If the three sides of a triangle are congruent to the three sides of another triangle, then the triangles are congruent.

Based on the diagram above, the theorem can be written as follows.

$$\begin{array}{c}\{\begin{array}{l}\overline{AB}\cong \overline{DE}\\ \overline{BC}\cong \overline{EF}\\ \overline{AC}\cong \overline{DF}\end{array}\Rightarrow △ABC\cong △DEF\end{array}$$

Proof

Side-Side-Side Congruence Theorem

This proof will be developed based on the given diagram, but it is valid for any pair of triangles.

The primary purpose of this proof is finding a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of them will be shown here.

1

Translate $△DEF$ So That Two Corresponding Vertices Match

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Translate $△DEF$ so that $D$ is mapped onto $A.$ If this translation maps $△DEF$ onto $△ABC,$ the proof is complete.

Since the image of the translation does not match $△ABC,$ at least one more transformation is needed.

2

Rotate $△A{E}^{\prime }{F}^{\prime }$ So That Two Corresponding Sides Match

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Rotate $△A{E}^{\prime }{F}^{\prime }$ counterclockwise about $A$ so that a pair of corresponding sides matches. If the image of this transformation is $△ABC,$ the proof is complete. Note that this rotation maps $E^{\prime }$ onto $B.$ Consequently, $\overline{A{E}^{\prime }}$ is mapped onto $\overline{AB}.$

As before, the image does not match $△ABC.$ Therefore, a third rigid motion is required.

3

Reflect $△AB{F}^{\prime \prime }$ So That Two More Corresponding Sides Match

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The points $C$ and $F^{\prime \prime }$ are on opposite sides of $\overleftrightarrow{AB}.$ Now, consider $\overline{C{F}^{\prime }}.$ Let $G$ denote the point of intersection between $\overleftrightarrow{AB}$ and $\overline{C{F}^{\prime \prime }}.$

It can be noted that $AC=A{F}^{\prime \prime }$ and $BC=B{F}^{\prime \prime }.$ By the Converse Perpendicular Bisector Theorem, $\overleftrightarrow{AB}$ is a perpendicular bisector of $\overline{C{F}^{\prime \prime }}.$ Points along the perpendicular bisector are equidistant from the endpoints of the segment, so $CG=G{F}^{\prime \prime }.$

Finally, $F^{\prime \prime }$ can be mapped onto $C$ by a reflection across $\overleftrightarrow{AB}$ by reflecting $△AB{F}^{\prime \prime }$ across $\overleftrightarrow{AB}.$ Because reflections preserve angles, $\overrightarrow{A{F}^{\prime \prime }}$ and $\overrightarrow{B{F}^{\prime \prime }}$ are mapped onto $\overrightarrow{AC}$ and $\overrightarrow{BC},$ respectively.

This time the image matches $△ABC.$

Consequently, the application of a sequence of rigid motions allows $△DEF$ to be mapped onto $△ABC.$ This means that $△DEF$ and $△ABC$ are congruent triangles. The proof is complete.

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Consider the triangles $△ABC$ and $△DEF,$ where If there exists a rigid motion that maps one of these onto the other, then they are congruent. As the sides $\overline{AB}$ and $\overline{DE}$ are congruent, there is a rigid motion that maps one of these onto the other. Performing this transformation for one of the triangles leads to the two congruent sides overlapping. The line $\overleftrightarrow{C{F}^{\prime }}$ can now be drawn, dividing the angle $\angle C$ into $\angle 1$ and $\angle 3,$ and $\angle F^{\prime }$ into $\angle 2$ and $\angle 4.$ Notice that the $△BC{F}^{\prime }$ is an isosceles triangle, leading to $\angle 1$ and $\angle 2$ being congruent. Similarly, $\angle 3$ and $\angle 4$ are congruent as $△AC{F}^{\prime }$ is also an isosceles triangle. This leads to which by construction means that Thus, $\angle C$ and $\angle F^{\prime }$ are congruent.

The triangles have two sides, and their included angle, that are congruent. Thus, by the SAS Congruence Theorem, the triangles are indeed congruent.