We will prove that the lines $AE$ and $CF$ divide the line segment $BH$ in the same ratio (it is easy to see that the points of intersection of the lines $AE$ and $CF$ with the line $BH$ will belong to this segment). Let $T$ be the point of intersection of the lines $AE$ and $BH$ (Fig. 52). Then

$$\frac{BT}{TH} = \frac{S_{\triangle BAT}}{S_{\triangle TAH}} = \frac{BA \cdot \sin \angle BAT}{AH \cdot \sin \angle TAH}.$$
The lines $AE$, $BE$ and $CE$ are concurrent and so by the trigonometric version of the Ceva's theorem we have:
$$\frac{\sin \angle BAT}{\sin \angle TAH} \cdot \frac{\sin \angle APE}{\sin \angle EPB} \cdot \frac{\sin \angle PBM}{\sin \angle MBA} = 1,$$
and so

Since $S_{\square ABM} = S_{\square CBM}$, we have that
$AB \sin \angle ABM = BC \sin \angle CBM$, hence
$$(*) = \frac{BA}{AH} \cdot \frac{\sin(90^\circ - \angle BAC)}{\sin \angle BAC} \cdot \frac{BC}{AB} = \frac{BC}{AH} \cdot \frac{\sin \angle ABH}{\sin \angle BAC} = (**).$$
Then using the sine theorem for the triangle $ABH$ and
taking into account that $\angle AHB = 180^\circ - \angle ACB$, we obtain:
$$\frac{AH}{\sin \angle ABH} = \frac{AB}{\sin \angle AHB} = \frac{AB}{\sin \angle ACB} = \frac{BC}{\sin \angle BAC},$$
that is $\frac{BT}{TH} = (**)=1$. So, we have proved that the line $AE$ passes through the midpoint of the line segment $BH$. One can similarly prove that the line $CF$ also passes through this point, which proves that the lines $AE$, $CF$ and $BH$ are concurrent.