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Difference between revisions of "Ptolemy's Inequality"

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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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'''Ptolemy's Inequality''' states that in for four points <math>A, B, C, D </math> in the plane,  
  
 
<center>
 
<center>
 
<math>
 
<math>
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\displaystyle AB \cdot CD + BC \cdot DA \ge AC \cdot BD
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AB \cdot CD + BC \cdot DA \ge AC \cdot BD
 
</math>,
 
</math>,
 
</center>
 
</center>
  
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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>.
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with equality [[iff]]. <math>ABCD </math> is a [[cyclic quadrilateral]] with [[diagonal]]s <math>AC </math> and <math>BD </math>.
  
 
== Proof ==
 
== Proof ==
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which is the desired inequality.  Equality holds iff. <math>A </math>, <math>P </math>, and <math>{C} </math> are [[collinear]].  But since the angles <math>BAP </math> and <math>BDC </math> are congruent, this would imply that the angles <math>BAC </math> and <math>BPC </math> are [[congruent]], i.e., that <math>ABCD </math> is a cyclic quadrilateral.
 
which is the desired inequality.  Equality holds iff. <math>A </math>, <math>P </math>, and <math>{C} </math> are [[collinear]].  But since the angles <math>BAP </math> and <math>BDC </math> are congruent, this would imply that the angles <math>BAC </math> and <math>BPC </math> are [[congruent]], i.e., that <math>ABCD </math> is a cyclic quadrilateral.
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[[Category:Theorems]]

Revision as of 13:42, 8 November 2007

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

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

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

Proof

We construct a point $P$ such that the triangles $APB, \; DCB$ 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 $ABD, \; PBC$ are also similar, which implies that

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

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

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

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

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