# 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

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

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

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

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 |