Menelaus's theorem
In Euclidean geometry, Menelaus's theorem, named for Menelaus of Alexandria, is a proposition about triangles in plane geometry. Suppose we have a triangle Template:Math, and a transversal line that crosses Template:Mvar at points Template:Mvar respectively, with Template:Mvar distinct from Template:Mvar. A weak version of the theorem states that
<math display=block>\left|\frac{\overline{AF}}{\overline{FB}}\right| \times \left|\frac{\overline{BD}}{\overline{DC}}\right| \times \left|\frac{\overline{CE}}{\overline{EA}}\right| = 1,</math>
where "| |" denotes absolute value (i.e., all segment lengths are positive).
The theorem can be strengthened to a statement about signed lengths of segments, which provides some additional information about the relative order of collinear points. Here, the length Template:Mvar is taken to be positive or negative according to whether Template:Mvar is to the left or right of Template:Mvar in some fixed orientation of the line; for example, <math>\tfrac{\overline{AF}}{\overline{FB}}</math> is defined as having positive value when Template:Mvar is between Template:Mvar and Template:Mvar and negative otherwise. The signed version of Menelaus's theorem states
<math display=block>\frac{\overline{AF}}{\overline{FB}} \times \frac{\overline{BD}}{\overline{DC}} \times \frac{\overline{CE}}{\overline{EA}} = - 1.</math>
Equivalently,<ref>Russell, p. 6.</ref>
<math display=block>\overline{AF} \times \overline{BD} \times \overline{CE} = - \overline{FB} \times \overline{DC} \times \overline{EA}.</math>
Some authors organize the factors differently and obtain the seemingly different relation<ref>Template:Citation</ref> <math display=block>\frac{\overline{FA}}{\overline{FB}} \times \frac{\overline{DB}}{\overline{DC}} \times \frac{\overline{EC}}{\overline{EA}} = 1,</math> but as each of these factors is the negative of the corresponding factor above, the relation is seen to be the same.
The converse is also true: If points Template:Mvar are chosen on Template:Mvar respectively so that <math display=block>\frac{\overline{AF}}{\overline{FB}} \times \frac{\overline{BD}}{\overline{DC}} \times \frac{\overline{CE}}{\overline{EA}} = -1,</math> then Template:Mvar are collinear. The converse is often included as part of the theorem. (Note that the converse of the weaker, unsigned statement is not necessarily true.)
The theorem is very similar to Ceva's theorem in that their equations differ only in sign. By re-writing each in terms of cross-ratios, the two theorems may be seen as projective duals.<ref>Template:Cite journal</ref>
ProofsEdit
A standard proofEdit
A proof given by John Wellesley Russell uses Pasch's axiom to consider cases where a line does or does not meet a triangle.<ref>Template:Cite book </ref> First, the sign of the left-hand side will be negative since either all three of the ratios are negative, the case where the line Template:Mvar misses the triangle (see diagram), or one is negative and the other two are positive, the case where Template:Mvar crosses two sides of the triangle.
To check the magnitude, construct perpendiculars from Template:Mvar to the line Template:Mvar and let their lengths be Template:Mvar respectively. Then by similar triangles it follows that <math display=block>\left|\frac{\overline{AF}}{\overline{FB}}\right| = \left|\frac{a}{b}\right|, \quad \left|\frac{\overline{BD}}{\overline{DC}}\right| = \left|\frac{b}{c}\right|, \quad \left|\frac{\overline{CE}}{\overline{EA}}\right| = \left|\frac{c}{a}\right|.</math>
Therefore, <math display=block>\left|\frac{\overline{AF}}{\overline{FB}}\right| \times \left|\frac{\overline{BD}}{\overline{DC}}\right| \times \left|\frac{\overline{CE}}{\overline{EA}}\right| = \left| \frac{a}{b} \times \frac{b}{c} \times \frac{c}{a} \right| = 1.</math>
For a simpler, if less symmetrical way to check the magnitude,<ref>Follows Template:Cite book</ref> draw Template:Mvar parallel to Template:Mvar where Template:Mvar meets Template:Mvar at Template:Mvar. Then by similar triangles <math display=block>\left|\frac{\overline{BD}}{\overline{DC}}\right| = \left|\frac{\overline{BF}}{\overline{CK}}\right|, \quad \left|\frac{\overline{AE}}{\overline{EC}}\right| = \left|\frac{\overline{AF}}{\overline{CK}}\right|,</math> and the result follows by eliminating Template:Mvar from these equations.
The converse follows as a corollary.<ref>Follows Russel with some simplification</ref> Let Template:Mvar be given on the lines Template:Mvar so that the equation holds. Let Template:Mvar be the point where Template:Mvar crosses Template:Mvar. Then by the theorem, the equation also holds for Template:Mvar. Comparing the two, <math display=block>\frac{\overline{AF}}{\overline{FB}} = \frac{\overline{AF'}}{\overline{F'B}}\ .</math> But at most one point can cut a segment in a given ratio so Template:Math
A proof using homothetiesEdit
The following proof<ref>Michèle Audin (1998) Géométrie, éditions BELIN, Paris: indication for exercise 1.37, page 273</ref> uses only notions of affine geometry, notably homotheties. Whether or not Template:Mvar are collinear, there are three homotheties with centers Template:Mvar that respectively send Template:Mvar to Template:Mvar, Template:Mvar to Template:Mvar, and Template:Mvar to Template:Mvar. The composition of the three then is an element of the group of homothety-translations that fixes Template:Mvar, so it is a homothety with center Template:Mvar, possibly with ratio 1 (in which case it is the identity). This composition fixes the line Template:Mvar if and only if Template:Mvar is collinear with Template:Mvar (since the first two homotheties certainly fix Template:Mvar, and the third does so only if Template:Mvar lies on Template:Mvar). Therefore Template:Mvar are collinear if and only if this composition is the identity, which means that the magnitude of the product of the three ratios is 1: <math display=block>\frac{\overrightarrow{DC}}{\overrightarrow{DB}} \times
\frac{\overrightarrow{EA}}{\overrightarrow{EC}} \times
\frac{\overrightarrow{FB}}{\overrightarrow{FA}} = 1,</math>
which is equivalent to the given equation.
HistoryEdit
It is uncertain who actually discovered the theorem; however, the oldest extant exposition appears in Spherics by Menelaus. In this book, the plane version of the theorem is used as a lemma to prove a spherical version of the theorem.<ref>Template:Cite book</ref>
In Almagest, Ptolemy applies the theorem on a number of problems in spherical astronomy.<ref name="rashed">Template:Cite book</ref> During the Islamic Golden Age, Muslim scholars devoted a number of works that engaged in the study of Menelaus's theorem, which they referred to as "the proposition on the secants" (shakl al-qatta'). The complete quadrilateral was called the "figure of secants" in their terminology.<ref name="rashed" /> Al-Biruni's work, The Keys of Astronomy, lists a number of those works, which can be classified into studies as part of commentaries on Ptolemy's Almagest as in the works of al-Nayrizi and al-Khazin where each demonstrated particular cases of Menelaus's theorem that led to the sine rule,<ref name="musa">Template:Cite journal</ref> or works composed as independent treatises such as:
- The "Treatise on the Figure of Secants" (Risala fi shakl al-qatta') by Thabit ibn Qurra.<ref name="rashed" />
- Husam al-Din al-Salar's Removing the Veil from the Mysteries of the Figure of Secants (Kashf al-qina' 'an asrar al-shakl al-qatta'), also known as "The Book on the Figure of Secants" (Kitab al-shakl al-qatta') or in Europe as The Treatise on the Complete Quadrilateral. The lost treatise was referred to by Sharaf al-Din al-Tusi and Nasir al-Din al-Tusi.<ref name="rashed" />
- Work by al-Sijzi.<ref name="musa" />
- Tahdhib by Abu Nasr ibn Iraq.<ref name="musa" />
- Roshdi Rashed and Athanase Papadopoulos, Menelaus' Spherics: Early Translation and al-Mahani'/al-Harawi's version (Critical edition of Menelaus' Spherics from the Arabic manuscripts, with historical and mathematical commentaries), De Gruyter, Series: Scientia Graeco-Arabica, 21, 2017, 890 pages. Template:ISBN
ReferencesEdit
External linksEdit
- Alternate proof of Menelaus's theorem, from PlanetMath
- Menelaus From Ceva
- Ceva and Menelaus Meet on the Roads
- Menelaus and Ceva at MathPages
- Demo of Menelaus's theorem by Jay Warendorff. The Wolfram Demonstrations Project.
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