EWD 1085

Proving the theorem of Menelaos

EWD1085 figure 1

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) + λμν = 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