Antipodal points on the nine-point conic

Traditionally, points X,Y on a conic C are antipodal if the line XY contains the centre of C.
We extend the concept to allow the centre to be replaced by the more general Z-centre.

Definition
Two points X,Y on a conic C are Z-antipodal if the line XY contains the Z-centre of C,
the pole of T(Z) in C.

Theorem 1
Two points X,Y on C(P,Q) are Z-antipodal if and only if
the P-Ceva conjugates of X, Y are conjugate in C(P,Z), or
the Q-Ceva conjugates of X, Y are conjugate in C(Z,Q).

Notes.
The conic C(W,W) is I(W), the inconic with perspector W.
The conic C(P,Z) can be replaced by any conic which meets T(Q) in the same points.
This emerges easily from Theorem 2 below. The family of such cubics contains just
one circumcubic, but the perspector is not very nice: it is the barycentric product of
the P-comp of Q and the crosspoint of P and Q.

This can be proved by Maple computation, but there is a barycentric proof below.
It also follows from another result which is more easily proved, even by hand.

Theorem 2
Suppose that U,V are fixed points, and that points X,Y are such that (U,V,X,Y) is harmonic.
Then the line L joining the P-Ceva conjugates of X and Y passes through a fixed point W.

Proof
let U = u:v:w, V = u':v':w', P= p:q:r.
First some notation to abbreviate common combinations.
let a = u(-u/p+v/q+w/r), b, c similarly.
let a' = u'(-u'/p+v'/q+w'/r), b', c' similarly.
let k = u(-u/p+'v'/q+w'/r)+u'(-u/p+v/q+w/r), l, m similarly.
As (U,V,X,Y) harmonic, we can write X = sU + tV, Y = sU-tV.
Now the P-ceva conjugates of X and Y are
X' = [s2a+t2a'+stk, s2b+t2b'+stl, s2c+t2c'+stm], and
Y' = [s2a+t2a' -stk, s2b+t2b' -stl, s2c+t2c' -stm].
From these it is clear that, when st ≠ 0, the line X'Y' contains the point W = k:l:m.
When st = 0, we have X = Y, and we appeal to continuity.

Proof of Theorem 1 from Theorem 2
Let U,V be the P-Ceva conjugates of the intersections of C(P,Q) and T(Z).
Then U,V are on T(Q) as the P-Ceva conjugates of points on C(P,Q).
If two points X,Y are on T(Q), then their P-Ceva conjugates X',Y' are on C(P,Q).
If ((U,V,X,Y) is harmonic, then Thereom 2 gives a point W which we identify.
As X tends to U, Y tends to U so the line X'Y' tends to the tangent to C(P,Q) at
the P-Ceva conjugate of U. Likewise, as X tends to V, X'Y' tends to the tangent
to C(P,Q) at the P-Ceva conjugate of V. The P-Ceva conjugates of U, V are just
the intersections of C(P,Q) and T(Z), so W is the pole of T(Z) in C(P,Q), i.e. W
is the Z-centre of C(U,P), and X',Y' are Z-antipodal.
The argument clearly reverses since, given a point X' on C(P,Q), there is a
unique Y' on C(P,Q) with X',Y' Z-antipodal. This can be found as above.
The condition (U,V,X,Y) harmonic is equivalent to the statement that they
are conjugate in any conic through U and V. The P-Ceva conjugate of T(Z)
is such a conic. It is C(P,Z).