Take $\{E\}\in BD\cap AC$, $BB'$ a diameter of the circle $(ABC)$. It is well known that $D,H,M, B'$ are collinear and $B'N\bot BN$, so $KH\parallel B'N\bot BN$. Easily, $H$ is the orthocenter of $\Delta AME$, so $K,H,E$ are collinear, done. Best regards, sunken rock

$M$ must be the midpoint of $KM$ so $K$ lies on $MN$ i.e. $BM$. Now $\angle AKC=\angle ANC=\angle AHC$ so $AHKC$ is cyclic, so an angle chase shows that $\angle KHA=\angle KBC$ and so $K$ lies on $(HDB)$. Then the three lines are concurrent at the radical centre of the three circles $(ABC),(HDB)$ and $(AHKC)$.

A note: Using sunken rock's notation, we obtain \[ MK\cdot MB=MN\cdot MB=MB'\cdot MD=MH\cdot MD \] which easily shows that $DBKH$ is cyclic for WakeUp's proof.