Ptolemy's Inequality

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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Ptolemy's Inequality

Proof