## Task

Suppose $\triangle ABC$ and $\triangle DEF$ share three corresponding congruent sides as pictured below:

Show that $\triangle ABC$ is congruent ot $\triangle DEF$ as follows:

- Apply a translation to move $\triangle ABC$ to $\triangle A^\prime B^\prime C^\prime$ with $A^\prime = D$.
- Apply a rotation to move $\triangle A^\prime B^\prime C^\prime$ to $A^{\prime \prime}B^{\prime \prime}C^{\prime \prime}$ with $A^{\prime \prime} = D$ and $B^{\prime \prime} = E$.
- Explain why $\left|A^{\prime \prime}C^{\prime \prime}\right| = \left|DF\right|$ and conclude that $\overleftrightarrow{DE}$ is the perpendicular bisector of $\overline{C^{\prime \prime}F}$.
- Show that reflection over $\overleftrightarrow{DE}$ maps $F$ to $C^{\prime \prime}$ and conclude that $\triangle ABC$ is congruent to $\triangle DEF$.

## IM Commentary

The goal of this task is to establish the SSS congruence criterion using rigid motions. The method used here relies on understanding that the perpendicular bisector of a segment $|PQ|$ is the set of points in the plane equidistant from $P$ and $Q$: see https://www.illustrativemathematics.org/illustrations/967. The argument presented here treats a specific triangle congruence but the steps outlined here can be adapted to show the congruence of any two triangles sharing three corresponding congruent sides:

- The translation is not necessary when $A = D$.
- The rotation is not necessary if $B^\prime = E$ (or if $A$ = $D$ and $B$ = $E$).
- The reflection is not necessary if $C^{\prime \prime}$ lies on the same side of $\overleftrightarrow{DE}$ as $F$.

Note, however, that in the case where no reflection is required, a further argument is needed to show the triangle congruence. One method which can be used is to reflect over $\overleftrightarrow{DE}$ and then apply the argument in the solution.

It is useful for teachers and students to compare the SSS congruence criterion to the SAS congruence criterion (see https://www.illustrativemathematics.org/illustrations/109). The SSS congruence criterion is fundamentally more difficult to prove than the SAS criterion. Another proof of the SSS criterion is presented in https://www.illustrativemathematics.org/illustrations/110 and it uses the SAS congruence criterion.