I have discovered something using Geogebra and I am positive it is true. I have tried to prove and my solution works but it is extremley convoluted. I'm hoping someone can provide a simple proof of the following:
Given triangle $ABC$ with altitudes $BB'$ and $CC'$ orthocenter $H$. Let the circumcircles of $B'HC'$ and $BHC$ intersect again at point $X$. Let $AX$ hit $BC$ at a point $M$.
Show that $M$ is the midpoint of side $BC$.
On
Construction: Extend AM to cut the circle BXHC at P. Join BP and join CP.
$\beta$ is the exterior angle of the cyclic quadrilateral $PCHX$. Therefore, $\alpha = \beta$.
$\gamma = \beta$ because they are the angles on the same segment of the cyclic quadrilateral $AHXC’$.
$\alpha = \gamma$ implies $AB // CP$.
Similarly, $\theta = \delta$ implies $BP // AC$
Therefore, $ABPC$ is a //gm. This means $BM = MC$
We have that the circumcircle of $B'HC'$ goes through $A$ and the circumcircle of $BHC$ goes through $A''$, the symmetric of $A$ with respect to the midpoint of $BC$. Moreover, $B,C,B',C'$ are concyclic points, hence if $Y=BC\cap B'C'$, we have that $HX$, i.e. the radical axis, goes through $Y$ (that is not crucial for our proof, but I think it is worth noticing). Since the centre of the circumcircle of $B'H C'$ is the midpoint of $AH$ and the centre of the circumcircle of $BHC$ is the midpoint of $AH''$, the radical axis is perpendicular to $AA''$.
Since $\widehat{HXA}=\widehat{HB'A}=\frac{\pi}{2}$, $AX\perp HX$, so $AX\cap BC=AA'\cap BC$ and the claim is proved.