CONCURRENT CIRCUMCIRCLES

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

B'a, C'a = the reflections of B',C' in AA', resp.

A'a = (perpendicular to BB' from B'a) /\ (perpendicular to CC' from C'a).

C'b, A'b = the reflections of C',A' in BB', resp.

B'b = (perpendicular to CC' from C'b) /\ (perpendicular to AA' from A'b).

A'c, B'c = the reflections of A',B' in CC', resp.

C'c = (perpendicular to AA' from A'c) /\ (perpendicular to BB' from B'c).

The circumcircles of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c are concurrent (?).

Let Ia,Ib,Ic be the incenters of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c, resp.

Are the triangles ABC, IaIbIc perspective?

Antreas P. Hatzipolakis, 27 Feb. 2013