AA Criterion: Applications

Deck 2 of 2 · HSG.SRT.A.3

This deck covers:

• Applying AA to identify similar triangles
• AA (similarity) vs ASA/AAS (congruence)
• Three real-world applications: shadows, map scaling, parallel segments

Learning Objectives: Complete Series

By the end of this two-deck series, you should be able to:

1. State the AA criterion for triangle similarity
2. Prove AA using properties of similarity transformations
3. Apply AA to determine if two triangles are similar
4. Explain why two angle pairs are sufficient
5. Distinguish between AA for similarity and ASA/AAS for congruence
6. Use AA to solve real-world problems

This deck focuses on Objectives 3, 5, and 6.

Recap: What Was Proved in Deck 1

AA Criterion (proven): If ∠A ≅ ∠D and ∠B ≅ ∠E, then △ABC ~ △DEF.

Why it works:

1. Angle sum → third angle matches automatically
2. Translate + rotate positions the triangle correctly
3. Dilation scales it to match size exactly

Now: Use AA as an efficient tool.

How to Apply the AA Criterion

Three-step application:

1. Identify corresponding angles in both triangles
2. Verify two pairs of corresponding angles are congruent
3. Conclude similarity — write the similarity statement in correspondence order

Sources of angle information:

• Given directly in the problem
• Parallel lines (corresponding, alternate interior angles)
• Right angles (shared 90°)
• Shared or vertical angles

Example A: Direct Angle Matching

△ABC: ∠A = 40°, ∠B = 70°
△DEF: ∠D = 40°, ∠E = 70°

Check:

• ∠A = ∠D ✓ (40°)
• ∠B = ∠E ✓ (70°)

Conclusion: △ABC ~ △DEF by AA ✓

The third angles match automatically — but we didn't need to check.

Quick Check: Confirming Example A

△ABC: ∠C = 180° − 40° − 70° = 70°
△DEF: ∠F = 180° − 40° − 70° = 70°

✓ Third angle pair also matches — exactly as AA predicts.

Key takeaway: Two pairs is enough. The third matches automatically. No need to calculate it before concluding similarity.

Example B: Parallel Lines Create Similar Triangles

Setup: Parallel lines ℓ₁ ∥ ℓ₂ cut by two transversals, forming △PQR (above) and △XYZ (below)

Question: Are △PQR and △XYZ similar?

Example B: Solution Using Parallel Line Properties

Using ℓ₁ ∥ ℓ₂:

• ∠PQR = ∠XYZ (alternate interior angles) ✓
• ∠QPR = ∠YXZ (corresponding angles) ✓

Two pairs match → △PQR ~ △XYZ by AA ✓

The parallel lines supply the angle equalities. We needed no other information.

Example C: Right Triangles with Equal Acute Angle

△ABC: ∠C = 90°, ∠A = 30° → ∠B = 60°
△DEF: ∠F = 90°, ∠D = 30° → ∠E = 60°

Check:

• ∠A = ∠D ✓ (30°)
• ∠C = ∠F ✓ (90°)

Conclusion: △ABC ~ △DEF by AA ✓

Insight: Why Trigonometry Works

All right triangles with the same acute angle θ are similar by AA.

Because they're similar, corresponding side ratios are equal — in every such triangle:

AA is the foundation of trigonometry — it's why these ratios depend only on the angle, not the triangle's size.

AA but NOT Congruent: A Concrete Example

△PQR: ∠P = 50°, ∠Q = 60°, PQ = 5
△XYZ: ∠X = 50°, ∠Y = 60°, XY = 10

Are they similar? ✓ — two angle pairs match → AA → △PQR ~ △XYZ

Are they congruent? ✗ — PQ = 5 ≠ XY = 10. Different sizes.

Key distinction: AA guarantees same shape, not same size.

Similarity vs Congruence: The Comparison

Correspondence Matters: Example

△ABC: ∠A = 50°, ∠B = 60°, ∠C = 70°
△DEF: ∠D = 70°, ∠E = 50°, ∠F = 60°

Correct correspondence: AE (50°), BF (60°), CD (70°)

Correct statement: △ABC ~ △EFD ✓

Not △ABC ~ △DEF — the angle values match, but the correspondence doesn't.

When setting up proportions, the order of vertices determines which sides correspond.

Quick Check: Similar, Congruent, or Both?

△ABC: ∠A = 45°, ∠B = 90°
△DEF: ∠D = 45°, ∠E = 90°

1. Are they similar? Why?
2. Are they congruent? What additional information is needed?
3. If AB = DE = 7 and AC = DF = 7√2, are they congruent?

Think about what AA alone tells you vs what AA + side information tells you.

AA Is for Triangles Only: Rectangle Counterexample

Rectangle 1: 2 × 3 (all angles = 90°)
Rectangle 2: 1 × 5 (all angles = 90°)

All four angle pairs match — but side ratios: 2/3 ≠ 1/5 → not similar.

Why AA fails for polygons with 4+ sides: Equal angles don't constrain side ratios. Triangles are unique because the angle sum theorem makes three angles interdependent — knowing two fixes the third and, combined with the transformation proof, forces the sides to be proportional.

Real-World Applications: Two-Step Process

AA + Proportions = Indirect Measurement

Step 1: Find two matching angle pairs → conclude AA → triangles are similar

Step 2: Corresponding sides are proportional → set up proportion → solve for unknown

This approach works for:

• Shadow and mirror height measurements
• Map scaling and distances
• Geometric figures with parallel lines

Problem 1: Shadow Measurement

Given:

• Tree: height = h, shadow = 15 m
• Pole: height = 2 m, shadow = 3 m
• Measured at the same time of day

Question: How tall is the tree?

Problem 1, Step 1: Verify AA

Triangle 1 (tree): right angle (tree vertical) + sun angle θ
Triangle 2 (pole): right angle (pole vertical) + same sun angle θ (same time of day)

Two pairs match → Triangles are similar by AA ✓

Notice: Both angle pairs were deduced from geometric properties — no measuring required. The right angle comes from vertical objects; the sun angle from same-time observation.

Problem 1, Step 2: Proportion and Solution

Corresponding sides of similar triangles are proportional:

Problem 2: Map Scaling

Setup:

• Two cities appear 5 cm apart on a map
• Map scale: 1 cm = 10 km in reality

The map and reality are similar figures with scale factor k.

Question: How far apart are the cities in reality?

Problem 2: Using the Scale Factor

Map and reality are similar figures with km/cm.

Note: The scale factor k can be any positive number. Here k = 10, but map scales often give non-integer k values (e.g., 1 cm : 7.5 km gives k = 7.5). The calculation is the same either way.

Your Turn: Map Problem

Given:

• Two cities are 3 cm apart on a map
• Map scale: 1 cm = 25 km

Question: How far apart are the cities in reality?

Set up the calculation using the scale factor and solve...

Problem 3: Parallel Segment in a Triangle

Setup: △ABC with segment DE where D is on AB, E is on AC, and DE ∥ BC

Claim: △ADE ~ △ABC

Problem 3, Step 1: Prove △ADE ~ △ABC

Finding angle pairs:

• ∠A is shared by both triangles → ∠A = ∠A ✓
• DE ∥ BC → ∠ADE and ∠ABC are corresponding angles → ∠ADE = ∠ABC ✓

Two pairs match → △ADE ~ △ABC by AA ✓

Problem 3, Step 2: Use Proportionality

Since △ADE ~ △ABC, corresponding sides are proportional:

This ratio k is the scale factor. Since D is between A and B, in this figure — but k can be any positive value in general.

Note: k = 2/3 is a perfectly valid scale factor. Non-integer values are normal.

Your Turn: Find the Unknown Length

Given: △ADE ~ △ABC with and

Find AE using the proportion:

Solve for AE before the next slide...

Check-In: The Two-Step Process

In your own words:

Step 1: How do you use AA to verify that two triangles are similar?

Step 2: Once similarity is confirmed, what do you set up to find unknown lengths?

Bonus: In the shadow problem, map problem, and triangle problem — where exactly did the angle information come from in each case?

Key Takeaways: Applying the AA Criterion

✓ AA in 3 steps: identify angles → verify two pairs → conclude similar
✓ AA + proportions = solve for unknown lengths
✓ Deduced angles count — no measuring required; geometric properties suffice

AA → similar, not congruent — to prove congruence, you need side information
Correspondence matters — match corresponding angles; order of vertices in the similarity statement matters
Triangles only — equal angles don't force similarity in other polygons
k can be any positive number — 2/3, 7.5, √2 are all valid scale factors

Next: Using AA to Prove Theorems

Coming up in the next lesson:

• Side-Splitter Theorem: A segment parallel to one side of a triangle divides the other two sides proportionally — a direct extension of Problem 3 in this deck
• Pythagorean Theorem: Proved via similar right triangles (AA at work again)
• Triangle Midsegment Theorem: Another consequence of AA and parallel segments

You've proven and applied AA. Next, you use it to derive other geometric results.