Proving the Pythagorean theorem geometrically | INFO 6205

A proof is a logical argument, proofs can be done without algebra.

A beautiful diagram is not yet a proof. Liam, in for Bear, reconstructs the Pythagorean theorem geometrically using Byrne-inspired graphics for INFO 6205 — Algorithms.

Follow the construction, justify the red and blue area equalities, then trace a 3–4–5 example. Congruent triangles, the twice-area rule, and partitioning do the logical work—not the colors alone. Carry the same discipline into algorithm correctness: state the precondition, justify each transformation, and distinguish a successful example from a general argument.

CHAPTERS 0:00 Introduction 0:30 Construction and the right-angle contract 1:51 Correctness: congruence and twice-area 2:58 The blue-side proof 3:49 Partition and substitution 4:24 Full worked trace: 3, 4, 5 5:16 Tests are not the theorem 5:36 Verdict 5:55 Your turn

SOURCES AND LIMITS Adapted from Euclid, Book I, Proposition 47, as presented in Oliver Byrne’s 1847 edition. This is a modern instructional reconstruction, not an exact facsimile. Vertex labels and color assignments are adapted for this film. Source edition: jemmybutton/byrne-euclid, https://github.com/jemmybutton/byrne-euclid (book text: CC BY-SA 4.0). This film adapts the construction, labels, colors, and explanation. Euclid I.47: https://mathcs.clarku.edu/~djoyce/elements/bookI/propI47.html Course: https://github.com/nikbearbrown/info-6205-algorithms 400 exact-arithmetic coordinate fixtures check the illustration’s geometry; they do not prove the theorem for every right triangle. The general argument uses congruence and Euclidean area relationships.

YOUR TURN Rebuild the construction for legs 5 and 12. Identify each congruent triangle and twice-area step. Make the angle non-right and locate the first failed justification. Record one correction in your Frictional log.

PRODUCTION Native 3840 × 2160 vector-scene render. Liam is an AI narrator using local Kokoro, not Bear’s recorded or cloned voice. Claude interface scenes are illustrative reconstructions. No burned-in captions. Independent educational production, not endorsed by Anthropic.

INFO 6205 — Algorithms playlist https://www.youtube.com/playlist?list=PLIU9y_cfCsac

What this video covers

• 0:00 — Introduction
• 0:30 — Construction and the right-angle contract
• 1:51 — Correctness: congruence and twice-area
• 2:58 — The blue-side proof
• 3:49 — Partition and substitution
• 4:24 — Full worked trace: 3, 4, 5
• 5:16 — Tests are not the theorem
• 5:36 — Verdict
• 5:55 — Your turn

Transcript

A lightly formatted transcript of the video, with timestamps so you can jump to any section on YouTube.

Introduction

[0:00] Hello, this is Liam in for Bear. For info 6205, ask Claude to explain Euclid's proposition 47 geometrically, build the construction, justify every area equality, then trace one example. We already have the buying graphics. Now, the job is to make the reasoning as visible as the colors. A convincing picture is not yet a proof. We will keep track of what each shape means, what each move preserves, and why the conclusion holds beyond the one triangle on screen. That is the same responsibility you have when explaining an algorithm.

Construction and the right-angle contract

[0:30] Part one, the construction. This is the algorithm section of our template, but the instructions build geometric objects rather than sort an array. First, state the contract. The triangle must have a right angle. Without that condition, the promised conclusion is not ours to claim. The two legs meet at the marked right angle. Blue and red identify those legs. Yellow identifies the opposite side, the hypotenuse. These are the colors of our adapted film, not a claim that every plate in Byan's book assigns the same colors. Small letter labels will keep repeated colors unambiguous. Construct an outward square on each side. A square is not just a four-sided patch. Its sides are equal, and its corners are right angles. Those

Facts will license the rotations later. Our claim concerns areas. The yellow square's area equals the blue square's area plus the red square's area. Here is the whole construction procedure. Take a right triangle. Build three outward squares. Through the right angle vertex, draw a line parallel to the outward sides of the yellow square. Extend it across that square. Then, join the auxiliary triangle vertices. Every line has a job. None is decoration. The new line partitions the yellow square into two rectangles. We tint one blue and the other red as destinations, but coloring them does not establish equality. We now owe two arguments. The red square equals the red destination in area, and the blue square equals the blue destination.

Correctness: congruence and twice-area

[1:51] Part two, correctness. Watch for two different licenses. A rigid rotation preserves a triangle's shape and area. A triangle and a parallelogram on the same base and between the same parallels have a two-to-one area relationship. The proof needs both. Rotation alone does not turn a square into a rectangle. Focus on the triangle joining the right angle corner, the red side end point, and the far yellow square corner. Rotate it a quarter turn about the red side end point. One edge lands on a side of the red square. The other lands on the hypotenuse. Equal square sides and their right angles make that landing exact, not approximate. Now compare the two larger regions to those congruent triangles. The red square is twice the area of its triangle, same base, same perpendicular height. The red destination rectangle is twice the area of the other triangle for the same reason. Equal triangle areas doubled on both sides give equal larger areas. Only now may the red destination fill. Its area has been connected to the red square by a chain of justified equalities. We did not stretch a shape until it looked right. We compared each region with twice the area of a triangle and proved those triangles congruent. Do

The blue-side proof

[2:58] The blue side explicitly. The triangle spanning a blue square side and the opposite vertex rotates a quarter turn about the upper triangle vertex. Its corresponding edges land on the blue leg and a yellow square side. Again, the square construction supplies the equal lengths and the right angle turn. The rotated triangles are congruent. The blue square is twice the area of its associated triangle. The blue destination rectangle is twice the area of the rotated triangle. Show the bases. Show the parallels. The two triangle areas are equal, so their doubles are equal. This pays the second obligation instead of hiding it behind the phrase similarly. The second destination now fills. Blue matches blue in area because of the argument we just completed, not because we reused the paint. Red has its own certificate. Each half of the yellow square has been accounted for, and neither argument depends on measuring pixels.

Partition and substitution

[3:49] Two destination rectangles cover the yellow square without a gap or an overlapping interior. Add their areas. Substitute the blue square for its equal rectangle and the red square for its equal rectangle. That establishes the claim for an arbitrary non-degenerate right triangle. The picture carries the objects. The equalities carry the proof. What if the angle is not right? Keep one endpoint at 0 -4 and move the other to 3 -1. The squared side lengths are 16, 10, and 18. 16 + 10 is not 18. The quarter turn correspondence also breaks. This is a failed precondition, not a counterexample to the theorem.

Full worked trace: 3, 4, 5

[4:24] Part three, a numerical check of the same construction. This is not a replacement proof. We choose legs of three and four units, then measure the areas by exact coordinate arithmetic. Predict the two destination areas before the numbers appear. Which region should have nine units of area and which should have 16? Red leg is three units, its square has nine unit cells. The blue leg is four units, its square has 16. Construct the yellow square by turning its side through a right angle. Its area, computed from the polygon coordinates, is 25. We are checking the drawing independently, not feeding the desired answer into a label. The red triangles each have area four and a half. Double either one and you get nine, matching the red square and its destination rectangle. The blue triangles each have area eight. Double either and you get 16. The whole destination is 9 + 16 or 25. Every numerical line corresponds to an earlier geometric reason. The Python checker

Tests are not the theorem

[5:17] Also tests 400 positive integer leg pairs with exact fractions. That can catch a wrong rotation or a misplaced dividing line. It cannot establish all right triangles by enumeration. The general argument comes from congruence, shared bases and parallels, and adding equal areas. Tests audit the implementation of our illustration.

Verdict

[5:36] Verdict construction proof example. The right angle is the contract. Congruent triangles and the twice area rule justify the substitutions. The final partition accounts for the whole square. For info 6205, carry this habit into code. Name what must be true, explain why a step preserves it, and distinguish one successful run from a correctness argument. Your turn. Rebuild this proof

Your turn

[5:57] For legs 5 and 12. Label every congruent triangle and every twice area step. Then make the angle non-right and identify the first failed justification. Separate the geometric proof from the Python checks. Keep a frictional log of one correction you made. Ask Claude to help draw and calculate, but you must be able to defend each equality. A polished animation with a missing reason is unfinished. Liam in forbear. >> Nick Bear Brown teaching Claude.