Let K=X(6)-Symmedian point of ABC. Antiparallel from K to BC intersect the AB, AC at Ac, Ab resp.
Define Ba, Bc, Ca, Cb cyclically.
KA: Antipode of K respect to circle (KBaCa). Define KB, KC cyclically.
* ABC and KAKB,KC are perspective triangles. Perspector is X(3527) = ISOGONAL CONJUGATE OF X(631)=perspector of 2nd Lemoine circle
