For non-degenerate 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) + 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
and from (1), (2), (3) we conclude
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
