Difference between revisions of "Ptolemy's Theorem"

(Proof)
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Given cyclic quadrilateral <math>\displaystyle ABCD,</math> extend <math>\displaystyle CD</math> to <math>\displaystyle P</math> such that <math>\angle BAC=\angle DAP.</math>  
 
Given cyclic quadrilateral <math>\displaystyle ABCD,</math> extend <math>\displaystyle CD</math> to <math>\displaystyle P</math> such that <math>\angle BAC=\angle DAP.</math>  
  
Since quadrilateral <math>\displaystyle ABCD,</math> is cyclic, <math>\displaystyle m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\displaystyle \angle ADP</math> is also supplementary to <math>\displaystyle \angle ADC,</math> so <math>\displaystyle \angle ADP=\angle ABC</math> Hence, <math>\displaystyle \triangle ABC \sim \triangle ADP</math> by <math>\displaystyle AA </math> similarity and <math>\displaystyle \frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.</math>
+
Since quadrilateral <math>\displaystyle ABCD,</math> is cyclic, <math>\displaystyle m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\displaystyle \angle ADP</math> is also supplementary to <math>\angle ADC,</math> so <math>\displaystyle \angle ADP=\angle ABC</math> Hence, <math>\displaystyle \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>\displaystyle \angle ABD=\angle ACD </math> (subtend the same arc) and <math>\displaystyle \angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAC=\andle CAP,</math> so <math>\displaystyle \triangle BAD\sim \triangle CAP.</math> This yields <math>\displaystyle \frac{AD}{AP}=\frac{BD}{CP}\implies CP=\frac{(AP)(BD)}{(AD)}.</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 BAC=\angle CAP,</math> so <math>\triangle BAD\sim \triangle CAP.</math> This yields <math>\frac{AD}{AP}=\frac{BD}{CP}\implies CP=\frac{(AP)(BD)}{(AD)}.</math>
  
However, <math>\displaystyle CP= CD+DP.</math> Substituting in our expressions for  <math>\displaystyle CP</math> and  <math>\displaystyle DP,</math>  <math>\displaystyle \frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.</math> Multiplying by <math>\displaystyle AB</math> yields  <math>\displaystyle (AC)(BD)=(AB)(CD)+(AD)(BC)</math>
+
However, <math>\displaystyle CP= CD+DP.</math> Substituting in our expressions for  <math>\displaystyle CP</math> and  <math>\displaystyle DP,</math>  <math> \frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.</math> Multiplying by <math>\displaystyle AB</math> yields  <math>\displaystyle (AC)(BD)=(AB)(CD)+(AD)(BC)</math>
  
 
--[[User:4everwise|4everwise]] 14:09, 22 June 2006 (EDT)
 
--[[User:4everwise|4everwise]] 14:09, 22 June 2006 (EDT)
 
  
 
=== Example ===
 
=== Example ===

Revision as of 19:19, 22 June 2006

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

$ac+bd=ef$.

Proof

Given cyclic quadrilateral $\displaystyle ABCD,$ extend $\displaystyle CD$ to $\displaystyle P$ such that $\angle BAC=\angle DAP.$

Since quadrilateral $\displaystyle ABCD,$ is cyclic, $\displaystyle m\angle ABC+m\angle ADC=180^\circ .$ However, $\displaystyle \angle ADP$ is also supplementary to $\angle ADC,$ so $\displaystyle \angle ADP=\angle ABC$ Hence, $\displaystyle \triangle ABC \sim \triangle ADP$ 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 $\angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAC=\angle CAP,$ so $\triangle BAD\sim \triangle CAP.$ This yields $\frac{AD}{AP}=\frac{BD}{CP}\implies CP=\frac{(AP)(BD)}{(AD)}.$

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

--4everwise 14:09, 22 June 2006 (EDT)

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.

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