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]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. | '''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]] <math>ABCD</math> with side lengths <math>\displaystyle {a},{b},{c},{d}</math> and [[diagonals]] <math>\displaystyle {e},{f}</math>: | Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>\displaystyle {a},{b},{c},{d}</math> and [[diagonals]] <math>\displaystyle {e},{f}</math>: | ||
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<math>\displaystyle ac+bd=ef</math>. | <math>\displaystyle ac+bd=ef</math>. | ||
| − | + | == Proof == | |
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> | ||
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--[[User:4everwise|4everwise]] 14:09, 22 June 2006 (EDT) | --[[User:4everwise|4everwise]] 14:09, 22 June 2006 (EDT) | ||
| − | + | == Example == | |
In a regular heptagon ''ABCDEFG'', prove that: ''1/AB = 1/AC + 1/AD''. | In a regular heptagon ''ABCDEFG'', prove that: ''1/AB = 1/AC + 1/AD''. | ||
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Now Ptolemy's theorem states that ''ab + ac = bc'', which is equivalent to ''1/a=1/b+1/c''. | Now Ptolemy's theorem states that ''ab + ac = bc'', which is equivalent to ''1/a=1/b+1/c''. | ||
| + | |||
| + | == See also == | ||
| + | * [[Geometry]] | ||
Revision as of 21:49, 24 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.
Contents
Definition
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 
--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.
