## Menelaus From Ceva

Ceva and Menelaus theorem go naturally together.

### Ceva's Theorem

Three Cevians AD, BE, and CF are concurrent iff

 (1) AF/FB  ·  BD/DC  ·  CE/EA = 1

holds.

### Menelaus Theorem

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:

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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:

 Identity Triangle Cevians Intersect At (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.

### 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.

### Menelaus and Ceva

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