Thales Projection Slides Thales' Parallel Projections
Exploring the Rigor of Proportionality
Unit: Similarity & Pythagoras
Lesson 01: Foundations
The Transversal Puzzle
Imagine three parallel lines spaced evenly.
If we drop a perpendicular transversal, it is cut into equal segments.
"Will any transversal, at any angle, also be cut into equal segments?"
L₁ L₂ L₃ A B
The Formal Claim
Thales' Theorem
If three or more parallel lines intersect two transversals, then they cut the transversals proportionally.
\[ \frac{AB}{BC} = \frac{DE}{EF} \]
A B C D E F
Proof by Auxiliary Construction
1
Translate
Construct a line through point A parallel to the second transversal \(DF\).
2
Parallelograms
Identify the resulting parallelograms to show segment congruence.
3
Area Ratios
Apply the lemma: Triangles with the same altitude have area ratios equal to base ratios.
"The power of a proof lies in constructing the bridge between the unknown and the known."
Synthesis Task
Using the Proportionality Worksheet, your objective is to reconstruct the formal proof of the Triangle Proportionality Theorem — a specific case of Thales' Theorem.
Consider: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those two sides proportionally.
Proportionality Proof Workshop Worksheet Proportionality Proof Workshop
Lesson 01: The Fundamental Theorem of Proportionality
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The Objective
To construct a formal, rigorous proof of the Triangle Proportionality Theorem : If a line parallel to one side of a triangle intersects the other two sides, then it divides those two sides proportionally.
1 Defining the Premises
A B C D E
Given: \(\triangle ABC\) with line \(DE\) parallel to \(BC\), where \(D\) lies on \(AB\) and \(E\) lies on \(AC\).
To Prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)
Strategy: We will use the relationship between area and altitude to establish these ratios.
Auxiliary Construction Instructions:
Draw segment \(BE\) and segment \(CD\).
Draw altitude \(h_1\) from \(E\) to side \(AB\) (label the foot \(F\)).
Draw altitude \(h_2\) from \(D\) to side \(AC\) (label the foot \(G\)).
Use this space to sketch the construction on the diagram above.
2 Establishing Area Ratios
Compare the areas of \(\triangle ADE\) and \(\triangle BDE\). Use their shared altitude from \(E\) to line \(AB\).
Expected result: \(\frac{\text{Area}(ADE)}{\text{Area}(BDE)} = \frac{AD}{DB}\)
Now, compare the areas of \(\triangle ADE\) and \(\triangle CDE\). Use their shared altitude from \(D\) to line \(AC\).
Expected result: \(\frac{\text{Area}(ADE)}{\text{Area}(CDE)} = \frac{AE}{EC}\)
3 The Parallel Equality
Analyze \(\triangle BDE\) and \(\triangle CDE\). Note that they share the same base, \(DE\), and their third vertices (\(B\) and \(C\)) lie on a line parallel to that base (\(BC \parallel DE\)).
Explain why \(\text{Area}(BDE) = \text{Area}(CDE)\). Cite specific geometric properties.
4 Q.E.D.
Synthesize the findings from parts 2 and 3 to complete the proof.
Similarity Mapping Slides Mapping Similarity
Criteria, Transformations, and Dilations
Lesson 02
What is Similarity?
"Two figures are similar if there exists a similarity transformation that maps one onto the other."
A similarity transformation is a composition of a dilation and one or more isometries (translation, rotation, reflection).
Pre-image Image
The AA Similarity Proof
STEP 1: DILATE
Dilate \(\triangle ABC\) by factor \(k = \frac{A'B'}{AB}\) about point \(A\).
This creates \(\triangle AB''C''\) where \(AB'' = A'B'\).
STEP 2: CONGRUENCE
Use the Parallel Postulate to show \(\angle AB''C'' \cong \angle ABC \cong \angle A'B'\).
By ASA or AAS, show \(\triangle AB''C'' \cong \triangle A'B'C'\).
STEP 3: COMPOSE
Since a dilation and a congruence map \(\triangle ABC\) to \(\triangle A'B'C'\), they are similar.
Q.E.D.
SAS Criterion
Ratio and Included Angle
\(\frac{AB}{A'B'} = \frac{AC}{A'C'}\) AND \(\angle A \cong \angle A'\)
If two sides of one triangle are proportional to two sides of another, and the included angles are congruent, then the triangles are similar.
Requires one rigid transformation
Requires one dilation
Side \(c\) Side \(b\) Side \(kc\) Side \(kb\)
Your Turn: The Transformation Proof Lab
You will now apply these concepts to construct a coordinate-independent transformation proof for the SAS and SSS Similarity Theorems.
Mission A
Construct the mapping sequence for SAS.
Mission B
Justify the SSS congruence of the dilated image.
Similarity Transformation Lab Worksheet Similarity Mapping Lab
Lesson 02: Transformation-Based Proofs
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The Transformation Challenge
In modern geometry, we define similarity through transformations. To prove that two triangles are similar, we must show that a sequence of rigid motions followed by a dilation maps the first triangle exactly onto the second.
1
SAS Similarity Construction
Given: \(\triangle ABC\) and \(\triangle DEF\) such that \(\angle A \cong \angle D\) and \(\frac{AB}{DE} = \frac{AC}{DF} = k\).
A B C
Pre-image (\(\triangle ABC\))
D E F
Target Image (\(\triangle DEF\))
Step A: Isometry Mapping
Identify the rigid motion(s) that map \(\angle A\) onto \(\angle D\).
Step B: Dilation
Apply a dilation to the image from Step A. Specify the center and scale factor.
Step C: Synthesis
Explain why the image of \(B\) and \(C\) under this sequence must coincide with \(E\) and \(F\).
2
SSS Similarity Strategy
To prove SSS Similarity, we dilate \(\triangle ABC\) by factor \(k\) to create \(\triangle A'B'C'\). We then must prove that \(\triangle A'B'C' \cong \triangle DEF\).
Outline the logical steps needed to establish congruence between the dilated triangle and the target triangle. Which congruence theorem (SSS, SAS, etc.) would you use as your final justification?
Critical Thinking: Why is it necessary to perform the dilation first in these proofs, rather than trying to map the triangles directly with a "similarity mapping" without defining its components?
Rigorous Geometry Sequence | similarity-proportionality-pythagoras
Geometric Mean Slides The Nested Altitude
Similarity in Right Triangles
Lesson 03: Geometric Means
One Triangle, Three Models
When we drop an altitude from the right angle to the hypotenuse, we "fracture" the large triangle into two smaller ones.
"Are these two new triangles similar to the original? Are they similar to each other?"
A B C D
The Theorem
The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the original triangle and to each other.
\[ \triangle ABD \sim \triangle BCD \]
\[ \triangle ABD \sim \triangle ACB \]
\[ \triangle BCD \sim \triangle ACB \ ]
How do we prove this rigorously?
HINT: AA CRITERION
Geometric Mean: Altitude
From the similarity \(\triangle ABD \sim \triangle BCD\), we derive:
\[ \frac{AD}{BD} = \frac{BD}{CD} \implies BD^2 = AD \cdot CD \]
"The altitude is the geometric mean between the segments of the hypotenuse."
Algebraic Vision
\(h^2 = x \cdot y\)
Where \(x\) and \(y\) are the base segments
Workshop Time
Grab the "Geometric Mean Workshop" sheet. Your task is to prove the Leg Rule: how each leg relates to the entire hypotenuse.
Find Similarity
Set Proportions
Solve Mean
Geometric Mean Workshop Worksheet Geometric Mean Workshop
Lesson 03: Proportions in the Right Triangle
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A B C D c a x y h
Notation Guide:
• \(h = BD\) (altitude)
• \(x = AD\), \(y = DC\) (segments)
• \(c = AB\), \(a = BC\) (legs)
• Hypotenuse \(b = x + y\)
1
The Geometric Mean Leg Rule
While the altitude is the geometric mean between the segments, each leg is also a geometric mean between the hypotenuse and the segment adjacent to that leg.
A. Identify the two triangles you must use to prove that \(c^2 = x \cdot b\). State the similarity criterion used.
B. Set up the proportion from the similarity above and perform the cross-multiplication to show the geometric mean relationship.
2
Analytical Problems
Problem 1: In a right triangle, the altitude to the hypotenuse divides it into segments of 4 cm and 9 cm. Calculate the altitude.
Problem 2: In the same triangle, calculate the length of the shorter leg using the Leg Rule.
3
The Pre-Pythagorean Glimpse
Consider the two leg equations: \(c^2 = x \cdot b\) and \(a^2 = y \cdot b\).
Add these two equations together. Factor out \(b\) from the right side. What fundamental theorem emerges from this algebraic manipulation of similarity ratios?
Pythagorean Proof Showdown Slides PROOF SHOWDOWN
The Pythagorean Theorem
> SIMILARITY VS. AREA <
1. The Similarity Method
Elegant & Algebraic
"By dropping one altitude, we create three similar triangles. The Pythagorean Theorem is merely the sum of two ratios."
The Logic:
Leg Rule A: \(a^2 = y \cdot c\)
Leg Rule B: \(b^2 = x \cdot c\)
Sum: \(a^2 + b^2 = c(y + x)\)
Substitute: \(y + x = c \implies a^2 + b^2 = c^2\)
A B C x y
2. The Windmill Proof
Euclid I.47
The Method of Shear
Euclid doesn't use similarity or algebra. He uses Area Preservation .
Shear squares into triangles.
Rotate triangles (SAS congruence).
Un-shear into rectangles in the large square.
Critique of Rigor
Similarity Proof
+ Extremely concise and efficient.
+ Connects measure to ratio.
- Relies on the Parallel Postulate.
- Abstract (algebraic vs. visual).
Euclidean Proof
+ Purely visual/spatial.
+ Doesn't require ratio theory.
- Highly complex construction.
- Harder to generalize quickly.
WHICH IS MORE RIGOROUS?
Does elegance define rigor, or does the avoidance of "advanced" theories like similarity make a proof more fundamental?
Team Ratio
Team Area
Pythagorean Proof Analysis Worksheet Pythagorean Proof Analysis
Lesson 04: Comparing Methodologies
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1
The Similarity Reconstruction
The most common "modern" proof of the Pythagorean Theorem utilizes the altitude-on-hypotenuse similarity.
A. Use the Geometric Mean Leg Rule to express \(a^2\) and \(b^2\) in terms of the hypotenuse segments \(x, y\) and the total hypotenuse \(c\).
B. Perform the summation and final substitution to complete the proof.
Critique: Why does this proof feel "circular" to some students? (Hint: Consider what theorems were used to prove the Leg Rule itself.)
2
The Euclidean "Windmill" Strategy
Euclid's proof (I.47) avoids similarity entirely, relying on area preservation through "shearing" and rotation.
Shearing Principle: Explain why a rectangle and a triangle with the same base and altitude relate such that \(\text{Area}(\text{Rect}) = 2 \cdot \text{Area}(\triangle)\).
Congruence Pivot: In Euclid's construction, two triangles are shown to be congruent using the SAS criterion. How does this "rotate" the area of a square into a portion of the larger square on the hypotenuse?
3
Comparative Rigor
Discussion Prompt: In undergraduate mathematics, we often prioritize "elegant" proofs. However, Euclid's proof is often considered more "foundational" because it does not require a theory of ratio and proportion (which historically were treated with great suspicion).
Which proof provides a better "explanation" for *why* the squares on the legs equal the square on the hypotenuse? Justify your choice based on mathematical intuition versus logical economy.
Circle Chord Power Slides Circle Chord Power
Similarity Beyond Polygons
Lesson 05: Final Application
Finding Similarity in Circles
When two chords intersect, they create vertically opposite angles and inscribed angles that subtend the same arc.
"Every intersection in a circle is a hidden pair of similar triangles."
A B C D P
Intersecting Chords
Proof Sketch:
By AA Similarity, \(\triangle APC \sim \triangle DPB\).
\(\frac{AP}{DP} = \frac{CP}{BP}\)
The Power Equation
\(AP \cdot BP = CP \cdot DP\)
"No matter how the chords are oriented, the product of the segments remains constant for a fixed point P."
Power of a Point (Outside)
Case 1: Secants
P
\(PA \cdot PB = PC \cdot PD\)
Case 2: Tangent-Secant
P
\(PT^2 = PA \cdot PB\)
Mastery Challenge
You are now equipped to prove the Power of a Point Theorem for any configuration. Grab the Circle Chord Mastery worksheet to synthesize polygon similarity with circle properties.
Circle Chord Mastery Worksheet Circle Chord Mastery
Lesson 05: Power of a Point Theorem
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1
The Intersecting Chord Proof
A B C D P
Proposition: Show that \(AP \cdot PB = CP \cdot PD\).
Instruction: Connect \(AC\) and \(DB\). Identify the inscribed angles that subtend the same arcs to establish triangle similarity.
2
The Secant-Secant Property
Two secants are drawn from an external point \(P\). The first secant intersects the circle at \(A\) and \(B\); the second intersects at \(C\) and \(D\).
A. Draw a diagram of this configuration. Label all points clearly.
B. Prove that \(\triangle PAD \sim \triangle PCB\). Be specific about which angles are shared or equal.
3
The Tangent Limit
As the secant \(PCD\) rotates such that \(C\) and \(D\) merge into a single point of tangency \(T\), the product \(PC \cdot PD\) becomes \(PT \cdot PT = PT^2\).
Problem: A tangent segment \(PT\) from external point \(P\) to circle \(O\) has length 12. A secant from \(P\) passes through the center and has an external part of length 8. Find the radius of the circle.
Critical Reflection: Why can the "Power of a Point" be considered a generalization of the Pythagorean Theorem? Consider a point outside a circle and the tangent segment to the radius.
Undergraduate Geometry Series | similarity-proportionality-pythagoras
