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.

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
• Desargues' Hexagon
• The Lepidoptera of the Circles

Menelaus and Ceva

• The Menelaus Theorem
• Menelaus Theorem: proofs ugly and elegant - A. Einstein's view
• Ceva's Theorem
• Ceva in Circumscribed Quadrilateral
• Ceva's Theorem: A Matter of Appreciation
• Ceva and Menelaus Meet on the Roads
• Menelaus From Ceva
• Menelaus and Ceva Theorems
• Ceva and Menelaus Theorems for Angle Bisectors
• Ceva's Theorem: Proof Without Words
• Cevian Cradle
• Cevian Cradle II
• Cevian Nest
• Cevian Triangle
• An Application of Ceva's Theorem
• Trigonometric Form of Ceva's Theorem
• Two Proofs of Menelaus Theorem
• Simultaneous Generalization of the Theorems of Ceva and Menelaus
  • Theorems of Ceva and Menelaus, an Illustrated Generalization
• Menelaus from 3D
• Terquem's Theorem
• Cross Points in a Polygon
• Two Cevians and Proportions in a Triangle, II
• Concurrence Not from School Geometry
• Two Triangles Inscribed in a Conic - with Elementary Solution
• From One Collinearity to Another
• Concurrence in Right Triangle
  

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