6.04 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 \overline{DFE}

Prove: \triangle ABC \cong \triangle DEF

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

1. There is a rigid transformation 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 transformations, 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 transformations

   • B maps to E using rigid transformations

   • C maps to F using rigid transformations

   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.

We will use two angles and and included side. To create a copy of \triangle ABC, we will first create a copy of \overline{AC}. The endpoints of this new segment are the two vertices of the triangle. Then we will create copies of \angle A on one of these vertices and a copy of \angle C on the other vertex. The intersection of the sides of the angles will determine the third vertex.

We will use GeoGebra to implement the steps.

We start by constructing a point D that will become one of the vertex of the triangles. We can do this using the Point tool as shown.

Next, we want to create a copy of \overline{AC}. To do this, we can use the Compass tool. Select vertices A and C first then select point D. This creates a circle centered at D with a radius of length AC.

We can then use the Point tool to create a new point E on the arc.

Connect D and E using the Segment tool. This is a copy of \overline{AC}. Because \overline{DE} is the radius of circle D it must be congruent to \overline{AC}.

To create a copy of \angle A and \angle C, we need to create an arc centered at A that intersects both \overline{AB} and \overline{AC}. To do this, use the Circle with Center through tool. Select point A and any other point on \overline{AB} which we will call F.

Find the point of intersection of the arc and \overline{AC} which we will call G and add it using the Point tool.

Use the Compass tool to create a copy of circle A centered at point D. To do this, select A and F, and then select D.

Locate the intersection of the arc of the smaller circle D and \overline{DE} using the Point tool. This point H is a copy of G.

Create an arc using the Compass tool. Select F and G, then select H.

Use the point tool to add the point I at the intersection of the arcs of circles D and H above \overline{DE}. We have created the point I to be the same distance from H as point F is from G. This means \overset{\large\frown}{FG} \cong \overset{\large\frown}{IH}.

Use the Line tool. Select I and then select D. \angle IDH is a copy of \angle A.

We need to repeat the previous steps to copy \angle C. Use the Circle with Center through tool. Select C and then another point J on \overline{BC}.

Locate K, the intersection of the arc and \overline{AC}, using the Point tool.

Use the Compass tool to create a copy of the arc on E. Select C and J, and then select E.

Locate the intersection of the arc and \overline{DE} using the Point tool. This point L is a copy of K.

Create an arc using the Compass tool. Select J and K, then select L.

Use the Line tool. Select M, the intersection of the arc centered at E and the arc centered at L, and then select E. \angle MEL is a copy of \angle C.

Locate N, the intersection of \overleftrightarrow{DI} and \overleftrightarrow{EM}, using the Point tool.

Finally, we can create \triangle DNE using the Polygon tool. This \triangle DNE is copy of \triangle ABC. We know the triangles are congruent by ASA because we constructed copies of two angles and their included side.

We can use the fact that we used the same setting of the compass when drawing the arcs to justify that some segments are congruent. Using SSS, ASA and CPCTC, we will prove that \overline{CD} is a bisector of \overline{AB} and that the segments are perpendicular, and therefore that \overline{CD} is a perpendicular bisector if \overline{AB}.

Start by marking point E, the intersection of \overline{CD} and \overline{AB}.

C and D lie on the arc centered at A, and so \overline{AC} \cong \overline{AD}. C and D also lie on the arc centered at B, and so \overline{BC} \cong \overline{BD}. Since by construction the two arcs have the same radius, we know that \overline{AC} \cong \overline{AD} \cong \overline{BC} \cong \overline{BD}. This means \triangle ACD \cong \triangle BCD by SSS congruence. Using CPCTC, \angle ACD \cong \angle BCD. This also means that \angle ACE \cong \angle BCE.

\triangle ACB is isosceles, so \angle CAE \cong \angle CBE. By ASA congruence, we know that \triangle ACE \cong \triangle BCE. By CPCTC, we have \overline{AE} \cong \overline{BE}. This means \overline{CD} is a bisector of \overline{AB}. Using the same congruence statement and CPCTC, we know that \angle AEC \cong \angle BEC. Since \angle AEC and \angle BEC are also linear pair, then m\angle AEC = m\angle BEC=90 \degree. This means that \overline{CE}, and by extension \overline{CD}, is perpendicular to \overline{AB}.

This proves that \overline{CD} is a perpendicular bisector of \overline{AB}.

It is important that the compass width is greater than half of AB to ensure that there will be two points of intersection. Setting the width to excatly half AB will only result in one intersection point and setting the width to less than half of AB will result in no intersection points.