Cut The Knot!An interactive column using Java appletsby Alex Bogomolny 
The Menelaus Theorem
November 1999
My previous column suffers from a conspicuous omission. I have quoted Dan Pedoe,
The theorems of Ceva and Menelaus naturally go together, since the one gives the conditions for lines through vertices of a triangle to be concurrent, and the other gives the condition for points on the sides of a triangle to be collinear, 
but then went on and discussed a fine proof of Ceva's theorem. Menelaus theorem was never mentioned again. This time, I'd like to give its due to the latter. Here are the two theorems side by side:
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 
At first sight, (1) and (2) express exactly the same fact that the product of three ratios of segments on the three sides of the triangle equals 1. Now, if (1) and (2) are the same, how can they be equivalent to the two essentially different facts? And the facts annunciated by the theorems of Ceva and Menelaus are indeed different. Therein lies a question, but also a clue to an answer.
What if applet does not run? 
Both theorems allow for points D, E, and F lying not only on the sides of ΔABC but also on their extensions. In fact, the theorem of Menelaus requires that at least one of the points lie on the extension of the corresponding side due to an obvious fact that a straight line can't cross internally all three sides of a triangle. (As an aside, in a 1945 Russian mathematics competition, one boy did not see that fact as obvious. For that insight, he was awarded the first prize although he did not solve a single problem.)
Of the three points D, E, or F, none, one, two or all three may lie externally to the triangle, on the side extensions. If 1 or 3 points lie on extensions, we have the Menelaus theorem. In the other two cases, the theorem is Ceva's. It then appears that the formulations of the theorems beg a question. For example, Menelaus' should have read
Let three points F, D, and E, lie respectively on the sides AB, BC, and AC of ΔABC or their extensions. Assume just 1 or all three of the points lie on side extensions. Then the points are collinear iff (2) holds. 
Although cumbersome, the latter formulation must be preferred unless of course there is a better way to rectify the situation. The commonly used approach avoids ambiguity by considering signed segments. By convention, for any two points P and Q, PQ and QP denote segments of different signs such that
For three points A, F, B, the ratio AF/FB is positive if F lies between A and B, and is negative if F lies on extension of the segment AB. It is also always true that
If we are bent on using the same product for both theorems, then there are two possibilities:


(When squared, the Ceva and Menelaus conditions coincide, naturally. Curiously, this is how they come out  squared  in a unified proof based on the 4 Travelers Problem. They come with their signs in plain view in a study of pole/polar relationships in a triangle.)
The lines AD, BE, and CF in Ceva's theorem are called Cevians. The theorem gives a necessary and sufficient condition for the concurrency of three Cevians. A straight line is often called a transversal to emphasize its relation to another shape. The Menelaus theorem gives a necessary and sufficient condition for three points  one on each side of a triangle  to lie on a transversal. What is a Cevian in one triangle is a transversal in another. For example, the Cevian BE serves as a transversal in ΔADC while CF is a transversal in ΔADB. Write condition (2) for the two triangles:
DB/CB · CE/AE · AK/DK = 1 and DC/BC · BF/AF · AK/DK = 1. 
Eliminate AK/DK from the two identities and recollect the sign convention (
There are great many proofs of the theorem of Menelaus. I'll give just two. One is the most economical in terms of required constructions (just one additional line), the other highlights an unexpected link between the theorem and other geometrical concepts.
What if applet does not run? 
Proof #1
Draw AP parallel to DE. Triangles ABP and BDF are similar as are triangles ACP and CDE. The first pair gives
Proof #2
From the vertices of ΔABC drop perpendiculars on the transversal. Consider three pairs of similar triangles AH_{a}F and BH_{b}F, CH_{c}D and BH_{b}D, and AH_{a}E and CH_{c}E. From these we get
AF/BF = AH_{a}/BH_{b}, BD/CD = BH_{b}/CH_{c}, CE/AE = CH_{c}/AH_{a}. 
Multiply the three to obtain (2).
The "backward" step is proven in a more or less standard manner very much as it was done for Ceva's theorem.
Let there be three points such that AF/BF·BD/CD·CE/AE = 1 holds. Assume on the contrary that the points are not colinear. Pick up any two. Say D and E. Draw the line DE and find its intersection F' with AB. Then by the "forward" step
The second proof is suggestive. Points D, E, and F serve as centers of homothety for pairs of similar shapes  segments in this case  BH_{b} and CH_{c}, AH_{a} and CH_{c}, and AH_{a} and BH_{b}. The Menelaus theorem then says that, given three shapes two of which were obtained from the third by central similarity transformaitions, then (naturally) there exists a homothety that transforms the first into the second, and centers of all three transformations are collinear.
As a bonus, we obtain a solution to the following problem of three circles:
Let there be three circles of different radii lying completely outside each other. To exclude a trivial case, assume also that their centers are not collinear, i.e. the three centers do not lie on the same straight line. Under these conditions, six external tangents to two of the three circles, taken pairwise, intersect at three points. Those three points are collinear. 
Furthermore, from the foregoing discussion it follows that two pairs of tangents may be taken internally.
(Note: A. Einstein used the Menelaus theorem in his correspondence on ugly and elegant proofs.)
The simplicity of the Menelaus theorem is deceptive. A few applications of (2) yield easily theorems of Desargues, Pappus, and Pascal. Another striking application can be found in [Honsberger].
At each vertex of a triangle there is a couple of angle bisectors: a bisector of the interior angle and a bisector of the exterior angle. It's well known that the bisectors taken one for each vertex are concurrent provided none or two of the bisectors are external. This is a particular case of Ceva's theorem. In general, every angle bisector crosses the opposite side of the triangle. It then follows from Menelaus' theorem, that every three such points are collinear provided none or two of the bisectors are internal.
What if applet does not run? 
References
 H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967
 C. W. Dodge, Euclidean Geometry and Transformations, Dover, 2004 (reprint of 1972 edition), p. 5.
 R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidian Geometry, MAA, 1995
 D. Pedoe, Geometry: A Comprehensive Course, Dover, 1970
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
 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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