Lesson 5
Points, Segments, and Zigzags
5.1: What's the Point? (5 minutes)
Warmup
This activity invites students to think carefully about the definition of congruence. Students can use any rigid motion to demonstrate all points are congruent.
Student Facing
If \(A\) is a point on the plane and \(B\) is a point on the plane, then \(A\) is congruent to \(B\).
Try to prove this claim by explaining why you can be certain the claim must be true, or try to disprove this claim by explaining why the claim cannot be true. If you can find a counterexample in which the “if” part (hypothesis) is true, but the “then” part (conclusion) is false, you have disproved the claim.
Student Response
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Activity Synthesis
Invite students to share. If students only suggest translation, ask if it is possible to use another transformation. If rotation is not mentioned by students, there is no need to continue to push for that solution.
5.2: What's the Segment? (20 minutes)
Activity
Students disprove a conjecture that all segments are congruent, introducing them to proof by contradiction. Then, students work to prove segments of the same length are congruent. This proof leads directly to the triangle congruence proofs in subsequent lessons.
Launch
Display the conjecture and invite students to prove or disprove it.
“If \(AB\) is a segment in the plane and \(CD\) is a segment in the plane, then \(AB\) is congruent to \(CD\).”
Give students 1 minute of quiet work time. Invite them to refer to the instructions in the warmup for how to prove or disprove an ifthen statement.
Select a student who has drawn a counterexample to share. Invite a student to rephrase the conjecture so it will always be true. (If \(AB\) is a segment in the plane and \(CD\) is a segment in the plane with the same length as \(AB\), then \(AB\) is congruent to \(CD\).)
Design Principle(s): Optimize output (for explanation); Cultivate conversation
Supports accessibility for: Memory; Language
Student Facing
Prove the conjecture: If \(AB\) is a segment in the plane and \(CD\) is a segment in the plane with the same length as \(AB\), then \(AB\) is congruent to \(CD\).
Student Response
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Student Facing
Are you ready for more?
Prove or disprove the following claim: “If \(EF\) is a piece of string in the plane, and \(GH\) is a piece of string in the plane with the same length as \(EF\), then \(EF\) is congruent to \(GH\).”
Student Response
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Anticipated Misconceptions
Students may forget to specify that the points will coincide since the segments are the same length. Remind them of the counterexample in the launch.
Activity Synthesis
Invite students to share pieces of the proof until the whole class agrees that the proof is sufficiently detailed and convincing. Help students determine when they should refer to rays versus segments to solidify the idea that the segments aren‘t congruent until they have used the fact that they are the same length.
Ask students to add this theorem to their reference charts as you add it to the class reference chart:
If two segments have the same length, then they are congruent. (Theorem)
5.3: Zig Then Zag (10 minutes)
Activity
In this partner activity, students take turns using their new theorem about congruent segments to prove figures made of congruent segments are congruent. As students trade roles listening and explaining their thinking, they have opportunities to explain their reasoning and critique the reasoning of others (MP3).
Launch
Arrange students in groups of 2.
Design Principle(s): Cultivate conversation
Supports accessibility for: Language
Student Facing
 Here are some statements about 2 zigzags. Put them in order to write a proof about figures \(QRS\) and \(XYZ\).
 1: Therefore, figure \(QRS\) is congruent to figure \(XYZ\).
 2: \(S'\) must be on ray \(YZ\) since both \(S'\) and \(Z\) are on the same side of \(XY\) and make the same angle with it at \(Y\).
 3: Segments \(QR\) and \(XY\) are the same length, so they are congruent. Therefore, there is a rigid motion that takes \(QR\) to \(XY\). Apply that rigid motion to figure \(QRS\).
 4: Since points \(S'\) and \(Z\) are the same distance along the same ray from \(Y\), they have to be in the same place.
 5: If necessary, reflect the image of figure \(QRS\) across \(XY\) to be sure the image of \(S\), which we will call \(S'\), is on the same side of \(XY\) as \(Z\).
 Take turns with your partner stating steps in the proof that figure \(ABCD\) is congruent to figure \(EFGH\).
Student Response
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Activity Synthesis
Invite students to write sentence frames for the new transformation they used in their proofs. Add them to the display of sentence frames for proofs. This display should be posted in the classroom for the remaining lessons within this unit. You will add to the display throughout the unit. An example template is provided in the blackline masters for this lesson.
Transformations:
 Segments \(\underline{\hspace{1in}}\) and \(\underline{\hspace{1in}}\) are the same length, so they are congruent. Therefore, there is a rigid motion that takes \(\underline{\hspace{1in}}\) to \(\underline{\hspace{1in}}\). Apply that rigid motion to \(\underline{\hspace{1in}}\).
Lesson Synthesis
Lesson Synthesis
Display two congruent zigzags with several segments each. Invite students to contribute one statement at a time to prove the zigzags are congruent. Refer them to the sentence frames when they get stuck.
5.4: Cooldown  Circular Logic (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
If 2 figures are congruent, then there is a sequence of rigid motions that takes one figure onto the other. We can use this fact to prove that any point is congruent to another point. We can also prove segments of the same length are congruent. Finally, we can put together arguments to prove entire figures are congruent.
These statements prove \(ABC\) is congruent to \(XYZ\).

Segments \(AB\) and \(XY\) are the same length, so they are congruent. Therefore, there is a rigid motion that takes \(AB\) to \(XY\). Apply that rigid motion to figure \(ABC\).

If necessary, reflect the image of figure \(ABC\) across \(XY\) to be sure the image of \(C\), which we will call \(C'\), is on the same side of \(XY\) as \(Z\).

\(C'\) must be on ray \(YZ\) since both \(C'\) and \(Z\) are on the same side of \(XY\) and make the same angle with it at \(Y\).

Since points \(C'\) and \(Z\) are the same distance along the same ray from \(Y\), they have to be in the same place.

Therefore, figure \(ABC\) is congruent to figure \(XYZ\).