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Difference between revisions of "Ptolemy's Inequality"
(reworded statement in a more general way)
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'''Ptolemy's Inequality''' states that in a [[quadrilateral]] <math> \displaystyle ABCD </math>,  
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'''Ptolemy's Inequality''' states that in for four points <math> \displaystyle A, B, C, D </math> in the plane,  
  
 
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<center>
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</center>
 
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with equality [[iff]]. <math> \displaystyle ABCD </math> is [[cyclic quadrilateral | cyclic]].
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with equality [[iff]]. <math> \displaystyle ABCD </math> is a [[cyclic quadrilateral]] with [[diagonal]]s <math> \displaystyle AC </math> and <math> \displaystyle BD </math>.
  
 
== Proof ==
 
== Proof ==

Revision as of 22:11, 24 November 2006

Ptolemy's Inequality states that in for four points $\displaystyle A, B, C, D$ in the plane,

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

with equality iff. $\displaystyle ABCD$ is a cyclic quadrilateral with diagonals $\displaystyle AC$ and $\displaystyle BD$.

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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