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

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

(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 =  D_{x}  D_{y}  1  
E_{x}  E_{y}  1  
F_{x}  F_{y}  1  . 
Writing D_{x} = λB_{x} + (1–λ)C_{x}, etc. we can factorize P = Q ·R where
Q =  λ  1–λ  0  R =  B_{x}  B_{y}  1  
0  μ  1–μ  C_{x}  C_{y}  1  
1–ν  0  ν  A_{x}  A_{y}  1 
From this factorization we conclude
(2)  Det. P = (Det. Q) · (Det. R) . 
Triangle ABC being nondegenerate 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 787121188
USA