Cut The Knot!An interactive column using Java applets
by Alex Bogomolny
My previous column suffers from a conspicuous omission. I have quoted Dan Pedoe,
but then went on and discussed only a fine proof of Ceva's theorem. Menelaus' theorem was never mentioned again. This time, I'd like to give the latter its due. Here are the two theorems side by side:
Three Cevians AD, BE, and CF are concurrent if and only if
Let three points F, D, and E, lie respectively on the sides AB, BC, and AC of ABC. Then the points are collinear if snf only if
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 enunciated by the theorems of Ceva and Menelaus are indeed different. Therein lies a question, but also a clue to an answer.
Both theorems allow the points D, E, and F to lie not only on the sides of ABC but also on the extensions of the sides. 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 all three sides of a triangle internally. (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 suggest 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 most common
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:
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 ADC while CF is a transversal in ADB. Write condition (2) for the two triangles:
Eliminate AK/DK from the two identities and recollect the sign
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.
Draw AP parallel to DE. Triangles ABP and BDF are similar as are
triangles ACP and CDE. The first pair gives
From the vertices of ABC drop perpendiculars on the transversal. Consider three pairs of similar triangles AHaF and BHbF, CHcD and BHbD, and AHaE and CHcE. From these we get
Multiply the three to obtain (2).
The latter 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 transformations, 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.
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 all or two of the bisectors are internal. 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 the Menelaus theorem, that every three such points are collinear provided all or two of the bisectors are external.
Alex Bogomolny has started and still maintains a popular Web site Interactive Mathematics Miscellany and Puzzles to which he brought more than 10 years of college instruction and, at least as much, programming experience. He holds M.S. degree in Mathematics from the Moscow State University and Ph.D. in Applied Mathematics from the Hebrew University of Jerusalem. He can be reached at email@example.com
Copyright ® 1997-1999 Alexander Bogomolny