|
|
| (40 intermediate revisions by 24 users not shown) |
| 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]].
| + | #REDIRECT[[Ptolemy's theorem]] |
| − | | |
| − | === Definition ===
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | === 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''
| |
