BMO 2010 Shortlist

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 $△KZL$ and $△BLN$ 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. □