# Difference between revisions of "Ptolemy's Theorem"

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== Problems == | == Problems == | ||

===2004 AMC 10B Problem 24=== | ===2004 AMC 10B Problem 24=== | ||

− | In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC= | + | In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC=9</math>. Point <math>D</math> is on the circumscribed circle of the triangle so that <math>AD</math> bisects angle <math>BAC</math>. What is the value of <math>AD/CD</math>? |

<math>\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}</math> | <math>\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}</math> |

## Revision as of 00:42, 16 August 2017

**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

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 .

## Problems

### 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