6.4: Ptolemy's inequality

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A quadrangle is defined as an ordered quadruple of distinct points in the plane. These 4 points are called vertexes of quadrangle. The quadrangle ABCD will be also denoted by $\square ABCD$.

Given a quadrangle $ABCD$, the four segments $[AB]$, $[BC]$, $[CD]$, and $[DA]$ are called sides of $\square ABCD$; the remaining two segments $[AC]$ and $[BD]$ are called diagonals of $\square ABCD$.

Theorem $\PageIndex{1}$ Ptolemy's inequality

In any quadrangle, the product of diagonals cannot exceed the sum of the products of its opposite sides; that is,

\[AC \cdot BD \le AB \cdot CD + BC \cdot DA\]

for any $\square ABCD$.

We will present a classical proof of this inequality using the method of similar triangles with an additional construction. This proof is given as an illustration — it will not be used further in the sequel.

Proof

$截屏2021-02-09 上午9.05.34.png$

Consider the half-line $[AX)$ such that $\measuredangle BAX = \measuredangle CAD$. In this case $\measuredangle XAD = \measuredangle BAC$ since adding $\measuredangle BAX$ or $\measuredangle CAD$ to the corresponding sides produces $\measuredangle BAD$. We can assume that

$AX = \dfrac{AB}{AC} \cdot AD.$

In this case we have

$\dfrac{AX}{AD} = \dfrac{AB}{AC}$, $\dfrac{AX}{AB} = \dfrac{AD}{AC}.$

Hence

$\triangle BAX \sim \triangle CAD$, $\triangle XAD \sim \triangle BAC$.

Therefore,

$\dfrac{BX}{CD} = \dfrac{AB}{AC}$, $\dfrac{XD}{BC} = \dfrac{AD}{AC}.$

or, equivalently

$AC \cdot BX = AB \cdot CD$, $AC \cdot XD = BC \cdot AD$.

Adding these two equalities we get

$AC \cdot (BX + XD) = AB \cdot CD + BC \cdot AD$.

It remains to apply the triangle inequality, $BD \le BX + XD$.

Using the proof above together with Corollary 9.3.2, one can show that the equality holds only if the vertexes $A, B, C$, and $D$ appear on a line or a circle in the same cyclic order; see also Theorem 10.4.1 for another proof of the equality case. Exercise 18.3.2 below suggests another proof of Ptolemy’s inequality using complex coordinates.