Skip to main content
Mathematics LibreTexts

2.5: Isosceles Triangles

  • Page ID
    34128
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    In Section 1.6, we defined a triangle to be isosceles if two of its sides are equal. Figure \(\PageIndex{1}\) shows an isosceles triangle \(\triangle ABC\) with \(AC=BC\). In \(\triangle ABC\) we say that \(\angle A\) is opposite side \(BC\) and \(\angle B\) is opposite side \(AC\).

    clipboard_e43133622825a973bd3b9c28191e3b46e.png
    Figure \(\PageIndex{1}\): \(\triangle ABC\) is isosceles with AC = BC.

    The most important fact about isosceles triangles is the following:

    Theorem \(\PageIndex{1}\)

    If two sides of a triangle are equal the angles opposite these sides are equal.

    Theorem \(\PageIndex{1}\) means that if \(AC = BC\) in \(\triangle ABC\) then \(\angle A = \angle B\).

    Example \(\PageIndex{1}\)

    Find \(x\):

    屏幕快照 2020-11-01 下午10.21.58.png

    Solution

    \(AC = BC\) so \(\angle A = \angle B\). Therefore, \(x = 40\).

    Answer: \(x = 40\).

    In \(\triangle ABC\) if \(AC = BC\) then side \(AB\) is called the base of the triangle and \(\angle A\) and \(\angle B\) are called the base angles. Therefore Theorem \(\PageIndex{1}\) is sometimes stated in the following way: "The base angles of an isosceles triangle are equal,"

    Proof of Theorem \(\PageIndex{1}\): Draw \(CD\), the angle bisector of \(\angle ACB\) (Figure \(\PageIndex{2}\)). The rest of the proof will be presented in double-column form. We have given that \(AC = BC\) and \(\angle ACD = \angle BCD\). We must prove \(\angle A = \angle B\).

    屏幕快照 2020-11-01 下午10.31.41.png
    Figure \(\PageIndex{2}\): Draw \(CD\), the angle bisector of \(\angle ACB\).
    Statements Reasons
    1. \(AC = BC\). 1. Given, \(\triangle ABC\) is isosceles.
    2. \(\angle ACD = \angle BCD\). 2. Given, \(CD\) is the angle bisector of \(\angle ACB\).
    3. \(CD = CD\). 3. Identity.
    4. \(\triangle ACD \cong \triangle BCD\). 4. \(SAS = SAS\): \(AC, \angle C, CD\) of \(\triangle ACD = BC\), \(\angle C, CD\) of \(\triangle BCD\).
    5. \(\angle A = \angle B\). 5. Corresponding angles of congruent trianglesare equal.
    Example \(\PageIndex{2}\)

    Find \(x, \angle A, \angle B\) and \(\angle C\):

    屏幕快照 2020-11-01 下午10.36.18.png

    Solution

    \(\angle B = \angle A = 4x + 5^{\circ}\) by Theorem \(\PageIndex{1}\). We have

    \[\begin{array} {rcl} {\angle A + \angle B + \angle C} & = & {180^{\circ}} \\ {4x + 5 + 4x + 5 + 2x - 10} & = & {180} \\ {10x} & = & {180} \\ {x} & = & {18} \end{array}\]

    \(\angle A = \angle B = 4x + 5^{\circ} = 4(18) + 5^{\circ} = 72 + 5^{\circ} = 77^{\circ}\).

    \(\angle C = 2x - 10^{\circ} = 2(18) - 10^{\circ} = 36 - 10^{\circ} = 26^{\circ}\).

    Check

    屏幕快照 2020-11-01 下午10.44.33.png

    Answer

    \(x = 18\), \(\angle A = 77^{\circ}\), \(\angle B = 77^{\circ}\), \(\angle C = 26^{\circ}\).

    In Theorem \(\PageIndex{1}\) we assumed \(AC = BC\) and proved \(\angle A = \angle B\). We will now assume \(\angle A = \angle B\) and prove \(AC = BC\). '1ihen the assumption and conclusion of a statement are interchanged the result is called the converse of the original statement.

    Theorem \(\PageIndex{2}\): The Converse of Theorem \(\PageIndex{1}\))

    If two angles of a triangle are equal the sides opposite these angles are equal.

    If Figure 4, if \(\angle A = \angle B\) then \(AC = BC\).

    屏幕快照 2020-11-01 下午10.48.05.png
    Figure \(\PageIndex{4}\). \(\angle A = \angle B\)
    Example \(\PageIndex{3}\)

    Find \(x\)

    屏幕快照 2020-11-01 下午10.49.04.png

    Solution

    \(\angle A = \angle B\) so \(x = AC = BC = 9\) by Theorem \(\PageIndex{2}\).

    Answer

    \(x = 9\).

    Proof of Theorem \(\PageIndex{2}\): Draw \(CD\) the angle bisector of \(\angle ACB\) (Figure \(\PageIndex{5}\)). We have \(\angle ACD = \angle BCD\) and \(\angle A = \angle B\). We must prove \(AC = BC\).

    屏幕快照 2020-11-01 下午10.50.45.png
    Figure \(\PageIndex{5}\). Draw \(CD\), the angle bisector of \(\angle ACB\).
    Statements Reasons
    1. \(\angle A = \angle B\). 1. Given.
    2. \(\angle ACD = \angle BCD\). 2. Given.
    3. \(CD = CD\). 3. Identity.
    4. \(\triangle ACD \cong \triangle BCD\). 4. \(AAS = AAS\): \(\angle A, \angle C, CD\) of \(\triangle ACD = \angle B\), \(\angle C\), \(CD\) of \(triangle BCD\).
    5. \(AC = BC\). 5. Corresponding sides of congruent triangles are equal

    The following two theorems are corollaries (immediate consequences) of the two preceding theorems:

    Theorem \(\PageIndex{3}\)

    An equilateral triangle is equiangular.

    In Figure \(\PageIndex{7}\), if \(AB = AC = BC\) then \(\angle A = \angle B = \angle C\).

    屏幕快照 2020-11-01 下午10.58.47.png
    Figure \(\PageIndex{7}\): \(\angle ABC\) is equilateral.
    Proof

    \(AC = BC\) so by Theorem \(\PageIndex{1}\) \(\angle B = \angle C\). Therefore \(\angle A = \angle B = \angle C\).

    Since the sum of the angle is \(180^{\circ}\) we must have in fact that \(\angle A = \angle B = \angle C = 60^{\circ}\).

    Theorem \(\PageIndex{4}\): The Converse of Theorem \(\PageIndex{3}\))

    An equiangular triangle is equilateral.

    In Figure \(\PageIndex{8}\), if \(\angle A = \angle B = \angle C\) then \(AB = AC = BC\).

    屏幕快照 2020-11-02 上午11.47.53.png
    Figure \(\PageIndex{8}\). \(\angle ABC\) is equiangular.
    Proof

    \(\angle A = \angle B\) so by Theorem \(\PageIndex{2}\), \(AC = BC\), \(\angle B = \angle C\) by Theorem \(\PageIndex{2}\), \(AB = AC\). Therefore \(AB = AC = BC\).

    Example \(\PageIndex{4}\)

    Find \(x\), \(y\) and \(AC\):

    屏幕快照 2020-11-02 上午11.51.13.png

    Solution

    \(\triangle ABC\) is equiangular and so by Theorem \(\PageIndex{4}\) is equilateral.

    Therefore \(\begin{array} {rcl} {AC} & = & {AB} \\ {x + 3y} & = & {7x - y} \\ {x - 7x + 3y + y} & = & {0} \\ {-6x + 4y} & = & {0} \end{array}\) and \(\begin{array} {rcl} {AB} & = & {BC} \\ {7x - y} & = & {3x + 5} \\ {7x - 3xy - y} & = & {5} \\ {4x - y} & = & {5} \end{array}\)

    We have a system of two equations in two unknowns to solve:

    屏幕快照 2020-11-02 上午11.55.24.png

    Check:

    屏幕快照 2020-11-02 上午11.56.09.png

    Answer: \(x = 2\), \(y = 3\), \(AC = 11\).

    Historical Note

    Theorem \(\PageIndex{1}\), the isosceles triangle theorem, is believed to have first been proven by Thales (c. 600 B,C,) - it is Proposition 5 in Euclid's Elements. Euclid's proof is more complicated than ours because he did not want to assume the existence of an angle bisector, Euclid's proof goes as follows:

    Given \(\triangle ABC\) with \(AC = BC\) (as in Figure \(\PageIndex{1}\) at the beginning of this section), extend \(CA\) to \(D\) and \(CB\) to \(E\) so that \(AD = BE\) (Figure \(\PageIndex{9}\)). Then \(\triangle DCB \cong \triangle ECA\) by \(SAS = SAS\). The corresponding sides and angles of the congruent triangles are equal, so \(DB = EA\), \(\angle 3 = \angle 4\) and \(\angle 1 + \angle 5 = \angle 2 + \angle 6\). Now \(\triangle ADB \cong \triangle BEA\) by \(SAS = SAS\). This gives \(\angle 5 = \angle 6\) and finally \(\angle 1 = \angle 2\).

    屏幕快照 2020-11-02 下午12.01.09.png
    Figure \(\PageIndex{9}\): The "bridge of fools".

    This complicated proof discouraged many students from further study in geometry during the long period when the Elements was the standard text, Figure \(\PageIndex{9}\) resembles a bridge which in the Middle Ages became known as the "bridge of fools," This was supposedly because a fool could not hope to cross this bridge and would abandon geometry at this point.

    Problems

    For each of the following state the theorem(s) used in obtaining your answer.

    1. Find \(x\):

    Screen Shot 2020-11-02 at 12.06.49 PM.png

    2. Find \(x\), \(\angle A\), and \(\angle B\):

    Screen Shot 2020-11-02 at 12.07.07 PM.png

    3. Find \(x\):

    Screen Shot 2020-11-02 at 12.07.27 PM.png

    4. Find \(x\), \(AC\), and \(BC\):

    Screen Shot 2020-11-02 at 12.07.40 PM.png

    5. Find \(x\):

    Screen Shot 2020-11-02 at 12.08.00 PM.png

    6. Find \(x\):

    Screen Shot 2020-11-02 at 12.08.15 PM.png

    7. Find \(x, \angle A, \angle B\), and \(\angle C\):

    Screen Shot 2020-11-02 at 12.08.31 PM.png

    8. Find \(x, \angle A, \angle B\), and \(\angle C\):

    Screen Shot 2020-11-02 at 12.08.45 PM.png

    9. Find \(x, AB, AC\), and \(BC\):

    Screen Shot 2020-11-02 at 12.09.04 PM.png

    10. Find \(x, AB, AC\), and \(BC\):

    Screen Shot 2020-11-02 at 12.09.43 PM.png

    11. Find \(x, y\), and \(AC\):

    Screen Shot 2020-11-02 at 12.10.02 PM.png

    12. Find \(x, y\), and \(AC\):

    Screen Shot 2020-11-02 at 12.10.22 PM.png

    13. Find \(x\):

    Screen Shot 2020-11-02 at 12.10.43 PM.png

    14. Find \(x, y\), and \(z\):

    Screen Shot 2020-11-02 at 12.10.58 PM.png


    This page titled 2.5: Isosceles Triangles is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Henry Africk (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.