Difference between revisions of "Ptolemy's Theorem"
(An example) |
IntrepidMath (talk | contribs) |
||
| Line 1: | Line 1: | ||
| − | '''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''' 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 === | === Definition === | ||
Revision as of 17:43, 18 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 
:
.
Example
In a regular heptagon ABCDEFG, prove that: 1/AB = 1/AC + 1/AD
Solution: Let ABCDEFG 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
