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# 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.

Keywords:

Quadrilateral, Circle, Cyclic, Ptolemy

Related:

CyclicQuadrilateral, ProofOfPtolemysTheorem, PtolemysTheorem, PythagorasTheorem, CrossedQuadrilateral

Type of Math Object:

Theorem

Major Section:

Reference

Groups audience:

## Mathematics Subject Classification

51-00*no label found*60K25

*no label found*18-00

*no label found*68Q70

*no label found*37B15

*no label found*18-02

*no label found*18B20

*no label found*

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new question: Prime numbers out of sequence by Rubens373

Oct 7

new question: Lorenz system by David Bankom

Oct 19

new correction: examples and OEIS sequences by fizzie

Oct 13

new correction: Define Galois correspondence by porton

Oct 7

new correction: Closure properties on languages: DCFL not closed under reversal by babou

new correction: DCFLs are not closed under reversal by petey

Oct 2

new correction: Many corrections by Smarandache

Sep 28

new question: how to contest an entry? by zorba

new question: simple question by parag

## Comments

## Ptolemy's thm

Maybe the word "Ptolemy" could go in the keywords; the search

engine currently finds no hits on that word.

We might mention some of the easy corollaries of Pt's thm. E.g.

if we specialize to the case in which $AC$ and

$BD$ are diameters of the given circle, we get Pythagoras's theorem.

LH