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High School Geometry
Domain Congruence
Cluster Understand congruence in terms of rigid motions
Standard Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions.
Task SSS Congruence Criterion

SSS Congruence Criterion


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Alignments to Content Standards: G-CO.B.8
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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:

  1. Apply a translation to move $\triangle ABC$ to $\triangle A^\prime B^\prime C^\prime$ with $A^\prime = D$.
  2. 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 $B^{\prime \prime} = E$. 
  3. Explain why $\left|A^{\prime \prime}C^{\prime \prime}\right| = \left|DF\right|$ and conclude that $\overleftrightarrow{DE}$ is the perpendicular bisector of $\overline{C^{\prime \prime}F}$.
  4. 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.

Solution

  1. The image of $\triangle ABC$ after translation by $\overline{AD}$ is pictured below, labeled $A^\prime B^\prime C^\prime$:


               

    The translation is chosen to send $A$ to $D$ so $A^\prime = D$ in the picture. Since $\triangle A^\prime B^\prime C^\prime$ is a translate of $\triangle ABC$ they are congruent and corresponding sides are congruent since translations preserve distance.

  2. If we rotate $\triangle A^\prime B^\prime C^\prime$ about  $A^\prime$ by $\angle B^\prime DE$ then $A^\prime = D$ does not move and $B^\prime$ maps to $E$. The image of $\triangle A^\prime B^\prime C^\prime$ after this rotation is shown below, labeled $\triangle A^{\prime \prime}B^{\prime \prime}C^{\prime \prime}$:


              

    Since $\triangle A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$ is a rotation of $\triangle A^\prime B^\prime C^\prime$ they are congruent and $\triangle A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$ is also congruent to $\triangle ABC$. Corresponding sides of these triangles are congruent since rotations (and translations) preserve distance.

  3. We know that $|AC| = |DF|$ by hypothesis. We saw in parts (a) and (b) that $\left|A^{\prime \prime} C^{\prime \prime} \right| = |AC|$ so we have $\left|A^{\prime \prime} C^{\prime \prime} \right| = |DF|$. Similar reasoning shows that $\left|B^{\prime \prime}C^{\prime \prime}\right| = |EF|$. The perpendicular bisector of $\overline{C^{\prime \prime}F}$ is the set of points equidistant from $C^{\prime \prime}$ and $F$ so this means that $D$ and $E$ both lie on the perpendicular bisector of $\overline{C^{\prime \prime}F}$. Thus $\overleftrightarrow{DE}$ is the perpendicular bisector of $\overline{C^{\prime \prime}F}$.
  4. We saw in part (c) that $\overleftrightarrow{DE}$ is the perpendicular bisector of $\overline{C^{\prime \prime}F}$. This means that reflection over $\overleftrightarrow{DE}$ maps $F$ to $C^{\prime \prime}$. This reflection maps $D$ and $E$ to themselves and so maps $\triangle A^{\prime \prime}B^{\prime \prime}C^{\prime \prime}$ to $\triangle DEF$, establishing the congruence of $\triangle ABC$ with $\triangle DEF$.

SSS Congruence Criterion

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:

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