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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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| − | == Problems ==
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| − | === Equilateral Triangle Identity ===
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| − | Let <math>\triangle ABC</math> be an equilateral triangle. Let <math>P</math> be a point on minor arc <math>AB</math> of its circumcircle. Prove that <math>PC=PA+PB</math>.
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| − | Solution: Draw <math>PA</math>, <math>PB</math>, <math>PC</math>. By Ptolemy's Theorem applied to quadrilateral <math>APBC</math>, we know that <math>PC\cdot AB=PA\cdot BC+PB\cdot AC</math>. Since <math>AB=BC=CA=s</math>, we divide both sides of the last equation by <math>s</math> to get the result: <math>PC=PA+PB</math>.
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| − | === Regular Heptagon Identity ===
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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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| − | === 1991 AIME Problems/Problem 14 ===
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| − | A hexagon is inscribed in a circle. Five of the sides have length <math>81</math> and the sixth, denoted by <math>\overline{AB}</math>, has length <math>31</math>. Find the sum of the lengths of the three diagonals that can be drawn from <math>A</math>.
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| − | [[1991_AIME_Problems/Problem_14#Solution|Solution]] | |
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| − | === Cyclic hexagon ===
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| − | A hexagon with sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle. Find the diameter of the circle.
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| − | Solution: Consider half of the circle, with the quadrilateral <math>ABCD</math>, <math>AD</math> being the diameter. <math>AB = 2</math>, <math>BC = 7</math>, and <math>CD = 11</math>. Construct diagonals <math>AC</math> and <math>BD</math>. Notice that these diagonals form right triangles. You get the following system of equations:
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| − | <math>(AC)(BD) = 7(AD) + 22</math> (Ptolemy's Theorem)
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| − | <math>\n(AC)^2 = (AD)^2 - 121</math>
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| − | <math>(BD)^2 = (AD)^2 - 4</math>
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| − | Solving gives <math>AD = 14</math>
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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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