Dear Alex,
I have found here one more concyclic point. It is centroid of equilateral triangle bounded by AX, CY and AC. Suppose D is intersection of AX, CY then this triangle is ACD and G is centroid of ACD. I use notations in your applet.
Proof:
- By my method: perpendicular with AC from A intersects circumcircle ABX at X', perpendicular with AC from C intersects circumcircle BCY at Y' then X'MY' are collinear. Intersection of X'Y' with AD is midpoint of X'Y'. This point is centroid of ACD because it collinear with N, A and C, P
- Another method: centroid G of ACD is intersection of AN, CP. Angle(ANP) = angle(CPB) = 90 therefore B, N, G, P are concyclic and our concyclic circle taken BG as diameter.

Generally, this fact can be generalized when XAB, YBC are any two similar triangles (by this order). In this case eight points:
- Point B
- Intersection of XC, YA
- Midpoints of AC, XC, YA, XB, YB
- Centroid G of triangle ACD (D = AX /\ CY)
are on one ellipse taken midpoint of GB as center.
Best regards,
Bui Quang Tuan