Movatterモバイル変換


[0]ホーム

URL:


Jump to content
WikipediaThe Free Encyclopedia
Search

Menelaus's theorem

From Wikipedia, the free encyclopedia
Geometric relation on line segments formed by a line cutting through a triangle
Menelaus's theorem, case 1: lineDEF passes inside triangleABC

InEuclidean geometry,Menelaus's theorem, named forMenelaus of Alexandria, is a proposition abouttriangles inplane geometry. Suppose we have a triangleABC, and atransversal line that crossesBC, AC, AB at pointsD, E, F respectively, withD, E, F distinct fromA, B, C. A weak version of the theorem states that

|AF¯FB¯|×|BD¯DC¯|×|CE¯EA¯|=1,{\displaystyle \left|{\frac {\overline {AF}}{\overline {FB}}}\right|\times \left|{\frac {\overline {BD}}{\overline {DC}}}\right|\times \left|{\frac {\overline {CE}}{\overline {EA}}}\right|=1,}

where "| |" denotesabsolute value (i.e., all segment lengths are positive).

The theorem can be strengthened to a statement aboutsigned lengths of segments, which provides some additional information about the relative order of collinear points. Here, the lengthAB is taken to be positive or negative according to whetherA is to the left or right ofB in some fixed orientation of the line; for example,AF¯FB¯{\displaystyle {\tfrac {\overline {AF}}{\overline {FB}}}} is defined as having positive value whenF is betweenA andB and negative otherwise. The signed version of Menelaus's theorem states

AF¯FB¯×BD¯DC¯×CE¯EA¯=1.{\displaystyle {\frac {\overline {AF}}{\overline {FB}}}\times {\frac {\overline {BD}}{\overline {DC}}}\times {\frac {\overline {CE}}{\overline {EA}}}=-1.}

Equivalently,[1]

AF¯×BD¯×CE¯=FB¯×DC¯×EA¯.{\displaystyle {\overline {AF}}\times {\overline {BD}}\times {\overline {CE}}=-{\overline {FB}}\times {\overline {DC}}\times {\overline {EA}}.}

Some authors organize the factors differently and obtain the seemingly different relation[2]FA¯FB¯×DB¯DC¯×EC¯EA¯=1,{\displaystyle {\frac {\overline {FA}}{\overline {FB}}}\times {\frac {\overline {DB}}{\overline {DC}}}\times {\frac {\overline {EC}}{\overline {EA}}}=1,}but as each of these factors is the negative of the corresponding factor above, the relation is seen to be the same.

Theconverse is also true: If pointsD, E, F are chosen onBC, AC, AB respectively so thatAF¯FB¯×BD¯DC¯×CE¯EA¯=1,{\displaystyle {\frac {\overline {AF}}{\overline {FB}}}\times {\frac {\overline {BD}}{\overline {DC}}}\times {\frac {\overline {CE}}{\overline {EA}}}=-1,}thenD, E, F arecollinear. 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 toCeva's theorem in that their equations differ only in sign. By re-writing each in terms ofcross-ratios, the two theorems may be seen asprojective duals.[3]

Proofs

[edit]
Menelaus's theorem, case 2: lineDEF is entirely outside triangleABC

A standard proof

[edit]

A proof given by John Wellesley Russell usesPasch's axiom to consider cases where a line does or does not meet a triangle.[4] First, the sign of theleft-hand side will be negative since either all three of the ratios are negative, the case where the lineDEF misses the triangle (see diagram), or one is negative and the other two are positive, the case whereDEF crosses two sides of the triangle.

To check the magnitude, construct perpendiculars fromA, B, C to the lineDEF and let their lengths bea, b, c respectively. Then bysimilar triangles it follows that|AF¯FB¯|=|ab|,|BD¯DC¯|=|bc|,|CE¯EA¯|=|ca|.{\displaystyle \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|.}

Therefore,|AF¯FB¯|×|BD¯DC¯|×|CE¯EA¯|=|ab×bc×ca|=1.{\displaystyle \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.}

For a simpler, if less symmetrical way to check the magnitude,[5] drawCK parallel toAB whereDEF meetsCK atK. Then by similar triangles|BD¯DC¯|=|BF¯CK¯|,|AE¯EC¯|=|AF¯CK¯|,{\displaystyle \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|,}and the result follows by eliminatingCK from these equations.

The converse follows as a corollary.[6] LetD, E, F be given on the linesBC, AC, AB so that the equation holds. LetF' be the point whereDE crossesAB. Then by the theorem, the equation also holds forD, E, F'. Comparing the two,AF¯FB¯=AF¯FB¯ .{\displaystyle {\frac {\overline {AF}}{\overline {FB}}}={\frac {\overline {AF'}}{\overline {F'B}}}\ .}But at most one point can cut a segment in a given ratio soF =F'.

A proof using homotheties

[edit]
Homothetie centers D, E F are colinear iff the composition is identity.

The following proof[7] uses only notions ofaffine geometry, notablyhomotheties.Whether or notD, E, F are collinear, there are three homotheties with centersD, E, F that respectively sendB toC,C toA, andA toB. The composition of the three then is an element of the group of homothety-translations that fixesB, so it is a homothety with centerB, possibly with ratio 1 (in which case it is the identity). This composition fixes the lineDE if and only ifF is collinear withD, E (since the first two homotheties certainly fixDE, and the third does so only ifF lies onDE). ThereforeD, E, F 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:DCDB×EAEC×FBFA=1,{\displaystyle {\frac {\overrightarrow {DC}}{\overrightarrow {DB}}}\times {\frac {\overrightarrow {EA}}{\overrightarrow {EC}}}\times {\frac {\overrightarrow {FB}}{\overrightarrow {FA}}}=1,}which is equivalent to the given equation.

History

[edit]

It is uncertain who actually discovered the theorem; however, the oldest extant exposition appears inSpherics by Menelaus. In this book, the plane version of the theorem is used as a lemma to prove a spherical version of the theorem.[8]

InAlmagest,Ptolemy applies the theorem on a number of problems in spherical astronomy.[9] During theIslamic 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'). Thecomplete quadrilateral was called the "figure of secants" in their terminology.[9]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'sAlmagest as in the works ofal-Nayrizi andal-Khazin where each demonstrated particular cases of Menelaus's theorem that led to thesine rule,[10] or works composed as independent treatises such as:

In the West the theorem was rediscovered by the Italian mathematicianGiovanni Ceva.[11]

References

[edit]
  1. ^Russell,p. 6.
  2. ^Johnson, Roger A. (2007) [1927],Advanced Euclidean Geometry, Dover, p. 147,ISBN 978-0-486-46237-0
  3. ^Benitez, Julio (2007)."A Unified Proof of Ceva and Menelaus' Theorems Using Projective Geometry"(PDF).Journal for Geometry and Graphics.11 (1):39–44.
  4. ^Russell, John Wellesley (1905). "Ch. 1 §6 "Menelaus' Theorem"".Pure Geometry. Clarendon Press.
  5. ^FollowsHopkins, George Irving (1902). "Art. 983".Inductive Plane Geometry. D.C. Heath & Co.
  6. ^Follows Russel with some simplification
  7. ^Michèle Audin (1998)Géométrie, éditions BELIN, Paris: indication for exercise 1.37, page 273
  8. ^Smith, D.E. (1958).History of Mathematics. Vol. II. Courier Dover Publications. p. 607.ISBN 0-486-20430-8.{{cite book}}:ISBN / Date incompatibility (help)
  9. ^abcdRashed, Roshdi (1996).Encyclopedia of the history of Arabic science. Vol. 2. London: Routledge. p. 483.ISBN 0-415-02063-8.
  10. ^abcMoussa, Ali (2011). "Mathematical Methods in Abū al-Wafāʾ's Almagest and the Qibla Determinations".Arabic Sciences and Philosophy.21 (1).Cambridge University Press:1–56.doi:10.1017/S095742391000007X.S2CID 171015175.
  11. ^O'Connor, John J.;Robertson, Edmund F.,"Giovanni Ceva",MacTutor History of Mathematics Archive,University of St Andrews

External links

[edit]
Wikimedia Commons has media related toMenelaos's theorem.
Mathematicians
(timeline)
Treatises
Concepts
and definitions
Results
InElements
Centers/Schools
Related
History of
Other cultures
Retrieved from "https://en.wikipedia.org/w/index.php?title=Menelaus%27s_theorem&oldid=1312268675"
Categories:
Hidden categories:

[8]ページ先頭

©2009-2025 Movatter.jp