Menelaus Theorem: Proofs Ugly and Elegant
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I adapt Einstein's second proof to the notations I used previously. It uses the principle that two triangles with a common (or supplementary) angle are related as the products of adjacent sides. (In the manner of the Ancients, Einstein means here the ratio of the areas of two triangles, and refers to the formula that gives the area as half the product of two sides and the sine of the angle between them.) We choose as these triangles those formed by the individual vertices of the triangle with the corresponding segments of the transversal:
AEF, and (those formed derived from it by cyclical transformation)
BFD,
CDE. We have
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area( |
If this equation is multiplied by those derived from it by cyclical transformation, we obtain
| 1 = AF·BD·CE / BF·CD·AE, |
as required.
Elegance, as beauty, is in the eye of the beholder. I agree with Einstein on one point and disagree on another. Line AP appears to be an artificial construct as it destroys the inner symmetry of the theorem. However, I think I mind less using auxiliary constructs. The additional lines in the second of the cited proofs helped relate (and that quite unexpectedly) the Menelaus Theorem to the homothety transformation and its properties. (The 4 travellers problem and one of its proofs serve as another example where a considerable benefit is drawn from an introduction of additional elements. The proof adds a 3-dimensional frame to a 2-dimensional problem making the whole picture a work of art.)
Yet another proof [F. G.-M., p. 84] adds elements to the original diagram. Like Proof #2 is does so in a symmetric manner. It renders the theorem virtually trivial.
Draw a line perpendicular to the transversal EDF. Let it intersect the latter at point K, and denote the projections of the vertices A, B, C on that line as Ka, Kb, and Kc, respectively. As is well known, the segments cut on two lines by a family of parallel lines are in the same ratio. Pairing the side lines of ΔABC with the just constructed line one at a time we get:
| AF/BF = KaK/KbK, BD/CD = KbK/KcK, CE/AE = KcK/KaK, |
so that obviously
| AF/BF · BD/CD · CE/AE = KaK/KbK · KbK/KcK · KcK/KaK = 1. |
Copyright © 1996-2008 Alexander Bogomolny
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