Skip to main content
Mathematics LibreTexts

4.3: Isoseles Triangles

  • Page ID
    23600
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    A triangle with two equal sides is called isosceles; the remaining side is called the base.

    Theorem \(\PageIndex{1}\)

    Assume \(\triangle ABC\) is an isosceles triangle with the base \([AB]\). Then

    \(\measuredangle ABC \equiv - \measuredangle BAC.\)

    Moreover, the converse holds if \(\triangle ABC\) is nondegenerate.

    The following proof is due to Pappus of Alexandria.

    Proof

    截屏2021-02-03 上午10.59.25.png

    Note that

    \(CA = CB\), \(CB = CA\), \(\measuredangle ACB \equiv -\measuredangle BCA\).

    By Axiom IV,

    \(\triangle CAB \cong \triangle CBA.\)

    Applying the theorem on the signs of angles of triangles (Theorem 3.3.1) And Axiom IV again, we get that

    \(\measuredangle BAC \equiv -\measuredangle ABC.\)

    To prove the converse, we assume that \(\measuredangle CAB \equiv - \measuredangle CBA\). By ASA condition (Theorem 4.2.1), \(\triangle CAB \cong \triangle CBA\). Therefore, \(CA = CB\).

    A triangle with three equal sides is called equilateral.

    Exercise \(\PageIndex{1}\)

    Let \(\triangle ABC\) be an equilateral triangle. Show that

    \(\measuredangle ABC = \measuredangle BCA = \measuredangle CAB.\)

    Hint

    Apply Theorem 4.3.1 twice


    This page titled 4.3: Isoseles Triangles 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.