Suppose that $AC=b,AB=c,BC=a$ Note that $\angle N'BX=\angle N'AC,\angle N'XB=\angle ACN$ Hence $\triangle N'XB\sim N'CA$ So $\dfrac{BX}{b}=\dfrac{NB}{NA}$ Note that $\dfrac{NB}{NA}=\dfrac{\sin \angle NCB}{\sin \angle NCA}=\dfrac{b}{a}\cdot \dfrac{AN}{NB}=\dfrac{b}{a}$ Which means that $BX=\dfrac{b^{2}}{a}$ Similarly $CY=\dfrac{c^{2}}{a}$ By cosine theorem $AX^{2}=\dfrac{b^{4}}{a^{2}}+c^{2}+\dfrac{2cb^{2}}{a}\cos B=\dfrac{b^{4}}{a^{2}}+c^{2}+\dfrac{b^{2}(a^{2}+c^{2}-b^{2})}{a^{2}}=\dfrac{a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}}{a^{2}}$ $AY^{2}=\dfrac{c^{4}}{a^{2}}+b^{2}+\dfrac{2bc^{2}}{a}\cos C=\dfrac{c^{4}}{a^{2}}+b^{2}+\dfrac{c^{2}(a^{2}+b^{2}-c^{2})}{a^{2}}=\dfrac{a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}}{a^{2}}$ Which means that $AX=AY$

Not sure what you mean by the title, but here's a synthetic solution. Let $Q$ be the reflection of $A$ across the perpendicular bisector of $BC$. Then we claim that $Q$ is the intersection of $XN'$ and $YM'$. Indeed this follows from the observation that $\angle QN'C = \angle N'XC + \angle N'CX = \angle ACN' + \angle N'CB = \angle ACB$ and similarly $\angle QM'B = \angle ABC$. Let $D$ be the foot of perpendicular from $A$ to $BC$ and $G$ be the centroid of $\triangle ABC$, then the homothety with centre $G$ and ratio $-1/2$ sends $Q$ to $D$ so $Q,G,D$ are collinear. Let the line $QD$ intersect $(ABC)$ for a second time at $P$. Now, we need to show that $D$ is the midpoint of $XY$. We know $QA \parallel BC$, so it suffices to prove that $(QA,QD; QX,QY) = -1$. For this, let the second intersection of the $A-$median with $(ABC)$ be $E$. Then $(AE,AQ; AB,AC) = -1 \implies (E,Q; B,C) = -1$. Now inverting at $G$ while preserving $(ABC)$ we get $(A,P; M',N') = -1$. But now projecting from $Q$ implies that $(QA, QD; QX,QY) = -1$ as needed. 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@above the title is for fun and the problem definetely has a synthetic solution and btw your solution consists of cross ratio stuff which at least I don't count it as a synthetic solution

Here is a synthetic solution using angle chasing/cyclic quad/similar triangles. Let $P,Q$ be midpoints of segments $BX,CY$, respectively. Since $\angle N'XB=\angle NCA$ and $\angle N'BX=\angle N'AC$, hence $\triangle N'XB\sim\triangle N'CA$. Similarly, $\triangle M'YC\sim\triangle M'BA$. Since $P,M$ are midpoints of sides $BX,AC$ of similar triangles $N'XB,N'CA$, then $\triangle N'PB\sim\triangle N'MA$. Similarly, $\triangle M'QC\sim\triangle M'NA$. Now, we have $\angle N'PB=\angle N'MA$ and $\angle M'QC=\angle M'NA$. This leads to the fact that $\square N'MCP$ and $\square M'NBY$ are both cyclic quadrilaterals. Furthermore, $\angle N'NM+\angle MM'N'=\angle N'NA+\angle ANM+\angle BM'N'=\angle BNC+\angle NBC+\angle NCB=180^\circ$, hence $M,N,N',M'$ are concyclic. Use some angle chasing, we have $$\angle MPC=\angle MN'C=\angle MN'N=\angle NM'M=\angle NM'B=\angle NQB$$Since $MN\parallel PQ$ and $\angle MPC=\angle NQB$, then $NMQP$ is an isosceles trapezoid. Finally, since $PN\parallel AX$ and $QM\parallel AY$, we have $$\angle AXY=\angle NPQ=\angle MQP=\angle AYX$$Therefore, $AX=AY$ as desired.

Attachments:

Another (first comes to mind) synthetic solution. Define $B', C'$ such that $ABCB'$ and $ACBC'$ form parallelogram, respectively. Then, one can find that the points $C', N', B, X$ and $B', M', C, Y$ are concyclic. Then, $C'XYB'$ will be an isosceles trapezoid. By using the fact that $AC' = BC = AB',$ $C'X = B'Y,$ and $\angle AC'X = \angle AB'Y,$ we conclude that $\triangle AC'X \cong \triangle AB'Y$ which gives the desired result.

Very Nice Geo : Let $B'$ and $C'$ be the reflections of $B$ and $C$ about $M$ and $N$ respectively. Claim: The point $C'$ lies on circle $(XBN')$ This is a simple angle-chase, $$\angle N'C'B=\angle CC'B=\angle C'CA=\angle NCA=\angle N'XB$$ Claim: The point $B'$ lies on circle $(YCM')$ This is a simple angle-chase, $$\angle M'B'C=\angle BB'C=\angle B'BA=\angle MBA=\angle M'YC$$ Claim: The points $C'$, $B'$, $Y$, and $X$ form an isosceles trapezoid It is easy to see that $B'C'||XY$ and that, $$\angle C'XB=\angle C'BA=\angle B'CA=\angle B'YC$$ Now since $A$ is the midpoint of $B'C'$ the result follows.

Attachments:

Let $S$, $T$ and $P$ be the midpoints of $CX$, $BY$ and $BC$. Let $D$ be the foot of perpendicular from $A$ to $XY$. We wish to show that $D$ is the midpoint of segment $XY$. In the 'flat' quadrilateral $XCBY$, $S$, $P$, and $T$ are the midpoints of the three other sides, so $D$ is the midpoint of segment $XY$ if and only if $SPTD$ is a parallelogram, or equivalently, $\overrightarrow{SD} = \overrightarrow{PT}$. Now note that by the given angle condition, $\triangle N'XC$ and $\triangle N'BA$ are spirally similar, and this sends $S$ to $N$, so $N'SBN$ is cyclic. Similarly, $M'TCM$ is also cyclic. Therefore, \[ \measuredangle NSD = \measuredangle NN'B = \measuredangle CN'B = \measuredangle CM'B = \measuredangle CM'M = - \measuredangle MTP. \]On the other hand, since $NDPM$ is an isosceles trapezoid, we get $\measuredangle NDS = -\measuredangle MPT$. Therefore, $\triangle NDS \cong \triangle MPT$ and they have opposite orientation which gives the desired result.

By Reim's theorem we get that if we let $A'$ a point in $(ABC)$ such that $AA' \parallel BC$ then $A', N', X$ and $A', M', Y$ are colinear triples. Now let $D$ in $BC$ such that $AD \perp BC$ and let $G$ be the centroid of $\triangle ABC$, by homothety we get that $D,G,A'$ are colinear, and let $A'D \cap (ABC)=W$ then, $-1=(B, C; A', AG \cap (ABC)) \overset{G}{=} (M', N'; W, A) \overset{A'}{=} (Y, X; D, \infty_{BC})$ therefore $XD=YD$ and thus $AX=AY$ as desired thus we are done .