The parallel chords theorem
If F is a family of (at least two) parallel chords of C(Π), then there is a
unique Πdiameter which bisects every member of F.
Observations
With the above notation, the Πcentre is ideal for Π if and only if the
plane conic C(Π) is a parabola. In such a case,
(1) the pline L(Π) is the tangent to C at the Πcentre,
(2) each Πdiameter cuts C(Π) exactly once, and
(3) the Πcentre corresponds to the direction of the axis of C(Π).
Proof
Let P' be the Πcentre 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 ppoints 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 Πdiameter 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 polarchord theorem, (A,B,R,Q) = 1 as Q is on the polar of R.
But Q' is ideal, so, by the midpoint theorem, R' is the midpoint of A'B'.
Thus the Πdiameter M' bisects A'B'.
Finally, as F contains at least two chords, there can be only one
such Πdiameter.

