Difference between revisions of "Ptolemy's Inequality"
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− | Ptolemy's Inequality | + | '''Ptolemy's Inequality''' states that in a [[quadrilateral]] <math> \displaystyle ABCD </math>, |
− | [ | + | <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. |
Revision as of 13:23, 24 November 2006
Ptolemy's Inequality states that in a quadrilateral ,
,
Proof
We construct a point such that the triangles
are similar and have the same orientation. In particular, this means that
.
But since this is a spiral similarity, we also know that the triangles are also similar, which implies that
.
Now, by the triangle inequality, we have . Multiplying both sides of the inequality by
and using
and
gives us
,
which is the desired inequality. Equality holds iff. ,
, and
are collinear. But since the angles
and
are congruent, this would imply that the angles
and
are congruent, i.e., that
is a cyclic quadrilateral.