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Cut The Knot!An interactive column using Java appletsby Alex Bogomolny |
The Menelaus Theorem
November 1999
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My previous column suffers from a conspicuous omission. I have quoted Dan Pedoe,
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 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
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.
Both theorems allow for points D, E, and F lying not only on the sides of 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
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
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.
Proof #1Draw AP parallel to DE. Triangles ABP and BDF are similar as are triangles ACP and CDE. The first pair gives Proof #2From the vertices of
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 - BHb and CHc, AHa and CHc, and AHa and BHb. 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:
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.
References
Menelaus and Ceva
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ABC. Then the points are collinear iff
ADB. Write condition (2) for the two triangles:| 37257071 |
