Editing Ceva's theorem (section)

==Proofs==

Several proofs of the theorem have been created.<ref name=r1>{{cite book |title=Pure Geometry
|first=John Wellesley|last=Russell|publisher=Clarendon Press|year=1905
|chapter= Ch. 1 §7 Ceva's Theorem
|url=https://books.google.com/books?id=r3ILAAAAYAAJ}}</ref><ref>[[Alfred S. Posamentier]] and Charles T. Salkind (1996), ''Challenging Problems in Geometry'', pages 177–180, Dover Publishing Co., second revised edition.</ref> 
Two proofs are given in the following. 

The first one is very elementary, using only basic properties of triangle areas.<ref name=r1 /> However, several cases have to be considered, depending on the position of the point {{mvar|O}}. 

The second proof uses [[Affine space#Barycentric coordinates|barycentric coordinates]] and [[vector (geometry)|vectors]], but is {{Vague|text=somehow|date=December 2024}} more natural and not case dependent. Moreover, it works in any [[affine plane]] over any [[field (mathematics)|field]].

===Using triangle areas===

First, the sign of the [[left-hand side]] is positive since either all three of the ratios are positive, the case where {{mvar|O}} is inside the triangle (upper diagram), or one is positive and the other two are negative, the case {{mvar|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 
: <math>\frac{|\triangle BOD|}{|\triangle COD|}=\frac{\overline{BD}}{\overline{DC}}=\frac{|\triangle BAD|}{|\triangle CAD|}.</math>
Therefore,
:<math>\frac{\overline{BD}}{\overline{DC}}=
\frac{|\triangle BAD|-|\triangle BOD|}{|\triangle CAD|-|\triangle COD|}
=\frac{|\triangle ABO|}{|\triangle CAO|}.</math>
(Replace the minus with a plus if {{mvar|A}} and {{mvar|O}} are on opposite sides of {{mvar|BC}}.)
Similarly,
: <math>\frac{\overline{CE}}{\overline{EA}}=\frac{|\triangle BCO|}{|\triangle ABO|},</math>
and
: <math>\frac{\overline{AF}}{\overline{FB}}=\frac{|\triangle CAO|}{|\triangle BCO|}.</math>
Multiplying these three equations gives
: <math>\left|\frac{\overline{AF}}{\overline{FB}}  \cdot \frac{\overline{BD}}{\overline{DC}} \cdot \frac{\overline{CE}}{\overline{EA}} \right|= 1,</math>
as required.

The theorem can also be proven easily using [[Menelaus's theorem]].<ref>Follows {{cite book |title=Inductive Plane Geometry|url=https://archive.org/details/inductiveplanege00hopkrich|first=George Irving|last=Hopkins|publisher=D.C. Heath & Co.|year=1902|chapter=Art. 986}}</ref> From the transversal {{mvar|BOE}} of triangle {{math|△''ACF''}},
: <math>\frac{\overline{AB}}{\overline{BF}} \cdot \frac{\overline{FO}}{\overline{OC}} \cdot \frac{\overline{CE}}{\overline{EA}} = -1</math>
and from the transversal {{mvar|AOD}} of triangle {{math|△''BCF''}},
: <math>\frac{\overline{BA}}{\overline{AF}} \cdot \frac{\overline{FO}}{\overline{OC}} \cdot \frac{\overline{CD}}{\overline{DB}} = -1.</math>
The theorem follows by dividing these two equations.

The converse follows as a corollary.<ref name=r1/> Let {{mvar|D, E, F}} be given on the lines {{mvar|BC, AC, AB}} so that the equation holds. Let {{mvar|AD, BE}} meet at {{mvar|O}} and let {{mvar|F'}} be the point where {{mvar|CO}} crosses {{mvar|AB}}. Then by the theorem, the equation also holds for {{mvar|D, E, F'}}. Comparing the two, 
: <math>\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 {{mvar|1=F = F’}}.

===Using barycentric coordinates===

Given three points {{mvar|A, B, C}} that are not [[collinearity|collinear]], and a point {{mvar|O}}, that belongs to the same [[plane (geometry)|plane]], the [[Affine space#Barycentric coordinates|barycentric coordinates]] of {{mvar|O}} with respect of {{mvar|A, B, C}} are the unique three numbers <math>\lambda_A, \lambda_B, \lambda_C</math> such that
:<math>\lambda_A + \lambda_B + \lambda_C =1,</math> 
and
:<math>\overrightarrow{XO}=\lambda_A\overrightarrow{XA} + \lambda_B\overrightarrow{XB} + \lambda_C\overrightarrow{XC},</math>
for every point {{mvar|X}} (for the definition of this arrow notation and further details, see [[Affine space]]).

For Ceva's theorem, the point {{mvar|O}} is supposed to not belong to any line passing through two vertices of the triangle. This implies that <math>\lambda_A \lambda_B \lambda_C\ne 0.</math>

If one takes for {{mvar|X}} the intersection {{mvar|F}} of the lines {{mvar|AB}} and {{mvar|OC}} (see figures), the last equation may be rearranged into
:<math>\overrightarrow{FO}-\lambda_C\overrightarrow{FC}=\lambda_A\overrightarrow{FA} + \lambda_B\overrightarrow{FB}.</math> 
The left-hand side of this equation is a vector that has the same direction as the line {{mvar|CF}}, and the right-hand side has the same direction as the line {{mvar|AB}}. These lines have different directions since {{mvar|A, B, C}} are not collinear. It follows that the two members of the equation equal the zero vector, and 
:<math>\lambda_A\overrightarrow{FA} + \lambda_B\overrightarrow{FB}=0.</math>
It follows that 
:<math>\frac{\overline{AF}}{\overline{FB}}=\frac{\lambda_B}{\lambda_A},</math>
where the left-hand-side fraction is the signed ratio of the lengths of the collinear [[line segment]]s {{mvar|{{overline|AF}}}} and {{mvar|{{overline|FB}}}}.

The same reasoning shows
:<math>\frac{\overline{BD}}{\overline{DC}}=\frac{\lambda_C}{\lambda_B}\quad \text{and}\quad \frac{\overline{CE}}{\overline{EA}}=\frac{\lambda_A}{\lambda_C}.</math>
Ceva's theorem results immediately by taking the product of the three last equations.