[1] Define the intersection point R of line PQ with side-line BC.

[2] Take the harmonic conjugate R* of R with respect to {B,C}.

[3] Line AR* passes through the fourth intersection point S of the two conics.

The property follows from Maclaurin's generation of the conics discussed in ConicsMaclaurin2.html . There it is shown that taking the

Applying this idea to the harmonic conjugates {P*,Q*} of the two pivots {P,Q} and the corresponding line P*Q* we obtain the intersection point S of lines {BC',CB'} belonging to both conics.

The rest is an easy consequence of this construction of S. By their definition lines {AR, PQ, BC, B'C'} make a harmonic bundle. It follows that the pole of line B'C' is the same point R* with respect to both conics. It follows also that line R*S passes through A and R* is harmonic conjugate to R with respect to {B,C}.

SteinerPoint.html

Maclaurin.html

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