Menelaus From CevaCeva and Menelaus theorem go naturally together. Ceva's TheoremThree Cevians AD, BE, and CF are concurrent iff
holds. Menelaus TheoremLet three points F, D, and E, lie respectively on the sides AB, BC, and AC of ΔABC. Then the points are collinear iff
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 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:
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
Desargues' Theorem
Menelaus and Ceva
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