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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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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

## 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