6.03 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.

To construct a triangle that is congruent to \triangle ABC, we will use geometric constructions to make a copy of each of its sides and angles.

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

Next, we want to create an arc with a radius of length AC. This will ensure the side we create is congruent to \overline{AC}.

To do this, we can make use of the Compass tool. Select vertices A and C first then select point D.

We can then use the Point tool to create a new point E on the arc of the circle we just drew. We know that the distance between points D and E is the same as the length of \overline{AC}.

We now want to create an arc with a radius of length AB. This will give us a corresponding congruent side to \overline{AB}.

Once again, we can make use of the Compass tool.

We won't place a point yet because we need to make sure we place it so that the third side will be the correct length. So first we will create an arc with a radius of length BC using the Compass tool. This time, select vertices B and C first to set the length. Then select point E. This will make sure the distance between point E and the new point is equal to the distance between points C and D while also making sure the distance between the new point and point D is the same as BA.

Next we can mark the point of intersection of the two circles we have created using the Point tool.

Finally, we can create \triangle DFE using the Polygon tool. \triangle DFE \cong \triangle ABC because all three pairs of corresponding sides are congruent.

If we want to confirm that the constructed triangle is a copy of the original triangle, we can look at the side lengths of both triangles in the Algebra view.

We can use the fact that we used the same setting of the compass when drawing the last two arcs to justify that some segments are congruent. We will use SSS and CPCTC to prove that \overleftrightarrow{AB} and \overline{CF} are perpendicular.

D and E lie on the same arc centered at C, so \overline{CD} \cong \overline{CE}.

F lies on both the arc centered at D and the arc centered at E, and both arcs have the same radius. This means \overline{DF} \cong \overline{EF}.

\overline{CF} is common to \triangle DFC and \triangle EFC. This means \triangle DFC \cong \triangle EFC by SSS congruence.

Using CPCTC, \angle DCF \cong \angle ECF. Since \angle DCF and \angle ECF are also linear pair, then {m\angle DCF = m\angle ECF = 90 \degree}. This means \overleftrightarrow{CF} is perpendicular to \overleftrightarrow{AB}.

We can use the fact that we used the same setting of the compass when drawing the last two arcs to justify that some segments are congruent. We will use SSS and CPCTC to prove that \angle ABC \cong \angle IDH.

G and F lie on the same arc centered at B and so \overline{BG} \cong \overline{BF}.

I and H lie on the same arc centered at D and so \overline{DI} \cong \overline{DH}.

Since, by construction the two arcs have the same radius, we know that {\overline{BG} \cong \overline{BF} \cong \overline{DI} \cong \overline{DH}}.

Also, \overline{FG} \cong \overline{HI} by construction. This means \triangle GBF \cong \triangle IDH.

Using CPCTC, \angle GBF \cong \angle IDH, so \angle ABC \cong \angle IDH.

The method for constructing an angle congruent to a given angle is very similar to the method for copying a triangle using the three sides of the triangle. The difference is that when constructing an angle, we draw an isosceles triangle first and copy that triangle.

1. There is a rigid transformation 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 transformations, 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 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, so \triangle ABC \cong \triangle DEF.

Another approach for a rigid transformation 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}.

We will use \overline{AB}, \overline{AC} and \angle A. To create a copy of \triangle ABC, we will first create a copy of \overline{AC}. Then we will create a copy of \angle A on one of the endpoints of the new segment. We will then create a copy of \overline{AB} along the side of the new angle.

We will use Geogebra to implement the steps.

We start by constructing a point D that will become one of the vertex of the triangle. 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 use the Point tool to create a new point E on the arc.

Connect D and E. 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, we need to create an arc centered at at A that intersects both \overline{AB} and \overline{AC}. To do this, use the Circle with Center through tool. Select 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.

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

Use the Point tool to add point J at the intersection of the arc of circle D and \overleftrightarrow{DI} above \overline{DE}. This point J is a copy of B.

Finally, we can create \triangle DJE using the Polygon tool. This \triangle DJE is a copy of \triangle ABC. We know that the triangles are congruent by SAS because we constructed copies of two sides and their included angle.

We already created a copy of \triangle ABC from the previous example by copying the three sides. Although the two methods are different, the resulting triangle are congruent.