# 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.

### Definition

Given a cyclic quadrilateral with side lengths and diagonals :

.

### Proof: Method I

Given cyclic quadrilateral extend to such that

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

### Proof: Method II

### 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*.