Turkey — Team Selection Test

Let the lines $EM$ and $BC$ meet at $K$, $HV$ and $BC$ meet at $P$. By the Power Rule in circumcircle of triangle $ABC$, we get $BK \cdot KC = EK \cdot KM$, and by the Power Rule in circumcircle of triangle $BHC$, we get $BP \cdot PC = HP \cdot PN$. If we show that $P = K$, then we obtain that $EK \cdot KM = HK \cdot KN$, which implies that by the Power Rule again, the points $H$, $E$, $M$, $N$ are concyclic. Using the Bisector Theorem, $P = K$ is equivalent to $BE/EC = BH/HC$.

W.L.O.G. let the point $E$ be on the smaller arc $AB$ of circumcircle of triangle $ABC$. Let the lines $CH$ and $AB$ meet at $D$, the lines $BH$ and $AC$ meet at $F$. It is clear that the points $D$ and $F$ are on the circle with diameter $[AH]$. By angle chasing, we obtain that $\angle ADE = \angle AFE$ and hence $\angle EDB = \angle EFC$. On the other hand, we get $\angle DBE = \angle ABE = \angle ACE = \angle FCE$. This means that $\triangle EDB \sim \triangle EFC$. Therefore, we have $BE/EC = BD/FC$ (1). Since $\angle BDH = \angle CFH = 90^\circ$ and $\angle DHB = \angle FHC$, we obtain that

$\Delta BDH \sim \Delta CFH$. Using the similarity, we conclude that $BD/FC = BH/HC$ (2). Using (1) and (2), we are done.