10.03 SSS and SAS congruence criteria

The best way to approach a proof is to label the given information and any information that can be concluded based on the given information. In this example, we can label \overline{DE} \cong \overline{EF} because E is a midpoint. Take a look at the labeled diagram:

We can see that these triangles have all three corresponding sides labeled as congruent so the triangles will be congruent by SSS congruence.

For the most part, the order of the statements in a proof is up to us. Just make sure that any statements based on a given piece of information (such as the definition of midpoint in this proof) come after that given statement.

We know by SSS congruency that \triangle{LKM} \cong \triangle{RQS} so by CPCTC, \overline{KM} \cong \overline{QS}, so by the definition of congruence, KM = QS. We will use this information to write and solve an equation to find the unknown variable.

1. There is a rigid motion that will map \overline{AB} onto \overline{DE} because \overline{AB}\cong \overline{DE}. First, we can rotate the triangle to get until \overline{AB}\parallel \overline{DE}. Then we can translate up or down and left or right until \overline{AB} is on top of \overline{DE}. Call this triangle \triangle A'B'C'. If C' and F are on the same side, reflect over \overline{DE} until the figure is as shown.

   

2. Since these transformations are all rigid motions, we have that: \overline{AC}\cong \overline{A'C'} \text{ and }\angle{BAC}\cong \angle{B'A'C'}
3. Using the given information and the transitive property of congruence, we have that:\overline{DF}\cong \overline{A'C'} \text{ and }\angle{B'A'C'}\cong \angle{EDF}
4. Since \overline{DF}\cong \overline{A'C'} \text{ and }\angle{B'A'C'}\cong \angle{EDF} we have that \overline{DE} must be the angle bisector of \angle C'DF using the definition of the angle bisector theorem since it breaks the angle into two congruent angles.

   

5. Since \overline{DE} is the angle bisector of \angle C'DF, we can use it as a line of reflection to map C' onto F.
6. Angle measures are preserved in reflections. We know that \overline{C'D} coincides with \overline{DF} and since these segments are congruent, point C' can be mapped on to point F by reflection across \overline{DE}.
7. We have now shown that:
   • A maps to D using rigid motions
   • B maps to E using rigid motions
   • C maps to F using rigid motions
   So we have that \triangle ABC can be mapped onto \triangle DEF, so \triangle ABC \cong \triangle DEF.

Another approach for a rigid motion transformation that could map \overline{AB} onto \overline{DE} could be a reflection across a diagonal line followed by a rotation and a translation.

Start by drawing any given information from the proof onto the diagram. We know that \angle ABD \cong \angle CBD, so we can label it appropriately.

Now that we see a shared angle between two congruent corresponding pairs of side lengths, we will use SAS to prove congruency between the triangles and prove the base angles theorem.

\triangle ABC \cong \triangle EBC because \overline{BC} is the same in both triangles. We are given that \overline{AC} = \overline{CE}. \overline{BC} is a perpendicular bisector of \overline{AE} since it is vertical to the base of the bridge, so m \angle BCA = m \angle BCE = 90 \degree. Since \angle BCA is the shared angle of \overline{BC} and \overline{AC}, and \angle BCE is the shared angle of \overline{BC} and \overline{CE}, the triangles are congruent by SAS.

We could use the same justification for \triangle {EFG} and \triangle {HFG}.