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| − | '''Ptolemy's Theorem''' gives a relationship between the side lengths and the diagonals of a [[cyclic quadrilateral]]; it is the [[equality condition | 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]] |
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| − | == Definition ==
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| − | Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and [[diagonal]]s <math>{e},{f}</math>:
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| − | <math>ac+bd=ef</math>.
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| − | == Proof ==
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| − | Given cyclic quadrilateral <math>ABCD,</math> extend <math>CD</math> to <math>P</math> such that <math>\angle BAC=\angle DAP.</math>
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| − | Since quadrilateral <math>ABCD</math> is cyclic, <math>m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\angle ADP</math> is also supplementary to <math>\angle ADC,</math> so <math>\angle ADP=\angle ABC</math>. Hence, <math>\triangle ABC \sim \triangle ADP</math> by AA similarity and <math>\frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.</math>
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| − | 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 BAD=\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>
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| − | However, <math>CP= CD+DP.</math> Substituting in our expressions for <math>CP</math> and <math>DP,</math> <math> \frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.</math> Multiplying by <math>AB</math> yields <math>(AC)(BD)=(AB)(CD)+(AD)(BC)</math>.
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| − | == Example ==
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| − | In a regular heptagon ''ABCDEFG'', prove that: ''1/AB = 1/AC + 1/AD''.
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| − | 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.
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| − | Now, Ptolemy's Theorem states that ''ab + ac = bc'', which is equivalent to ''1/a=1/b+1/c''.
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| − | == See also ==
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| − | * [[Geometry]]
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| − | [[Category:Geometry]]
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| − | [[Category:Theorems]]
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