Nine-point centre (anticentre or Euler centre) & Maltitudes of Cyclic Quadrilateral

1) The perpendiculars from the midpoints of the sides of a cyclic quadrilateral ABCD to the opposite sides (called the maltitudes) are concurrent in P (called the anticentre, nine-point centre or Euler centre by different authors).

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. More-over, the centres of these 4 circles are concyclic, with the centre of the formed circle at P, and all 5 these circles are congruent, with radii half that of the circumcircle of ABCD.

Nine-point centre (anticentre or Euler centre) & Maltitudes of Cyclic Quadrilateral

Challenge
a) Can you prove the maltitudes concurrency result? Also show (prove) that the nine-point centre (anticentre or Euler centre) and the circumcentre are symmetric with respect to the centroid of cyclic quadrilateral ABCD. If stuck, see my paper at Generalizations involving maltitudes.
b) Can you prove the nine-point (or Euler) centre result involving the concurrency of the four 9-point circles & generalize further?
The beautiful result in b) 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.