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# 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.

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

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

$$\measuredangle A'C''B' \equiv -\measuredangle A'C'B'.$$

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.