Menelaus From Ceva
Three Cevians AD, BE, and CF are concurrent iff
|(1)||AF/FB · BD/DC · CE/EA = 1|
Let three points F, D, and E, lie respectively on the sides AB, BC, and AC of ΔABC. Then the points are collinear iff
|(2)||AF/BF · BD/CD · CE/AE = 1|
The theorems are in fact equivalent. That Ceva's theorem is implied by that of Menelaus has already been established. Below, I prove that the converse is also true: Ceva's theorem implies the theorem of Menelaus. The proof requires introduction of several additional lines:
|What if applet does not run?|
Let X, Y, Z be the intersections of BE and AD, AD and CF, and BE and CF, respectively. There are several triangles with three concurrent cevians to which we may apply Ceva's theorem:
|(1)||AF/FB · BX/XE · EC/CA = 1||ABE||BC, AX, EF||D|
|(2)||CE/EA · AY/YD · DB/BC = 1||ACD||AB, CY, DE||F|
|(3)||BD/DC · CZ/ZF · FA/AB = 1||BCF||AC, BZ, DF||E|
|(4)||EX/XB · BC/CD · DF/FE = 1||BDE||BF, CE, DX||A|
|(5)||DY/YA · AB/BF · FE/ED = 1||ADF||AE, BD, FY||C|
|(6)||FZ/ZC · CA/AE · ED/DF = 1||CEF||CD, AF, EZ||B|
Now multiply (1)-(6), taking into account the sign convention for directed segments. The product reduces to
|(7)||(AF/BF · BD/CD · CE/AE)2 = 1.|
But the straight line DEF is a transversal of ΔABC. It could be argued that the points D, E, F either all lie on the extensions of the sides of ΔABC, or only one does, while the other two lie on its sides. A simple verification of signs then shows that the product in (7) is positive:
|(8)||AF/BF · BD/CD · CE/AE = 1,|
which proves Menelaus' theorem.
There are two additional proofs of the equivalency of the two theorems. One is derived from a solution of the 4-travelers problem and the other from the observation of poles and polars with respect to a triangle and the Desargues' theorem.
- C. W. Dodge, Euclidean Geometry and Transformations, Dover, 2004 (reprint of 1972 edition), pp. 22-23.
- J. R. Silvester, Ceva = (Menelaus)2, The Math. Gazette, v 84, N 5 (2000), pp 268-271.
Copyright © 1996-2018 Alexander Bogomolny