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capm #"=rM cg ӗ( Dtb `H 8opxyp*BTheorem: Sylvester-Gallai The following result, which plays a useful role in the theory of 'harmonic separation', is particularly interesting because, after its enunciation by Sylvester in 1893, it remained unproved for about forty years. Then T. Gallai proved it by an ingenious argument using parallel lines. The projective proof given here is due to R. Steibnerg.
Theorem
If n given points are not all on one line, there exists a line containing exactly two of them.
Proof:
Suppose, if possible, that every join of two of the n points contains a third. Since the points are not all collinear, they must include three forming a triangle PQR. Let p be a line through P that contains no other point of the set. All the joins of pairs of the n points meet p in a certain set of at least two points: P itself, one on QR, and possibly others. These points occur in a certain cyclic order. Let A be consecutive to P in this order, so that one of the segments AP is not met by any of the joins. This point A is not one of the n but lies on a line containing at least three of them, say B, C, D, so named that AB//CD. Since P and B are two of the n points, their join must contain a third, say O. Suppose ABCD /\O==APC'D'. Then AP//C'D'; that is, the joins OC and OD each meet one of the two segments AP, contrary to our definition of A. Hence, in fact, one of the joins must contain only two points of the set.t.t.t.t.M ӗ( P0!D? \
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