We begin the proof with a lemma.
Lemma 1. Triangle $ABC$ and point $X'$ are given such that $\triangle BX'A \sim \triangle AX'C$ and $X'$ doesn't lie on line $BC$. The reflection of $A$ with respect to $X'$ lies on the circumcircle of $ABC$.
Proof. We denote by $O$ the circumcenter of $ABC$ and by $AB, AC$ the midpoints of $M, N$, respectively. It is enough to show that $AX'O = 90^\circ$. Observe that points $B$ and $C$ are on different sides of line $AX'$, otherwise from $\angle AX'B = \angle AX'C$ it follows that $X'$ would lie on $BC$. Therefore, based on the given similarity we can conclude that $\angle X'MA = \angle X'NC$. This implies that the circumcircle of $AMN$ passes through $X'$ and $O$, as evident. Thus, we have $\angle AMO = \angle AX'O = 90^\circ$, which is equivalent to the desired result.
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Now, we proceed our proof. Let's denote the reflections of points $B$ and $C$ with respect to points $F$ and $E$ as $K$ and $L$, respectively. It is evident that $K$ and $L$ are the feet of the altitudes from vertices $C$ and $B$ in triangle $ABC$, and the quadrilateral $BKLC$ is cyclic. Moreover, according to the lemma, the quadrilaterals $CXLY$ and $BXKY$ are also cyclic. If these three circles are not congruent, their pairwise radical axes must be concurrent. These radical axes are $BK$, $CL$, and $XY$. Therefore, $XY$ must pass through point $A$. However, this is not possible since $X$ and $Y$ lie on different sides of the line $CE$, and considering the fact that $CXLY$ is cyclic, line $XY$ must intersect line $CL$. Furthermore, since $\triangle ABC$ is an acute triangle, point $A$ lies on the extension of the line $CL$ from point $L$. Therefore, $XY$ cannot pass through point $A$. The resulting contradiction shows that the six points $C, X, L, K, Y$, and $B$ are concyclic; i.e., lie on a circle with center $M$.
Note that points $X$ and $Y$ are on the same side of $BC$ (the segment $XY$ intersects the segments $BK$ and $CL$). Thus, $\angle XCY$ is an acute angle. This implies that point $E$ is inside $\triangle CXY$, as otherwise we would have
$$180^\{\circ\} \leq \angle EXC + \angle EYC + \angle XCY = 2\angle XCY$$
which is a contradiction. Therefore, we can write:
$$\angle XEY = \angle EXC + \angle EYC + \angle XCY = 2\angle XCY = \angle XMY$$

Thus, quadrilateral XEMY is cyclic. Similarly, it can be proven that quadrilateral FMXY is also cyclic. Moreover, it is evident that points A, F, M, and E lie on a circle. So, the hexagon FMEXAY is cyclic. Since AM passes through the center of this circle and MY = MX, it follows that AX is also equal to AY. Finally, we conclude that $AM \perp XY$.