# Ptolemy’s theorem

If $ABCD$ is a cyclic quadrilateral^{}, then the product of the two diagonals is equal to the sum of the products of opposite sides.

$$AC\cdot BD=AB\cdot CD+AD\cdot BC.$$ |

When the quadrilateral^{} is not cyclic we have the following inequality

$$AB\cdot CD+BC\cdot AD>AC\cdot BD$$ |

An interesting particular case is when both $AC$ and $BD$ are diameters^{}, since we get another proof of Pythagoras’ theorem.

Title | Ptolemy’s theorem |

Canonical name | PtolemysTheorem |

Date of creation | 2013-03-22 11:43:13 |

Last modified on | 2013-03-22 11:43:13 |

Owner | drini (3) |

Last modified by | drini (3) |

Numerical id | 18 |

Author | drini (3) |

Entry type | Theorem |

Classification | msc 51-00 |

Classification | msc 60K25 |

Classification | msc 18-00 |

Classification | msc 68Q70 |

Classification | msc 37B15 |

Classification | msc 18-02 |

Classification | msc 18B20 |

Related topic | CyclicQuadrilateral |

Related topic | ProofOfPtolemysTheorem |

Related topic | PtolemysTheorem |

Related topic | PythagorasTheorem |

Related topic | CrossedQuadrilateral |