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• ## Re: [EMHL] Re: S, H, M_a are collinear

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• Dear Vladimir! 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
Message 1 of 12 , May 18, 2011
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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
of PbPc.

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.

Sincerely Alexey

[Non-text portions of this message have been removed]
• Dear Alexey, ... Right! Thank you. Please see my private letter. Regards, Vladimir
Message 1 of 12 , May 18, 2011
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Dear Alexey,

> Your proof don't use the condition that APbPPc is cyclic. So we obtain nice theorem which is true for an arbitrary point P.

Right! Thank you. Please see my private letter.

Regards,