Menelaus

1. Menelaus

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 ratios of lengths of segments,like (FB/FA) are negative
for points F between A and B and positive for locations of F outside the segment AB.

A proof can be given by projecting segments AA*, BB* and CC* on line e and replacing the ratios entering the relation with the
ratios of the projecting segments, e.g. DA/DC = AA*/CC*, etc. ... The product becomes:
                       (DA/DC)(EB/EA)(FC/FB) = (A*A/C*C)(B*B/A*A)(C*C/B*B) = 1,
proving the necessity. The sufficiency can be proved by contradiction and use of the necessity. In fact, assuming the condition
and that line DE passes through another point F* of BC, different from F, one deduces that FC/FB=F*C/F*B, which implies the
contradiction F=F*.

Remarks
1) For a proof of Menelaus on the basis of Ceva's theorem look at the file MenelausFromCeva.html .
2) This theorem implies (in fact, it is equivalent to) the theorem of Ceva, discussed in the file Ceva.html .
3) For a nice application of Menelaus to a property of parallelogramms look at the file MenelausApp.html .
4) An other application of the theorem can be found in Newton.html .
5) One of the most interesting applications of Menelaus is to the proof that the product of homotheties is again a homothety     (or
translation in some cases), see the file Homotheties_Composition.html for a discussion on this.
6) The formulation used here for the ratios puts the dividing point at the beginning of segments e.g. DA/DC instead    of using
AD/DA as is done in other references e.g. [Bottema, p. 85], where the product is then -1 instead of +1.

2. Projective version

The condition of Menelaus theorem can be generalized to cross ratios (see CrossRatio0.html ) by introducing an arbitrary line L,
which together with the other four lines builds a set in general position i.e. no three lines of the set pass through the same
point. The figure below shows such a figure. The condition of Menelaus is then equivalent to ([Green, p. 354])
(ABDJ)*(BCFI)*(CAEG) = 1.

The proof can be given as in section-4 in the case of Ceva's theorem (see Ceva.html ). Define the projective transformation which
maps the side-lines of the triangle, each to itself and the line L to the line at infinity. Then the cross ratios are preserved and
transfer to ratios in the case of line at infinity. Then apply the Menelaus theorem in its usual form.

Bibliography

[Bottema] O. Bottema Topics in Elementary Geometry Springer Verlag Heidelberg 2007
[Green] Green H.G. On the theorems of Ceva and Menlelaus American Math. Monthly, Vol. 64(1957)

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1. Menelaus

2. Projective version

See Also

Bibliography