Let Ma, Mb, Mc be the midpoints of the sides of ABC. The perpendicular from Ma to the B-bisector (resp. the C-bisector) cuts MbMc at Ab (resp. Ac). Define analogously Bc, Ba, Ca, Cb. The triangles MaAbAc, MbBcBa, McCaCb have as common circumcenter the Spieker point of ABC.