# Difference between revisions of "Ptolemy's Theorem"

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− | '''Ptolemy's | + | '''Ptolemy's Theorem''' gives a relationship between the side lengths and the diagonals of a [[cyclic quadrilateral]]; it is the [[equality condition | equality case]] of [[Ptolemy's Inequality]]. Ptolemy's Theorem frequently shows up as an intermediate step in problems involving inscribed figures. |

− | == | + | == Statement == |

− | Given a cyclic quadrilateral <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and | + | Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and [[diagonal]]s <math>{e},{f}</math>: |

− | < | + | <cmath>ac+bd=ef.</cmath> |

+ | |||

+ | == Proof == | ||

+ | |||

+ | Given cyclic quadrilateral <math>ABCD,</math> extend <math>CD</math> to <math>P</math> such that <math>\angle BAD=\angle CAP.</math> | ||

+ | |||

+ | Since quadrilateral <math>ABCD</math> is cyclic, <math>m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\angle ADP</math> is also supplementary to <math>\angle ADC,</math> so <math>\angle ADP=\angle ABC</math>. Hence, <math>\triangle ABC \sim \triangle ADP</math> by AA similarity and <math>\frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.</math> | ||

+ | |||

+ | Now, note that <math>\angle ABD=\angle ACD </math> (subtend the same arc) and <math>\angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAD=\angle CAP,</math> so <math>\triangle BAD\sim \triangle CAP.</math> This yields <math>\frac{AB}{AC}=\frac{BD}{CP}\implies CP=\frac{(AC)(BD)}{(AB)}.</math> | ||

+ | |||

+ | However, <math>CP= CD+DP.</math> Substituting in our expressions for <math>CP</math> and <math>DP,</math> <math> \frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.</math> Multiplying by <math>AB</math> yields <math>(AC)(BD)=(AB)(CD)+(AD)(BC)</math>. | ||

+ | |||

+ | == Problems == | ||

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

+ | 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> | ||

+ | |||

+ | Solution: Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>CD</math>'s length must also be <math>x</math> since <math>\angle BAD</math> and <math>\angle DAC</math> intercept arcs of equal length(because <math>\angle BAD =\angle DAC</math>). Using Ptolemy's Theorem, <math>7x+8x=9(AD)</math>. The ratio is <math>\boxed{\frac{5}{3}}\implies(B)</math> | ||

+ | |||

+ | === Equilateral Triangle Identity === | ||

+ | Let <math>\triangle ABC</math> be an equilateral triangle. Let <math>P</math> be a point on minor arc <math>AB</math> of its circumcircle. Prove that <math>PC=PA+PB</math>. | ||

+ | |||

+ | Solution: Draw <math>PA</math>, <math>PB</math>, <math>PC</math>. By Ptolemy's Theorem applied to quadrilateral <math>APBC</math>, we know that <math>PC\cdot AB=PA\cdot BC+PB\cdot AC</math>. Since <math>AB=BC=CA=s</math>, we divide both sides of the last equation by <math>s</math> to get the result: <math>PC=PA+PB</math>. | ||

+ | |||

+ | === Regular Heptagon Identity === | ||

+ | In a regular heptagon <math> ABCDEFG </math>, prove that: <math> \frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD} </math>. | ||

+ | |||

+ | Solution: Let <math> ABCDEFG </math> be the regular heptagon. Consider the quadrilateral <math> ABCE </math>. If <math> a </math>, <math> b </math>, and <math> c </math> represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of <math> ABCE </math> are <math> a </math>, <math> a </math>, <math> b </math> and <math> c </math>; the diagonals of <math> ABCE </math> are <math> b </math> and <math> c </math>, respectively. | ||

+ | |||

+ | Now, Ptolemy's Theorem states that <math> ab + ac = bc </math>, which is equivalent to <math> \frac{1}{a}=\frac{1}{b}+\frac{1}{c} </math> upon division by <math> abc </math>. | ||

+ | |||

+ | === 1991 AIME Problems/Problem 14 === | ||

+ | A hexagon is inscribed in a circle. Five of the sides have length <math>81</math> and the sixth, denoted by <math>\overline{AB}</math>, has length <math>31</math>. Find the sum of the lengths of the three diagonals that can be drawn from <math>A</math>. | ||

+ | |||

+ | [[1991_AIME_Problems/Problem_14#Solution|Solution]] | ||

+ | |||

+ | === 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 <math>ABCD</math>, <math>AD</math> being the diameter. <math>AB = 2</math>, <math>BC = 7</math>, and <math>CD = 11</math>. Construct diagonals <math>AC</math> and <math>BD</math>. Notice that these diagonals form right triangles. You get the following system of equations: | ||

+ | |||

+ | <math>(AC)(BD) = 7(AD) + 22</math> (Ptolemy's Theorem) | ||

+ | |||

+ | <math>\text(AC)^2 = (AD)^2 - 121</math> | ||

+ | |||

+ | <math>(BD)^2 = (AD)^2 - 4</math> | ||

+ | |||

+ | Solving gives <math>AD = 14</math> | ||

+ | |||

+ | == See also == | ||

+ | * [[Geometry]] | ||

+ | |||

+ | [[Category:Geometry]] | ||

+ | [[Category:Theorems]] |

## Latest 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