[alogo] Menelaus from Ceva

Consider a triangle ABC and three points D on AC, E on BA and F on CB. Then a necessary and sufficient condition for these points to be collinear on a line e is (DA/DC)(EB/EA)(FC/FB) = 1. Here we take into account the orientation of the segments, so that the ratios like (FB/FA) are negative for points F between A and B and positive for locations of F outside the segment [AB]. Here we prove this theorem by deducing it from Ceva's theorem (look at Ceva.html ).

Since in Ceva.html we proved Ceva's theorem by deducing it from Menelaus, the two theorems are equivalent.

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The proof, orginally given by John R. Silvester, may proceed as follows:
Apply Ceva's theorem to triangles and respective [Cevians]:
BCE and the lines BA, CX, ED, through point F = > (AC/AE)(XE/XB)(DB/DC)=1,
CAF and the lines CB, AY, FE, through point D = > (BA/BF)(YF/YC)(EC/EA)=1 ,
ABD and the lines AC, BZ, DF, through point E = > (CB/CD)(ZD/ZA)(FA/FB)=1 ,
BEF and the lines BD, EA, FX, through point C = > (DE/DF)(AF/AB)(XB/XE)=1 ,
CFD and the lines CE, FB, DY, through point A = > (EF/ED)(BD/BC)(YC/YF)=1 ,
ADE and the lines AF, DC, EZ, through point B = > (FD/FE)(CE/CA)(ZA/ZD)=1 .
Multiply the equations to find:
(DB/DC)²(EC/EA)²(FA/FB)² = 1.

But (DB/DC)(EC/EA)(FA/FB) is different from -1, otherwise by Ceva's theorem AD, BE and CF would be concurrent (or parallel). Hence the previous relation yields (DB/DC)(EC/EA)(FA/FB) = 1.

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