My proof also uses harmonic fours, but together with the Lemma on
similarity of circles given in my 1st post in this thread.
So, let U be the 2nd intersection of SPa and o, the circumcircle of
ABC, and let AU meet PbPc at M. We must prove that M is the midpoint
1. Let Q = BC/\PbPc and let W be the 2nd intersection of SQ and o.
Points B,Pa,C,Q are a harmonic four; this remains true after their
projection from center S to o (into B,U,C,W) and then, after the
projections from center A to PbPc (into Pc,M,Pb,W'). So it suffices to
prove that AW || PbPc.
2. Now consider the direct similarity s with center S that takes o to
p, the circle APbPc (I've already used s in the post mentioned above).
It takes B, C to Pc, Pb, hence Q'=s(Q) lies on the line PbPc. On the
other hand, by the Lemma, W'=s(W) lies on the line WA. But the
triangles SQQ' and SWW' are similar and even homothetic (because Q, S,
W are collinear). It follows that line WW' (= line AW) || line QQ'
(=line PbPc). QED
Your proof don't use the condition that APbPPc is cyclic. So we obtain nice theorem which is true for an arbitrary point P.
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