Skip to main content
Mathematics LibreTexts

10.2: Cross-ratio

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    The following theorem gives some quantities expressed in distances or angles that do not change after inversion.

    Theorem \(\PageIndex{1}\)

    Let \(ABCD\) and \(A'B'C'D'\) be two quadrangles such that the points \(A',B',C'\), and \(D'\) are the inverses of \(A,B,C\), and \(D\) respectively.



    \(\dfrac{AB \cdot CD}{BC \cdot DA} = \dfrac{A'B' \cdot C'D'}{B'C' \cdot D'A'}\).


    \(\measuredangle ABC + \measuredangle CDA \equiv -(\measuredangle A'B'C' + \measuredangle C'D'A')\).

    (c) If the quadrangle \(ABCD\) is inscribed, then so is \(\square A'B'C'D'\).


    (a). Let \(O\) be the center of the inversion. According to Lemma 10.1.1, \(\triangle AOB \sim \triangle B'OA'\). Therefore,

    \(\dfrac{AB}{A'B'} = \dfrac{OA}{OB'}.\)


    \(\dfrac{BC}{B'C'} = \dfrac{OC}{OB'}\), \(\dfrac{CD}{C'D'} = \dfrac{OC}{OD'}\), \(\dfrac{DA}{D'A'} = \dfrac{OA}{OD'}\).


    \(\dfrac{AB}{A'B'} \cdot \dfrac{B'C'}{BC} \cdot \dfrac{CD}{C'D'} \cdot \dfrac{D'A'}{DA} = \dfrac{OA}{OB'} \cdot \dfrac{OB'}{OC} \cdot \dfrac{OC}{OD'} \cdot \dfrac{OD'}{OA}.\)

    Hence (a) follows.

    (b). According to Lemma 10.1.1,

    \[\begin{array} {l} {\measuredangle ABO \equiv -\measuredangle B'A'O, \measuredangle OBC \equiv -\measuredangle OC'B',} \\ {\measuredangle CDO \equiv -\measuredangle D'C'O, \measuredangle ODA \equiv -\measuredangle OA'D'.} \end{array}\]

    By Axiom IIIb,

    \(\measuredangle ABC \equiv \measuredangle ABO + \measuredangle OBC\), \(\measuredangle D'C'B' \equiv \measuredangle D'C'O + \measuredangle OC'B'\),
    \(\measuredangle CDA \equiv \measuredangle CDO + \measuredangle ODA\), \(\measuredangle B'A'D' \equiv \measuredangle B'A'O + \measuredangle OA'D'\),

    Therefore, summing the four identities in 10.2.1, we get that

    \(\measuredangle ABC +\measuredangle CDA \equiv -(\measuredangle D'C'B' + \measuredangle B'A'D')\).

    Applying Axiom IIIb and Exercise 7.4.5, we get that

    \(\begin{array} {rcl} {\measuredangle A'B'C' + \measuredangle C'D'A'} & \equiv & {-(\measuredangle B'C'D' + \measuredangle D'A'B') \equiv} \\ {} & \equiv & {\measuredangle D'C'B' + \measuredangle B'A'D'.} \end{array}\)

    Hence (b) follows.

    (c). Follows from (b) and Corollary 9.3.2.

    This page titled 10.2: Cross-ratio is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.