ProofOfPtolemysInequality 2002-06-09 05:21:11 2006-10-02 04:59:56 Proof Ptolemy's theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) \usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Looking at the quadrilateral $ABCD$ we construct a point $E$, such that the triangles $ACD$ and $AEB$ are similar ($\angle ABE=\angle CDA$ and $\angle BAE=\angle CAD$). \begin{center} \includegraphics{ptolomaeus.eps} \end{center} This means that: $$\frac{AE}{AC}=\frac{AB}{AD}=\frac{BE}{DC},$$ from which follows that $$BE=\frac{AB\cdot DC}{AD}.$$ Also because $\angle EAC=\angle BAD$ and $$\frac{AD}{AC}=\frac{AB}{AE}$$ the triangles $EAC$ and $BAD$ are similar. So we get: $$EC=\frac{AC\cdot DB}{AD}.$$ Now if $ABCD$ is cyclic we get $$\angle ABE+\angle CBA=\angle ADC+\angle CBA=180^\circ.$$ This means that the points $C$, $B$ and $E$ are on one line and thus: $$EC=EB+BC$$ Now we can use the formulas we already found to get: $$\frac{AC\cdot DB}{AD}=\frac{AB\cdot DC}{AD}+BC.$$ Multiplication with $AD$ gives: $$AC\cdot DB=AB\cdot DC+BC\cdot AD.$$ Now we look at the case that $ABCD$ is not cyclic. Then $$\angle ABE+\angle CBA=\angle ADC+\angle CBA\neq 180^\circ,$$ so the points $E$, $B$ and $C$ form a triangle and from the triangle inequality we know: $$EC<EB+BC.$$ Again we use our formulas to get: $$\frac{AC\cdot DB}{AD}<\frac{AB\cdot DC}{AD}+BC.$$ From this we get: $$AC\cdot DB<AB\cdot DC+BC\cdot AD.$$ Putting this together we get Ptolomy's inequality: $$AC\cdot DB\leq AB\cdot DC+BC\cdot AD,$$ with equality iff $ABCD$ is cyclic.