# Difference between revisions of "Ptolemy's Theorem"

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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''; the diagonals of ''ABCE'' are ''b'' and ''c'', respectively. | 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''; the diagonals of ''ABCE'' are ''b'' and ''c'', respectively. | ||

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

== See also == | == See also == | ||

* [[Geometry]] | * [[Geometry]] |

## Revision as of 13:54, 26 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*; 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*.