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Below is a “meta-walkthrough” of how one might think, step-by-step, when searching for (and finally writing down) a proof of the Pythagorean theorem. I’ll deliberately expose the mental moves rather than jump straight to a finished argument, so you can see the kind of questions and observations that naturally lead to several classical proofs. I conclude each line of thought with a concise, formal proof that crystallizes the discoveries.

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1. Start from the statement and translate it into geometric/combinatorial language.

• Goal: For a right triangle with legs $a,b$ and hypotenuse $c$, show $a^2 + b^2 = c^2$.

• Observation: “Squares on the sides” reminds us that the theorem can be rephrased in terms of areas of squares. So we may try to turn a relationship about lengths into one about areas.

2. Ask: How can we make squares, or areas, appear in a diagram that contains only a single right triangle to start with?

• Squares attached to each side is one idea.

• Duplicating or rearranging copies of the triangle is another—maybe we can tile something whose area is easy to compute in two different ways.

• Using similarity might let us equate ratios involving squared lengths.

3. Branch into three promising strategies:

A. Rearrangement / dissection arguments
B. Similar-triangle arguments (Euclid I.47 style)
C. Coordinate/analytic approach (drop the triangle onto a grid)

Below I follow each branch from “search” to “finished proof.”

A. Rearrangement strategy (think “Lego bricks”)

a) If I take 4 copies of the right triangle and arrange them into a larger square, do I know the side length of that square? (Classic puzzle vision.)

b) For a right triangle, a neat arrangement is to put the right angles in the corners of a square. The outer perimeter then forms a square whose side length is $a+b$. That square’s area is $(a+b)^2$.

c) The four triangles leave a central gap. What shape is the gap? It’s another square, and its side equals $c$.

d) Therefore area of large square = 4 × (area of triangle) + area of inner square. Solve for $c^2$.

Formal proof A (dissection):

• Place 4 congruent copies of the right triangle so their right angles meet at the corners of a large square of side $a+b$. (Picture a pinwheel.)

• The interior gap is a square of side $c$ (adjacent legs $a,b$ force the hypotenuse edges to form a square).

• Area comparison: $(a+b)^2 = 4(\frac12 ab) + c^2$. Simplify:

 $$a^2 + 2ab + b^2 = 2ab + c^2 \quad\Longrightarrow\quad a^2 + b^2 = c^2.$$

QED.

B. Similar-triangle strategy (Euclid style)

a) Highlight the altitude from the right angle to the hypotenuse—it splits the original triangle into two smaller right triangles. Each is similar to the original (because they share acute angles).

b) Therefore ratios of corresponding sides yield equalities involving squares (products of legs and projections).

c) Specifically:
$a^2 = c \cdot m$ and $b^2 = c \cdot n$, where $m,n$ are the projections of $a,b$ on $c$ and $m+n=c$.

d) Adding gives $a^2 + b^2 = c(m+n) = c^2$.

Formal proof B (Euclid I.47 modernized):

• Draw altitude from right-angle vertex $P$ to hypotenuse $\overline{AB}$; let foot be $D$. Let $AP = b$, $PB = a$, $AB = c$, $AD = n$, $DB = m$.

• $\triangle PAD \sim \triangle PBA$ (AA)
⇒ $\frac{b}{c} = \frac{n}{b}$ ⇒ $b^2 = cn$.

• $\triangle PBD \sim \triangle PAB$ (AA)
⇒ $\frac{a}{c} = \frac{m}{a}$ ⇒ $a^2 = cm$.

• Add: $a^2 + b^2 = c(m+n) = c^2.$

QED.

C. Coordinate geometry strategy (algebra meets geometry)

a) Place the right triangle with its right angle at the origin, legs along the axes: $(0,0)$ to $(a,0)$ and $(0,b)$.

b) The hypotenuse is then the segment joining $(a,0)$ and $(0,b)$.

c) Use the distance formula to compute its length.

Formal proof C (analytic):

• Hypotenuse length:
$c = \sqrt{(a-0)^2 + (0-b)^2} = \sqrt{a^2 + b^2}.$

• Square both sides, obtain $c^2 = a^2 + b^2.$

QED.

What “thinking very hard” meant in practice:

1. Translate length relation → area or algebra.
2. Look for self-similar pieces you can copy or rearrange.
3. Keep an eye out for right angles (they often create similar triangles).
4. If stuck in pure geometry, impose coordinates; if stuck in algebra, hunt for an area interpretation.
5. After you find a rough idea, strip all unnecessary constructions, rename points sensibly, and write the terse formal argument.

Any of these three proofs can be presented in under ten lines, yet each came from a slightly different mode of thought: combinatorial tiling, similarity/ratios, or analytic geometry. Practicing these mental moves on simpler lemmas builds the instinct that eventually produces “aha!” moments for the Pythagorean theorem and many other results.