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