Ceva's theorem

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′.

Given three points A, B, C, that are not collinear, and a point O, that belongs to the same plane, the barycentric coordinates of O with respect of A, B, C are the unique three numbers ${\displaystyle \lambda _{A},\lambda _{B},\lambda _{C}}$ such that

${\displaystyle \lambda _{A}+\lambda _{B}+\lambda _{C}=1,}$

and

${\displaystyle {\overrightarrow {XO}}=\lambda _{A}{\overrightarrow {XA}}+\lambda _{B}{\overrightarrow {XB}}+\lambda _{C}{\overrightarrow {XC}},}$

for every point X (for the definition of this arrow notation and further details, see Affine space).

For Ceva's theorem, the point O is supposed to not belong to any line passing through two vertices of the triangle. This implies that ${\displaystyle \lambda _{A}\lambda _{B}\lambda _{C}\neq 0.}$

If one takes for X the intersection F of the lines AB and OC (see figures), the last equation may be rearranged into

${\displaystyle {\overrightarrow {FO}}-\lambda _{C}{\overrightarrow {FC}}=\lambda _{A}{\overrightarrow {FA}}+\lambda _{B}{\overrightarrow {FB}}.}$

The left-hand side of this equation is a vector that has the same direction as the line CF, and the right-hand side has the same direction as the line AB. These lines have different directions since A, B, and C are not collinear. It follows that the two members of the equation equal the zero vector, and

${\displaystyle \lambda _{A}{\overrightarrow {FA}}+\lambda _{B}{\overrightarrow {FB}}=0.}$

It follows that

${\displaystyle {\frac {AF}{FB}}={\frac {\lambda _{B}}{\lambda _{A}}},}$

where the left-hand-side fraction is the signed ratio of the lengths of the collinear line segments AF and FB.

The same reasoning shows

${\displaystyle {\frac {BD}{DC}}={\frac {\lambda _{C}}{\lambda _{B}}}\quad {\text{and}}\quad {\frac {CE}{EA}}={\frac {\lambda _{A}}{\lambda _{C}}}.}$

Ceva's theorem results immediately by taking the product of the three last equations.