BMO Shortlist, 2010

From the point $G$ we draw a parallel to the line $AC$, which intersects $BZ$ and $AB$ at the points $V$ and $S$ respectively.

Therefore, since $AM = MC$ from the construction hypothesis we have that $SV = VG$, that is $V$ is the midpoint of the segment $SG$. If $T$ is the midpoint of the chord $EG$ we have $ES \parallel TV$ and therefore $\angle BEG = \angle VTG$ and since $\angle BEG = \angle BZG$ we will have $\angle VTG = \angle VZG$. Therefore, the quadrilateral $VTZG$ is cyclic, so $\angle TZV = \angle TGV$.

Since $GS \parallel AH$, we have $\angle TGV = \angle THA$. Therefore, we conclude that the quadrilateral $MTZH$ is cyclic and since $\angle MTH = 90^\circ$ it implies that $\angle MZH = 90^\circ$, so $BZ$ is the height of the triangle $KBH$, that is the point $L$ is the orthocenter of the triangle $KBH$, since from our hypotheses $KN \perp BH$.
In addition, since the quadrilateral $LZHN$ is cyclic it is known that the second point of intersection $P$ of the circumscribed circles of the triangles $\triangle KZL$ and $\triangle BLN$ will be located on the $KB$. It suffices now to prove that the points $P$, $L$ and $H$ are collinear. It is true that from the cyclic quadrilaterals $PLNB$, $NLZH$ and $BKZN$ follows
$$\angle KLP = \angle KBN, \angle KLZ = \angle BLN = \angle BHK, \angle ZLH = \angle ZNH = \angle BKH.$$
From the later relations we have $\angle KLP + \angle KLZ + \angle ZLH = 180^\circ$, so the points $P$, $L$ and $H$ are collinear, therefore the lines $BZ$, $KN$ and $HP$ are concurrent. □