Menelaus From Ceva

Ceva and Menelaus theorem go naturally together.

Ceva's Theorem

Three Cevians AD, BE, and CF are concurrent iff

holds.

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:

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:

Now multiply (1)-(6), taking into account the sign convention for directed segments. The product reduces to

But the straight line DEF is a transversal of ABC. It could be argued that the points D, E, F either all lie on the extensions of the sides of ABC, 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:

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.

References

1. C. W. Dodge, Euclidean Geometry and Transformations, Dover, 2004 (reprint of 1972 edition), pp. 22-23.
2. J. R. Silvester, Ceva = (Menelaus)², The Math. Gazette, v 84, N 5 (2000), pp 268-271.

Desargues' Theorem

• Desargues' Theorem
• 2N-Wing Butterfly Problem
• Cevian Triangle
• Do You Speak Mathematics?
• Desargues in the Bride's Chair (with Pythagoras)
• Menelaus From Ceva
• Monge from Desargues
• Monge via Desargues
• Nobbs' Points, Gergonne Line
• Soddy Circles and David Eppstein's Centers
• Pascal Lines: Steiner and Kirkman Theorems II
• Pole and Polar with Respect to a Triangle
• The Lepidoptera of the Circles

Copyright © 1996-2008 Alexander Bogomolny