MGU Mathematics Series — Geometry
Part III: Similarity, Right Triangles, and Trigonometry
Similar Figures · Scale Factors · AA · SSS · SAS · Midsegment · Indirect Measurement
Chapter 6 of 14
MGU Mathematics Series  ·  Geometry  ·  Library Textbook Edition

Chapter 6:
Similarity and
Proportional Reasoning

Similar figures, scale factors, corresponding parts, AA, SSS, and SAS similarity, the Triangle Proportionality Theorem, the Midsegment Theorem, scale drawings, and indirect measurement — the proportional structure that connects geometry to algebra, right triangles, and trigonometry

■ Math Level: Geometry ■

■ 14 Sections  •  Part III of 5 ■

Follows Chapters 1–5: Foundations, Logic, Lines, Triangles  ·  Prepares students for Chapters 7 and 8: Right Triangles and Trigonometry

Chapter 6 — Front Matter

Purpose and Overview

This chapter connects geometry to one of the most powerful tools in algebra: proportional reasoning. You have already studied congruence, which is the relationship between figures that are exactly the same size and shape. Similarity is a related but broader idea. Similar figures share the same shape but are not required to be the same size. A photograph and its enlargement are similar. A scale model and the building it represents are similar. A map and the territory it describes are related by similarity.

Understanding similarity means understanding how ratios and proportions describe geometric relationships. It means being able to find unknown lengths by reasoning from what is known. It means seeing how the same shape, at different scales, preserves angles while stretching or shrinking distances by a consistent factor.

This chapter builds directly on the ratio and proportion work you did in Algebra I. It also prepares you for the next two chapters, which depend heavily on similarity: right triangle geometry and right-triangle trigonometry. The entire foundation of trigonometry rests on the fact that similar right triangles have equal ratios of corresponding sides. Work through this chapter carefully — the ideas here reappear in every chapter that follows, and in physics, engineering, architecture, maps, photography, design, and computing.

Opening Question

A surveyor cannot walk across a river to measure its width. Standing on one bank, she marks two points and measures angles to a tree on the opposite bank. From those measurements alone, she calculates the river's width. How is this possible? The answer lies in similarity. If two triangles have the same angles, their sides are proportional — even if the triangles are very different in size. By constructing a smaller triangle she can measure and a larger triangle that matches it in shape, she can use proportions to find the unknown distance. By the end of this chapter, you will be able to use the same reasoning.

14Sections
12Objectives
17Key Terms
15+Worked Examples

Why This Chapter Matters

Similarity is one of the most widely applied ideas in all of mathematics. Every map is a similar figure of the region it represents. Blueprints are similar to the structures they describe. Enlarging a photograph produces a similar image. Astronomers calculate distances to stars using similar triangles formed by angles of observation. The sine, cosine, and tangent ratios of trigonometry are defined using similarity of right triangles. Images in computer graphics are scaled using dilations — the transformation version of similarity. Organ proportions, body ratios, and imaging scales in biology and medicine all involve similarity reasoning.

Similarity is also the geometric concept that connects most directly to algebra. When you write a proportion and solve for an unknown, you are using algebra to answer a geometric question. That connection runs through the rest of this course.

Learning Objectives

  1. 1Explain what it means for two figures to be similar.
  2. 2Identify and write scale factors between similar figures.
  3. 3Name corresponding sides and angles in similar figures.
  4. 4Determine whether two polygons are similar.
  5. 5Set up and solve proportions to find unknown lengths in similar figures.
  6. 6Apply the AA, SSS, and SAS triangle similarity shortcuts.
  7. 7Use the Triangle Proportionality Theorem and its converse.
  8. 8Apply the Triangle Midsegment Theorem.
  9. 9Read and interpret scale drawings and maps.
  10. 10Solve measurement problems using indirect measurement and similarity.
  11. 11Recognize and avoid common similarity mistakes.
  12. 12Connect similarity reasoning to right triangles and trigonometry.
Chapter 6 — Contents

Sections at a Glance

Sections 6.1 – 6.5
Foundations: What Similarity Means
Similar Figures · Scale Factors · Corresponding Parts · Similar Polygons · Proportions
Section 6.1

What Similarity Means

The Idea of Same Shape, Different Size

When you look at a photograph and a small thumbnail of the same image, the two pictures look the same in shape but different in size. The proportions of every feature are preserved between the two images. This is the intuitive meaning of similarity.

In geometry, similar figures are figures that have the same shape but are not necessarily the same size. Two conditions must both be satisfied:

Definition — Similar Figures

Condition 1: All corresponding angles are congruent.

Condition 2: All corresponding sides are proportional — that is, their ratios are all equal.

If either condition fails, the figures are not similar. Both conditions together define similarity.

It is important to understand that satisfying only one condition is not enough. You can construct quadrilaterals where all corresponding angles are equal but sides are not proportional, and vice versa. Both conditions together define similarity.

The symbol for similarity is ∼. We write △ABC ∼ △DEF to say that triangle ABC is similar to triangle DEF.

The Order of the Similarity Statement

When writing a similarity statement, the order of letters matters. The letters must be written so that corresponding vertices match. If △ABC ∼ △DEF, then vertex A corresponds to vertex D, vertex B to vertex E, and vertex C to vertex F. This means:

From △ABC ∼ △DEFCongruent AnglesProportional Sides
A ↔ D∠A ≅ ∠DAB/DE = BC/EF = AC/DF
B ↔ E∠B ≅ ∠E(same common ratio)
C ↔ F∠C ≅ ∠F(same common ratio)

Similarity vs. Congruence

Congruence is a special case of similarity. If two figures are congruent, they are also similar — but their scale factor is 1. Congruent figures must be identical in both shape and size. Similar figures must match in shape, and their sizes must be proportional, but the sizes can differ.

Example 6.1.1
Recognizing similarity from angle measures
  • ProblemTriangle PQR has angles measuring 50°, 60°, and 70°. Triangle STU has angles measuring 60°, 70°, and 50°. Are the triangles similar?
  • ThinkSimilarity requires all corresponding angles to be congruent. List and match the angle measures from each triangle.
  • Represent△PQR: 50°, 60°, 70°  ·  △STU: 60°, 70°, 50°
    Match by measure: ∠P = 50° = ∠U, ∠Q = 60° = ∠S, ∠R = 70° = ∠T
  • CheckEvery angle in one triangle has a matching angle of equal measure in the other triangle. ✓
Answer: Yes. △PQR ∼ △UST (vertices written in the order that pairs corresponding angles).
Reading the Similarity Statement

When matching angles, do not assume letters appear in the same left-to-right order. Always match angles by measure first, then write the statement so that corresponding vertices align. The order of letters in the statement is the correspondence.

Practice 6.1Concept Check
  1. In your own words, state the two conditions that make two figures similar.
  2. How does the symbol ∼ differ from the symbol ≅? Write an example of each using triangles.
  3. △ABC ∼ △DEF. List all six pairs of corresponding parts (three angle pairs and three side pairs).
  4. Triangle XYZ has angles 40°, 75°, and 65°. Triangle LMN has angles 75°, 65°, and 40°. Write a correct similarity statement matching corresponding vertices.
  5. Explain why congruent figures are also similar, but similar figures are not always congruent.
Section 6.2

Scale Factors

Defining the Scale Factor

When two figures are similar, the ratios of all corresponding side lengths are equal. This common ratio is called the scale factor from the first figure to the second. If △ABC ∼ △DEF, the scale factor from △ABC to △DEF is:

Scale Factor

k = DE/AB = EF/BC = DF/AC

If k > 1, the second triangle is larger. If k < 1, the second triangle is smaller. If k = 1, the triangles are congruent. The scale factor from △DEF back to △ABC is 1/k.

Example 6.2.1
Finding the scale factor from corresponding sides
  • Problem△ABC ∼ △DEF. The sides of △ABC are AB = 4, BC = 6, AC = 8. The sides of △DEF are DE = 10, EF = 15, DF = 20. Find the scale factor from △ABC to △DEF.
  • ThinkCompute each ratio of a side in △DEF to the corresponding side in △ABC and verify they are all equal.
  • Solve k = DE/AB = 10/4 = 5/2 k = EF/BC = 15/6 = 5/2 k = DF/AC = 20/8 = 5/2
  • CheckAll three ratios equal 5/2. Each side of △DEF is 5/2 times the corresponding side of △ABC. ✓
Answer: The scale factor from △ABC to △DEF is 5/2, or 2.5.
Example 6.2.2
Using the scale factor to find missing lengths
  • Problem△GHI ∼ △JKL with scale factor 3/4 from △GHI to △JKL. If GH = 12 and HI = 20, find JK and KL.
  • ThinkMultiply each side of △GHI by the scale factor to get the corresponding side of △JKL.
  • Solve JK = (3/4) · GH = (3/4) · 12 = 9 KL = (3/4) · HI = (3/4) · 20 = 15
  • Check9/12 = 3/4 ✓ and 15/20 = 3/4 ✓
Answer: JK = 9 and KL = 15.
Student Warning — Direction of the Scale Factor

The scale factor from figure A to figure B is not the same as the scale factor from figure B to figure A. Always identify which figure is the original and which is the image before computing. If the scale factor from A to B is k, the scale factor from B to A is 1/k.

Practice 6.2Scale Factor Practice
  1. △ABC ∼ △DEF. AB = 5, BC = 7, AC = 9, DE = 10, EF = 14, DF = 18. Find the scale factor from △ABC to △DEF.
  2. △PQR ∼ △STU. PQ = 6, ST = 9. Find the scale factor from △PQR to △STU.
  3. Scale factor from △RST to △UVW is 3/5. If RS = 15, find UV. If VW = 12, find ST.
  4. Scale factor from figure A to figure B is 2.5. What is the scale factor from figure B to figure A?
  5. Two similar rectangles. Shorter one is 12 cm long. Scale factor is 7/4. Find the length of the larger rectangle.
  6. △MNO ∼ △XYZ. MN = 8, XY = 12, NO = 6. Find YZ and the scale factor.
Section 6.3

Corresponding Sides and Angles

In any similarity statement, the letters give you the correspondence directly. In △ABC ∼ △DEF, side AB corresponds to side DE, side BC corresponds to side EF, side AC corresponds to side DF. Corresponding sides are the sides connecting corresponding vertices.

When two figures are similar, corresponding sides form equal ratios. These equal ratios can be used to write proportions that solve for unknown side lengths. If △ABC ∼ △DEF:

Proportions from Corresponding Sides

AB/DE = BC/EF = AC/DF

Any two of these equal ratios can be set equal to form a proportion, which can then be solved for an unknown. Keep one figure's sides consistently in the numerators.

Example 6.3.1
Finding missing sides using proportions
  • Problem△RST ∼ △UVW. Given RS = 9, ST = 12, RT = 15, UV = 6. Find VW and UW.
  • ThinkFind the scale factor, then write proportions pairing corresponding sides.
  • DiagramDraw △RST (larger) and △UVW (smaller) side by side. Mark RS↔UV, ST↔VW, RT↔UW with matching tick marks.
  • Solve Scale factor (RST → UVW): UV/RS = 6/9 = 2/3 RS/UV = ST/VW 9/6 = 12/VW → 9 · VW = 72 → VW = 8 RS/UV = RT/UW 9/6 = 15/UW → 9 · UW = 90 → UW = 10
  • Check(2/3) · 12 = 8 = VW ✓   (2/3) · 15 = 10 = UW ✓
Answer: VW = 8 and UW = 10.
Practice 6.3Corresponding Parts and Proportions
  1. △ABC ∼ △DEF. AB = 10, DE = 6, BC = 15. Find EF. Verify by computing the scale factor.
  2. △PQR ∼ △STU. PQ = x, ST = 9, QR = 14, TU = 21. Set up and solve a proportion for x.
  3. △ABC ∼ △DEF. AB = 2x − 1, DE = 5, BC = 12, EF = 15. Find x and the length of AB.
  4. From the similarity △XYZ ∼ △LMN, list all three pairs of corresponding sides and all three pairs of corresponding angles.
Section 6.4

Similar Polygons

Triangles are not the only figures that can be similar. Any two polygons can be similar if both conditions hold: all corresponding angles are congruent and all corresponding sides are proportional. For polygons with more than three sides, both conditions must be verified separately. Unlike triangles — where knowing just the angles is often enough — for quadrilaterals and higher polygons, neither condition alone guarantees the other.

Student Warning — Quadrilaterals and Higher

A rectangle with dimensions 3 × 6 and a rectangle with dimensions 3 × 9 have all right angles in common, but their sides are not proportional (6/9 ≠ 3/3), so they are not similar. Always check both angle congruence and side proportionality for polygons with four or more sides.

Checking Polygon Similarity — Four Steps

Step 1. Check that the polygons have the same number of sides. Step 2. Set up a correspondence of vertices. Step 3. Verify that all pairs of corresponding angles are congruent. Step 4. Verify that all pairs of corresponding sides have the same ratio. If all conditions are satisfied, state the similarity using the correct vertex correspondence.

Example 6.4.1
Checking polygon similarity
  • ProblemQuadrilateral ABCD has angles 80°, 100°, 80°, 100° and sides AB = 4, BC = 6, CD = 4, DA = 6. Quadrilateral EFGH has angles 80°, 100°, 80°, 100° and sides EF = 6, FG = 9, GH = 6, HE = 9. Are they similar?
  • ThinkCheck angles first (match A↔E, B↔F, C↔G, D↔H), then compute all four side ratios.
  • Solve Angles: 80°=80°, 100°=100°, 80°=80°, 100°=100° ✓ EF/AB = 6/4 = 3/2 FG/BC = 9/6 = 3/2 GH/CD = 6/4 = 3/2 HE/DA = 9/6 = 3/2 ✓ All equal.
  • CheckBoth conditions satisfied. The scale factor is 3/2.
Answer: ABCD ∼ EFGH with scale factor 3/2.
Practice 6.4Similar Polygons
  1. Two squares: one has side length 5, the other has side length 9. Are they similar? Explain.
  2. A rectangle with dimensions 4 × 10 and a rectangle with dimensions 6 × 15. Are they similar? If so, find the scale factor.
  3. A rhombus with angles 60°/120° and a square. Are they similar? Explain why or why not.
  4. Describe a pair of non-congruent rectangles that are similar. What condition must their dimensions satisfy?
Section 6.5

Ratios in Similar Figures

Proportions and Cross Multiplication

A proportion is a statement that two ratios are equal: a/b = c/d. In this proportion, a and d are called the extremes, and b and c are called the means. The fundamental property of proportions — cross multiplication — states: a · d = b · c. This is the primary algebraic tool for solving proportion equations.

Setting Up Proportions Correctly

Correct: (side from figure 1)/(side from figure 2) = (side from figure 1)/(side from figure 2)

Incorrect: (side from figure 1)/(side from figure 2) = (side from figure 2)/(side from figure 1)

Keep one figure's sides consistently in the numerators and the other figure's sides consistently in the denominators. Mixing figures within a proportion leads to wrong answers.

Example 6.5.1
Solving a proportion for an unknown side
  • Problem△ABC ∼ △DEF. Given AB = x, BC = 18, DE = 10, EF = 15. Find x.
  • RepresentAB/DE = BC/EF
  • Solve x/10 = 18/15 15x = 10 · 18 = 180 x = 12
  • Check12/10 = 6/5 and 18/15 = 6/5. ✓
Answer: AB = 12.
Example 6.5.2
Proportion with a variable on both sides
  • ProblemIn similar triangles, one pair of corresponding sides has lengths x and x + 4. Another pair has lengths 6 and 10. Find x.
  • Representx/6 = (x + 4)/10
  • Solve 10x = 6(x + 4) 10x = 6x + 24 4x = 24 x = 6
  • CheckSides: 6 and 10, and 6+4=10 and 10. Scale factor = 1. The triangles are congruent. ✓
Answer: x = 6.
Practice 6.5Proportion Problems
  1. △ABC ∼ △DEF. AB = 10, DE = 6, BC = 15. Find EF.
  2. △PQR ∼ △STU. PQ = x, ST = 9, QR = 14, TU = 21. Find x.
  3. △ABC ∼ △DEF. AB = 2x−1, DE = 5, BC = 12, EF = 15. Find x and AB.
  4. Proportion: 3/x = 7/21. Solve for x.
  5. Two corresponding sides are (2x+1) and (3x−2); a second pair of corresponding sides are 8 and 12. Write and solve a proportion for x.
  6. Perimeter of △ABC = 36 and scale factor from △ABC to △DEF is 5/3. Find the perimeter of △DEF.
Sections 6.6 – 6.8
Triangle Similarity Shortcuts
AA Similarity · SSS Similarity · SAS Similarity
Section 6.6

AA Triangle Similarity

Why Two Angles Are Enough for Triangles

For general polygons, you must check both angles and side ratios to confirm similarity. For triangles, there are shortcuts. The most powerful comes from a basic fact: the angle sum of any triangle is always 180°. If two triangles share two pairs of congruent angles, the third pair must also be congruent (since 180° minus two equal values gives equal third values). You do not need to check all three angle pairs.

Postulate
AA Similarity Postulate

If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

The reason AA works: once the angles of a triangle are fixed, the shape is determined. Any two triangles with the same angles will have the same shape, and their sides will be proportional.

Example 6.6.1
Applying AA Similarity from angle measures
  • ProblemIn △ABC: ∠A = 50°, ∠B = 70°. In △DEF: ∠D = 50°, ∠F = 60°. Are the triangles similar?
  • ThinkFind the third angle in each triangle, then match all three pairs.
  • Solve △ABC: ∠C = 180° − 50° − 70° = 60° △DEF: ∠E = 180° − 50° − 60° = 70° ∠A = 50° = ∠D ∠B = 70° = ∠E ∠C = 60° = ∠F (two pairs congruent → AA applies)
Answer: △ABC ∼ △DEF by AA Similarity.
Example 6.6.2
AA Similarity with parallel lines
  • ProblemTwo triangles share vertex P. Line AB is parallel to line CD; points A and C lie on one transversal through P, and B and D lie on another. Show △PAB ∼ △PCD.
  • DiagramDraw two lines from point P extending downward. Mark points A and B on the upper pair of lines (closer to P). Mark C and D further out. Draw AB ∥ CD.
  • PlanUse parallel line angle relationships to find two pairs of congruent angles.
  • Solve AB ∥ CD: ∠PAB ≅ ∠PCD (corresponding angles, transversal PA) ∠PBA ≅ ∠PDC (corresponding angles, transversal PB) Two pairs of angles congruent → AA Similarity.
Answer: △PAB ∼ △PCD by AA Similarity.
Practice 6.6AA Similarity
  1. △ABC: ∠A = 55°, ∠B = 80°. △DEF: ∠D = 55°, ∠F = 45°. Determine whether the triangles are similar by AA. Write the correct similarity statement.
  2. In a diagram, two triangles share a vertex. One angle of each triangle measures 90°. A second angle of each measures 32°. Show that the triangles are similar by AA.
  3. △ABC: ∠A = 3x°, ∠B = 2x°, ∠C = x°. △PQR: ∠P = 90°, ∠Q = 60°. Find x and determine whether the triangles are similar.
  4. Explain why AA Similarity works for triangles but not for quadrilaterals. Give a counterexample showing that two quadrilaterals with matching angles need not be similar.
Section 6.7

SSS Triangle Similarity

Theorem
SSS Similarity Theorem

If the ratios of all three pairs of corresponding sides of two triangles are equal, then the triangles are similar.

That is, if AB/DE = BC/EF = AC/DF, then △ABC ∼ △DEF.

Note the difference from SSS Congruence, which required all three pairs of sides to be equal. SSS Similarity only requires the three ratios to be equal.

Example 6.7.1
Applying SSS Similarity
  • Problem△ABC has sides 6, 8, and 10. △PQR has sides 9, 12, and 15. Are the triangles similar?
  • ThinkMatch sides in order (smallest to smallest, etc.) and check whether all three ratios are equal.
  • Solve 9/6 = 3/2 12/8 = 3/2 15/10 = 3/2 ✓ All equal.
Answer: △ABC ∼ △PQR by SSS Similarity, with scale factor 3/2.
Example 6.7.2
Checking SSS — triangles that are not similar
  • Problem△XYZ has sides 4, 7, and 9. △LMN has sides 8, 14, and 20. Are they similar?
  • Solve 8/4 = 2 14/7 = 2 20/9 ≈ 2.22 ✗ Third ratio differs.
Answer: The triangles are NOT similar. The side ratios are not all equal.
Student Warning — Match Sides in Corresponding Order

When applying SSS Similarity, match sides in the correct corresponding order: smallest to smallest, middle to middle, largest to largest (or as specified by the problem). Matching them out of order can give false ratios and a false conclusion.

Practice 6.7SSS Similarity
  1. Sides 3, 4, 5 and sides 9, 12, 15. Similar by SSS?
  2. Sides 6, 8, 10 and sides 9, 12, 15. Similar by SSS?
  3. Sides 5, 8, 11 and sides 10, 16, 23. Similar by SSS?
  4. Sides 4, 6, 8 and sides 6, 9, 12. Similar by SSS? Find the scale factor.
  5. Sides 7, 7, 10 and sides 14, 14, 20. Similar by SSS? Find the scale factor.
  6. △GHI: sides 4, 7, 9. △JKL: sides 8, 14, 20. Similar? Explain.
Section 6.8

SAS Triangle Similarity

Theorem
SAS Similarity Theorem

If two sides of one triangle are proportional to two sides of another triangle, and the included angles between those sides are congruent, then the triangles are similar.

That is, if AB/DE = AC/DF and ∠A ≅ ∠D (the included angle), then △ABC ∼ △DEF.

The angle must be the included angle — the angle formed between the two sides being compared. If the angle is not included, SAS Similarity does not apply.

Example 6.8.1
Applying SAS Similarity
  • ProblemIn △ABC and △DEF: AB = 6, AC = 8, ∠A = 40°, DE = 9, DF = 12, ∠D = 40°. Are the triangles similar?
  • ThinkCheck the ratio of the two given sides from each triangle, then verify the included angle.
  • Solve AB/DE = 6/9 = 2/3 AC/DF = 8/12 = 2/3 ✓ Equal ratios. ∠A = ∠D = 40° ✓ Included angles congruent.
Answer: △ABC ∼ △DEF by SAS Similarity.
Example 6.8.2
SAS with nested triangles — finding a missing side
  • Problem△PQR and △PST share vertex P. PQ = 4, PS = 10, PR = 6, PT = 15, ∠P is shared. Are the triangles similar? If so, find QR if ST = 30.
  • DiagramDraw △PST with vertex P at top. Draw △PQR inside it so that Q is on PS and R is on PT. The shared angle at P is the included angle for both triangles.
  • Solve PS/PQ = 10/4 = 5/2 PT/PR = 15/6 = 5/2 ✓ Equal ratios. ∠P is the included angle shared by both triangles. △PQR ∼ △PST by SAS Similarity. Scale factor (PQR → PST) = 5/2. ST/QR = 5/2 30/QR = 5/2 → 5·QR = 60 → QR = 12
  • CheckScale factor check: (5/2) · 12 = 30 = ST. ✓
Answer: △PQR ∼ △PST by SAS Similarity, and QR = 12.
Practice 6.8SAS Similarity
  1. △ABC: AB = 8, AC = 12, ∠A = 40°. △DEF: DE = 10, DF = 15, ∠D = 40°. Are the triangles similar by SAS?
  2. △PQR and △PST share vertex P. PQ = 5, PS = 10, PR = 7, PT = 14. Is ∠P sufficient to confirm similarity by SAS? Explain.
  3. △ABC: AB = 6, BC = 9, ∠B = 60°. △XYZ: XY = 4, YZ = 6, ∠Y = 60°. Confirm SAS similarity and find the scale factor.
  4. Explain why the angle must be the included angle in SAS Similarity. Give an example where two sides are proportional and a non-included angle is congruent, but the triangles are not similar.
Similarity Proof PracticeProofs Using AA, SSS, SAS
Proof Example 6.8.3
Proving △ADE ∼ △ABC using AA

Given: In △ABC, DE ∥ BC with D on AB and E on AC.
Prove: △ADE ∼ △ABC.

StatementReason
DE ∥ BCGiven
∠ADE ≅ ∠ABCCorresponding angles, DE ∥ BC (transversal AB)
∠AED ≅ ∠ACBCorresponding angles, DE ∥ BC (transversal AC)
△ADE ∼ △ABCAA Similarity Postulate
  1. Proof: △GHI has sides 6, 10, 14. △JKL has sides 9, 15, 21. Prove △GHI ∼ △JKL by writing out all three ratio checks and stating the theorem used.
  2. Proof: In △PQR, S is the midpoint of PQ and T is the midpoint of PR. Prove △PST ∼ △PQR using SAS Similarity. Then conclude ST ∥ QR and ST = (1/2)QR.
  3. Proof: △ABC and △ADE share vertex A. Given ∠BAC ≅ ∠DAE (vertical angles) and ∠B ≅ ∠D. Prove △ABC ∼ △ADE and state the shortcut used.
Sections 6.9 – 6.10
Proportional Parts and the Midsegment
Triangle Proportionality Theorem · Converse · Midsegment Theorem
Section 6.9

Proportional Parts in Triangles

Theorem
Triangle Proportionality Theorem

If a line is parallel to one side of a triangle and intersects the other two sides at distinct points, then it divides those two sides proportionally.

In △ABC, if DE ∥ BC with D on AB and E on AC, then: AD/DB = AE/EC

Equivalently: AD/AB = AE/AC   (comparing each part to the whole side)

Diagram Instruction

Draw triangle ABC with A at top, B at lower left, C at lower right. Draw a line parallel to BC that cuts AB at point D and AC at point E. Label the four segments: AD (upper part of AB), DB (lower part of AB), AE (upper part of AC), EC (lower part of AC). These four segments satisfy AD/DB = AE/EC.

Example 6.9.1
Applying the Triangle Proportionality Theorem
  • ProblemIn △ABC, DE ∥ BC, D on AB, E on AC. Given AD = 5, DB = 10, AE = 7. Find EC.
  • RepresentAD/DB = AE/EC
  • Solve 5/10 = 7/EC 5 · EC = 70 EC = 14
  • Check5/10 = 1/2 and 7/14 = 1/2. ✓
Answer: EC = 14.
Converse
Converse of the Triangle Proportionality Theorem

If a line divides two sides of a triangle proportionally, then the line is parallel to the third side. This converse is used to prove that two lines are parallel by showing proportional segments on two sides of a triangle.

Example 6.9.2
Using the Converse to prove lines parallel
  • ProblemIn △PQR, S is on PQ and T is on PR. PS = 4, SQ = 8, PT = 5, TR = 10. Is ST ∥ QR?
  • Solve PS/SQ = 4/8 = 1/2 PT/TR = 5/10 = 1/2 ✓ Equal ratios.
Answer: By the Converse of the Triangle Proportionality Theorem, ST ∥ QR.
Practice 6.9Proportional Parts
  1. DE ∥ BC in △ABC. AD = 6, DB = 9, AE = 8. Find EC.
  2. DE ∥ BC in △ABC. AD = 4, AB = 10, AE = 6. Find EC and AC.
  3. ST ∥ QR in △PQR? PS = 5, SQ = 15, PT = 7, TR = 21. Check and justify.
  4. ST ∥ QR in △PQR? PS = 4, SQ = 6, PT = 6, TR = 8. Check and justify.
  5. DE ∥ BC. AD = 3x, DB = 6, AE = x+2, EC = 4. Find x.
  6. DE ∥ BC. AD = 5, AB = 20, AE = 7. Find AC.
Section 6.10

Triangle Midsegment Theorem

What a Midsegment Is

A midsegment of a triangle is a segment that connects the midpoints of two sides of the triangle. Every triangle has three midsegments — one for each pair of sides.

Theorem
Triangle Midsegment Theorem

The midsegment connecting the midpoints of two sides of a triangle is:

1. Parallel to the third side.

2. Half as long as the third side.

If M is the midpoint of AB and N is the midpoint of AC in △ABC, then: MN ∥ BC and MN = (1/2) · BC.

Diagram Instruction

Draw triangle ABC. Mark the midpoint of AB as M and the midpoint of AC as N. Draw segment MN. Use a double tick mark on MN and a single tick mark on BC to show the parallel and half-length relationship. Add parallel arrows (arrows pointing in the same direction) on MN and BC to emphasize they are parallel.

Why the Midsegment Theorem works: Since M and N are midpoints, AM = (1/2)AB and AN = (1/2)AC. The angle at A is shared by both △AMN and △ABC. The ratios AM/AB = AN/AC = 1/2 with included angle ∠A congruent. By SAS Similarity, △AMN ∼ △ABC with scale factor 1/2. Therefore MN = (1/2)BC, and corresponding sides of similar triangles are parallel.

Example 6.10.1
Applying the Midsegment Theorem — find the midsegment
  • ProblemIn △XYZ, M is the midpoint of XY and N is the midpoint of XZ. If YZ = 28, find MN.
  • SolveMN = (1/2) · YZ = (1/2) · 28 = 14
Answer: MN = 14.
Example 6.10.2
Finding the third side from the midsegment
  • ProblemIn △ABC, DE is a midsegment connecting midpoints of AB and BC. If DE = 11, find AC.
  • Solve DE = (1/2) · AC 11 = (1/2) · AC AC = 22
Answer: AC = 22.
Example 6.10.3
Midsegment with algebra
  • ProblemIn △PQR, midsegment ST is parallel to QR. If ST = 2x + 1 and QR = 5x − 4, find x and the length of QR.
  • ThinkST = (1/2)QR, so 2 · ST = QR.
  • Solve 2x + 1 = (1/2)(5x − 4) 2(2x + 1) = 5x − 4 4x + 2 = 5x − 4 6 = x QR = 5(6) − 4 = 26
  • CheckST = 2(6) + 1 = 13. And (1/2) · 26 = 13. ✓
Answer: x = 6 and QR = 26.
Student Warning — Midsegment is HALF the Third Side

The midsegment is half the length of the parallel side, not equal to it. Writing MN = BC instead of MN = (1/2)BC is one of the most common midsegment errors. Remember: the midsegment connects midpoints, so the corresponding triangle is smaller by a factor of 2.

Practice 6.10Midsegment Problems
  1. Midsegment MN in △ABC. YZ = 34. Find MN.
  2. Midsegment DE = 15. Find the parallel third side.
  3. Midsegment = 3x + 2, third side = 8x − 4. Find x and the midsegment length.
  4. Midsegment = 4n − 1, third side = 6n + 8. Find n and the third side.
  5. In △PQR, S is mid of PQ and T is mid of PR. ST = 7.5. Find QR.
  6. PQ ∥ midsegment RS. RS = 13. What is PQ?
Sections 6.11 – 6.14
Applications, Mistakes, and Connections
Scale Drawings · Indirect Measurement · Common Mistakes · Preparing for Trigonometry
Section 6.11

Scale Drawings and Maps

A scale drawing is a drawing in which all lengths are proportional to the corresponding lengths of the actual object. The scale is the ratio of the drawing's length to the actual length. Scale is often written as a ratio such as 1:100, meaning every 1 unit in the drawing represents 100 units in reality. It may also be written with units: "1 inch = 50 feet."

Scale Drawing Formulas

To find an actual length from a drawing: Actual length = Drawing length × Scale factor

To find a drawing length from an actual measurement: Drawing length = Actual length ÷ Scale factor

Always identify the units. A scale of "1 inch = 50 feet" uses two different units — do not mix them in a single calculation.

Example 6.11.1
Drawing to actual
  • ProblemA floor plan has scale 1 inch = 12 feet. A room measures 3.5 inches wide on the plan. How wide is the actual room?
  • SolveActual width = 3.5 × 12 = 42 feet
Answer: The actual room is 42 feet wide.
Example 6.11.2
Actual to drawing
  • ProblemAn architect draws a room that is 30 feet long. Scale: 1 inch = 10 feet. How long should the room appear on paper?
  • SolveDrawing length = 30 ÷ 10 = 3 inches
Answer: The room should be drawn 3 inches long.
Example 6.11.3
Map distance
  • ProblemOn a map, the scale is 1 cm = 50 km. Two cities are 7.4 cm apart. What is the actual distance?
  • SolveActual distance = 7.4 × 50 = 370 km
Answer: The cities are 370 km apart.
Practice 6.11Scale Drawings and Maps
  1. Scale: 1/4 inch = 1 foot. Room is 4.5 in × 3 in on the plan. Actual dimensions?
  2. Scale: 1 cm = 75 km. Two airports are 8.4 cm apart. Actual distance?
  3. Building is 240 feet tall. Scale: 1 inch = 40 feet. Height on paper?
  4. Photo: 4 in × 6 in. Enlarged so width = 10 in. New height?
  5. Scale model: 1:200. Model is 2.5 feet long. Actual length in feet?
  6. Scale: 1 inch = 25 miles. Two towns are 6.5 inches apart. Actual distance?
Section 6.12

Similarity in Measurement Problems

One of the most practical applications of similarity is indirect measurement — finding lengths that cannot be measured directly by using similar triangles and proportions. Two objects casting shadows at the same time in the same sunlight create similar right triangles: the sun's rays are parallel, so both triangles have the same angle of elevation. Since corresponding angles match (AA Similarity), the heights and shadow lengths are proportional.

Example 6.12.1
Shadow measurement
  • ProblemA 5-foot person casts a 3-foot shadow. At the same time, a tree casts a 21-foot shadow. How tall is the tree?
  • ThinkBoth the person and the tree are vertical. The sun's angle is the same. By AA Similarity (shared sun angle + right angles at the base), the triangles are similar.
  • DiagramDraw two right triangles side by side. Left triangle: height 5, base 3. Right triangle: height h (unknown), base 21. Mark equal angles at the top (sun angle).
  • Representperson's height / person's shadow = tree's height / tree's shadow
  • Solve 5/3 = h/21 3h = 105 h = 35
Answer: The tree is 35 feet tall.
Example 6.12.2
Mirror measurement
  • ProblemA student places a mirror flat on the ground 8 feet from the base of a wall. Standing 2 feet from the mirror, she can just see the top of the wall reflected in the mirror. Her eye level is 5.5 feet. How tall is the wall?
  • ThinkWhen light reflects off a flat mirror, the angle of incidence equals the angle of reflection. This creates two similar right triangles: student-to-mirror and mirror-to-wall.
  • DiagramDraw two right triangles sharing a common point at the mirror. Left triangle: height 5.5 (eye level), base 2 (student to mirror). Right triangle: height h (wall), base 8 (mirror to wall). Mark equal reflection angles at the mirror.
  • Solve 5.5/2 = h/8 2h = 5.5 × 8 = 44 h = 22
Answer: The wall is 22 feet tall.
Practice 6.12Measurement Applications

For each problem: draw a diagram, set up a proportion, and solve.

  1. A 6-foot person casts a 4-foot shadow. A flagpole casts a 30-foot shadow at the same time. How tall is the flagpole?
  2. A student places a mirror 10 feet from a building. Standing 3 feet from the mirror, she sees the top of the building. Her eye is 5.4 feet high. How tall is the building?
  3. A 5.5-foot person casts a 2.2-foot shadow. A cell tower casts a 36-foot shadow at the same time. How tall is the tower?
  4. A surveyor sets up two similar triangles. One triangle has legs 20 m and 15 m. The corresponding leg of the similar triangle is 48 m. Find the other leg, which represents a pond's width.
  5. A building that is 18 feet wide is drawn on an architect's plan with a width of 3 inches. The same plan shows a room that is 4.5 inches wide. What is the actual width of the room?
  6. A 30-foot tree casts a shadow, and at the same time a 5-foot person casts an 8-foot shadow. How long is the tree's shadow?
Section 6.13

Common Similarity Mistakes

Mistake 1 — Incorrect similarity statement (wrong vertex order)
△ABC ∼ △DEF where ∠A=50°, ∠B=60°, ∠C=70°, ∠D=60°, ∠E=50°, ∠F=70°

The correct statement must pair corresponding vertices: ∠A(50°)=∠E(50°), ∠B(60°)=∠D(60°), ∠C(70°)=∠F(70°). Correct: △ABC ∼ △EDF.

Mistake 2 — Mixing figures in a proportion
AB/EF = DE/BC for △ABC ∼ △DEF

One figure's sides must stay consistently on top (or bottom) throughout the proportion. Correct: AB/DE = BC/EF.

Mistake 3 — Confusing similarity with congruence
Setting AB = DE instead of AB/DE = BC/EF

Similar figures have proportional sides (equal ratios), not necessarily equal sides. Only congruent figures have equal corresponding sides. Use ∼ for similarity and set up a proportion, not an equation of equality.

Mistake 4 — Applying SSS Similarity with only two ratios checked
"Two ratios equal 3/2, so the triangles must be similar by SSS."

SSS Similarity requires all three ratios to be equal. Always compute and check all three pairs of corresponding sides.

Mistake 5 — Using a non-included angle for SAS Similarity
Two proportional sides and a non-included congruent angle → concluding SAS Similarity

The angle in SAS Similarity must be the included angle — the angle between the two sides being compared. A non-included angle does not justify the theorem.

Mistake 6 — Writing midsegment = third side (forgetting the factor of 1/2)
Midsegment MN = BC = 18, so MN = 18.

The Midsegment Theorem says MN = (1/2) · BC. If BC = 18, then MN = 9.

Mistake 7 — Inverting the scale factor when converting
Scale: 1 cm = 10 m. Room is 30 m. Drawing length = 30 × 10 = 300 cm.

To go from actual to drawing, divide by the scale: 30 ÷ 10 = 3 cm. Multiplying converts in the wrong direction.

Mistake 8 — Concluding polygon similarity from angles alone (non-triangles)
"Both rectangles have four right angles, so they must be similar."

For polygons with four or more sides, both conditions — angle congruence and side proportionality — must be verified. AA works for triangles only.

Practice 6.13Error Analysis

Each problem contains an error. Identify it and write the correct answer.

  1. △ABC ∼ △DEF. A student sets up: AB/EF = DE/BC. What is wrong? Write the correct proportion.
  2. △PQR has sides 5, 8, 11. △STU has sides 10, 16, 23. A student claims they are similar by SSS because 10/5 = 16/8 = 2. What is wrong?
  3. △ABC ∼ △DEF with scale factor 3/2. A student writes QR = 10 + 3 = 13, saying "the sides increased by 3." If BC = 10 and EF = 10 · (3/2) = 15, what is the actual error and the correct value?
  4. In △ABC, midsegment MN connects midpoints of AB and AC. BC = 20. A student writes MN = 20. Identify the error and write the correct value.
  5. A student writes △ABC ∼ △DEF for two triangles where ∠A = 50°, ∠B = 60°, ∠D = 60°, ∠E = 50°. What is wrong with this similarity statement? Write the correct statement.
Section 6.14

Preparing for Right-Triangle Trigonometry

The Connection Between Similarity and Trigonometry

Everything in Chapter 8 (Right-Triangle Trigonometry) rests on one key idea from this chapter: similar right triangles have equal ratios of corresponding sides.

Consider any right triangle with an acute angle of measure θ. Now consider a different right triangle that also has an acute angle of measure θ. Because both are right triangles (they share the 90° angle) and both have angle θ, they share two pairs of congruent angles. By AA Similarity, the two triangles are similar. Since the triangles are similar, their corresponding side ratios are equal. In particular:

Ratio NameRatioNote
Sine of θopposite / hypotenuseSame value for every right triangle with angle θ
Cosine of θadjacent / hypotenuseSame value for every right triangle with angle θ
Tangent of θopposite / adjacentSame value for every right triangle with angle θ

These ratios depend only on the angle θ — not on the size of the triangle. That is precisely why we can define sine, cosine, and tangent as ratios that belong to an angle. These ratios are consistent because similar triangles are consistent. Similarity is the reason trigonometry works.

Preview — Chapters 7 and 8

In Chapter 7, you will study right triangles, the Pythagorean Theorem, special right triangles (45-45-90 and 30-60-90), and radical lengths. In Chapter 8, you will use similarity to define and apply the trigonometric ratios. Whenever you see a sine, cosine, or tangent, you are looking at a ratio of two sides in a right triangle — and the reason that ratio is reliable is that all right triangles with the same acute angle are similar.

Practice 6.14Readiness Check
  1. Two right triangles each have an acute angle of 35°. Without measuring, explain why their corresponding side ratios must be equal.
  2. A right triangle has legs 3 and 4, and hypotenuse 5. A similar right triangle has hypotenuse 10. Find the two legs.
  3. Right triangles A and B both have a 60° acute angle. In triangle A, the hypotenuse is 8 and the side opposite the 60° angle is 4√3. In triangle B, the hypotenuse is 12. Find the side opposite the 60° angle in triangle B.
  4. Explain in your own words why the sine of a 30° angle is the same regardless of how large the right triangle is.
MGU Library Connection
  • Appendix E: Triangle Congruence and Similarity Reference — AA, SSS, SAS shortcuts and comparison with congruence shortcuts
  • Appendix G: Trigonometry Ratio Reference — how similarity establishes the consistency of sine, cosine, and tangent
  • Chapter 7: Right Triangles and the Pythagorean Theorem — similarity reappears in geometric mean and similar right triangles
  • Chapter 8: Right-Triangle Trigonometry — the direct application of this chapter's similarity results
  • Chapter 14: Transformations — dilations are the transformation version of similarity
  • MGU Dictionary entries: similar figures, scale factor, proportion, midsegment, indirect measurement
Chapter 6 — Back Matter

Chapter Summary

Key Terms Review

similar figures
Figures with congruent corresponding angles and proportional corresponding sides. Written as △ABC ∼ △DEF.
similarity statement
A statement such as △ABC ∼ △DEF that identifies corresponding vertices by their position in the statement.
scale factor
The common ratio of corresponding side lengths between two similar figures. Going from figure B back to A uses the reciprocal.
corresponding sides
Sides that connect corresponding vertices in a similarity statement and form equal ratios.
corresponding angles
Angles at corresponding vertices in a similarity statement; these angles are congruent in similar figures.
proportion
An equation stating that two ratios are equal: a/b = c/d.
cross multiplication
The method of solving a proportion a/b = c/d by computing a · d = b · c.
AA Similarity
Triangle similarity established by two pairs of congruent angles. Accepted as a postulate.
SSS Similarity
Triangle similarity established by three equal ratios of corresponding sides.
SAS Similarity
Triangle similarity established by two proportional sides and the congruent included angle.
Triangle Proportionality Theorem
A line parallel to one side of a triangle divides the other two sides proportionally. Its converse also holds.
midsegment
A segment connecting the midpoints of two sides of a triangle. Every triangle has three midsegments.
Triangle Midsegment Theorem
A midsegment is parallel to the third side and equals half its length.
scale drawing
A drawing in which all lengths are proportional to actual lengths by a fixed scale ratio.
scale
The ratio of a drawing length to the actual length it represents, such as 1 inch = 10 feet.
indirect measurement
Finding a length that cannot be measured directly, using similar triangles and proportions.
similarity ratio
Another name for the scale factor; the common ratio of all corresponding side lengths in similar figures.

Concept Review Questions

Answer in complete sentences. Use correct geometric language.

  1. What two conditions must be satisfied for two polygons to be similar?
  2. Why is it not sufficient to check only angles when determining similarity for a quadrilateral?
  3. What does the scale factor tell you about the relationship between two similar figures?
  4. Explain the difference between △ABC ≅ △DEF and △ABC ∼ △DEF.
  5. Why does AA Similarity work for triangles but not for quadrilaterals?
  6. State the Triangle Proportionality Theorem in your own words.
  7. A midsegment connects the midpoints of two sides of a triangle. What two things does the Midsegment Theorem say about every midsegment?
  8. How does similarity make indirect measurement possible?
  9. Why do all right triangles with the same acute angle have equal ratios of corresponding sides?
  10. A student claims that two rectangles are similar because they both have four right angles. Is the student correct? Explain and provide a counterexample.

Skill Practice

Part AScale Factors
  1. △ABC: sides 5, 7, 9. △DEF: sides 10, 14, 18. Scale factor?
  2. △PQR: sides 6, 8, 10. △STU: sides 9, 12, 15. Scale factor?
  3. Scale factor = 3/2; smaller triangle has sides 4, 6, 8. Find larger triangle's sides.
  4. △RST ∼ △UVW. RS = 10, UV = 15, ST = 8. Find VW.
  5. Scale factor from A to B is 2.5. What is the scale factor from B to A?
  6. △MNO: sides 3, 5, 7. △XYZ: sides 6, 10, 15. Are they similar?
Part BProportions and Missing Sides
  1. △ABC ∼ △DEF. AB = 10, DE = 6, BC = 15. Find EF.
  2. △PQR ∼ △STU. PQ = x, ST = 9, QR = 14, TU = 21. Find x.
  3. DE ∥ BC; AD = 6, DB = 9, AE = 8. Find EC.
  4. DE ∥ BC; AD = 4, AB = 10, AE = 6. Find EC and AC.
  5. Midsegment = 13. Find the parallel third side.
  6. Midsegment = 3x + 2; third side = 8x − 4. Find x and QR.
Part CTriangle Similarity — Name the Shortcut

For each pair, determine if similar. If yes, name the shortcut (AA, SSS, or SAS) and write the similarity statement.

  1. ∠A = 45°, ∠B = 65°; ∠D = 45°, ∠F = 70°.
  2. Sides 4, 6, 8 and sides 6, 9, 12.
  3. PQ = 5, QR = 7, ∠Q = 50°; ST = 10, TU = 14, ∠T = 50°.
  4. ∠A = 90°, ∠B = 32°; ∠D = 90°, ∠E = 58°.
  5. AB = 6, AC = 8, ∠A = 40°; DE = 9, DF = 12, ∠D = 40°.
  6. Sides 4, 7, 9 and sides 8, 14, 20.
Part DScale Drawings and Indirect Measurement
  1. Scale: 1 inch = 12 feet. Room is 3.5 inches wide. Actual width?
  2. Building: 240 feet tall. Scale: 1 in = 40 ft. Height on paper?
  3. Person: 6 ft, shadow: 4 ft. Tree shadow: 30 ft. Tree height?
  4. Mirror 10 ft from building. Student 3 ft from mirror. Eye: 5.4 ft. Wall height?
  5. Map scale: 1 cm = 50 km. Cities are 7.4 cm apart. Actual distance?
  6. Photo: 4 × 6 inches. Width enlarged to 10 inches. New height?

Word Problems

For each problem: identify the similar triangles or figures, set up the proportion, and solve. Show all steps.

  1. A photograph is 4 inches wide and 6 inches tall. It is enlarged so that the width becomes 10 inches. What is the new height?
  2. Two similar triangles have perimeters 24 and 36. If a side of the smaller triangle is 8, find the corresponding side of the larger.
  3. A surveyor needs to find the width of a river. She sets up similar triangles using stakes on her side of the bank. One triangle has legs 20 m and 15 m. The corresponding leg of the similar triangle is 48 m. Find the width represented by the other leg.
  4. A room that is 18 feet wide is drawn on a plan with a width of 3 inches. A second room is 4.5 inches wide on the same plan. What is the actual width of the second room?
  5. △ABC ∼ △PQR. AB = 14, BC = 21, CA = 28. Scale factor from △ABC to △PQR is 3/7. Find the perimeter of △PQR.
  6. A model bridge is built at a scale of 1:200. The model is 2.5 feet long. How long is the actual bridge?
  7. In △PQR, DE is a midsegment parallel to QR. If DE = 4n − 1 and QR = 6n + 8, find n and the length of QR.
  8. A 5.5-foot person casts a 2.2-foot shadow. A nearby cell tower casts a 36-foot shadow at the same time. How tall is the tower?

Error Analysis

Each problem shows a student's work. Identify the error and write the correct solution.

Problem 1
△ABC ∼ △DEF. AB=6, DE=9, BC=10. Student computes EF = 10+3 = 13.

What is the error? What is the correct value of EF?

Problem 2
Two rectangles both have four right angles. Student concludes they are similar.

Is this correct? Give a counterexample using specific dimensions.

Problem 3
In △ABC, midsegment MN connects midpoints of AB and AC. BC = 18. Student writes MN = 18.

Identify the error and find the correct value of MN.

Problem 4
Proportion for △ABC ∼ △DEF: AB/EF = DE/BC

What is wrong with this proportion? Write a correct proportion using these sides.

Problem 5
∠A=50°, ∠B=60°, ∠C=70°; ∠D=60°, ∠E=50°, ∠F=70°. Student writes △ABC ∼ △DEF.

Is this correct? Write the correct similarity statement.

Problem 6
Scale: 1 cm = 10 m. Room is 30 m wide. Student computes drawing width = 30 × 10 = 300 cm.

Identify the error. What is the correct drawing width?

Chapter 6 Checkpoint

Chapter 6 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 6
Part A — Definitions and Concepts
  1. In your own words, explain what it means for two figures to be similar.
  2. State the AA Similarity Postulate.
  3. State the Triangle Midsegment Theorem.
  4. What is the difference between the scale factor from figure A to figure B, and the scale factor from figure B to figure A?
Part B — Scale Factors and Proportions
  1. △RST ∼ △UVW. RS = 10, UV = 15, ST = 8. Find the scale factor and VW.
  2. △ABC ∼ △DEF. AB = x + 3, DE = 10, BC = 8, EF = 12. Find x.
  3. In △PQR, ST ∥ QR, S on PQ, T on PR. PS = 4, SQ = 8, PT = 5. Find TR.
Part C — Triangle Similarity

For each pair, determine whether the triangles are similar. If yes, name the shortcut and write the similarity statement. If no, explain why not.

  1. △ABC: ∠A = 45°, ∠B = 65°.   △DEF: ∠D = 45°, ∠F = 70°.
  2. △GHI: sides 4, 6, 8.   △JKL: sides 6, 9, 12.
  3. △PQR: PQ = 5, QR = 7, ∠Q = 50°.   △STU: ST = 10, TU = 14, ∠T = 50°.
Part D — Midsegment Theorem
  1. In △ABC, M is the midpoint of AB and N is the midpoint of AC. If MN = 13, find BC.
  2. In △XYZ, PQ is a midsegment parallel to XZ. If PQ = 4n − 1 and XZ = 6n + 8, find n and XZ.
Part E — Applications
  1. A 5.5-foot person casts a 2.2-foot shadow. A nearby cell tower casts a 36-foot shadow at the same time. How tall is the tower?
  2. A map has a scale of 1 inch = 25 miles. Two towns are 6.5 inches apart. What is the actual distance in miles?
  3. A room that is 18 feet by 24 feet is drawn on an architect's plan. If the 24-foot side measures 4 inches on the plan, what is the scale? How long is the 18-foot side on the plan?
Part F — Proof
  1. Given: ∠BAC ≅ ∠DAE and ∠B ≅ ∠D. Prove that △ABC ∼ △ADE. State the similarity shortcut used and write a two-column proof.
  2. In △PQR, S is the midpoint of PQ and T is the midpoint of PR. Use SAS Similarity to prove △PST ∼ △PQR. Then conclude ST ∥ QR and ST = (1/2)QR.
Bridge Note — Toward Right Triangles and Trigonometry
Chapter 7: Right Triangles and the Pythagorean Theorem

Chapter 6 has established the reasoning you will need in the next two chapters. In Chapter 7, you will work with right triangles in depth — computing missing side lengths using the Pythagorean Theorem, working with special right triangles (45-45-90 and 30-60-90), and using the distance formula on the coordinate plane. Similarity appears again when you study similar right triangles formed by the altitude drawn to the hypotenuse. In Chapter 8, the entire framework rests on the similarity result from Section 6.14: all right triangles with the same acute angle are similar, so their corresponding side ratios are equal. The sine, cosine, and tangent functions are those ratios. The skills you practiced here — identifying corresponding parts, writing proportions, using cross multiplication, and applying similarity shortcuts — are the same skills you will use to solve for missing sides and angles in right triangles. Carry these ideas forward. Similarity is not only a topic in this chapter; it is a thread that runs through the rest of this book and into every science course that uses measurement, ratio, and angle reasoning.