Let $E'F'$ be the common tangent closer to N such that $E'$ on $O_1$. It is easy to see that $MN$ intersects the midpoint of $AB$. Note that $\triangle CMD$ is congruent to $\triangle AMB$, and can be obtained via a 180 degree rotation about $M$. The line $MN$ maps to $MN$ about this rotation. Clearly, $CDAB$ is a parallelogram. Thus $CB//MN//AD//BF//AE$. It is easy to see that since $EFBA$ is an isosceles trapezoid, $EFCD$ is also an isosceles trapezoid. Thus it is cyclic. It suffices to show that M lies on the circumcircle, as $\angle E'MF'=\angle E'MN+\angle NMF'=\angle F'E'N+\angle E'F'N = 180^\circ -\angle E'NF'$, showing that the circumcircles have equal radius. Clearly, the circumradiuses of $EMF$, $ANB$, $AMB$, $CMD$ are equal. We consider the circumcentre of $EFCD$, $O$. Clearly, $EO = FO = MO = CO = DO$, thus, $M$ lies on the circumcircle.

Using sine-law it is enough to prove that $\measuredangle EMF=180-\measuredangle ENF$. Let $a=\measuredangle MAB, b=\measuredangle MBA, c=\measuredangle MDA$. Segments $AC$ and $BD$ have the same midpoint so $ABCD$ is a parallelogram, and therefore we have: $\measuredangle MDC=b, \measuredangle MBC=c, \measuredangle MCD=a, \measuredangle AMB=\measuredangle CMD=180-(a+b), \measuredangle BMC=\measuredangle AMD=a+b, \measuredangle DAC=\measuredangle BDA=180-(a+b+c)$ Since $\measuredangle MEA=\measuredangle MAB=a, \measuredangle MED=\measuredangle MCD=a$ $A,D,E$ are collinear. In the same way we prove that $B,C,F$ are collinear. Then $\measuredangle MEF=180-\measuredangle MCF=\measuredangle MCB=180-(a+b+c)$ Moreover $\measuredangle MFE=180-\measuredangle MDE=\measuredangle DMA=c$ Hence $\measuredangle EMF=a+b$. But we have $\measuredangle ENF=360-\measuredangle ENF_M=360-\measuredangle ENM+\measuredangle FNM=360-(180-\measuredangle EAM +180-\measuredangle FBM)=\measuredangle EAM +\measuredangle FBM=180-(a+b)=180-\measuredangle EMF$, $Q.E.D$

Let α = ∠$BAC$, β = ∠$ABD$. Because $M$ is the midpoint of $AC$ and $BD$, $ABCD$ is a parallelogram, α = ∠$ACD$, β = ∠$BDC$ and $AD || BC$. Extend line $NM$ to meet $AB$ at $I$. Since $AB$ tangents to the circles, $IA² = IM×IN = IB²$, or $IA = IB$. Thus $IM || AD || BC$, or $MN || AD || BC$. Furthermore, in circle $(CDM)$, α = ∠$MCD$ = ∠$MED$, and in circle $(O1)$, α = ∠$BAM$ = ∠$AEM$ to imply that the three points $A, D$ and $E$ are collinear. With the same argument, the same can be said of the three points $B, C$ and $F$. Therefore, $AE || MN || BF$ and $AM = EN, BM = FN$. In other words, triangles $ENF$ is congruent to triangle $AMB$ and is also congruent to triangle $CMD$. Because the two congruent triangles $ENF$ and $CMD$ are inscribed in two different circles, these two circles must have equal radii.

Solution: Claim. $A,D,E$ and $B,C,F$ are collinear. (Proof) \[\angle AEM = \angle BAM = \angle MCD = \angle MED\]thus $A,E,D$ are collinear, similarly $B,C,F$ are also collinear. Claim.$MN||AE$ and $MN||BF$. (Proof) $MN$ bisects $AB$ so $MN||AE||BF$. Thus $E$ is the reflection of $A$ over $O_1O_2$ and $F$ is the reflection of $B$ over $O_1O_2$ which means $EF$ is the other tangent to $(O_1)$ and $(O_2)$. Extend $MN$ to meet $(DCM)$ at $R$. We have \[ \angle NFE =\angle NMF = \angle REF \implies NF||ER.\]Therefore $NFER$ is parallelogram or $\angle FNE = \angle FRE = 180^{\circ} - \angle FME$ and we are done.

My solution: Let $P$ be the midpoint of $AB$. As $M$ is the $N$-Humpty point of $\triangle NAB$, we have that $P$ lies on $MN$. Also, $MP$ is the $A$-medial line in $\triangle ACB$, and so we have $BC \parallel MN$. Similarly, $AD \parallel MN$. Now, let $XY$ be the other external common tangent of $(O_1)$ and $(O_2)$, with $X$ on $(O_1)$. Then the pairs $\{X,A\},\{Y,B\},\{M,N\}$ are reflections about the line joining $O_1$ and $O_2$, which gives $AX \parallel BY \parallel MN$. Thus, $C$ lies on $BY$, and $D$ lies on $AX$. Now, $ABCD$ is a parallelogram, and so $CD=AB=XY$, where the last statement follows from the fact that length of external common tangents to any two circles is same. This means that $CDXY$ is an isosceles trapezoid, which in turn gives that it is cyclic. Also, $\angle MCD=\angle MAB=\angle MXD$, which means that $M$ also lies on this circle. Thus, we have $X=E$ and $Y=F$. Finally, Note that, by an easy angle chase, we have that $\angle EMF=180^{\circ}-\angle ENF$, and so, after applying Sine Law, we get that the circumradii of $\triangle MEF$ and $\triangle NEF$ are equal.

Let the common tangent to two circles nearer to $N$ touch $O_1$ and $O_2$ at $E'$ and $F'$, respectively. I contend that the reflection of $N$ over $\overline{EF}$ lies on $\odot(MCD)$, which yields the result. Let $N'$ be the reflection of $N$ over $\overline{E'F'}$.[asy][asy]import olympiad;import geometry; size(8cm);defaultpen(fontsize(10pt)); path o1=circle( (0,0),0.75 ); path o2=circle( (1,0),0.55 ); pair M=intersectionpoints(o1,o2)[0]; pair N=intersectionpoints(o1,o2)[1]; pair S=(3.75,0); pair B=intersectionpoints(o2,circle((2.375,0), 1.375))[0]; pair F=intersectionpoints(o2,circle((2.375,0), 1.375))[1]; pair A=intersectionpoints(o1,circle((1.875,0), 1.875))[0]; pair E=intersectionpoints(o1,circle((1.875,0), 1.875))[1]; pair C=2M-A; pair D=2M-B; pair N1=2foot(N,E,F)-N; draw(o1);draw(o2);draw(circumcircle(M,E,F), gray+dashed); draw(A--B--C--D--cycle,grey+ linewidth(.4)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); dot("$B$",B,dir(B)); dot("$F$",F,dir(F)); dot("$A$",A,dir(A)); dot("$E$",E,dir(E)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$N'$",N1,dir(N1)); [/asy][/asy] Claim: $ME'F'N'$, $MDF'N'$ and $MCE'N'$ are cyclic. Proof. Firstly, \begin{align*} \measuredangle E'MF'=\measuredangle E'MN+\measuredangle NMF'=\measuredangle F'E'N+ \measuredangle NF'E'=\measuredangle F'NE'=\measuredangle E'N'F' , \end{align*}which yields the first cyclicity. For the other two, note that $MDF'N'$ is an isosceles trapezoid (and similarly $MCE'N'$ is an isosceles trapezoid). Indeed, $DM=MB=NF'=N'F'$ and \begin{align*} \measuredangle DMF'=\measuredangle BMF'= \measuredangle NBF'+\measuredangle MBF'=\measuredangle NF'E'+\measuredangle MF'E'=\measuredangle MF'N'. \end{align*} Hence, $E\equiv E'$ and $F\equiv F'$. Moreover, $N'$, the reflection of $N$ over $\overline{EF}$ lies on $\odot(MCD)$, as desired.