The figure below shows a triangle ABC
so that H is the
orthocenter, BD is the internal bisector, and M is the midpoint of AC. Line EHF is perpendicular to
BD. The circumcircle of the triangle BEF cuts the circumcircle of the
triangle ABC and BD at G and N, respectively. Prove that the points G, H,
N, and M are collinear.

See also

Conformal Mapping or Transformation of Problem
1382