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Mathematics LibreTexts

6.4: Ptolemy's inequality

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    23615
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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.