Difference between revisions of "Ptolemy's Theorem"
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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>. | 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>. | ||
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== Example == | == Example == | ||
Revision as of 17:04, 15 February 2007
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 
.
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.
