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2.6: The SSS Theorem

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    34129
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    We now consider the case where the side of two triangles are known to be of the same length.

    Theorem \(\PageIndex{1}\): Side-Side-Side (SSS) Theorem

    Two triangles are congruent if three sides of one are equal respectively to three sides of the other (\(SSS = SSS\)).

    Theorem \(\PageIndex{1}\) is demonstrated in Figure \(\PageIndex{1}\): if \(a=d, b=e,\) and \(c=f\) then \(\triangle ABC \cong \triangle DEF\)

    clipboard_e421ec307516542d1734d0a6ddb05ea41.png
    Figure \(\PageIndex{1}\): \(\triangle ABC \cong \triangle DEF\) because \(SSS = SSS\).
    Example \(\PageIndex{1}\)

    Find \(x, y, z:\)

    clipboard_e89a56f3010d2114e1c007fecb808f43e.png

    Solution

    \(AB = 7 = DF\). Therefore \(\angle C,\) the angle opposite AB must correspond to \(\angle E\), the angle opposite \(DF\). In the same way \(\angle A\) corresponds to \(\angle F\) and \(\angle B\) corresponds to \(\angle D\). We have \(\triangle ABC \cong \triangle FDE\) by \(SSS = SSS\), so

    \(x^{\circ}=\angle D=\angle B=44^{\circ}\)

    \(y^{\circ}=\angle F=\angle A=57^{\circ}\)

    \(z^{\circ}=\angle E=\angle C=79^{\circ}\)

    Answer: \(x = 44, y=57, z=79\)

    Proof of Theorem \(\PageIndex{1}\)

    In Figure \(\PageIndex{1}\), place \(\triangle ABC\) and \(\triangle DEF\) so that their longest sides coincide, in this case \(AB\) and \(DE\). This can be done because \(AB = c= r = DE.\) Now draw \(CF\), forming angles \(1,2,3,\) and 4 (Figure \(\PageIndex{2}\)). The rest of the proof will be presented in double-column form:

    clipboard_ed00c957957c7f639aaf53f40aa954d11.png
    Figure \(\PageIndex{2}\): Place \(\triangle ABC\) and \(\triangle DEF\) so that \(AB\) and DE coincide and draw \(CF\).
    Statement Reasons
    1. \(\angle 1 = \angle 2\). 1. The base angles of isosceles triangle \(CAF\) are equal (Theorem \(\PageIndex{1}\), section 2.5).
    2. \(\angle 3 = \angle 4\). 2. The base angles of isosceles triangle \(CBF\) are equal.
    3. \(\angle C = \angle F\). 3. \(\angle C = \angle 1 + \angle 3 = \angle 2 + \angle 4 = \angle F\).
    4. \(AC = DF\). 4. Given, \(AC = b = e = DF\).
    5. \(BC = EF\). 5. Given, \(BC = a = d = EF\).
    6. \(\triangle ABC \cong \triangle DEF\). 6. \(SAS = SAS\): \(AC, \angle C, BC\) of \(\triangle ABC = DF\), \(\angle F\), \(EF\) of \(\triangle DEF\).
    Example \(\PageIndex{2}\)

    Given \(AB = DE, BC = EF,\) and \(AC = DF\). Prove \(\angle C = \angle F\)

    clipboard_ee0c0f0d6ef166db1034093a3a6e507fe.png

    Solution

    Statements Reasons
    1. \(AB = DE\). 1. Given.
    2. \(BC = EF\). 2. Given.
    3. \(AC = DF\). 3. Given.
    4. \(\triangle ABC \cong \triangle DEF\). 4. \(SSS = SSS\): \(AB, BC, AC\) of \(\triangle ABC = DE\), \(EF, DF\) of \(\triangle DEF\).
    5. \(\angle C = \angle F\). 5. Corresponding angles of congruent triangles are equal.
    Application: Triangular Bracing

    The SSS Theorem is the basis of an important principle of construction engineering called triangular bracing. Imagine the line segments in Figure \(\PageIndex{3}\) to be beans of wood or steel joined at the endpoints by nails or screws. If pressure is applied to one of the sides, \(ABCD\) will collapse and look like \(A'B'C'D'\).

    clipboard_edb18c1069fe3bcc823b78ff19e5e82b2.png
    Figure \(\PageIndex{3}\): \(ABCD\) collapses into \(A'B'C'D'\), when pressure is applied.

    Now suppose points \(A\) and \(C\) are joined by a new beam, called a brace (Figure \(\PageIndex{4}\)). The structure will not collapse as long as the beans remain unbroken and joined together. It is impossible to deform \(ABCD\) into any other shape \(A'B'C'D'\) because if \(AB = A'B'\), \(BC = B'C'\), and \(AC = A'C'\) then \(\triangle ABC\) would be congruent to \(\triangle A'B'C'\) by \(SSS = SSS\).

    clipboard_e4bdc1e76fefcc81a8df578b81d2d2606.png
    Figure \(\PageIndex{4}\): \(ABCD\) cannot collapse into \(A'B'C'D'\) as long as the beams remain unbroken and Jotned together.

    We sometimes say that a triangle is a rigid figure; once the sides of a triangle are fixed the angles cannot be changed. Thus in Figure \(\PageIndex{4}\), the shape of \(\triangle ABC\) cannot be changed as long as the lengths of its sides remain the same.

    Problems

    1 - 8. For each of the following (1) write the congruence statement,

    (2) given the reason for (1) (\(SAS\), \(ASA\), \(AAS\), or \(SSS\) Theorem), and

    (3) find \(x\), or \(x\) and \(y\), or \(x, y\), and \(z\).

    1.

    Screen Shot 2020-11-02 at 5.19.40 PM.png

    2.

    Screen Shot 2020-11-02 at 5.20.08 PM.png

    3.

    Screen Shot 2020-11-02 at 5.20.26 PM.png

    4.

    Screen Shot 2020-11-02 at 5.28.04 PM.png

    5.

    Screen Shot 2020-11-02 at 5.28.23 PM.png

    6.

    Screen Shot 2020-11-02 at 5.28.39 PM.png

    7.

    Screen Shot 2020-11-02 at 5.29.05 PM.png

    8.

    Screen Shot 2020-11-02 at 5.29.18 PM.png

    9. Given \(AB = DE\), \(BC = EF\), and \(AC = DF\). Prove \(\angle A = \angle D\).

    Screen Shot 2020-11-02 at 5.29.36 PM.png

    10. Given \(AC = BC\). \(AD = BD\). Prove \(\angle ADC = \angle BDC\).

    Screen Shot 2020-11-02 at 5.29.53 PM.png

    11. Given \(AB = AD\), \(BC = DC\). Prove \(\angle BAC = \angle CAD\).

    Screen Shot 2020-11-02 at 5.30.25 PM.png

    12. Given \(AB = CD\), \(BC = DA\). Prove \(\angle BAC = \angle DCA\).

    Screen Shot 2020-11-02 at 5.31.23 PM.png

    13. Given \(AE = CE\), \(BE = ED\). Prove \(AB = CD\).

    Screen Shot 2020-11-02 at 5.31.39 PM.png

    14. Given \(AB||CD\), \(AD||BC\). Prove \(AB = CD\).

    Screen Shot 2020-11-02 at 5.31.52 PM.png


    This page titled 2.6: The SSS Theorem 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.