Difference between revisions of "Ptolemy's Theorem"

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'''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]].
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'''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.
  
=== Definition ===
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== Statement ==
  
Given a cyclic quadrilateral <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and diagonals <math>{e},{f}</math>:
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Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and [[diagonal]]s <math>{e},{f}</math>:
  
<math>ac+bd=ef</math>
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<cmath>ac+bd=ef.</cmath>
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== Proof ==
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Given cyclic quadrilateral <math>ABCD,</math> extend <math>CD</math> to <math>P</math> such that <math>\angle BAD=\angle CAP.</math>
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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>
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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>
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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>.
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== Problems ==
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===2004 AMC 10B Problem 24===
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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>?
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<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>
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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>
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=== Equilateral Triangle Identity ===
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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>.
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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>.
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=== Regular Heptagon Identity ===
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In a regular heptagon <math> ABCDEFG </math>, prove that: <math> \frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD} </math>.
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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.
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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>.
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=== 1991 AIME Problems/Problem 14 ===
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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>.
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[[1991_AIME_Problems/Problem_14#Solution|Solution]]
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=== Cyclic Hexagon ===
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A hexagon with sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.  Find the diameter of the circle.
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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:
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<math>(AC)(BD) = 7(AD) + 22</math> (Ptolemy's Theorem)
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<math>\text(AC)^2 = (AD)^2 - 121</math>
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<math>(BD)^2 = (AD)^2 - 4</math>
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Solving gives <math>AD = 14</math>
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== See also ==
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* [[Geometry]]
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[[Category:Geometry]]
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[[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.

Statement

Given a cyclic quadrilateral $ABCD$ with side lengths ${a},{b},{c},{d}$ and diagonals ${e},{f}$:

\[ac+bd=ef.\]

Proof

Given cyclic quadrilateral $ABCD,$ extend $CD$ to $P$ such that $\angle BAD=\angle CAP.$

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

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

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

Problems

2004 AMC 10B Problem 24

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

$\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}$

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

Equilateral Triangle Identity

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

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

Regular Heptagon Identity

In a regular heptagon $ABCDEFG$, prove that: $\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD}$.

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 $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$ upon division by $abc$.

1991 AIME Problems/Problem 14

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

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

$(AC)(BD) = 7(AD) + 22$ (Ptolemy's Theorem)

$\text(AC)^2 = (AD)^2 - 121$

$(BD)^2 = (AD)^2 - 4$

Solving gives $AD = 14$

See also

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