# trigonometric version of Ceva’s theorem

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

 $\frac{\sin ACZ}{\sin ZCB}\cdot\frac{\sin BAX}{\sin XAC}\cdot\frac{\sin CBY}{% \sin YBA}=1.$

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

 $\frac{\sin ACZ}{\sin ZCB}\cdot\frac{\sin BAX}{\sin XAC}\cdot\frac{\sin CBY}{% \sin YBA}=1$

then $AD,BE,CF$ are concurrent.

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 Related topic Median Related topic Centroid Related topic Orthocenter Related topic OrthicTriangle Related topic Cevian Related topic Incenter Related topic GergonnePoint Related topic MenelausTheorem Related topic ProofOfVanAubelTheorem Related topic VanAubelTheorem