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| [[File:Menelaus' theorem 1.svg|right|thumb|Menelaus' theorem, case 1: line DEF passes inside triangle ABC]]
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| '''Menelaus' theorem''', named for [[Menelaus of Alexandria]], is a theorem about [[triangle]]s in [[plane geometry]]. Given a triangle ''ABC'', and a [[Transversal (geometry)|transversal]] line that crosses ''BC'', ''AC'' and ''AB'' at points ''D'', ''E'' and ''F'' respectively, with ''D'', ''E'', and ''F'' distinct from ''A'', ''B'' and ''C'', then
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| : <math>\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = -1.</math>
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| or simply
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| : <math>AF \times BD \times CE= - FB \times DC \times EA .</math>
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| This equation uses signed lengths of segments, in other words the length ''AB'' is taken to be positive or negative according to whether ''A'' is to the left or right of ''B'' in some fixed orientation of the line. For example, ''AF''/''FB'' is defined as having positive value when ''F'' is between ''A'' and ''B'' and negative otherwise.
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| The [[Theorem#Converse|converse]] is also true: If points ''D'', ''E'' and ''F'' are chosen on ''BC'', ''AC'' and ''AB'' respectively so that
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| : <math>\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = -1,</math>
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| then ''D'', ''E'' and ''F'' are collinear. The converse is often included as part of the theorem.
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| The theorem is very similar to [[Ceva's theorem]] in that their equations differ only in sign.
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| == Proof == | |
| [[File:Menelaos' theorem 2.svg|right|thumb|Menelaus' theorem, case 2: line DEF is entirely outside triangle ABC]]
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| A standard proof is as follows:<ref>Follows Russel</ref>
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| 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 DEF misses the triangle (lower diagram), or one is negative and the other two are positive, the case where DEF crosses two sides of the triangle. (See [[Pasch's axiom]].)
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| To check the magnitude, construct perpendiculars from ''A'', ''B'', and ''C'' to the line ''DEF'' and let their lengths be ''a, b,'' and ''c'' respectively. Then by [[Similarity (geometry)|similar]] triangles it follows that |''AF''/''FB''| = |''a''/''b''|, |''BD''/''DC''| = |''b''/''c''|, and |''CE''/''EA''| = ''c''/''a''. So
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| : <math>\left|\frac{AF}{FB}\right| \cdot \left|\frac{BD}{DC}\right| \cdot \left|\frac{CE}{EA}\right| = \left| \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} \right| = 1.</math>
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| For a simpler, if less symmetrical way to check the magnitude,<ref>Follows {{cite book |title=Inductive Plane Geometry|first=George Irving|last=Hopkins|publisher=D.C. Heath & Co.|year=1902|chapter=Art. 983}}</ref> draw ''CK'' parallel to ''AB'' where ''DEF'' meets ''CK'' at ''K''. Then by similar triangles
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| : <math>\left|\frac{BD}{DC}\right| = \left|\frac{BF}{CK}\right|,\,\left|\frac{AE}{EC}\right| = \left|\frac{AF}{CK}\right|</math> | |
| and the result follows by eliminating ''CK'' from these equations.
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| The converse follows as a corollary.<ref>Follows Russel with some simplification</ref> Let ''D'', ''E'' and ''F'' be given on the lines ''BC'', ''AC'' and ''AB'' so that the equation holds. Let ''F''′ be the point where ''DE'' crosses ''AB''. Then by the theorem, the equation also holds for ''D'', ''E'' and ''F''′. Comparing the two,
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| : <math>\frac{AF}{FB} = \frac{AF'}{F'B}.</math>
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| But at most one point can cut a segment in a given ratio so ''F''=''F''′.
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| === A non-computational proof using homothecies ===
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| The following proof<ref>See Michèle Audin, Géométrie, éditions BELIN, Paris 1998: indication for exercice 1.37, p. 273</ref> uses only notions of [[affine geometry]], notably [[Homothetic transformation|homothecies]].
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| Whether or not ''D'', ''E'', ''F'' are collinear, there are three homothecies with centers ''D'', ''E'', ''F'' that respectively send ''B'' to ''C'', ''C'' to ''A'', and ''A'' to ''B''. The composition of the three then is an element of the group of homothecy-translations that fixes ''B'', so it is a homothecy with center ''B'', possibly with ratio 1 (in which case it is the identity). This composition fixes the line ''DE'' if and only if ''F'' is collinear with ''D'' and ''E'' (since the first two homothecies certainly fix ''DE'', and the third does so only if ''F'' lies on ''DE''). Therefore ''D'', ''E'', ''F'' are collinear if and only if this composition is the identity, which means that the product of the three ratios is 1:
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| : <math>\frac{\overrightarrow{DC}}{\overrightarrow{DB}} \times
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| \frac{\overrightarrow{EA}}{\overrightarrow{EC}} \times
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| \frac{\overrightarrow{FB}}{\overrightarrow{FA}} = 1,</math>
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| which is equivalent to the given equation.
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| ==History==
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| 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>{{cite book|
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| last=Smith|first=D.E.|title=History of Mathematics|isbn=0-486-20430-8
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| |publisher=Courier Dover Publications|year=1958|volume=II|page=607}}</ref>
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| In [[Almagest]], [[Ptolemy]] applies the theorem on a number of problems in spherical astronomy.<ref name="rashed">{{cite book|last=Rashed|first=Roshdi|title=Encyclopedia of the history of Arabic science|volume=2|page=483|year=1996|publisher=Routledge|location=London|isbn=0-415-02063-8}}</ref> During the [[Islamic Golden Age]], Muslim scholars devoted a number of works that engaged in the study of Menelaus' 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' theorem that led to the [[sine rule]],<ref name="musa">{{cite journal|last=Moussa|first=Ali|title=Mathematical Methods in Abū al-Wafāʾ's Almagest and the Qibla Determinations|journal=Arabic Sciences and Philosophy|year=2011|volume=21|issue=1|publisher=[[Cambridge University Press]]|doi=10.1017/S095742391000007X}}</ref> or works composed as independent treatises such as:
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| * The "Treatise on the Figure of Secants" (''Risala fi shakl al-qatta''') by [[Thabit ibn Qurra]].<ref name="rashed" />
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| * [[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 [[Al-Tusi]] and [[Nasir al-Din al-Tusi]].<ref name="rashed" />
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| * Work by [[al-Sijzi]].<ref name="musa" />
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| * ''Tahdhib'' by [[Abu Nasr ibn Iraq]].<ref name="musa" />
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| ==References==
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| {{Reflist}}
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| * {{cite book |title=Pure Geometry
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| |first=John Wellesley|last=Russell|publisher=Clarendon Press|year=1905
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| |chapter= Ch. 1 §6 "Menelaus' Theorem"
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| |url=http://books.google.com/books?id=r3ILAAAAYAAJ}}
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| == External links ==
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| * [http://planetmath.org/?op=getobj&from=objects&id=3092 Alternate proof] of Menelaus' theorem, from [[PlanetMath]]
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| * [http://www.cut-the-knot.org/Curriculum/Geometry/MenelausFromCeva.shtml Menelaus From Ceva]
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| * [http://www.cut-the-knot.org/4travelers/CevaAndMenelaus.shtml Ceva and Menelaus Meet on the Roads]
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| * [http://www.mathpages.com/home/kmath442/kmath442.htm Menelaus and Ceva] at MathPages
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| * [http://demonstrations.wolfram.com/MenelausTheorem/ Menelaus' Theorem] by Jay Warendorff. [[The Wolfram Demonstrations Project]].
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| * {{MathWorld |title=Menelaus' Theorem |urlname=MenelausTheorem}}
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| [[Category:Affine geometry]]
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| [[Category:Triangle geometry]]
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| [[Category:Articles containing proofs]]
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| [[Category:Theorems in plane geometry]]
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