Show that the bissectors of $\angle BHC$ and $\angle BFC$ meet on $BC$ Given an acute triangle $\triangle ABC$ with orthocenter $H$. Let $D = BH \cap AC, E = CH \cap AB$ and $F = (AEDH) \cap (ABC) \neq A$. Show that the inner angle bisectors of $\angle BFC$ and $\angle BHC$ meet on the side $BC$.

It is easy to prove that the dotted ray passes through the midpoint of $BC$ and the antipode of $A$ in the $(ABC)$.
My first idea was to find $\angle CBF$ but that is a bit tough. So maybe drawing the perpendiculars from the meeting of the bissector of $\angle CHB$ with $BC$ to $FB$ and $FC$ may lead to something but it seems I need the forementioned angles.
 A: 
We use this fact that the bisector of $\angle BFC$ and Perpendicular bisector of BC meet on circumcircle of triangle BFC and the bisector of $\angle BHC$ and Perpendicular bisector of BC meet on circumcircle of triangle BHC. These two circle have common chord BC and the meeting points are both on perpendicular bisector of BC, that means bisectors of angles $\hat {BFC}$ and $\hat{BHC}$meet at G which is on BC.
A: Here is a possible solution which is chasing proportions. Using the given notation, the solution will live in the following picture:

We use the inversion in $A$ with power $AD\cdot AC = AE\cdot AB$, denoted by a star as upper decoration. So $B=E^*$ and $C=D^*$. (And the orthocenter $H$ is mapped to $H^*$, the projection of $A$ on $BC$.) The point $F$ which is on $(AED)\cap (ABC)$ is mapped to the finite point in the transformed intersection, which is an intersection of lines
$$ 
(AED)^*\cap(ABC)^*=(A^*E^*D^*)\cap(A^*B^*C^*)=(\infty BC)\cap(\infty ED)
\ ,
$$
so $F^*$ is the intersection of the lines $DE$ and $BC$. Now we will write some relations involving proportions, obtained from similar triangles related to inverted points, and Menelaus:
$$
\begin{aligned}
\frac{FB}{F^*E} &=
\frac{AF}{AE} =
\frac{AB}{AF^*} 
&&
\text{ from }
&\Delta AFB &\sim \Delta AEF^*\ ,
\\ 
\frac{FC}{F^*D} &=
\frac{AF}{AD} =
\frac{AC}{AF^*} 
&&
\text{ from }
&\Delta AFC &\sim \Delta ADF^*\ ,
\\
\frac{HB}{HC} &=
\frac{HE}{HD} =
\frac{BE}{CD} 
&&
\text{ from }
&\Delta HBE &\sim \Delta HCD\ ,
\\
1 &=\frac{F^*E}{F^*D}\cdot\frac{CD}{CA}\cdot\frac{BA}{BE}
&&
\text{ from Menelaus in }
&\Delta AED &\text{ w.r.t. the line }F^*CB\ ,
\end{aligned}
$$
and get
$$
\frac{FB}{FC}
=
\frac{FB}{F^*E}\cdot
\frac{F^*E}{F^*D}\cdot
\frac{F^*D}{FC}
=\frac{AB}{AF^*}
\cdot\left(
\frac{BE}{BA}\cdot\frac{CA}{CD}
\right)\cdot
\frac{AF^*}{AC}
=\frac{BE}{CD}
=
\frac{HB}{HC}\ .
$$
Now use the angle bisector theorem in $\Delta FBC$ and $\Delta HBC$ to see that the bisectors in them from $F$, respectively $H$ hit $BC$ in the same point $X$.
$\square$

We are done, but here is a hint to the idea behind the proof. (Or consider it a "second proof", starting with a longer introductory story, and a shorter computation.)
For a (capital letter) point $Z$ in the plane denote by (its lower case pendant) $z\in\Bbb C$ its affix in the complex plane. (Do so for all letters, copy decorations.) The cross ratio "cr" of the four points $B,C;H,F$ has modulus $\frac{HB}{HC}:\frac{FB}{FC}$, so we want in fact to show that
$$
\operatorname{cr}(b,c;h,f):=\frac{h-b}{h-c}:\frac{f-b}{f-c}
$$
has modulus one. The cross ratio is invariant w.r.t. translations, rescaling, and inversion (i.e. w.r.t. homographic transformations), so we may replace the four points $(B,C;H,F)$ by "transformed" points obtained by some geometric inversion, e.g. the inversion w.r.t. $A$ used in the proof. So we pass to the tuple
$
(B^*,C^*;H^*,F^*)=(E,D;H^*,F^*)
$ which turns out to be simpler to handle.
Using the expression for $\frac{F^*E}{F^*D}$ from Menelaus, we are also done in some few lines:
$$
\begin{aligned}
\frac{HB}{HC}\cdot\frac{FC}{FB}
&=|\operatorname{cr}(b,c;h,f)|\\
&=|\operatorname{cr}(e,d;h^*,f^*)|=
\frac{H^*E}{H^*D}\cdot
\frac{F^*D}{F^*E}
=
\frac{AC\cdot\frac{BE}{BC}}{AB\cdot\frac{CD}{BC}}
\cdot
\frac{CD}{CA}\cdot\frac{BA}{BE}
\\
&=1\ .
\end{aligned}
$$
A: 
With the ideas of sirous, we use barycentric calculus. The coordinates of $D$ and $E$ are
\begin{equation}
D\bigg(\frac{a^2+b^2-c^2}{2b^2},0,\frac{-a^2+b^2+c^2}{2b^2}\bigg),\quad E\bigg(\frac{a^2-b^2+c^2}{2c^2},\frac{-a^2+b^2+c^2}{2c^2},0\bigg)
\end{equation}
The circumcircle $(ABC)$ has equation $-a^{2}yz-b^{2}xz-c^{2}xy=0$ and $(AED)$ has $-a^{2}yz-b^{2}xz-c^{2}xy+\Big(\frac{a^2-b^2+c^2}{2}y+\frac{a^2+b^2-c^2}{2}z\Big)(x+y+z)=0$ whose intersection $F\neq A$ is
\begin{equation}
F\Big(a^2\big(a^2+b^2-c^2\big)\big(a^2-b^2+c^2\big):-\big(-b^2+c^2\big)\big(a^2+b^2-c^2\big)\big(-a^2+b^2+c^2\big):\big(-b^2+c^2\big)\big(a^2-b^2+c^2\big)\big(-a^2+b^2+c^2\big)\Big)
\end{equation}
The bisector of $BC$ is $y-z=\frac{-b^2+c^2}{a^2}x$ whose below intersection with $(ABC)$ is $U\big(-a^2:b(b+c):c(b+c)\big)$.
The lines $BC$ and $UF$ intersects on $G_1\Big(0:\big(a^2+b^2-c^2\big)c:\big(a^2-b^2+c^2\big)b\Big)$.
The orthocenter is $H\Big(\big(a^2+b^2-c^2\big)\big(c^2+a^2-b^2\big):\big(b^2+c^2-a^2\big)\big(a^2+b^2-c^2\big):\big(c^2+a^2-b^2\big)\big(b^2+c^2-a^2\big)\Big)$ and $(HBC)$ is $-a^{2}yz-b^{2}xz-c^{2}xy+\big(b^2+c^2-a^2\big)x(x+y+z)=0$ and the intersection of this circle with the bisector of $BC$ is $V\big(-a^2:a^2+bc-c^2:a^2+bc-b^2\big)$.
The lines $BC$ and $HV$ intersects on $G_2\Big(0:\big(a^2+b^2-c^2\big)c:\big(a^2-b^2+c^2\big)b\Big)$.
We have $G_1=G_2$ and the demostration is done.
A: Note that $\angle FBE = \angle FCD$ and $\angle BEF = 180^\circ - \angle FEA = 180^\circ - \angle FDA = \angle CDF$. It follows that $\triangle BEF \sim \triangle CDF$. Therefore $$\frac{BF}{CF} = \frac{BE}{CD}.$$
On the other hand, $\triangle BEH \sim \triangle CDH$, hence $$\frac{BE}{CD} = \frac{BH}{CH}.$$
Therefore $$\frac{BF}{CF} = \frac{BH}{CH}.$$
We are done by angle bisector theorem: the bisectors of angles $BFC, BHC$ intersect the segment $BC$ at the point $X$ satisfying $$\frac{BX}{CX} = \frac{BF}{CF} = \frac{BH}{CH}.$$
A: Here is yet on other proof, essentially using the observations (qualified as simple) of the OP regarding the mid point of $BC$ and the antipode of $A$ to conclude. These points and some related points should be introduced first, so in order to separate notations from arguments, i will restate, then give the quick proof in a minimal picture. After the proof, some interesting bonus properties in the given constellation of points may be (or may be not) interesting. A list of these properties with hints to show them is provided, and for the convenience of the reader an involved picture is then also given. Well, please ignore this bonus if it feels off-topic.

We restate and introduce notations.
Proposition: Let $\triangle ABC$ be a triangle with orthocenter $H$, circumcenter $O$. Let $H^*=AH\cap BC$, $D = BH \cap AC$, $E = CH \cap AB$ be the feet of the heights in the given triangle.
Let $F\ne A$ be the second point of intersection of the circles $(AEDH)$ and  $(ABC)$.
We furthermore construct the following points:
Let $M$ be the mid point of the side $BC$, let $K\in AH\cap (ABC)$ be the reflection of $H$ w.r.t. $BC$, let the side bisector $OM$ of the side $BC$ intersect $(ABC)$ in the points $Q$ and $V$, so that $A,V$ are the same side w.r.t. $BC$ and $AQ$ bisects $\hat A$. Let $A'$ be the antipode of $A$ in the circle $(O)=(ABC)$, so $A'=AO\cap FHM$.
Let $X$ be the intersection of the
inner angle bisectors of $\sphericalangle BFC$, which is $FQ$, and of $\sphericalangle BKC$, which is $KV$.
Show that $X\in BC$ .
Note: Since $\Delta BKC$ and $\Delta BHC$ are reflected in $BC$, the angle bisector of $\sphericalangle BKC$ also passes through $X$, which is mot-a-mot is the posted question.
Note: As mentioned by the OP, $A'$ is on $HM$, since for instance $H^*M$ is half of and parallel to $KA'$, so it is the mid line in $\Delta HKA'$. The point $F$ is also on the line $HA'$ since $\widehat{AFH}=90^\circ$ in the circle $(AFEHD)$ with diameter $AH$ and $\widehat{AFA'}=90^\circ$ in the circle $(ABC)$ with diameter $AA'$.


The picture is as above, and now we can start the answer:
Proof: Apply Pascal's theorem in the (degenerated) hexagon $A'FQQVK$.
$\square$
Details: The three colinear points from Pascal's Theorem are explicitly in this special case $A'F\cap QV=M$, $FQ\cap VK=X$, and $QQ\cap KA'$. The latter point is the point at infinity lying on the two parallel lines $QQ$ (tangent in $Q$ at the circle $(ABC)$) and $KA'$, and thus on each further line having their common direction. The line $BC$ is such a line, and it contains $M$. So it contains also $X$.
