# Ceva's theorem

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

Ceva's theorem is a theorem about triangles in Euclidean plane geometry. Given a triangle ABC, let the lines AO, BO and CO be drawn from the vertices to a common point O (not on one of the sides of ABC), to meet opposite sides at D, E and F respectively. (The segments AD, BE, and CF are known as cevians.) Then, using signed lengths of segments,

${\displaystyle {\frac {AF}{FB}}\cdot {\frac {BD}{DC}}\cdot {\frac {CE}{EA}}=1.}$

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.

A slightly adapted converse is also true: If points D, E and F are chosen on BC, AC and AB respectively so that

${\displaystyle {\frac {AF}{FB}}\cdot {\frac {BD}{DC}}\cdot {\frac {CE}{EA}}=1,}$

then AD, BE and CF are concurrent, or all three parallel. The converse is often included as part of the theorem.

The theorem is often attributed to Giovanni Ceva, who published it in his 1678 work De lineis rectis. But it was proven much earlier by Yusuf Al-Mu'taman ibn Hűd, an eleventh-century king of Zaragoza.[1]

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.)

The theorem is very similar to Menelaus' theorem in that their equations differ only in sign.

## Proof of the theorem

(Here directed segments are not used, except in the case of proving using Menelaus' Theorem)

A standard proof is as follows;[2] Posamentier and Salkind[3]:177–180 give four proofs.

First, the sign of the left-hand side is positive since either all three of the ratios are positive, the case where O is inside the triangle (upper diagram), or one is positive and the other two are negative, the case O is outside the triangle (lower diagram shows one case).

To check the magnitude, note that the area of a triangle of a given height is proportional to its base. So

${\displaystyle {\frac {|\triangle BOD|}{|\triangle COD|}}={\frac {BD}{DC}}={\frac {|\triangle BAD|}{|\triangle CAD|}}.}$

Therefore,

${\displaystyle {\frac {BD}{DC}}={\frac {|\triangle BAD|-|\triangle BOD|}{|\triangle CAD|-|\triangle COD|}}={\frac {|\triangle ABO|}{|\triangle CAO|}}.}$

(Replace the minus with a plus if A and O are on opposite sides of BC.) Similarly,

${\displaystyle {\frac {CE}{EA}}={\frac {|\triangle BCO|}{|\triangle ABO|}},}$

and

${\displaystyle {\frac {AF}{FB}}={\frac {|\triangle CAO|}{|\triangle BCO|}}.}$

Multiplying these three equations gives

${\displaystyle \left|{\frac {AF}{FB}}\cdot {\frac {BD}{DC}}\cdot {\frac {CE}{EA}}\right|=1,}$

as required.

The theorem can also be proven easily using Menelaus' theorem.[4] From the transversal BOE of triangle ACF,

${\displaystyle {\frac {AB}{BF}}\cdot {\frac {FO}{OC}}\cdot {\frac {CE}{EA}}=-1}$

and from the transversal AOD of triangle BCF,

${\displaystyle {\frac {BA}{AF}}\cdot {\frac {FO}{OC}}\cdot {\frac {CD}{DB}}=-1.}$

The theorem follows by dividing these two equations.

The converse follows as a corollary.[2] Let D, E and F be given on the lines BC, AC and AB so that the equation holds. Let AD and BE meet at O and let F′ be the point where CO crosses AB. Then by the theorem, the equation also holds for D, E and F′. Comparing the two,

${\displaystyle {\frac {AF}{FB}}={\frac {AF'}{F'B}}}$

But at most one point can cut a segment in a given ratio so F=F′.

## 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.[5][6]

Routh's theorem gives the area of the triangle formed by three cevians in the case that they are not concurrent. Ceva's theorem can be obtained from it by setting the area equal to zero and solving.

The analogue of the theorem for general polygons in the plane has been known since the early nineteenth century.[7] The theorem has also been generalized to triangles on other surfaces of constant curvature.[8]

## References

1. ^ Holme, Audun (2010). Geometry: Our Cultural Heritage. Springer. p. 210. ISBN 3-642-14440-3.
2. ^ a b Follows Russell, John Wellesley (1905). "Ch. 1 §7 Ceva's Theorem". Pure Geometry. Clarendon Press.
3. ^ Alfred S. Posamentier and Charles T. Salkind (1996), Challenging Problems in Geometry, Dover Publishing Co., second revised edition.
4. ^ Follows Hopkins, George Irving (1902). "Art. 986". Inductive Plane Geometry. D.C. Heath & Co.
5. ^ Landy, Steven (December 1988). "A Generalization of Ceva's Theorem to Higher Dimensions". The American Mathematical Monthly. 95 (10): 936–939. doi:10.2307/2322390. JSTOR 2322390.
6. ^ Wernicke, Paul (November 1927). "The Theorems of Ceva and Menelaus and Their Extension". The American Mathematical Monthly. 34 (9): 468–472. doi:10.2307/2300222. JSTOR 2300222.
7. ^ Grünbaum, Branko; Shephard, G. C. (1995). "Ceva, Menelaus and the Area Principle". Mathematics Magazine. 68 (4): 254–268. doi:10.2307/2690569. JSTOR 2690569.
8. ^ Masal'tsev, L. A. (1994). "Incidence theorems in spaces of constant curvature". Journal of Mathematical Sciences. 72 (4): 3201–3206. doi:10.1007/BF01249519.