# 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

$$\frac{AZ}{ZB}\cdot \frac{BX}{XC}\cdot \frac{CY}{YA}=1.$$ |

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

$$\frac{AZ}{ZB}\cdot \frac{BX}{XC}\cdot \frac{CY}{YA}=1$$ |

then $AX,BY,CZ$ are concurrent^{}.

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 infinity^{}) 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 | Orthocenter^{} |

Related topic | OrthicTriangle |

Related topic | Cevian |

Related topic | Incenter^{} |

Related topic | GergonnePoint |

Related topic | MenelausTheorem |

Related topic | ProofOfVanAubelTheorem |

Related topic | VanAubelTheorem |

Related topic | BisectorsTheorem |

Related topic | DirectedSegment |