Let ABC be an acute triangle. Let H denote its orthocenter and D,E and F the feet of its altitudes from A,B and C, respectively. Let the common point of DF and the altitude through B be P. The line perpendicular to BC through P intersects AB in Q. Furthermore, EQ intersects the altitude through A in N. Prove that N is the midpoint of AH.Proposed by Karl Czakler geometrymidpointperpendicular