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# proof of Ptolemy’s inequality

Looking at the quadrilateral $ABCD$ we construct a point $E$, such that the triangles $ACD$ and $AEB$ are similar ($\angle ABE=\angle CDA$ and $\angle BAE=\angle CAD$).

This means that:

$\frac{AE}{AC}=\frac{AB}{AD}=\frac{BE}{DC},$ |

from which follows that

$BE=\frac{AB\cdot DC}{AD}.$ |

Also because $\angle EAC=\angle BAD$ and

$\frac{AD}{AC}=\frac{AB}{AE}$ |

the triangles $EAC$ and $BAD$ are similar. So we get:

$EC=\frac{AC\cdot DB}{AD}.$ |

Now if $ABCD$ is cyclic we get

$\angle ABE+\angle CBA=\angle ADC+\angle CBA=180^{\circ}.$ |

This means that the points $C$, $B$ and $E$ are on one line and thus:

$EC=EB+BC$ |

Now we can use the formulas we already found to get:

$\frac{AC\cdot DB}{AD}=\frac{AB\cdot DC}{AD}+BC.$ |

Multiplication with $AD$ gives:

$AC\cdot DB=AB\cdot DC+BC\cdot AD.$ |

Now we look at the case that $ABCD$ is not cyclic. Then

$\angle ABE+\angle CBA=\angle ADC+\angle CBA\neq 180^{\circ},$ |

so the points $E$, $B$ and $C$ form a triangle and from the triangle inequality we know:

$EC<EB+BC.$ |

Again we use our formulas to get:

$\frac{AC\cdot DB}{AD}<\frac{AB\cdot DC}{AD}+BC.$ |

From this we get:

$AC\cdot DB<AB\cdot DC+BC\cdot AD.$ |

Putting this together we get Ptolomy’s inequality:

$AC\cdot DB\leq AB\cdot DC+BC\cdot AD,$ |

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