Notice $$\measuredangle BXC=\measuredangle BXA=\measuredangle XAB=\measuredangle BHC,$$so $X$ lies on $(BHC).$ Similarly, $Y$ lies on $(BHC)$ so $F$ is the $A$-HM point of $\triangle ABC.$ $\square$

@above directed angles are bad In this solution, we denote segment $AB$ as $(AB),$ and line $AB$ as $\overline{AB}.$ [asy][asy] unitsize(25); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.57104904356743, xmax = 9.29800414774593, ymin = -4.398668189419848, ymax = 3.6041014066115533; /* image dimensions */ /* draw figures */ draw((1.421975246673722,2.7712339825901857)--(0,0), linewidth(0.65)); draw((0,0)--(4,0), linewidth(0.65)); draw((4,0)--(1.421975246673722,2.7712339825901857), linewidth(0.65)); draw((0,0)--(2.1442872828261756,1.9947885050838443), linewidth(0.65)); draw((4,0)--(0.8336695185024892,1.6247071145070373), linewidth(0.65)); draw(circle((1.421975246673722,2.0470348664316598), 0.724199116158526), linewidth(0.65)); draw(circle((2,-0.7241991161585263), 2.12707883254119), linewidth(0.65)); draw((1.421975246673722,2.7712339825901857)--(2,0), linewidth(0.65)); draw((0,0)--(2.866599318978629,1.218343027577503), linewidth(0.65)); draw((0.24536379033125622,0.4781802464238889)--(4,0), linewidth(0.65)); draw((0,0)--(0.24536379033125622,0.4781802464238889), linewidth(0.65)); /* dots and labels */ dot((1.421975246673722,2.7712339825901857),dotstyle); label("$A$", (1.3767107127595135,2.8617630504185954), NE * labelscalefactor); dot((0,0),dotstyle); label("$B$", (-0.17133634710640336,-0.03516712009050919), NW * labelscalefactor); dot((4,0),dotstyle); label("$C$", (4.1016356543948405,-0.06232584043903204), NE * labelscalefactor); dot((1.4219752466737219,1.322835750273134),linewidth(4pt) + dotstyle); label("$H$", (1.3586048991938302,1.3951921515983612), NE * labelscalefactor); dot((0.8336695185024892,1.6247071145070373),linewidth(4pt) + dotstyle); label("$E$", (0.6615310769150256,1.5400386601238163), W * labelscalefactor); dot((2.1442872828261756,1.9947885050838443),linewidth(4pt) + dotstyle); label("$D$", (2.1824194164324178,2.0470014399629095), NE * labelscalefactor); dot((1.7114871257381217,1.3832222183709362),linewidth(4pt) + dotstyle); label("$F$", (1.8021973315530697,1.2322398295072239), NE * labelscalefactor); dot((2,0),linewidth(4pt) + dotstyle); label("$M$", (2.0194670943412683,-0.08252066808642318), S * labelscalefactor); dot((0.24536379033125622,0.4781802464238889),dotstyle); label("$Y$", (0.07309213603032036,0.489901473314266), NW * labelscalefactor); dot((2.866599318978629,1.218343027577503),dotstyle); label("$X$", (2.9066519590597473,1.3046630837699515), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The configuration reminds us greatly of the $A$-HM point, denoted as $X_A=(ADHE) \cap (BHC) \neq H,$ which lies on the $A$-median. Thus, it suffices to show $X$ and $Y$ lie on $(BHC),$ as this would imply $F=X_A,$ which the required property. We proceed by angle chasing. Consider $X$ first. It suffices to show $\angle BXC = \angle BHC.$ Notice \begin{align*} \angle BXC &= 180^\circ - \angle BXA \\ &= 180^\circ-(90^\circ - \angle XBD) \\ &=90^\circ+ \angle XBD \\ &= 90^\circ + \angle ABD \\ &= 180^\circ-(90^\circ- \angle ABD) \\ &=180^\circ - \angle EHB \\ &= \angle BHC. \end{align*}Therefore, $X$ lies on $(BHC).$ A similar argument can be made to show $Y$ also lies on $(BHC).$ However, notice this configuration only takes care of the case when $X \in (AC), Y \in (AB).$ In certain configurations, it is possible for $X \in \overline{AC} \setminus \{(AC)\}$ or $Y \in \overline{AB} \setminus \{AB\},$ or both to occur. In these cases, the argument will be the same as before, but instead we consider angles $\angle XCB$ and $\angle YBC.$ Exhausting all cases, we are done, as $F=X_A.$

It suffices to show that $X$ and $Y$ lie on $(BCH)$ as this would imply that $F$ is the $A$-humpty point which is known to be on the $A$-median. Toss on the complex plane with $(ABC)$ as the unit circle with $A=a,B=b,C=c,H=a+b+c$. We have $E=e=\frac12(c+a+b-ab/c)$ so $Y=y=2e-a=c+b-ab/c$. To show that $BCHY$ is a cyclic quadrilateral, it suffices to show \begin{align*}\frac{h-b}{h-c}:\frac{y-b}{y-c}&=\frac{a+c}{a+b}:\frac{c-ab/c}{b-ab/c}\\&= \frac{a+c}{a+b}:\frac{c^2-ab}{bc-ab}\\ &= \frac{a+c}{a+b}\cdot \frac{bc-ab}{c^2-ab}\\ &= \frac{bc^2-ba^2}{bc^2-ba^2+ac^2-ab^2}\in\mathbb{R}.\end{align*}But observe \begin{align*}\overline{\left(\frac{bc^2-ba^2}{bc^2-ba^2+ac^2-ab^2}\right)}&=\frac{\frac{1}{bc^2}-\frac{1}{ba^2}}{\frac{1}{bc^2}-\frac{1}{ba^2}+\frac{1}{ac^2}-\frac{1}{ab^2}}\\&=\frac{-1}{-1}\cdot\frac{\frac{ba^2-bc^2}{(abc)^2}}{\frac{ba^2-bc^2}{(abc)^2}+\frac{ab^2-ac^2}{(abc)^2}}\\&=\frac{bc^2-ba^2}{bc^2-ba^2+ac^2-ab^2}\end{align*}so $Y$ lies on $(BCH)$. Similarly, $X$ lies on $(BCH)$ are we are done!

We will use barycentric coordinates. Reminder that we will use Convay's notation: $S_A= \frac{-a^2+b^2+c^2}{2}$ and $S_{AB}= S_A \cdot S_B$ and $S=S_{AB}+S_{BC}+S_{AC}$ Let $(ABC)$ be the reference triangle hence $A(1,0,0),B(0,1,0),C(0,0,1)$. It is well known that $H(S_{BC}:S_{AC}:S_{AB}),E(2S_B:2S_A:0),D(2S_C:0:2S_A)$. Now we will derive the points $Y,X$. We can use reflection formula $2E-A=Y$ etc. But we have to normalize coordinates'sum. It is not hard to see that the sum of coordinates $E$ is $2c^2$. Hence write $A=(2c^2:0:0)$. Now, $Y=(4S_B-2c^2:4S_A:0)$. Now we will derive equation for cirlce $BHC$. Reminder that equation for circle: $ -a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$. then \begin{align*} B(0,1,0): v=0& \\ C(0,0,1): w=0& \\ H(S_{BC}:S_{AC}:S_{AB}):u= \frac{S_{ACB}(a^2S_A+b^2S_B+c^2S_C)}{S\cdot S_{BC}} \end{align*}Now we have equation for our circle, we have to test $Y \in (BHC)$: $$-c^2(4S_B-2c^2)(4S_A)+\frac{S_{ACB}(a^2S_A+b^2S_B+c^2S_C)}{S\cdot S_{BC}} \cdot (4S_B-c^2))(4S_B+4S_A-2c^2)=0\rightarrow$$$$-4c^2S_A+\frac{S_{ACB}(a^2S_A+b^2S_B+c^2S_C)}{S\cdot S_{BC}} )(2c^2)=0 \rightarrow $$$$-4c^2+ \frac{S_{CB}(a^2S_A+b^2S_B+c^2S_C)}{S\cdot S_{BC}}(2c^2)=0$$$$ \frac{(a^2S_A+b^2S_B+c^2S_C)}{S}=2$$Which is true after expanding! Hence we have proven $Y \in (BHC)$. similiarly, we can prove $X \in (BHC)$. Therefore we can finish the problem with $A-HM$ point. But, let's assume you are not good at syntetic geometry like me, and want to finish the problem full analytically.Then lets derife point $F$. Since $circle(BYXHC)=circle(HXY)$, we can derive the formula for circle $(ADE)$, and find their intersection. Note that we already know formula of $(BHXYC)$. Let's plug in values of $A,D,E$ to the circle formula: \begin{align*} A(1,0,0): u=0& \\ E(2S_B:2S_A:0):v=S_B& \\ D(2S_C:0:2S_A):w=S_C& \\ \end{align*} Now let $Q \in \overline{AM}$, where $M=(0,\frac{1}{2},\frac{1}{2})$. We will parametrize the coordinates of $Q$. Since area between them is $0$ then: $$\det\begin{vmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} \\ x & y & z \end{vmatrix}=0\rightarrow y=z$$Hence $Q=(1-2q,q,q)$. Now lets check is there any point $Q$ in intersection of our circles. We have to observe $$\frac{S_{A}(a^2S_A+b^2S_B+c^2S_C)}{S}(1-2x)-x(S_B+S_C)=0$$$$\frac{S_{A}(a^2S_A+b^2S_B+c^2S_C)}{S}(1-2x)-xa^2=0$$$$\frac{S_{A}(a^2S_A+b^2S_B+c^2S_C)}{S}= \frac{xa^2}{1-2x}$$$$\frac{S_{A}(a^2S_A+b^2S_B+c^2S_C)}{Sa}=\frac{x}{1-2x}$$Call the final $LHS=N$ Then $Q=(Ny,y,y)$. where $ny+y+y=1$, which is intersection of the circles. Observe that $H \neq Q$ meaning there is a point on line $AM$, intersecting $BC$ and circleS $(HXY),(ADE)$, since circle's intersection is unique, we get $Q=F$ hence $AF$ intersects middle point of $BC$ and we are done! Remark: Solution can look long, actually it is short as complex bash solution above but showing equations step + extra part makes solution long

It is obvious, $ABX$ is isosceles triangle and also $ACY$. From there u can use ; $\angle DBX=\angle DBA$, $\angle ACE=\angle YCE$. From angle chasing you can easily find that X and Y points are on the $(BHC)$

You can config bash this. First, $A$ is the orthocentre of $BHC$, so by the Reflections of the Orthocentre Lemma we get $X,Y\in (BHC)$, which gives $(HXY)=(BHC)$. Hence $F$ is the magic point of the median of $A$ in $ABC$ (the so-called $A$ HM point).