
# 6.4: Ptolemy's inequality


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

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.