# 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 Ptolemy's Inequality. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures.

## Contents

## Statement

Given a cyclic quadrilateral with side lengths and diagonals :

## Proof 1

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 .

# Proof 2 (inversion)

We provide a proof for the general case of Ptolemy's theorem, Ptolemy's Inequality.

Let be four points in the Euclidean plane. Taking an inversion centered at (the point doesn't matter, it can be any of the four) with radius , we have that by the Triangle Inequality, with equality holding when are collinear, i.e. when lie on a circle containing Additionally, by the Inversion Distance Formula, we may express the inequality as the following:

Dividing by and multiplying everything by we get and thus the desired.

## Problems

## 2023 AIME I Problem 5

Square is inscribed in a circle. Point is on this circle such that , and . What is the area of the square?

Solution: We may assume that is between and . Let , , , , and . We have , because is a diagonal. Similarly, . Therefore, . Similarly, .

By Ptolemy's Theorem on , , and therefore . By Ptolemy's on , , and therefore . By squaring both equations, we obtain

Thus, , and . Plugging these values into , we obtain , and . Now, we can solve using and (though using and yields the same solution for ).

The answer is .

### 2004 AMC 10B Problem 24

In triangle we have , , . Point is on the circumscribed circle of the triangle so that bisects angle . What is the value of ?

Solution: Set 's length as . 's length must also be since and intercept arcs of equal length(because ). Using Ptolemy's theorem, . The ratio is

### Equilateral Triangle Identity

Let be an equilateral triangle. Let be a point on minor arc of its circumcircle. Prove that .

Solution: Draw , , . By Ptolemy's theorem applied to quadrilateral , we know that . Since , we divide both sides of the last equation by to get the result: .

### Regular Heptagon Identity

In a regular heptagon , prove that: .

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

Now, Ptolemy's theorem states that , which is equivalent to upon division by .

### 1991 AIME Problems/Problem 14

A hexagon is inscribed in a circle. Five of the sides have length and the sixth, denoted by , has length . Find the sum of the lengths of the three diagonals that can be drawn from .

### Cyclic Hexagon

A hexagon with sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle. Find the diameter of the circle.

Solution: Consider half of the circle, with the quadrilateral , being the diameter. , , and . Construct diagonals and . Notice that these diagonals form right triangles. You get the following system of equations:

(Ptolemy's theorem)

Solving gives