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Mathematics LibreTexts

4.4: Side-Side-Side condition

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    Theorem \(\PageIndex{1}\) SSS condition

    \(\triangle ABC \cong \triangle A'B'C'\) if

    \(A'B' = AB\), \(B'C' = BC\) and \(C'A' = CA\).

    Note that this condition is valid for degenerate triangles as well.


    Choose \(C''\) so that \(A'C'' = A'C'\) and \(\measuredangle B'A'C'' = \measuredangle BAC\). According to Axiom IV,

    \(\triangle A'B'C'' \cong \triangle ABC.\)

    It will suffice to prove that

    \[\triangle A'B'C' \cong \triangle A'B'C''.\]

    The condition 4.4.1 trivially holds if \(C'' = C'\). Thus, it remains to consider the case \(C'' \ne C'\).

    截屏2021-02-03 下午3.10.08.png

    Clearly, the corresponding sides of \(\triangle A'B'C'\) and \(\triangle A'B'C''\) are equal. Hence the triangles \(\trianggle C'A'C''\) and \(\triangle C'B'C''\) are isosceles. By Theorem 4.3.1, we have

    \(\begin{array} {l} {\measuredangle A'C''C' \equiv -\measuredangle A'C'C''} \\ {\measuredangle C'C''B' \equiv -\measuredangle C''C'B'.} \end{array}\)

    Adding them, we get

    \(\measuredangle A'C''B' \equiv -\measuredangle A'C'B'.\)

    Applying Axiom IV again, we get 4.4.1

    Corollary \(\PageIndex{1}\)

    If \(AB + BC = AC\), then \(B \in [AC]\).


    We may assume that \(AB > 0\) and \(BC > 0\); otherwise \(A = B\) or \(B = C\).

    Arguing by contradiction, suppose \(AB + BC = AC\). Choose \(B' \in [AC]\) such that \(AB = AB'\); note that \(BC = B'C\) and \(\measuredangle AB'C = \pi\).

    By \(SSS\),

    \(\triangle ABC \cong \triangle AB'C.\)

    Therefore \(\measuredangle ABC = \pi\). By Theorem 2.4.1, \(B\) lies between \(A\) and \(C\).

    Advanced Exercise \(\PageIndex{1}\)

    Let \(M\) be the midpoint of the side \([AB]\) of \(\triangle ABC\) and \(M'\) be the midpoint of the side \([A'B']\) of \(\triangle A'B'C'\). Assume \(C'A' = CA\), \(C'B' = CB\), and \(C'M' = CM\). Prove that

    \(\triangle A'B'C' \cong \triangle ABC\).


    Consider the points \(D\) and \(D'\), such that \(M\) is the midpoint of \([CD]\) and \(M'\) is the midpoint of \([C'D']\). Show that \(\triangle BCD \cong \triangle B'C'D'\) and use it to prove that \(\triangle A'B'C' \cong \triangle ABC\).

    Exercise \(\PageIndex{2}\)

    Let \(\triangle ABC\) be an isosceles triangle with the base \([AB]\). Suppose that \(CA' = CB'\) for some points \(A' \in [BC]\) and \(B' \in [AC]\). Show that

    (a) \(\triangle AA'C \cong \triangle BB'C\);

    (b) \(\triangle ABB' \cong \triangle BAA'\).

    截屏2021-02-03 下午4.01.29.png


    (a) Apply SAS.

    (b) Use (a) and apply SSS.

    Exercise \(\PageIndex{3}\)

    Let \(\triangle ABC\) be a nondegenerate triangle and let \(f\) be a motion of the plane such that

    \(f(A) = A\), \(f(B) = B\) and \(f(C) = C\).

    Show that \(f\) is the identity map; that is, \(f(X) = X\) for any point \(X\) on the plane.


    Without loss of generally, we may assume that \(X\) is distinct from \(A, B,\) and \(C\). Set \(f(X) = X'\); assume \(X' \ne X\).

    Note that \(AX = AX'\), \(BX = BX'\), and \(CX = CX'\). By SSS we get that \(\angle ABX = \pm \measuredangle ABX'\). Since \(X \ne X'\), we get that \(\measuredangle ABX \equiv - \measuredangle ABX'\). The same way we get that \(\measuredangle CBX \equiv -\measuredangle CBX'\). Subtracting these two identities from each other, we get that \(\measuredangle ABC \equiv - \measuredangle ABC\). Conclude that \(\measuredangle ABC = 0\) or \(\pi\). That is, \(\triangle ABC\) is degenerate -- a contradiction.

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