4.4: Side-Side-Side condition

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.

Proof

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 $C'' \ne C'$.

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 $B \in [AC]$.

Proof

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

$\triangle A'B'C' \cong \triangle ABC$.

Hint

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) $\triangle ABB' \cong \triangle BAA'$.

Hint

(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.

Hint

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.