Proving the theorem of Menelaos For non-degenerate triangle ABC and D, E, F collinear as in above figures the theorem of Menelaos states

 BD––– DC
·
 CE––– EA
·
 AF––– FB
=  –1       .

In terms of λ, μ, ν, defined by

 D = λB + (1–λ)C,     E = μC + (1–μ)A,     F = νA + (1–ν)B

we have to conclude

 1–λ––– λ
·
 1–μ––– μ
·
 1–ν––– ν
=  –1       or

 (0) (1–λ) · (1–μ) · (1–ν) + λ·μ·ν = 0

from the fact that D, E, and F are collinear. This is expressed by

 (1) Det. P = 0   where   P = Dx Dy 1 Ex Ey 1 Fx Fy 1 .

Writing Dx = λBx + (1–λ)Cx, etc. we can factorize   P = Q ·R   where

 Q = λ 1–λ 0 R = Bx By 1 0 μ 1–μ Cx Cy 1 1–ν 0 ν Ax Ay 1

From this factorization we conclude

 (2) Det. P = (Det. Q) · (Det. R)           .

Triangle ABC being non-degenerate is expressed by

 (3) Det. R ≠ 0           ,

and from (1), (2), (3) we conclude

 Det. Q = 0           ,

which, in view of Q’s definition, equivales (0).

I designed the above proof in reaction to a very classical argument in which the above two figured had to be dealt with separately. I like the proof for the way R enters the picture. (And just in case you feel tempted to send me shorter proofs of Menelaos’s theorem: I know all about barycentric coordinates.)

Austin, 10 October 1990

prof.dr. Edsger W.Dijkstra
Department of Computer Sciences
The University of Texas at Austin
Austin, TX 78712-1188
USA