1) The perpendiculars from the midpoints of the sides of a cyclic quadrilateral to the opposite sides (called the maltitudes) are concurrent in P.
2) The four nine-point circles of triangles ABC, BCD, CDA and DAB are also concurrent in the same point P, and is therefore often called the nine-point (or Euler) centre of a cyclic quadrilateral.
Nine-point (or Euler) centre & Maltitudes of Cyclic Quadrilateral
a) Can you prove the maltitudes result? If stuck, see my paper at Generalizations involving maltitudes
b) Can you prove the nine-point (or Euler) centre result? This beautiful result generalizes to any cyclic polygon and a proof is given in Yaglom, I.M. (1968). Geometric Transformations II. Washington, DC: The Mathematical Association of America, p. 24.
Michael de Villiers, 6 April 2010.