|
|
| (29 intermediate revisions by 18 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]]. Ptolemy's Theorem frequently shows up as an intermediate step in problems involving inscribed figures.
| + | #REDIRECT[[Ptolemy's theorem]] |
| − | | |
| − | == Definition ==
| |
| − | | |
| − | Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>\displaystyle {a},{b},{c},{d}</math> and [[diagonals]] <math>\displaystyle {e},{f}</math>:
| |
| − | | |
| − | <math>\displaystyle ac+bd=ef</math>.
| |
| − | | |
| − | == Proof ==
| |
| − | | |
| − | Given cyclic quadrilateral <math>\displaystyle ABCD,</math> extend <math>\displaystyle CD</math> to <math>\displaystyle P</math> such that <math>\angle BAC=\angle DAP.</math>
| |
| − | | |
| − | Since quadrilateral <math>\displaystyle ABCD</math> is cyclic, <math>\displaystyle m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\displaystyle \angle ADP</math> is also supplementary to <math>\angle ADC,</math> so <math>\displaystyle \angle ADP=\angle ABC</math>. Hence, <math>\displaystyle \triangle ABC \sim \triangle ADP</math> by AA similarity and <math>\frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.</math>
| |
| − | | |
| − | Now, note that <math>\angle ABD=\angle ACD </math> (subtend the same arc) and <math>\angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAC=\angle CAP,</math> so <math>\triangle BAD\sim \triangle CAP.</math> This yields <math>\frac{AD}{AP}=\frac{BD}{CP}\implies CP=\frac{(AP)(BD)}{(AD)}.</math>
| |
| − | | |
| − | However, <math>\displaystyle CP= CD+DP.</math> Substituting in our expressions for <math>\displaystyle CP</math> and <math>\displaystyle DP,</math> <math> \frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.</math> Multiplying by <math>\displaystyle AB</math> yields <math>\displaystyle (AC)(BD)=(AB)(CD)+(AD)(BC)</math>.
| |
| − | | |
| − | --[[User:4everwise|4everwise]] 14:09, 22 June 2006 (EDT)
| |
| − | | |
| − | == Example ==
| |
| − | | |
| − | 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''.
| |
| − | | |
| − | == See also ==
| |
| − | * [[Geometry]]
| |
