Awesome problem! Set $G=DE \cap BC$. Let $I$ be on $AB$ with $BIGE$ cyclic. Then, $\measuredangle ADE=\measuredangle BGE=\measuredangle BIE=\measuredangle AIE$ so $ADIE$ is cyclic and $\measuredangle XFE=\measuredangle ADG=\measuredangle AIE=\measuredangle XIE$ so $XIEF$ is cyclic. Moreover, $\angle DEI=\angle ABC=\angle AEB$ and subtracting $\angle AIE$ gives $\measuredangle IGB=\measuredangle IEB=\measuredangle DEA=\measuredangle DIA$ and $\measuredangle BIE=\measuredangle BGE=\measuredangle CGD$ so $\angle DGI=\angle DIE$. Thus $DI^2=DG\cdot DE=DF^2$. Similarly, $J=(CGE)\cap AC$ has $DF=DI=DJ$, $EJFY$ cyclic, and $AIEJD$ cyclic. Now, \[\angle XEY=\angle FEY-\angle FEX = \angle FJY-\angle FIX=\angle FJA-\angle FIA.\]Let $Q=\overline{JCA} \cap (D)$ and $P=\overline{IAX}\cap (D)$ where $(D)$ is the circle with center $D$ through $F,I,J$. So the desired angle becomes \[\angle FJA-\angle FIA=\angle FJQ-\angle FIP =\frac{\angle PDQ}{2}\]since the angle is the angle that arc $PQ$ inscribes. This is the cool part: as $\triangle ABC$ is isosceles, lines $AB$ and $AC$ are reflections about $AD$. But $AD$ cuts $(D)$ in half! Thus reflecting over $AD$, $Q$ goes to $I$ and $P$ goes to $J$. So $\angle IDJ = \angle IAJ=\angle BAC$ from $AIEJD$ cyclic. Thus $\frac{\angle PDQ}{2}=\frac{\angle BAC}{2}$ so we conclude that $\angle XEY=\frac{\angle BAC}{2}$ as desired. Remark. YAOaops pointed out that we can just say $\overarc{PQ}=\angle PAQ=\angle BAC$ by symmetry oops

We claim $\angle XEY=\frac12\angle BAC$. Let $\omega_1=(ABC)$ and $\omega_2=(DEF)$ which are both tangent to $AD$. We claim the $X$-Apollonius circle of $XAD$ passes through $E$. We have \begin{align*} \frac{EA}{ED}&=\frac{\sin\angle EDA}{\sin\angle EAD}\\ &=\frac{EDR_{\omega_1}}{EAR_{\omega_2}}\\ &=\sqrt{\frac{R_{\omega_1}}{R_{\omega_2}}} \end{align*} and \begin{align*} \frac{XA}{XD}&=\frac{\sin\angle XDA}{\sin\angle XAD}\\ &=\frac{DFR_{\omega_1}}{ABR_{\omega_2}}\\ &=\frac{XDR_{\omega_1}}{XAR_{\omega_2}}\\ &=\sqrt{\frac{R_{\omega_1}}{R_{\omega_2}}}. \end{align*} Similarly, $\frac{YA}{YD}=\frac{XA}{XD}$. Therefore, $E$, $X$, and $Y$ all lie on this circle. If the reflection of $X$ over $AD$ is $X'$, then $X'$ also lies on this circle, and $\angle YX'X=\frac12\angle A$, so $\angle XEY=\frac12\angle BAC$, which is fixed.

Let $DE \cap BC=G$ and let $(BGE) \cap AB=K$ and $(CGE) \cap AC=J$, by Miquel $K,G,J$ are colinear. Claim 1: $AKEJD$ is cyclic. Proof: On line $BC$ apply reim's on $(BGE), (CGE)$ to finish. Claim 2: $D$ is center of $(JKF)$ Proof: Note $\angle CJE=\angle BKE= \angle DGF=\angle ADG=\angle DFE$ so $DG \cdot DE=DF^2$ and $XKEF, YEJF$ are cyclic, now $\angle KJD= \angle KEG=\angle ABC=\angle ACB=\angle DEJ=\angle DKJ$, from here $DK$ is tangent to $(BGE)$ and $DJ$ is tangent to $(CGE)$ therefore by PoP the claim is true. Finish: $\angle YEX=\angle AKF-\angle AJF=\angle BAC-\angle KFJ=\angle BAC-\frac{\angle BAC}{2}=\frac{\angle BAC}{2}$ which finishes the problem as RHS is fixed thus we are done .

Very pretty! Set $P$ to be $(ABCE) \cap (EDF)$. Set $Q$ to be $(EXY) \cap XA$. We have a lemma, or more particularly a subproblem: Quote: Let $AD$ be a common tangent of two circles $\omega_1$ and $\omega_2$, denote by $B$ one of the intersections of the circles. Let point $P$ be variable on the $B$-apollonian circle of $\triangle BAD$. Then if $PA$ and $PD$ meet $\omega_1$ and $\omega_2$ at $S$ and $T$ respectively, then $ST \parallel AD$. The converse also holds by an easy case check. To show this, note forgotten coaxiality gives $\frac{PS \cdot PA}{PT \cdot PD}$ is fixed, but we know $\frac{PA}{PD}$ is fixed, whence we find $\frac{PS}{PT}$ is fixed. Then taking the limiting case as $P$ lies on $AB$ yields the result. $\square$ Notice this immediately implies that $(EXY)$ is the $E$-apollonian circle of $\triangle EAD$. From then, let $Y'$ be the reflection of $Y$ over $AD$, so $\measuredangle YEX = \measuredangle YY'X = 90^\circ - \measuredangle BCA$, which is fixed.

Here is a solution using inversion:

Let $XY$ intersect $BC$ at $I$ and let $(EFD)$ intersect $BC$ again at $G$. Notice that $I$ is the Miquel's point of complete quadrilateral $IXAB$. Claim: Triangle $ABC$ and $DXY$ are similar $$\angle DXY=\angle DAY=\angle CBA$$$$\angle DYX=\angle DAC=\angle BCA$$Claim: $I$, $F$, and $D$, are collinear $$\angle IFX=\angle ICX=\angle BCA=\angle XYD=180^{\circ}-\angle XFD$$Claim: $XYGE$ is concyclic $$IY\cdot IX=ID\cdot IF=IG\cdot IE$$ Now to finish notice that $DX=DY$ and $DG=DE$ so $D$ is the center of $(XYGE)$, finally $$\angle AXE+\angle AYE=\angle XAY-\angle XEY=\angle XAY-\frac{1}{2}\angle XDY=\frac{1}{2}\angle XAY=\frac{1}{2}\angle BAC$$

Notice that there are two choices for $F$, which motivates the following: Let $F_1,X_1,Y_1$ be the points resulting from one of the choices, and $F_2,X_2,Y_2$ be the other. We see that $X_2$ is the reflection of $Y_1$ over $DA$ (and the symmetric). But also, by power ratio lemma and $$\frac{Y_1A\cdot Y_1C}{Y_1D\cdot Y_1F_1}=\left(\frac{Y_1A}{Y_1D}\right)^2=\left(\frac{\sin Y_1DA}{\sin DAY_1}\right)^2=\text{const}$$because while the sins aren't constant, their squares are, we get the circle $(X_1Y_1EX_2Y_2)$. Now finish by noticing $\angle X_1EY_1=\angle X_1X_2Y_1=\angle(AB,\overline{BC})=90^\circ-\angle ABC$.