6.05 Right triangle congruence

1. There is a rigid transformation that will map \angle XYZ to \angle RST because \angle RST \cong \angle XYZ. First, we can rotate \triangle XYZ clockwise about Y until \overline{YZ} \parallel \overline{ST}.

2. Since the transformation is a rigid transformation, we have that: \angle XYZ \cong \angle X'Y'Z' \text{, } \overline{XZ} \cong \overline {X'Z'} \text{, and } \overline{YZ} \cong \overline{Y'Z'}

3. Using the given information and the transitive property of congruence, we have that: \angle RST \cong \angle X'Y'Z' \text{, } \overline{RT} \cong \overline {X'Z'} \text{, and } \overline{ST} \cong \overline{Y'Z'}

4. Finally, draw a vector from X to R, and translate \triangle XYZ along the vector. A translation preserves angle measures and side lengths.

We have now shown that:

• X maps to R using rigid transformations
• Y maps to S using rigid transformations
• Z maps to T using rigid transformations

So we have that \triangle XYZ can be mapped onto \triangle RST, so \triangle RST \cong \triangle XYZ.

When proving triangles congruent, start by labeling and identifying the given information as well as anything that we can conclude directly from the diagram (such as vertical angles or reflexively congruent parts). Since we are trying to prove two triangles congruent, we need to decide which triangle congruence theorem best fits the given information. Then, when writing the proof we always start with the given information, build any conclusions based on the given information, and end with the conclusion we are trying to prove.

We are given that \overline{LK} \perp \overline{IJ} which means that \angle{IKL} and \angle{JKL} are both right angles by the definition of perpendicular lines. That means \triangle{ILK} and \triangle{JLK} are both right triangles by the definition of right triangles. We are also given that \overline{IL}\cong \overline{JL} which means we have two congruent hypotenuses, and we know that \overline{LK}\cong \overline{LK} by the reflexive property so we have a pair of congruent legs as well. This is enough information to conclude that \triangle{ILK}\cong \triangle{JLK} by the HL theorem.

Triangle proofs can be written in any style - just make sure that each congruent part needed to justify the triangle congruence theorem has its own statement and reason included.

Since m \angle ABC and m \angle CDE = 90 \degree, we know that \triangle ABC and \triangle CDE are both right triangles.

We know that if the hypotenuse and leg of one triangle are congruent to the hypotenuse and leg of another triangle, then the triangles are congruent. We need to choose values for x and y that will make the hypotenuses and the legs congruent.

We can use that \overline{AC} \cong \overline{CE} and \overline{AB} \cong \overline{CD} to solve for the variables. By the definition of congruence, since \overline{AC} \cong \overline{CE} and \overline{AB} \cong \overline{CD}, the two pairs of segments will be equal in length.

To construct a triangle congruent to \triangle ABC, we will use the Hypotenuse-Leg congruence criteria and construct a congruent hypotenuse and leg.

We will first copy one of the legs of the right triangle. Let's choose \overline {AB}.

Start by creating a horizontal line \overleftrightarrow{DE} using the Line tool.

Next, we create a point F somewhere on the line by using the Point tool.

Then create an arc with a radius of length \overline{AB} (one of the legs). Once again, we can use the Compass tool. Select vertices A and B first then select point F. This ensures that the radius of the circle is the same as the length of \overline {AB}.

Next, we mark the points of intersection of the line and the arcs using the Point tool. \overline {FH} and \overline {FG} are both congruent to \overline {AB}.

We now create two arcs with a radius of length \overline{BC} (the hypotenuse) using the Compass tool. For the first arc, select vertices B and C first then select point G. For the second arc, select vertices B and C first then select point H.

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

Finally, we can create \triangle FHI using the Polygon tool. This \triangle FHI is congruent to \triangle ABC.

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.