# Ptolemy's Theorem

**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 :

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## 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*.