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.
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)
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.