Difference between revisions of "Ptolemy's Theorem"
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== Definition == | == Definition == | ||
| − | Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>\displaystyle {a},{b},{c},{d}</math> and [[ | + | Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>\displaystyle {a},{b},{c},{d}</math> and [[diagonal]]s <math>\displaystyle {e},{f}</math>: |
| − | <math>\displaystyle ac+bd=ef</math>. | + | <math>\displaystyle ac+bd=ef</math>. |
== Proof == | == Proof == | ||
Revision as of 11:47, 10 July 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; 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.
