# Lesson 11

Splitting Triangle Sides with Dilation, Part 2

## 11.1: Notice and Wonder: Parallel Segments (5 minutes)

### Warm-up

The purpose of this warm-up is to elicit the idea that lines parallel to the base of a triangle make interesting shapes, including similar triangles, which will be useful when students prove these conjectures in a later activity. While students may notice and wonder many things about these images, triangles and angles formed by parallel lines are the important discussion points.

This prompt gives students opportunities to see and make use of structure (MP7). The specific structure they might notice is that given parallel lines, we can draw conclusions about angle measure, which can then lead to additional conclusions.

### Launch

Display the image for all to see. Ask students 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.

### Student Facing

What do you notice? What do you wonder?

### Activity Synthesis

Ask students to share the things they noticed and wondered. Record and display their responses for all to see. If possible, record the relevant reasoning on or near the image. After all responses have been recorded without commentary or editing, ask students, “Is there anything on this list that you are wondering about now?” Encourage students to respectfully disagree, ask for clarification, or point out contradicting information.

## 11.2: Prove It: Parallel Segments (15 minutes)

### Activity

The goal of this activity is for students to prove the converse of the theorem they developed in a previous lesson. In that lesson, they proved that a line that splits two sides of a triangle proportionally is parallel to the third side. In this activity, they prove that a line parallel to one side of a triangle creates similar triangles, and thus splits the other two sides proportionally.

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

### Launch

Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time. Use this routine to help students improve their written responses for the proof that a line parallel to one side of a triangle always creates similar triangles. Give students time to meet with 2–3 partners to share and receive 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 $$\angle NMB \cong \angle CAB$$?”, and “How do you know that triangle $$BMN$$ and triangle $$BAC$$ are similar?” Invite students to go back and revise or refine their written responses based on the feedback from peers. This will help students justify why a line parallel to one side of a triangle creates a similar triangle.
Design Principle(s): Optimize output (for justification); Cultivate conversation
Representation: Internalize Comprehension. Use color coding and annotations to highlight connections between representations in a problem. Students may benefit from highlighting the two triangles in different colors so that they can discern which angles and sides are relevant.
Supports accessibility for: Visual-spatial processing, conceptual processing

### Student Facing

Does a line parallel to one side of a triangle always create similar triangles?

1. Create several examples. Decide if the conjecture is true or false. If it’s false, make a more specific true conjecture.
2. Find any additional information you can be sure is true.
Label it on the diagram.

3. Write an argument that would convince a skeptic that your conjecture is true.

### Anticipated Misconceptions

If students are stuck with what to label on the diagram, suggest they look in the reference chart for properties of parallel lines.

If students are stuck starting their proof, ask about how we can prove that two triangles are similar. Suggest they look in the reference chart. (Find a sequence of rigid motions and a dilation that takes one triangle to the other. Use a dilation to make the triangles congruent, and then use one of the triangle congruence theorems to finish the proof. Use the Angle-Angle Triangle Similarity Theorem.)

### Activity Synthesis

The goal of this discussion is to connect similarity to the ability to calculate side lengths. Ask students to name ratios that must be equivalent, now that we have proven that the triangles are similar. ($$\frac{AM}{MB}=\frac{CN}{NB}$$$$\frac{AC}{MN}=\frac{CB}{NB}=\frac{AB}{MB}$$)

Record these as theorems for all to see (though students do not need to copy them into their reference chart, as they are unlikely to be referred to later):

• A line parallel to one side of a triangle creates a similar triangle.
• A line parallel to one side of a triangle divides the other two sides proportionally.

## 11.3: Preponderance of Proportional Relationships (15 minutes)

### Activity

Because the previous activity ended with students identifying equivalent ratios, this activity asks students to calculate using equivalent ratios. This builds a bridge from knowing that a segment is parallel to one side of a triangle to knowing that equivalent ratios can be written.

Monitor for students who:

• identify a scale factor from one triangle to the other
• use the Pythagorean Theorem
• use equivalent ratios based on ratios of corresponding sides across the two triangles (For example, the shorter sides in each triangle are 3 units long and 7.5 units long. The lengths of the medium sides should have the same ratio. The medium side in the small triangle is 4 units. $$3:7.5$$ is equivalent to $$4:10$$, so $$DE$$ is 10.)
• use proportional relationships based on ratios of side lengths within one triangle (For example, the ratio of these sides in the smaller triangle is $$3:4$$, so in the larger triangle, the ratio should also be $$3:4$$. Since $$3:4$$ is equivalent to $$7.5:10$$$$DE$$ is 10.)

### Launch

Representing, Conversing: MLR7 Compare and Connect. Use this routine to prepare students for the whole-class discussion about strategies for calculating the side lengths of similar triangles. After students calculate the side lengths of each triangle, invite them to create a visual display of their work. Then ask students to quietly circulate and observe at least two other visual displays in the room. Give students quiet think time to consider what is the same and what is different about their strategies. Next, ask students to find a partner to discuss what they noticed. Listen for and amplify the language students use to compare and contrast the various strategies for calculating the side lengths of similar triangles.
Design Principle(s): Cultivate conversation

### Student Facing

Find the length of each unlabelled side.

1. Segments $$AB$$ and $$EF$$ are parallel.
2. Segments $$BD$$ and $$FG$$ are parallel. Segment $$EG$$ is 12 units long. Segment $$EB$$ is 2.5 units long.

### Student Facing

#### Are you ready for more?

Find the lengths of sides $$CE,CB$$, and $$CA$$ in terms of $$x, y,$$ and $$z$$. Explain or show your reasoning.

### Anticipated Misconceptions

Ask students to think about what type of triangle they see if they forget to use the Pythagorean Theorem for the third side.

Students who struggle to visualize the similar triangles can trace each triangle separately onto tracing paper and label the corresponding sides using colored pencils.

### Activity Synthesis

Select previously identified students to share these strategies in this order:

• using equivalent ratios based on ratios of corresponding sides across the two triangles
• identifying a scale factor from one triangle to the other
• using the Pythagorean Theorem
• using ratios of side lengths within one triangle

Make sure all students understand how the first strategy follows directly from the theorem proved in the previous activity, and that if line $$FG$$ was not parallel to line $$BD$$, we couldn’t solve the problem.

Encourage students to look for and discuss connections between finding a scale factor, using equivalent ratios across the two triangles, and using equivalent ratios within the two triangles. (Ratios of corresponding sides across triangles are all equivalent because the second component in each ratio is the scale factor times the first. A ratio within one triangle is equivalent to the corresponding ratio within the other triangle because both components get multiplied by the scale factor.)

## Lesson Synthesis

### Lesson Synthesis

Invite one or more students to state the main theorems proved in this lesson:

• A line parallel to one side of a triangle creates a similar triangle.
• A line parallel to one side of a triangle divides the other two sides proportionally.

Invite additional students to consult their reference charts and identify other theorems we have proven about the same diagram. (If a line divides two sides of a triangle proportionally, the line must be parallel to the third side of the triangle.) Remind students of the definition of converse. Now we know that a line is parallel to one side of a triangle if and only if it splits the other two sides proportionally.

## 11.4: Cool-down - More Proportional Relationships? (5 minutes)

### Cool-Down

In triangle $$ABC$$, segment $$FG$$ is parallel to segment $$AC$$. We can show that corresponding angles in triangle $$ACB$$ and triangle $$GFB$$ are congruent, so the triangles are similar by the Angle-Angle Triangle Similarity Theorem. There must be a dilation that sends triangle $$GFB$$ to triangle $$ACB$$, and so pairs of corresponding side lengths are in the same proportion. Then we can show that segment $$GF$$ divides segments $$AB$$ and $$CB$$ proportionally. In other words, $$\frac{BG}{GA}$$=$$\frac{BF}{FC}$$.
For example, suppose $$G$$ is $$\frac23$$ of the way from $$A$$ to $$B$$ and $$F$$ is $$\frac23$$ of the way from $$C$$ to $$B$$. Then if $$BA=9$$ and $$BC=12$$, we know that $$GA=6$$ and $$FC=8$$. What will $$BG$$ and $$BF$$ equal? Since $$BG=3$$ and $$BF=4$$, we know that $$\frac36=\frac48$$ and can show that $$\frac{BG}{GA}$$=$$\frac{BF}{FC}$$.