Problems of Ukrainian Authors

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

$$\frac{BT}{TH}=\frac{S_{△BAT}}{S_{△TAH}}=\frac{BA\cdot \sin \angle BAT}{AH\cdot \sin \angle TAH}.$$ 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 $$(\ast )=\frac{BA}{AH}\cdot \frac{\sin (90^{\circ }-\angle BAC)}{\sin \angle BAC}\cdot \frac{BC}{AB}=\frac{BC}{AH}\cdot \frac{\sin \angle ABH}{\sin \angle BAC}=(\ast \ast ).$$ 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 $\frac{BT}{TH}=(\ast \ast )=1$. 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.