Difference between revisions of "Ptolemy's Theorem"
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=== Definition === | === Definition === | ||
| − | Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and [[diagonals]] <math>{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>: |
| − | <math>ac+bd=ef</math>. | + | <math>\displaystyle ac+bd=ef</math>. |
=== Proof === | === Proof === | ||
Revision as of 21:30, 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 
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.
