Two points P and Q lying on side BC of triangle ABC and their distance from the midpoint of BC are equal.The perpendiculars from P and Q to BC intersect AC and AB at E and F,respectively.M is point of intersection PF and EQ.If H1 and H2 be the orthocenters of triangles BFP and CEQ, respectively, prove that AM⊥H1H2.Author:Mehdi E'tesami Fard , Iran geometryanalytic geometrygraphing linesslopeparallelogramsymmetrygeometry proposed