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Difference between revisions of "Ptolemy's Inequality"

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Ptolemy's Inequality for a [[quadrilateral]] ABCD states that AB·CD + BC·DA ≥ AC·BD with [[equal]]ity [[iff]] ABCD is a [[cyclic quadrilateral]].
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'''Ptolemy's Inequality''' states that in a [[quadrilateral]] <math> \displaystyle ABCD </math>,
  
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[http://planetmath.org/encyclopedia/ProofOfPtolemysInequality.html A proof of Ptolemy's inequality.]
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<center>
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<math>
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\displaystyle AB \cdot CD + BC \cdot DA \ge AC \cdot BD
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</math>,
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</center>
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with equality [[iff]]. <math> \displaystyle ABCD </math> is [[cyclic quadrilateral | cyclic]].
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== Proof ==
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We construct a point <math> \displaystyle P </math> such that the [[triangles]] <math> \displaystyle APB, \; DCB </math> are [[similar]] and have the same [[orientation]].  In particular, this means that
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<center>
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<math>
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BD = \frac{BA \cdot DC }{AP} \; (*)
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</math>.
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</center>
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But since this is a [[spiral similarity]], we also know that the triangles <math> \displaystyle ABD, \; PBC </math> are also similar, which implies that
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<center>
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<math>
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BD = \frac{BC \cdot AD}{PC} \; (**)
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</math>.
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</center>
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Now, by the [[triangle inequality]], we have <math> \displaystyle AP + PC \ge AC </math>.  Multiplying both sides of the inequality by <math> \displaystyle BC </math> and using <math> \displaystyle (*) </math> and <math> \displaystyle (**) </math> gives us
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<center>
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<math>
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BA \cdot DC + BC \cdot AD \ge AC \cdot BC
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</math>,
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</center>
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which is the desired inequality.  Equality holds iff. <math> \displaystyle A </math>, <math> \displaystyle P </math>, and <math> \displaystyle {C} </math> are [[collinear]].  But since the angles <math> \displaystyle BAP </math> and <math> \displaystyle BDC </math> are congruent, this would imply that the angles <math> \displaystyle BAC </math> and <math> \displaystyle BPC </math> are [[congruent]], i.e., that <math> \displaystyle ABCD </math> is a cyclic quadrilateral.

Revision as of 12:23, 24 November 2006

Ptolemy's Inequality states that in a quadrilateral $\displaystyle ABCD$,

$\displaystyle AB \cdot CD + BC \cdot DA \ge AC \cdot BD$,

with equality iff. $\displaystyle ABCD$ is cyclic.

Proof

We construct a point $\displaystyle P$ such that the triangles $\displaystyle APB, \; DCB$ are similar and have the same orientation. In particular, this means that

$BD = \frac{BA \cdot DC }{AP} \; (*)$.

But since this is a spiral similarity, we also know that the triangles $\displaystyle ABD, \; PBC$ are also similar, which implies that

$BD = \frac{BC \cdot AD}{PC} \; (**)$.

Now, by the triangle inequality, we have $\displaystyle AP + PC \ge AC$. Multiplying both sides of the inequality by $\displaystyle BC$ and using $\displaystyle (*)$ and $\displaystyle (**)$ gives us

$BA \cdot DC + BC \cdot AD \ge AC \cdot BC$,

which is the desired inequality. Equality holds iff. $\displaystyle A$, $\displaystyle P$, and $\displaystyle {C}$ are collinear. But since the angles $\displaystyle BAP$ and $\displaystyle BDC$ are congruent, this would imply that the angles $\displaystyle BAC$ and $\displaystyle BPC$ are congruent, i.e., that $\displaystyle ABCD$ is a cyclic quadrilateral.

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