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Difference between revisions of "Ptolemy's Inequality"

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Revision as of 16:10, 15 December 2007

Ptolemy's Inequality is a famous inequality attributed to the Greek mathematician Ptolemy.

Theorem

The 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 if and only if $ABCD$ 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.

See Also

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