484)); draw((-2.1878044470715396,6.550740570496484)--(12.759664485425345,-4.150158239143364)); draw((8.151516483487773,2.3089678810362515)--(-4.032054729327774,-3.850717430101127)); draw((2.981856018208117,4.429854225766368)--(4.363804878048786,-4.000437834622246)); draw(circle((4.285595556339402,-8.386175952017743), 9.473842424205992), dotted); draw(circle((4.0897416332288925,7.1303254123795075), 6.304244869866325), dotted); draw((2.770308337468983,0.9657005935218326)--(-4.032054729327774,-3.850717430101127)); draw((2.770308337468983,0.9657005935218326)--(3.6985717072242785,0.05767898958177634)); draw((2.770308337468983,0.9657005935218326)--(4.363804878048786,-4.000437834622246)); draw((-2.1878044470715396,6.550740570496484)--(2.770308337468983,0.9657005935218326)); draw((2.770308337468983,0.9657005935218326)--(8.151516483487773,2.3089678810362515)); draw((2.770308337468983,0.9657005935218326)--(12.759664485425345,-4.150158239143364)); draw((2.770308337468983,0.9657005935218326)--(2.981856018208117,4.429854225766368)); draw(circle((14.492725326566756,1.9883764485512387), 11.766942082408164)); draw(circle((-0.6600317978286934,2.9137195386012835), 3.944871513018204), linewidth(2.8) + ccqqqq); draw(circle((0.1894933877272876,-2.6011352904860123), 4.402604232526042), linewidth(2.8) + ccqqqq); draw((12.759664485425345,-4.150158239143364)--(24.401680836436284,-4.357766212481088)); draw((8.151516483487773,2.3089678810362515)--(24.401680836436284,-4.357766212481088)); /* dots and labels */ dot((-2.1878044470715396,6.550740570496484),dotstyle); label("$B$", (-2.8316449445788967,6.702232452262923), NE * labelscalefactor); dot((8.151516483487773,2.3089678810362515),dotstyle); label("$C$", (8.681738069670304,2.1574759992698165), NE * labelscalefactor); dot((12.759664485425345,-4.150158239143364),dotstyle); label("$D$", (13.112875611338582,-4.6217857097782336), NE * labelscalefactor); dot((-4.032054729327774,-3.850717430101127),dotstyle); label("$A$", (-4.763166437100967,-4.091564123595705), NE * labelscalefactor); dot((2.981856018208117,4.429854225766368),dotstyle); label("$N$", (3.152284385195359,4.657092048416025), NE * labelscalefactor); dot((4.363804878048786,-4.000437834622246),dotstyle); label("$M$", (4.51571132109329,-3.788580360062831), NE * labelscalefactor); dot((3.6985717072242785,0.05767898958177634),dotstyle); label("$K$", (4.06123567579398,-0.4178859907596104), NE * labelscalefactor); dot((3.27531080367777,2.639694242340432),dotstyle); label("$L$", (3.417395178286623,2.877062437660392), NE * labelscalefactor); dot((-0.7320875677395589,14.76089410920631),dotstyle); dot((5.564353253060242,1.0009678644880278),dotstyle); label("$O$", (5.424662611691912,0.2259545067477463), NE * labelscalefactor); dot((2.770308337468983,0.9657005935218326),dotstyle); label("$T$", (1.9782223015054732,0.6425571816054477), NE * labelscalefactor); dot((24.401680836436284,-4.357766212481088),dotstyle); label("$S$", (24.436893773379737,-5.038388384635935), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Hello! My solution. [asy][asy] import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.1457979222845776, xmax = 12.266476492244193, ymin = -3.2063481389724027, ymax = 7.21112291135303; /* image dimensions */ pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); draw((0.38,3.64)--(0.,0.)--(6.,0.)--(2.0907699367980856,4.506099179603401)--cycle, cqcqcq); /* draw figures */ draw((0.38,3.64)--(0.,0.), sqsqsq); draw((0.,0.)--(6.,0.), sqsqsq); draw((6.,0.)--(2.0907699367980856,4.506099179603401), sqsqsq); draw((2.0907699367980856,4.506099179603401)--(0.38,3.64), sqsqsq); draw((1.2353849683990428,4.073049589801701)--(3.,0.)); draw((0.38,3.64)--(6.,0.)); draw((0.,0.)--(2.0907699367980856,4.506099179603401)); draw(circle((1.5,1.3357495541274977), 2.008538491379194), linewidth(1.2) + linetype("2 2")); draw((0.3082288761040996,2.952508181628744)--(1.8298294582557977,2.700964550168843)); draw((0.3082288761040996,2.952508181628744)--(1.5513998576512036,3.34363025924552)); draw((0.3082288761040996,2.952508181628744)--(3.,0.)); draw((1.2353849683990428,4.073049589801701)--(0.3082288761040996,2.952508181628744)); /* dots and labels */ dot((0.38,3.64),linewidth(3.pt) + dotstyle); label("$A$", (0.04930944651363667,3.6982547664758494), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.26217639884493277,-0.22992783887942173), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (5.950235739139869,-0.3510612231855314), NE * labelscalefactor); dot((2.0907699367980856,4.506099179603401),linewidth(3.pt) + dotstyle); label("$D$", (2.160491287277274,4.615407533364966), NE * labelscalefactor); dot((1.2353849683990428,4.073049589801701),linewidth(3.pt) + dotstyle); label("$M$", (1.2606432895747401,4.269312149633223), NE * labelscalefactor); dot((3.,0.),linewidth(3.pt) + dotstyle); label("$N$", (3.0084249774200464,-0.4029755307452927), NE * labelscalefactor); dot((1.8298294582557977,2.700964550168843),linewidth(3.pt) + dotstyle); label("$K$", (1.9009197494784662,2.7984067687733205), NE * labelscalefactor); dot((1.5513998576512036,3.34363025924552),linewidth(3.pt) + dotstyle); label("$L$", (1.8143959035455302,3.4213784594904557), NE * labelscalefactor); dot((0.3082288761040996,2.952508181628744),linewidth(3.pt) + dotstyle); label("$P$", (0.04930944651363667,2.936844922266017), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $P\equiv AB\cap \odot (BLN)$.It suffices to show that $APKM$ is cyclic. We will show that $\triangle{MLP}\simeq \triangle{ABC}$.Since $\angle{MLP}=\angle{ABC}$ it suffices to show that $\frac{AB}{CB}=\frac{ML}{PL} \ (1)$. $\bullet$ The sine law gives $\frac{PL}{\sin \hat{PNL}}=\frac{NL}{\sin \hat{NPL}}\Rightarrow PL=\frac{NL\cdot \sin \hat{ABD}}{\sin \hat{CBD}}$. Thus,$(1)$ becomes $\frac{AB}{CB}=\frac{ML\cdot \sin \hat{CBD}}{NL\cdot \sin \hat{ABD}} \ (2)$. $\bullet$ The sine law gives $\frac{ML}{\sin \hat{MDL}}=\frac{MD}{\sin \hat{MLD}}\Rightarrow ML=\frac{MD\cdot \sin \hat{ADB}}{\sin \hat{MLD}}$ and $\frac{NL}{\sin \hat{NBL}}=\frac{BN}{\sin \hat{BLN}}\Rightarrow NL=\frac{BN\cdot \sin \hat{CBD}}{\sin \hat{BLN}}$. Thus,since $\angle{MLD}=\angle{BLN}$,$(2)$ becomes $\frac{AB}{CB}=\frac{MD\cdot \sin \hat{ADB}}{BN\cdot \sin \hat{CBD}}\cdot \frac{\sin \hat{CBD}}{\sin \hat{ABD}}\Leftrightarrow $ $\Leftrightarrow \frac{AB}{CB}=\frac{AD\cdot \sin \hat{ADB}}{BC\cdot \sin \hat{ABD}}\Leftrightarrow \frac{AB}{AD}=\frac{\sin \hat{ADB}}{\sin \hat{ABD}}$which is true from the sine law. Thus $\triangle{MLP}\simeq \triangle{ABC}$ as we wanted.This gives $\angle{PML}=\angle{BAC}\Leftrightarrow \angle{PMK}=\angle{PAK}$ which gives that $APMK$ is cyclic and we are done.

Let S be midpoint of AB.Circle MNS intersects AB second time in point W.Actually this point is common point of our circles(which can be proved by counting angles).

Misha57:It is an incrediblle solution!

Here is outline of another interesting approach. Let $Y=K \infty_{BC} \cap L \infty_{AD}$. It is well known that $\frac{AK}{KC}=\frac{DL}{LB}$, so $Y$ is on $AB$. Let $X$ be other intersection of $(KLY)$ with $AB$. Then from easy angle chase we are done.

Let \(X\) be the center of spiral similarity sending \(\overline{AMD}\) to \(\overline{CNB}\); then from \(\triangle XMA\sim\triangle XNC\), point \(K=\overline{AC}\cap\overline{MN}\) lies on \((XMA)\), and from \(\triangle XNB\sim\triangle XMD\), point \(L=\overline{BD}\cap\overline{MN}\) lies on \((XNB)\). [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair A,B,C,D,EE,M,NN,K,L,X,Y; A=(0.17479,-2.73); B=(5.07399,-2.73); C=(6.66487,3.17233); D=(-2.51,1.08); EE=extension(A,C,B,D); M=(A+D)/2; NN=(B+C)/2; K=extension(A,C,M,NN); L=extension(B,D,M,NN); X=reflect(circumcenter(EE,A,D),circumcenter(EE,B,C))*EE; Y=reflect(circumcenter(A,K,M),circumcenter(B,L,NN))*X; draw(circumcircle(X,D,M),gray+dashed); draw(circumcircle(X,C,NN),gray+dashed); draw(circumcircle(A,K,M)); draw(circumcircle(B,L,NN)); draw(A--C,gray); draw(B--D,gray); draw(A--B--C--D--cycle,linewidth(1)); dot("\(A\)",A,SW); dot("\(B\)",B,SE); dot("\(C\)",C,NE); dot("\(D\)",D,NW); dot("\(X\)",X,S); dot("\(Y\)",Y,S); dot("\(E\)",EE,N); dot("\(M\)",M,SW); dot("\(N\)",NN,E); dot("\(K\)",K,NE); dot("\(L\)",L,NW); [/asy][/asy] Finally if the circles \((XMA)\) and \((XNB)\) intersect again at \(Y\), then again by \(\triangle XMA\sim\triangle XNC\) we have \[\measuredangle XYA=\measuredangle XMA=\measuredangle XNC=\measuredangle XNB=\measuredangle XYB,\]so \(Y\) lies on \(\overline{AB}\).

The bare bones of @above would be: By gliding principle spiral similarities $AM\mapsto CN$ and $DM\mapsto BN$ have common center $X$. So we see that $X\in (EKL)$. And result follows by Miquel on $ABE$.

I did not see this idea in the above solutions. We can restate the original problem in the following way. With the same notations, let $ X$ be the point that satisfies $ \angle XMN=\angle BAK$ and $ \angle MNX=\angle ABL$ and $ X$ is in the same half-plane, determined by the line $ MN,$ in which are also the points $ A$ and $ B.$ Prove that $ X$ lies on $ AB.$ We fix the points $ A,B,D$ and begin to move the point $ C',$ that initially coincides with $ C,$ along the ray $ AC.$ Then $ N'$, as a midpoint of $ BC'$ moves along the line through the initial point $ N$ and parallel to $ AC.$ Note that $ \displaystyle k:=\frac{MX'}{MN'}$ is constant, because the angles $ \alpha, \beta$ are fixed. The point $ X'$ is obtained from the point $ N'$ by rotating $ N'$ around the point $ M$ (which is fixed) at an angle $ \alpha$ and then by homothety with ratio $ k.$ This means that $ X'$ moves along a line that is parallel to $ AB$ because it is obtained by rotating the line $ QN$ at $ \alpha$ and then by homothety. So, to prove that $ X'\in AB$ it suffices to point out only one particular case for which $ X'\in AB.$ Now, let's see what happens when $ N'\equiv Q.$ Because $ \angle MQA=\beta,$ this implies $ N'X'$ is on $ AB,$ and in this particular case indeed $ X'\in AB.$ Therefore, $ X'\in AB$ for all $ C'\in AC$ and the result follows. Remark. I put a few more comments on my blog.

Here is a purely angle chasing solution. Let $P$ be the midpoint of $AB$ and $Q$ be the second intersection point of the circumcircle of $MNP$ with $AB$. Then $\angle KMQ = \angle NMQ = \angle NQB = \angle KAB = \angle KAQ$, so $AMKQ$ is cyclic. Analogously $BNLQ$ is cyclic and we are done.

Denote by $E$ midpoint of $AB,$ by $F$ second point of intersection of $AB,\odot (MEN).$ By Reim's both $AMKF,BNLF$ are cyclic, so done.