trigonometric version of 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 AD,BE,CF are concurrentMathworldPlanetmath.

Remarks: All the angles are directed angles (counterclockwise is positive), and the intersection points may lie in the prolongation of the segments.

Title trigonometric version of Ceva’s theorem
Canonical name TrigonometricVersionOfCevasTheorem
Date of creation 2013-03-22 14:49:17
Last modified on 2013-03-22 14:49:17
Owner drini (3)
Last modified by drini (3)
Numerical id 6
Author drini (3)
Entry type Theorem
Classification msc 51A05
Related topic Triangle
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Related topic Centroid
Related topic OrthocenterMathworldPlanetmath
Related topic OrthicTriangle
Related topic Cevian
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