Let ABC be a triangle. Let O be the center of the excircle opposite A. Let the excircle opposite B intersect the extensions of BC and BA at D and E, respectively, and let the excircle opposite C intersect the extensions of CB and CA at G and F, respectively. Let DE and GF intersect at I. Let H be the contact point of the excircle O with BC. Then AOHI is a parallelogram.