The parallel chords theorem
If F is a family of (at least two) parallel chords of C(Π), then there is a
unique Π-diameter which bisects every member of F.
Observations
With the above notation, the Π-centre is ideal for Π if and only if the
plane conic C(Π) is a parabola. In such a case,
(1) the p-line L(Π) is the tangent to C at the Π-centre,
(2) each Π-diameter cuts C(Π) exactly once, and
(3) the Π-centre corresponds to the direction of the axis of C(Π).
Proof
Let P' be the Π-centre of C(Π), and A'B' a member of F.
Since F is a family of parallel segments on Π the common direction
defines an ideal point Q' for Π.
Note that, by the above observations, if C(Π) is a parabola, then we
cannot have Q' = P', since there are no chords in the axial direction.
Of course, if C(Π) is not a parabola, then P' is not ideal, so we cannot
have Q' = P' in this case either.
Let A,B,P,Q be the p-points corresponding to A',B',P',Q' respectively.
Note that, as Q' is ideal for Π, Q lies on L(Π), the polar of P.
Let M be the polar of Q with respect to C. By La Hire's theorem,
P lies on M, so M embeds as a Π-diameter M' on Π.
Let M cut AB at R. Since Q' ≠ P', M' is not the ideal line for Π.
M' does contain the ideal point Q' so it contains no other ideal point.
Thus, R' is a point on Π, and hence on A'B'.
Again by La Hire's theorem, Q is on the polar of R.
By the polar-chord theorem, (A,B,R,Q) = -1 as Q is on the polar of R.
But Q' is ideal, so, by the mid-point theorem, R' is the mid-point of A'B'.
Thus the Π-diameter M' bisects A'B'.
Finally, as F contains at least two chords, there can be only one
such Π-diameter.
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