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 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

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

Therefore,

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

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

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

and

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

Multiplying these three equations gives

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

as required.

The theorem can also be proven easily using Menelaus' theorem.^[5] From the transversal BOE of triangle △ACF,

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

and from the transversal △AOD of triangle △BCF,

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

The theorem follows by dividing these two equations.

The converse follows as a corollary.^[3] Let D, E, F be given on the lines BC, AC, AB so that the equation holds. Let AD, 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, F'. Comparing the two,

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

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

Using barycentric coordinates

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, 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

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

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

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

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