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Below is a “walk-through” of three of the most celebrated proofs of the Pythagorean Theorem, written so that you can see not only the finished argument but also the thought-process that produces it. (If you only need one proof, feel free to stop after Method 1.)

Notation: We work in a right triangle ΔABC with right angle at C, legs a = BC and b = AC, and hypotenuse c = AB. The goal is to show a² + b² = c².

──────────────────────────────── METHOD 1. Euclid’s original proof (Elements, I.47) ────────────────────────────────

0. Big idea: Compare areas of squares built on each side of the right triangle by cutting and rearranging them into equal shapes.

1. Construct squares externally on all three sides:
   • Square on AB has area c².
   • Square on AC has area b².
   • Square on BC has area a².

2. Draw the two lines through C that are respectively parallel to AB and to the sides of the square on AB; they subdivide the big square (on AB) into two rectangles.

3. Show that each of those rectangles has exactly the same area as one of the smaller squares (on AC or BC). The key geometric fact is that triangles you get in the construction are congruent to pieces of the original right triangle, so the rectangles can be dissected and slid onto the smaller squares.

4. Therefore, area(c²-square) = area( b²-square) + area( a²-square ), proving a² + b² = c².

Why it works: Euclid realized that a right angle makes each leg parallel to a side of the square on the hypotenuse, so dissection lines fall into place “for free.”

──────────────────────────────── METHOD 2. Similar-triangle proof (a favorite in high-school texts) ────────────────────────────────

0. Big idea: Drop an altitude from the right-angle vertex to the hypotenuse and watch how similar triangles encode the necessary proportions.

1. Let CD ⟂ AB meet AB at D. Call AD = m and DB = n, so m+n = c.

2. Observe the three triangles: • ΔABC (original)
   • ΔACD (shares ∠A with ABC)
   • ΔBCD (shares ∠B with ABC)

Each is similar, because they have two angles in common (AA similarity).

3. Translate similarity into proportions:

For ΔACD ~ ΔABC:
AC / AB = CD / AC ⇒ b / c = h / b ⇒ b² = c·h.

For ΔBCD ~ ΔABC:
BC / AB = CD / BC ⇒ a / c = h / a ⇒ a² = c·h.

(Here h = CD.)

4. Add the two equalities: a² + b² = c·h + c·h = 2c·h.

5. Note that h = CD is the geometric mean of m and n (from similarity, h² = mn), but you don’t even need that: just rewrite

a² + b² = c·(m + n) because m+n = c, so

a² + b² = c · c = c².

Why it works: Similarity compresses “shape” information, so lengths along corresponding sides must be in the same ratio; the altitude splits the big triangle into two mini-copies whose legs become the geometric means we need.

──────────────────────────────── METHOD 3. Rearrangement / “Chinese” proof (often shown as a moving-tiles animation) ────────────────────────────────

0. Big idea: Place four congruent right triangles inside a square in two different patterns and compare the uncovered area.

1. Build a big square of side length a+b. Place four copies of ΔABC inside so that their right angles meet at the center (hypotenuses form a small tilted square). Empty area = c².

2. Same big square. Place the four triangles with legs flush to the outer edges (right angles at the corners). Empty area = a² + b² (the two little corner squares).

3. Since the coverings only differ by rearrangement of identical triangles, the vacant areas must be equal: a² + b² = c².

Why it works: Congruent shapes tiled inside the same bounding region leave identical leftover area, so you translate a geometry statement about equality of areas into the algebraic equation we want.

──────────────────────────────── Meta-thinking: How do you “come up” with these? ────────────────────────────────

1. Symmetry & dissection: If the statement to be proved is purely about lengths squared (areas of squares), try to force a literal square-area comparison (Euclid, Chinese).

2. Similar shapes: Many right-triangle results use the dropped altitude because it replicates the right angle and spawns similar triangles, delivering nice proportional equations.

3. Coordinate or vector methods: Put the triangle on a Cartesian grid (say, C = (0,0), A = (a,0), B = (0,b)). Then AB = √(a² + b²), square both sides, and you immediately get the theorem. This is “cheating” in purely Euclidean terms because you smuggle the distance formula in, but it shows how analytic geometry packages the Pythagorean Theorem right into its metric.

4. Differential/inner-product viewpoint: In any inner-product space, “orthogonality” and the norm satisfy ⟨v,v⟩ = ∑(coordinates)², so the theorem generalizes to much higher dimensions. But historically, people noticed the planar case first.

Take-away: No single proof is canonical; the theorem is so central that over 300 proofs exist. Each approach illuminates a different aspect—area comparisons, similarity, algebraic rearrangement, analytic geometry, or even differential geometry. Choose the one that best matches the tools you’re comfortable with (or the tools you wish to showcase).