Let ABC be a triangle where \angle ACB\equal{}90^{\circ}. Let D be the midpoint of BC and let E, and F be points on AC such that CF\equal{}FE\equal{}EA. The altitude from C to the hypotenuse AB is CG, and the circumcentre of triangle AEG is H. Prove that the triangles ABC and HDF are similar. geometryparallelogramgeometric transformationrotationgeometry proposed