Let ABC be an non- isosceles triangle, Ha, Hb, and Hc be the feet of the altitudes drawn from the vertices A,B, and C, respectively, and Ma, Mb, and Mc be the midpoints of the sides BC, CA, and AB, respectively. The circumscribed circles of triangles AHbHc and AMbMc intersect for second time at point A′. The circumscribed circles of triangles BHcHa and BMcMa intersect for second time at point B′. The circumscribed circles of triangles CHaHb and CMaMb intersect for second time at point C′. Prove that points A′,B′ and C′ lie on the same line. geometrymidpointsaltitudesisoscelesUkraine Correspondence