Let the vertices of our basic triangle, points {a,b,c}, be on the parabola 4ky-x2=0 (k is the focus-vertex distance). Consider also points {a'=b+c-a, b'=c+a-b, c'=a+b-c} which are the vertices of the anticomplementary triangle.
The parallels from {a',b',c'} to the axis of the parabola meet the opposite sides of the anticomplementary triangle on points of the parabola.
In fact, the parallel line L1 from a'=b+c-a, is the line having constant x-coordinate and equal to (b1+c1-a1).
The line L2 through {b',c'} in parametric form is b'+t(c'-b) and meets L1 at a point such that b'1+t(c'1-b1)=b1+c1-a1. From which follows immediately that
t=(b1-a1)/(b1-c1).
Thus, the y-coordinate of the intersection point is b'2+t(c'2-b'2)=a2+(2t-1)(b2-c2) , which replacing y2=x12/(4k) reduces to (1/(4k))(b1+c1-a1)2, thus proving the claim.
Remark-1 It follows that for every tripple of non-collinear points {A,B,C} and every direction L, there is conic passing through the six points {A,B,C,A'',B'',C''}, where the three last points are the intersections of the parallels to L from the vertices and the opposite sides to these vertices of the anticomplementary triangle.
Further, for all directions not coinciding with the sides of ABC the conic is a parabola whereas for directions coinciding with a side of the triangle the corresponding conic degenerates to a set of two parallel lines. Remark-2 Non degnerate parabolas touch the line at infinity at their "point" corresponding to the direction of the axis. This is used in the reference given below.