## Ceva and Menelaus Meet on the Roads

Ceva's theorem gives a condition for concurrency of three lines, while Menelaus's theorem provides a similar condition for collinearity of three points. Stuart Anderson made a brilliant observation that an analogous dichotomy holds for a lemma that underlies his solution to the 4 Travelers Problem. Pursuing the analogy he came up with a unified proof of Ceva's and Menelaus' theorems which is presented below.

As usual, let ABC be a triangle, and let D lie on BC, E on CA, F on AB. Let travelers d, e, and f start at time t = 0 from D, E, and F along lines AD, BE, and CF. Assume that they all meet each other (pairwise). Focus on two of the travelers, say d and e. Start two other travelers d' and e' from D and E along BC and CA, also at t = 0, and give them speeds so that d' meets e at B and e' meets d at A. You can always do this by choosing the correct speeds.

Now the 4 travelers problem shows that d' meets e' at C. Let's look at the times of travel (Vx means velocity of traveler x):

 (1) EA / Ve' = AD / Vd because d and e' meet at A, (2) BD / Vd' = BE / Ve because e and d' meet at B, (3) CE / Ve' = DC / Vd' because d' and e' meet at C.

Combining these gives

 (4) CE / EA · BD / DC · AD / BE = Vd / Ve,

and multiplying this together with the two other equations we get by cyclic permutation gives

 (5) (CE / EA · AF / FB · BD / DC)2 = 1,

which, quite appropriately, is the square of the Ceva and Menelaus conditions.

Now let's look at the quantities in (5) as signed segments. It's a simple matter to verify that, for the concurrency condition, the ratios in (5) are either all positive, or exactly two of them are negative. Thus the product of the ratios is always positive, which leads to Ceva's theorem. For the collinearity condition, the ratios in (5) are either all negative, or exactly two of them are positive. The product, therefore, is always negative, which leads to Menelaus' theorem.

Conversely, in the three cyclic equations like (4), we can choose Vd arbitrarily, then choose Ve to satisfy the first of the equations, then choose Vf compatible with Ve in the second equation. In the third equation, we have no more choices, but (5) ensures that the third equation is equivalent the product of the first two, so it is automatically satisfied. Similarly, (4) and its permutations make it possible to choose Vd', Ve', and Vf ' to satisfy (1)-(3) and their permutations. (Note: because they drop out of (4), Vd', Ve' and Vf ' can be chosen independently when focusing on each pair of travelers. The Vd' used with d and e need not equal the Vd' used with d and f.) The timings then imply that each traveler meets each of the others.

Therefore the condition that travelers d, e, and f all meet pairwise is equivalent to the hypothesis of Ceva and Menelaus. Now by the lemma I used in proving the 4 travelers problem, the travelers d, e, and f are all collinear (Menelaus) or their paths are concurrent (Ceva).

This is quite an interesting (to me, at least) connection, because in the original 4 travelers problem as posed, the parallelism of the travelers or the concurrency of their paths is in neither the statement nor the conclusion, but occur in the body of the proof. Further, the original form of the 4 travelers problem excluded the Ceva-like case by requiring the lines to be in general position. It is only when you look at both the case of general position lines and the general position travelers that the connection to Ceva and Menelaus appears.

Note that I don't think this is the "best" way to prove Ceva and Menelaus, because other ways are simpler, but it makes a connection I haven't seen anywhere.

(AB: Neither have I. However, Maj/Cpt/Lt Pestich pointed out that the close relationship between the two theorems can be derived by observing the pole/polar relationships in a triangle. Lt Pestich has also mentioned as his source the Russian edition of I. M. Yaglom's Geometric Transformations, v. 2 (1956), problem 165. The relevant part of the book is available in English as Geometric Transformations, v. III, MAA, 1973, problem 66. Our proof is different and more transparent.) 68246600