Let $ ABC $ be an acutangle triangle and $ D $ any point on the $ BC $ side. Let $ E $ be the symmetrical of $ D $ in $ AC $ and $ F $ is the symmetrical $ D $ relative to $ AB $. $ A $ straight $ ED $ intersects straight $ AB $ at $ G $, while straight $ F D $ intersects the line $ AC $ in $ H $. Prove that the points $ A, E, F, G$ and $ H $ are on the same circumference.