Let ABC be a triangle and A', B', and C' be the midpoints of the sides opposite A, B, and C. Let A", B", and C" be the points of tangency of the incircles of AB'C' with B'C', BA'C' with A'C', and CA'B' with A'B'. Then A'A", B'B", and C'C" are concurrent and meet at the Nagel point.