Difference between revisions of "Ptolemy's Inequality"

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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-Ptolemy's Inequality for a [[quadrilateral]] ABCD states that AB·CD + BC·DA ≥ AC·BD with [[equal]]ity [[iff]] ABCD is a [[cyclic quadrilateral]].
+'''Ptolemy's Inequality''' states that in a [[quadrilateral]] <math> \displaystyle ABCD </math>,
-[http://planetmath.org/encyclopedia/ProofOfPtolemysInequality.html A proof of Ptolemy's inequality.]
+<center>
+<math>
+\displaystyle AB \cdot CD + BC \cdot DA \ge AC \cdot BD
+</math>,
+</center>
+with equality [[iff]]. <math> \displaystyle ABCD </math> is [[cyclic quadrilateral | cyclic]].
+== Proof ==
+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
+<center>
+<math>
+BD = \frac{BA \cdot DC }{AP} \; (*)
+</math>.
+</center>
+But since this is a [[spiral similarity]], we also know that the triangles <math> \displaystyle ABD, \; PBC </math> are also similar, which implies that
+<center>
+<math>
+BD = \frac{BC \cdot AD}{PC} \; (**)
+</math>.
+</center>
+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
+<center>
+<math>
+BA \cdot DC + BC \cdot AD \ge AC \cdot BC
+</math>,
+</center>
+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.

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Difference between revisions of "Ptolemy's Inequality"

Revision as of 12:23, 24 November 2006

Proof