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Ceva's theorem

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Image:Ceva's theorem 1.svg
Ceva's theorem, case 1: the three lines are concurrent at a point O inside ABC
Image:Ceva's theorem 2.svg
Ceva's theorem, case 2: the three lines are concurrent at a point O outside ABC

Ceva's theorem is a well-known theorem in elementary geometry. Given a triangle ABC, and points D, E, and F that lie on lines BC, CA, and AB respectively, the theorem states that lines AD, BE and CF are concurrent if and only if

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.


There is also an equivalent trigonometric form of Ceva's Theorem, that is, AD,BE,CF concur if and only if

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{\sin\angle BAD}{\sin\angle CAD}\times\frac{\sin\angle ACF}{\sin\angle BCF}\times\frac{\sin\angle CBE}{\sin\angle ABE}=1.


The theorem was proved by Giovanni Ceva in his 1678 work De lineis rectis, but it was also proved much earlier by Al-Mu'taman ibn Hűd, an eleventh-century king of Saragossa.

Associated with the figures are several terms derived from Ceva's name: cevian (the lines AD, BE, CF are the cevians of O), cevian triangle (the triangle DEF is the cevian triangle of O); cevian nest, anticevian triangle, Ceva conjugate. (Ceva is pronounced Chay'va; cevian is pronounced chev'ian.)

Contents

Proof of the theorem

Suppose Failed to parse (Missing texvc executable; please see math/README to configure.): AD , Failed to parse (Missing texvc executable; please see math/README to configure.): BE

and Failed to parse (Missing texvc executable; please see math/README to configure.): CF
intersect at a point Failed to parse (Missing texvc executable; please see math/README to configure.): O

. Because Failed to parse (Missing texvc executable; please see math/README to configure.): \triangle BOD

and Failed to parse (Missing texvc executable; please see math/README to configure.): \triangle COD
have the same height, we have
Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{|\triangle BOD|}{|\triangle COD|}=\frac{BD}{DC}.


Similarly,

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{|\triangle BAD|}{|\triangle CAD|}=\frac{BD}{DC}.


From this it follows that

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{BD}{DC}= \frac{|\triangle BAD|-|\triangle BOD|}{|\triangle CAD|-|\triangle COD|} =\frac{|\triangle ABO|}{|\triangle CAO|}.


Similarly,

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{CE}{EA}=\frac{|\triangle BCO|}{|\triangle ABO|},


and

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{AF}{FB}=\frac{|\triangle CAO|}{|\triangle BCO|}.


Multiplying these three equations gives

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1,


as required. Conversely, suppose that the points Failed to parse (Missing texvc executable; please see math/README to configure.): D , Failed to parse (Missing texvc executable; please see math/README to configure.): E

and Failed to parse (Missing texvc executable; please see math/README to configure.): F
satisfy the above equality. Let Failed to parse (Missing texvc executable; please see math/README to configure.): AD
and Failed to parse (Missing texvc executable; please see math/README to configure.): BE
intersect at Failed to parse (Missing texvc executable; please see math/README to configure.): O

, and let Failed to parse (Missing texvc executable; please see math/README to configure.): CO

intersect Failed to parse (Missing texvc executable; please see math/README to configure.): AB
at Failed to parse (Missing texvc executable; please see math/README to configure.): F '

. By the direction we have just proven,

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{AF '}{F 'B} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.


Comparing with the above equality, we obtain

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{AF '}{F 'B}=\frac{AF}{FB}.


Adding 1 to both sides and using Failed to parse (Missing texvc executable; please see math/README to configure.): AF '+F 'B=AF + FB = AB , we obtain

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{AB}{F 'B}=\frac{AB}{FB}.


Thus Failed to parse (Missing texvc executable; please see math/README to configure.): F 'B=FB , so that Failed to parse (Missing texvc executable; please see math/README to configure.): F

and Failed to parse (Missing texvc executable; please see math/README to configure.): F '
coincide (recalling that the distances are directed). Therefore Failed to parse (Missing texvc executable; please see math/README to configure.): AD

, Failed to parse (Missing texvc executable; please see math/README to configure.): BE

and Failed to parse (Missing texvc executable; please see math/README to configure.): CF = CF '
intersect at Failed to parse (Missing texvc executable; please see math/README to configure.): O

, and both implications are proven.

For the trigonometic form of the theorem, one approach is to view the three cevians, concurrent at point O, as partitioning the triangle Failed to parse (Missing texvc executable; please see math/README to configure.): \triangle ABC

into three smaller triangles: 

Failed to parse (Missing texvc executable; please see math/README to configure.): \triangle AOB ,Failed to parse (Missing texvc executable; please see math/README to configure.): \triangle BOC , and Failed to parse (Missing texvc executable; please see math/README to configure.): \triangle COA .

Applying the law of sines to each triangle we get:

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{\sin\angle OAB}{\sin\angle OBA}=\frac{OB}{OA} \text{  ; } \frac{\sin\angle OBC}{\sin\angle OCB}=\frac{OC}{OB}\text{  ; } \frac{\sin\angle OCA}{\sin\angle OAC}=\frac{OA}{OC}


When we mutiply the three equations, the right side will equal 1. The six sines on the left side, when rearranged, will yield the expression given in the theorem.

Generalizations

The theorem can be generalized to higher dimensional simplexes using barycentric coordinates. Define a cevian of an n-simplex as a ray from each vertex to a point on the opposite (n-1)-face (facet). Then the cevians are concurrent if and only if a mass distribution can be assigned to the vertices such that each cevian intersects the opposite facet at its center of mass. Moreover, the intersection point of the cevians is the center of mass of the simplex. (Landy. See Wernicke for an earlier result.)

The analogue of the theorem for general polygons in the plane has been known since the early nineteenth century (Grünbaum & Shephard 1995, p. 266). The theorem has also been generalized to triangles on other surfaces of constant curvature (Masal'tsev 1994).

See also

External links

References

  • Grünbaum, Branko & Shephard, G. C. (1995), "Ceva, Menelaus and the Area Principle", Mathematics Magazine 68 (4): 254–268, <http://links.jstor.org/sici?sici=0025-570X(199510)68%3A4%3C254%3ACMATAP%3E2.0.CO%3B2-0>.
  • J. B. Hogendijk, "Al-Mutaman ibn Hűd, 11the century kin of Saragossa and brilliant mathematician," Historia Mathematica 22 (1995) 1-18.
  • Landy, Steven. A Generalization of Ceva's Theorem to Higher Dimensions. The American Mathematical Monthly, Vol. 95, No. 10 (Dec., 1988), pp. 936-939
  • Masal'tsev, L. A. (1994) "Incidence theorems in spaces of constant curvature." Journal of Mathematical Sciences, Vol. 72, No. 4
  • Wernicke, Paul. The Theorems of Ceva and Menelaus and Their Extension. The American Mathematical Monthly, Vol. 34, No. 9 (Nov., 1927), pp. 468-472de:Satz von Ceva

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