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Ceva's theorem
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===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.
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