Ptolemy's Theorem

Revision as of 16:43, 18 June 2006 by IntrepidMath (talk | contribs)

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 $ABCD$ with side lengths ${a},{b},{c},{d}$ and diagonals ${e},{f}$:

$ac+bd=ef$.

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