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.


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.
