Difference between revisions of "Ptolemy's Theorem"

m (Ptolemy's heorem moved to Ptolemy's Theorem: making a proper noun)
Line 6: Line 6:
  
 
<math>ac+bd=ef</math>.  
 
<math>ac+bd=ef</math>.  
 +
 +
=== Proof: Method I===
 +
 +
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> Hence, <math>\displaystyle \angle ADP=\angle ABC</math> and <math>\displaystyle \triangle ABC \sim\triangle ADP</math> by <math>\displaystyle AA </math> similarity.
 +
 +
 +
 +
===Proof: Method II===
 +
  
 
=== Example ===
 
=== Example ===

Revision as of 13:49, 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: Method I

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 $\displaystyle \angle ADC.$ Hence, $\displaystyle \angle ADP=\angle ABC$ and $\displaystyle \triangle ABC \sim\triangle ADP$ by $\displaystyle AA$ similarity.


Proof: Method II

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.

Invalid username
Login to AoPS