Ceva’s theorem

Let ABC be a given triangleMathworldPlanetmath and P any point of the plane. If X is the intersectionMathworldPlanetmath point of AP with BC, Y the intersection point of BP with CA and Z is the intersection point of CP with AB, then


Conversely, if X,Y,Z are points on BC,CA,AB respectively, and if


then AX,BY,CZ are concurrentMathworldPlanetmath.

Remarks: All the segments are directed segments (that is AB=-BA), and so theorem is valid even if the points X,Y,Z are in the prolongations (even at the infinityMathworldPlanetmath) and P is any point on the plane (or at the infinity).

Title Ceva’s theorem
Canonical name CevasTheorem
Date of creation 2013-03-22 11:57:15
Last modified on 2013-03-22 11:57:15
Owner drini (3)
Last modified by drini (3)
Numerical id 16
Author drini (3)
Entry type Theorem
Classification msc 51A05
Related topic Triangle
Related topic Median
Related topic Centroid
Related topic OrthocenterMathworldPlanetmath
Related topic OrthicTriangle
Related topic Cevian
Related topic IncenterMathworldPlanetmath
Related topic GergonnePoint
Related topic MenelausTheorem
Related topic ProofOfVanAubelTheorem
Related topic VanAubelTheorem
Related topic BisectorsTheorem
Related topic DirectedSegment