6.4: Ptolemy's inequality
- Page ID
- 23615
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(\square ABCD\).
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.