Difference between revisions of "Ptolemy's Theorem"
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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>{a},{b},{c},{d}</math> and [[diagonals]] <math>{e},{f}</math>: | ||
| − | <math>ac+bd=ef</math> | + | <math>ac+bd=ef</math>. |
| + | |||
| + | === 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'' | ||
Revision as of 16:30, 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.
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
