# Difference between revisions of "Ptolemy's Theorem"

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'''Ptolemy's theorem''' gives a relationship between the side lengths and the diagonals of a [[cyclic quadrilateral]]; it is the equality case of the [[Ptolemy inequality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. | '''Ptolemy's theorem''' gives a relationship between the side lengths and the diagonals of a [[cyclic quadrilateral]]; it is the equality case of the [[Ptolemy inequality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. | ||

− | + | == Definition == | |

Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>\displaystyle {a},{b},{c},{d}</math> and [[diagonals]] <math>\displaystyle {e},{f}</math>: | Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>\displaystyle {a},{b},{c},{d}</math> and [[diagonals]] <math>\displaystyle {e},{f}</math>: | ||

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<math>\displaystyle ac+bd=ef</math>. | <math>\displaystyle ac+bd=ef</math>. | ||

− | + | == Proof == | |

Given cyclic quadrilateral <math>\displaystyle ABCD,</math> extend <math>\displaystyle CD</math> to <math>\displaystyle P</math> such that <math>\angle BAC=\angle DAP.</math> | Given cyclic quadrilateral <math>\displaystyle ABCD,</math> extend <math>\displaystyle CD</math> to <math>\displaystyle P</math> such that <math>\angle BAC=\angle DAP.</math> | ||

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--[[User:4everwise|4everwise]] 14:09, 22 June 2006 (EDT) | --[[User:4everwise|4everwise]] 14:09, 22 June 2006 (EDT) | ||

− | + | == Example == | |

In a regular heptagon ''ABCDEFG'', prove that: ''1/AB = 1/AC + 1/AD''. | In a regular heptagon ''ABCDEFG'', prove that: ''1/AB = 1/AC + 1/AD''. | ||

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Now Ptolemy's theorem states that ''ab + ac = bc'', which is equivalent to ''1/a=1/b+1/c''. | Now Ptolemy's theorem states that ''ab + ac = bc'', which is equivalent to ''1/a=1/b+1/c''. | ||

+ | |||

+ | == See also == | ||

+ | * [[Geometry]] |

## Revision as of 21:49, 24 June 2006

**Ptolemy's theorem** gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of the Ptolemy inequality. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures.

## Contents

## Definition

Given a cyclic quadrilateral with side lengths and diagonals :

.

## Proof

Given cyclic quadrilateral extend to such that

Since quadrilateral is cyclic, However, is also supplementary to so Hence, by AA similarity and

Now, note that (subtend the same arc) and so This yields

However, Substituting in our expressions for and Multiplying by yields

--4everwise 14:09, 22 June 2006 (EDT)

## Example

In a regular heptagon *ABCDEFG*, prove that: *1/AB = 1/AC + 1/AD*.

Solution: Let *ABCDEFG* be the regular heptagon. Consider the quadrilateral *ABCE*. If *a*, *b*, and *c* represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of *ABCE* are *a*, *a*, *b* and *c*; and the diagonals of *ABCE* are *b* and *c*, respectively.

Now Ptolemy's theorem states that *ab + ac = bc*, which is equivalent to *1/a=1/b+1/c*.