5.04 ASA and AAS congruence criteria

We can set up a figure so we can do the proof in terms of a particular diagram. Since we can't use the ASA congruency theorem to prove itself, we must use another strategy. In this case, we need to identify a series of rigid transformations that will map \triangle ABC to \triangle DEF. Our rigid transformations are:

• Translation
• Rotation
• Reflection

Given:

• \angle{ABC}\cong \angle{DEF}
• \overline{BC}\cong \overline{EF}
• \angle{ACB}\cong \angle{DFE}

Prove: \triangle ABC \cong \triangle DEF

We need to consider that the two triangles could be any orientation.

1. There is a rigid motion that will map \overline{BC} onto \overline{EF} because \overline{BC}\cong \overline{EF}. First, we can rotate and reflect the triangle to get \overline{BC}\parallel \overline{EF} so that and \angle B and \angle E are oriented in the same direction. Call this triangle \triangle A'B'C'. If A' and D are on the same side, reflect over \overline{EF} until the figure is as shown.

   

2. Since these transformations are all rigid motions, we have that: \angle{ABC}\cong \angle{A'B'C'} \text{ and }\angle{ACB}\cong \angle{A'C'B'}
3. Using the given information and the transitive property of congruence, we have that:\angle{DEF}\cong \angle{A'B'C'} \text{ and }\angle{DFE}\cong \angle{A'C'B'}

4. Point A' can be mapped on to point D by a reflection across \overline{EF}.

   We can justify that point A' coincides with point D after the reflection as follows:

   If the reflection of \triangle{A'FE} is \triangle{A''FE}, points E and F will remain fixed during the reflection. \overrightarrow{EA''} will coincide with \overrightarrow{ED}, and \overrightarrow{FA''} will coincide with \overrightarrow{FD} because reflections preserve angles, and rays coming from the same point at the same angle will coincide. We have that\overrightarrow{EA''} and \overrightarrow{FA''} can only intersect at point A'', and \overrightarrow{ED} and \overrightarrow{FD} can only intersect at point D. Since the rays coincide, their intersection points, A'' and D, will also coincide.

5. 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 using rigid motions, so: \triangle ABC \cong \triangle DEF

Notice that one triangle is equilateral, while the other is equiangular. The corollary to the base angles theorem and its converse tells us that both triangles will be equilateral and equiangular.

Both triangles are equiangular, which means that the two triangles share three common angle measures. Since they both also have a corresponding side of length 8, we can use the AAS test to justify that they are congruent.

We could also show that the triangles are congruent by stating that both triangles are equilateral, which means that both must have three sides of length 8. We can then use the SSS test to justify that they are congruent.

We want to find as much information as we can in order to satisfy one of the congruence tests.

Since \overline{AD} and \overline{BC} are straight line segments, we can find vertically opposite angles, and since \overline{AB}\parallel\overline{DC} we can find alternate angles on parallel lines.

We could also use ASA to prove that \triangle{ABX} \cong \triangle{DCX}, ignoring vertical angles in a proof like the one that follows:

The triangles are congruent by ASA, so we can identify corresponding parts and create an equation to solve for x.

Based on the diagram, we see that\angle E \cong \angle Y and \angle D \cong \angle X, so the included side for the first triangle is \overline{DE} and the included side for the second triangle is \overline{XY}. By the definition of congruence, if \overline{DE}\cong \overline{XY}, then the two segments will be equal in length. Therefore, we need to find x so that 15=4x-1.