# bisectors theorem

Bisectors theorem.

Let $ABC$ a triangle, and $AP$ the angle bisector^{} of $\mathrm{\angle}CAB$. Then

$$\frac{BP}{PC}=\frac{BA}{CA}.$$ |

Generalization of bisectors theorem.

Let $ABC$ a triangle, and $P$ any point on the side $BC$. Then

$$\frac{BP}{PC}=\frac{BA\mathrm{sin}BAP}{CA\mathrm{sin}PAC}.$$ |

The latest theorem even holds when $P$ is not on the segment $BC$ but anywhere on the line (possibly at the infnity), but in this case, the ratio $BP/PC$ must be done with directed segments and the angles considered as directed angles.

Title | bisectors theorem |
---|---|

Canonical name | BisectorsTheorem |

Date of creation | 2013-03-22 14:49:23 |

Last modified on | 2013-03-22 14:49:23 |

Owner | drini (3) |

Last modified by | drini (3) |

Numerical id | 9 |

Author | drini (3) |

Entry type | Theorem |

Classification | msc 51A05 |

Synonym | angle bisector theorem |

Synonym | generalization of bisectors theorem |

Related topic | CevasTheorem |

Related topic | HarmonicDivision |