# Lesson 8

The Perpendicular Bisector Theorem

## 8.1: Which One Doesn’t Belong: Intersecting Lines (5 minutes)

### Warm-up

This activity prompts students to compare four images. It gives students a reason to use language precisely (MP6). It gives the teacher an opportunity to hear how students use terminology and talk about characteristics of the items in comparison to one another. In particular, students will be focused on the characteristics of perpendicular lines.

### Launch

Arrange students in groups of 2–4. Display the figures for all to see. Give students 1 minute of quiet think time and then time to share their thinking with their small group. In their small groups, ask each student to share their reasoning why a particular item does not belong, and together, find at least one reason each item doesn‘t belong.

### Student Facing

Which one doesn’t belong?

### Student Response

Teachers with a valid work email address can click here to register or sign in for free access to Student Response.

### Activity Synthesis

Ask each group to share one reason why a particular item does not belong. Record and display the responses for all to see. After each response, ask the class if they agree or disagree. Since there is no single correct answer to the question of which one does not belong, attend to students’ explanations and ensure the reasons given are correct.

During the discussion, ask students to explain the meaning of any terminology they use, such as perpendicular. Also, press students on unsubstantiated claims.

## 8.2: Lots of Lines (15 minutes)

### Activity

The proof that if a point \(C\) is the same distance from \(A\) as it is from \(B\), then \(C\) must be on the perpendicular bisector of \(AB\) is challenging. Students may struggle to make sense of what it means to prove that a point must be on a line. This proof, like the proof of the Isosceles Triangle Theorem, requires thinking carefully about how to define an auxiliary line.

Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).

### Launch

Arrange students in groups of 2. Use the Notice and Wonder instructional routine with this image to launch the activity.

Display the image for all to see. Ask student to think of at least one thing they notice and at least one thing they wonder. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice with their partner, followed by a whole-class discussion.

Things students may notice:

- It’s a triangle.
- Segment \(PA\) is congruent to segment \(PB\).
- Angle \(A\) is congruent to angle \(B\) (by the Isosceles Triangle Theorem).
- There’s a dotted line.

Things students may wonder:

- Is the dotted line the line of symmetry?
- Is the dotted line the angle bisector of angle \(APB\)?
- Is the dotted line the perpendicular bisector of segment \(AB\)?

The purpose of this launch is to elicit the idea that special lines through \(P\) when segments \(PA\) and \(PB\) are congruent have many overlapping properties, which will be useful when students draw auxiliary lines later in this activity. While students may notice and wonder many things about these images, wondering about the line through \(P\) is an important focus of the discussion.

If students have access to GeoGebra Geometry from Math Tools, suggest that it might be a helpful tool in this activity.

*Reading, Listening, Conversing: MLR6 Three Reads.*Use a modified version of this routine to support reading comprehension of this problem. Use the first read to orient students to the situation. Ask students to describe what the situation is about (Diego, Jada, and Noah want to prove that if a point \(C\) is the same distance from \(A\) as it is from \(B\), then \(C\) must be on the perpendicular bisector of \(AB\).). Use the second read to identify geometric relationships stated by Diego, Jada, or Noah. Listen for and amplify important relationships such as “In Noah’s approach, \(CD\) is the angle bisector of angle \(ACB\)” and “In Jada’s approach, \(D\) is the midpoint of \(AB\).” After the third read, ask students to brainstorm possible questions to help Jada and Noah prove that \(CD\) must be the perpendicular bisector of \(AB\). This routine will help students make sense of the language in the word problem and the reasoning needed to solve the problem.

*Design Principle(s): Support sense-making*

*Representation: Internalize Comprehension.*Provide appropriate reading accommodations and supports to ensure students access to written directions, word problems, and other text-based content.

*Supports accessibility*

*for:*

*Language; Conceptual processing*

### Student Facing

Diego, Jada, and Noah were given the following task:

Prove that if a point \(C\) is the same distance from \(A\) as it is from \(B\), then \(C\) must be on the perpendicular bisector of \(AB\).

At first they were really stuck. Noah asked, “How do you prove a point is on a line?” Their teacher gave them the hint, “Another way to think about it is to draw a line that you know \(C\) is on, and prove that line has to be the perpendicular bisector.”

They each drew a line and thought about their pictures. Here are their rough drafts.

Diego’s approach: “I drew a line through \(C\) that was perpendicular to \(AB\) and through the midpoint of \(AB\). That line is the perpendicular bisector of \(AB\) and \(C\) is on it, so that proves \(C\) is on the perpendicular bisector.”

Jada’s approach: “I thought the line through \(C\) would probably go through the midpoint of \(AB\) so I drew that and labeled the midpoint \(D\). Triangle \(ACB\) is isosceles, so angles \(A\) and \(B\) are congruent, and \(AC\) and \(BC\) are congruent. And \(AD\) and \(DB\) are congruent because \(D\) is a midpoint. That made two congruent triangles by the Side-Angle-Side Triangle Congruence Theorem. So I know angle \(ADC\) and angle \(BDC\) are congruent, but I still don’t know if \(DC\) is the perpendicular bisector of \(AB\).”

Noah’s approach: “In the Isosceles Triangle Theorem proof, Mai and Kiran drew an angle bisector in their isosceles triangle, so I’ll try that. I’ll draw the angle bisector of angle \(ACB\). The point where the angle bisector hits \(AB\) will be \(D\). So triangles \(ACD\) and \(BCD\) are congruent, which means \(AD\) and \(BD\) are congruent, so \(D\) is a midpoint and \(CD\) is the perpendicular bisector.”

- With your partner, discuss each student’s approach.
- What do you notice that this student understands about the problem?
- What question would you ask them to help them move forward?

- Using the ideas you heard and the ways you think each student could make their explanation better, write your own explanation for why \(C\) must be on the perpendicular bisector of \(A\) and \(B\).

### Student Response

Teachers with a valid work email address can click here to register or sign in for free access to Student Response.

### Student Facing

#### Are you ready for more?

Elena has another approach: “I drew the line of reflection. If you reflect across \(C\), then \(A\) and \(B\) will switch places, meaning \(A'\)coincides with \(B\), and \(B'\) coincides with \(A\). \(C\) will stay in its place, so the triangles will be congruent.”

1. What feedback would you give Elena?

2. Write your own explanation based on Elena‘s idea.

### Student Response

Teachers with a valid work email address can click here to register or sign in for free access to Extension Student Response.

### Anticipated Misconceptions

Remind students who struggle with their critique or edits to make use of the tips from the display, such as asking for 3 statements and 3 reasons, or looking for congruent triangles.

### Activity Synthesis

Focus discussion on what students proved and the implications. Display the image from the launch of this activity, and ask what must be true about this image, based on what they just proved. (The dotted line is the perpendicular bisector of \(AB\). The dotted line is the line of reflection of \(AB\).)

Display this image of the construction of a perpendicular bisector and ask students how what they just proved explains why this construction works. (Both circles have radius \(AB\), so the intersection points of the circles are the same distance from \(A\) and \(B\). Therefore, both intersection points are on the perpendicular bisector of \(AB\).)

Ask students to add part 1 of the Perpendicular Bisector Theorem to their reference charts as you add it to the class reference chart:

If a point \(C\) is the same distance from \(A\) as it is from \(B\), \(C\) must be on the **perpendicular bisector** of \(AB\). (Theorem)

## 8.3: Not Too Close, Not Too Far (15 minutes)

### Activity

In a previous activity, students had a chance to practice giving feedback on proofs. In this activity, they will put their proof-writing and feedback-giving skills to work as they draft proofs and then give feedback to a partner. Students have already drawn diagrams of this situation in an earlier lesson, so the focus of this activity is writing a proof that is clear and easy to understand (MP3). Students will make revisions to the proofs they write in the cool-down, so focus discussion on common concerns.

Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).

### Launch

Arrange students in groups of 2.

Display one of the correct, complete diagrams from the cool-down of the Side-Angle-Side Triangle Congruence lesson. Ask students which statement this image illustrates and why. (If a point is on the perpendicular bisector of a line segment, then that point must be the same distance from each endpoint of the segment.)

*Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time.*Use this routine to help students improve their written responses for the proof of the conjecture: If \(P\) is a point on the perpendicular bisector of \(AB\), then \(PA\) is congruent to \(PB\). Give students time to meet with 2–3 partners to share and get feedback on their responses. Display feedback prompts that will help students strengthen their ideas and clarify their language. For example, “How do you know that \(M\) is the midpoint of \(AB\)?”, “How do you know that angle \(AMP\) is a right angle?”, and “How do you know that triangle \(AMP\) is congruent to triangle \(BMP\)?” In the cool down, students will revise and refine their written explanation based on the feedback from peers. This will help students justify their thinking for each step of their proof.

*Design Principle(s): Optimize output (for explanation); Cultivate conversation*

*Engagement: Internalize Self Regulation.*Demonstrate giving and receiving constructive feedback. Use a structured process and display sentence frames to support productive feedback. For example, “This method works (or doesn’t work) because…,” “Another strategy could be _____ because….”

*Supports accessibility for: Social-emotional skills; Organization; Language*

### Student Facing

- Work on your own to make a diagram and write a rough draft of a proof for the statement:
If \(P\) is a point on the perpendicular bisector of \(AB\), prove that the distance from \(P\) to \(A\) is the same as the distance from \(P\) to \(B\).

- With your partner, discuss each other’s drafts. Record your partner‘s feedback for your proof.
- What do you notice that your partner understands about the problem?
- What question would you ask them to help them move forward?

### Student Response

Teachers with a valid work email address can click here to register or sign in for free access to Student Response.

### Anticipated Misconceptions

Students may think this is the same proof as the previous activity. Ask students what the difference is between this claim and that one. (This one is backwards.) Inform students this is the converse, and the converse of a true statement is not always true, so they will need to write a new proof for this claim.

### Activity Synthesis

Students will make revisions to the proofs they write in the cool-down, so focus discussion on common concerns about the draft proofs. Invite a few students to share their progress.

## Lesson Synthesis

### Lesson Synthesis

Display these statements for all to see:

- If a point \(C\) is the same distance from \(A\) as it is from \(B\), prove that \(C\) must be on the perpendicular bisector of \(AB\).
- If \(P\) is a point on the perpendicular bisector of \(AB\), prove that the distance from \(P\) to \(A\) is the same as the distance from \(P\) to \(B\).

Ask students what is the same about the two statements and what is different. (They are the same thing—only the ifs and thens are switched.)

Tell students that when two statements are switched like this, we call them **converses**. Ask student if these statements are converses: “If you practice, then you’ll get better,” and “If you don’t practice, then you won’t get better.” (No; one is the negation of the other, but they don’t switch the order of what is given and what is concluded.)

Ask students for the converse of this statement: “If I won the lottery, then I am rich.” (If I am rich, then I won the lottery.)

Ask students if the converse of a true statement has to be true. (No; you could get rich by receiving an inheritance or having an insanely popular YouTube channel.)

“Today, we proved these converse statements about perpendicular bisectors are *both* true, which means that a point being the same distance from two endpoints of a segment and a point being on the perpendicular bisector of that segment is the exact same thing.” A statement and its converse will be true if they are from a definition, such as: “If a quadrilateral has four equal sides and four equal angles, then it is a square.” and “If a quadrilateral is a square, then it has four equal sides and four equal angles.”

Display one of the correct, complete images from the cool-down of the Side-Angle-Side Triangle Congruence lesson. Ask students what they can conclude about this image, and which statement from the Perpendicular Bisector Theorem they are using to draw the conclusion. (Point \(P\) is the same distance from points \(A\) and \(B\). Part 2, the converse.)

Ask students to add part 2 of the Perpendicular Bisector Theorem (the converse of part 1) to their reference charts as you add it to the class reference chart:

If \(C\) is a point on the perpendicular bisector of \(AB\), the distance from \(C\) to \(A\) is the same as the distance from \(C\) to \(B\). (Theorem)

## 8.4: Cool-down - Reflect and Revise (5 minutes)

### Cool-Down

Teachers with a valid work email address can click here to register or sign in for free access to Cool-Downs.

## Student Lesson Summary

### Student Facing

The perpendicular bisector of a line segment is exactly those points that are the same distance from both endpoints of the line segment. This idea can be broken down into 2 statements:

- If a point is on the perpendicular bisector of a segment, then it must be the same distance from both endpoints of the line segment.
- If a point is the same distance from both endpoints of a line segment, then it must be on the perpendicular bisector of the segment.

These statements are **converses** of one another. Two statements are converses if the “if” part and the “then” part are swapped. The converse of a true statement isn‘t always true, but in this case, both statements are true parts of the Perpendicular Bisector Theorem.

A line of reflection is the perpendicular bisector of segments connecting points in the original figure with corresponding points in the image. Therefore, these 3 lines are all the same:

- The perpendicular bisector of a segment.
- The set of points equidistant from the 2 endpoints of a segment.
- The line of reflection that takes the 2 endpoints of the segment onto each other, and the segment onto itself.

It is useful to know that the perpendicular bisector of a line segment is also all the points which are the same distance from both endpoints of the line segment, because then:

- If 2 points are both equidistant from the endpoints of a segment, then the line through those points must be the perpendicular bisector of the segment (because 2 points define a unique line).
- If 2 points are both equidistant from the endpoints of a segment, then the line through those must be the line of reflection that takes the segment to itself and swaps the endpoints.
- If a point is on the line of reflection, then it is the same distance from that point to a point in the original figure and to its corresponding point in the image.
- If a point is on the perpendicular bisector of a segment, then it is the same distance from that point to both endpoints of the segment.