 # Cut The Knot!

An interactive column using Java applets
by 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.

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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 PQ = - QP. (It is also possible to assign a direction on the line PQ thus endowing each of the segments with a proper sign. Regardless of the choice of direction, we shall have PQ = - QP, the only important relation for the present.) In general, two segments (on one and the same line) looking into the same direction have the same sign, those looking into opposite directions have different signs.

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 AF/FB = -AF/BF. For Ceva's theorem, either all three points D, E, and F lie on the corresponding segments BC, CA, and AB internally, or some two of them lie on segment extensions, i.e. externally. Accordingly, in (1), either all three ratios are positive or exactly two of them are negative. In both cases the product is 1. For the Menelaus theorem, the product in (1) is negative, while the product in (2) is negative for Ceva's theorem.

If we are bent on using the same product for both theorems, then there are two possibilities:

Ceva: AF/FB · BD/DC · CE/EA = 1 AF/FB · BD/DC · CE/EA = -1

Ceva: AF/BF · BD/CD · CE/AE = -1 AF/BF · BD/CD · CE/AE = 1

(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 (CB = -BC, etc.) The effort should lead to (1) which shows that the Menelaus theorem implies that of Ceva. (More accurately, this shows that (1) holds for three intersecting Cevians. The converse follows in a standard manner by the reductio ad absurdum argument. The necessary and sufficient conditions in the Menelaus theorem also follow from each other in a similar manner.) The theorems are indeed equivalent. But this fact deserves a separate discussion. 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.

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### Proof #1

Draw AP parallel to DE. Triangles ABP and BDF are similar as are triangles ACP and CDE. The first pair gives AF/BF = PD/BD, the second AE/CE = PD/CD. Combine the two to get (2).

### Proof #2

From the vertices of ΔABC 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

 AF/BF = AHa/BHb, BD/CD = BHb/CHc, CE/AE = CHc/AHa.

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 AF'/BF'·BD/CD·CE/AE = 1. From which AF'/BF' = AF/BF. By subtracting 1 from both sides one gets AB/AF' = AB/AF, from which F' = F.

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:

 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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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### References

1. H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967
2. C. W. Dodge, Euclidean Geometry and Transformations, Dover, 2004 (reprint of 1972 edition), p. 5.
3. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidian Geometry, MAA, 1995
4. D. Pedoe, Geometry: A Comprehensive Course, Dover, 1970 70386994