Since $AB = DE$, the triangles $NAB$ and $NED$ are congruent, so $NA = NE$ and $NB = ND$. Let $K$ and $L$ be the antipodes of $N$ in $\gamma_1$ and $\gamma_2$, respectively, and notice that they are the midpoints of the arcs $AE$ and $BD$, respectively. Notice further that the angles $KME$ and $LMD$ have equal measures, to deduce that the four arcs $KA, KE, LB$ and $LD$ all have the same measure. The arcs $EMN, AFN$ and $BMN$ have equal measures as well. Consequently, the inscribed angles $AFN$ and $DCN$ subtend arcs of equal measures on the respective circles, so they are congruent; that is, the points $A, F, C, D$ are co-cyclic.

We now show that the center of the circle through $A, F, C, D$ is the midpoint $O$ of the segment $KL$. To this end, we show that $O$ lies on the perpendicular bisector of any segment $X_1X_2$ through $N$, where $X_1$ is on $\gamma_1$ and $X_2$ is on $\gamma_2$; in particular, $O$ lies on the perpendicular bisectors of both segments $AC$ and $DF$, whence the conclusion.

Let $O_1$ and $O_2$ be the centers of the circles $\gamma_1$ and $\gamma_2$, respectively, and notice that $OO_1NO_2$ is a parallelogram to deduce that the segments $NO$ and $O_1O_2$ cross each other at their common midpoint $P$. Let further $X, X'_1$ and $X'_2$ be the midpoints of the segments $X_1X_2, NX_1$ and $NX_2$, respectively, and notice that the segments $NX$ and $X'_1X'_2$ have the same midpoint $X'$, to infer that $OX$ is parallel to $PX'$. Since the latter is perpendicular to $X'_1X'_2$, it follows that $OX$ is perpendicular to $X_1X_2$, so $OX$ is indeed the perpendicular bisector of the segment $X_1X_2$.