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.

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.

### See Also

MenelausFromCeva.html

Ceva.html

MenelausApp.html

Newton.html

Homotheties_Composition.html

CrossRatio0.html

### 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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