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Below is a “guided tour” through three of the cleanest classical proofs of the Pythagorean Theorem. Reading them side-by-side helps illuminate why the result is so fundamental: it can be reached by pure geometry (similarity), by area computations (dissection), or by algebraic/coordinate reasoning.

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0. Statement

For every right triangle with legs of lengths $a,b$ and hypotenuse $c$,

$a^{2}+b^{2}=c^{2}.$

We assume right triangle means $\angle C = 90^{\circ}$ opposite side $c$.

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1. Proof by Similar Triangles (Euclid I.47)

Idea: Drop an altitude from the right‐angle vertex to the hypotenuse. Similarity splits the original triangle into two smaller right triangles, each similar to the whole as well as to each other. The similarity ratios translate into the desired square-of-length relations.

1. Let $\triangle ABC$ be right with $\angle C=90^{\circ}$. Draw altitude $CD$ to hypotenuse $\overline{AB}$, meeting at $D$.

2. Triangles $\triangle ACD$ and $\triangle ABC$ are similar (both right, and share $\angle A$).
   Likewise, $\triangle BCD$ is similar to $\triangle ABC$.

3. Using similarity:

   • From $\triangle ACD \sim \triangle ABC$: $\frac{CD}{BC} = \frac{AC}{AB} \;\;\Longrightarrow\;\; CD\cdot AB = AC\cdot BC = ab.$

   • From $\triangle BCD \sim \triangle ABC$: $\frac{CD}{AC} = \frac{BC}{AB} \;\;\Longrightarrow\;\; CD\cdot AB = BC\cdot AC = ab.$

   Those equalities give
   $AD = \frac{AC^{2}}{AB} = \frac{a^{2}}{c}, \qquad BD = \frac{BC^{2}}{AB} = \frac{b^{2}}{c}.$

4. Now $AB = AD + DB$ implies
   $c = \frac{a^{2}}{c} + \frac{b^{2}}{c} \quad\Longrightarrow\quad a^{2}+b^{2}=c^{2}.$

Key take-away: The theorem drops out of the proportionality of similar triangles; no area arguments are necessary.

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2. Proof by Area Dissection (Preserved-Shape Rearrangement)

Idea: Build squares on each side of the right triangle, then cut the two smaller squares into pieces that reassemble to cover the large square.

1. Draw right triangle $ABC$ with legs $a,b$ and hypotenuse $c$.

2. Construct squares externally on each side.

3. One elegant dissection, often attributed to the Bhāskara proof, is:

   • Copy the original triangle four times.
   • Arrange them inside the $c \times c$ square so their hypotenuses form a smaller, central tilted square.
   • Two possible arrangements exist: one leaves a central square of side $(b-a)$, the other fills the big square completely. Comparing the leftover areas shows that the sum of the areas of the two $a$- and $b$-squares equals that of the $c$-square.

Formally:

Total area of $4$ triangles: $4\cdot\frac12 ab = 2ab$.
Area of large square: $c^{2}$.
Area of leftover central square: $(b-a)^{2}$.

So
$c^{2} = 2ab + (b-a)^{2} = 2ab + (a^{2}+b^{2}-2ab) = a^{2}+b^{2}.$

The beauty here is purely area preservation: cut, slide, re-attach—no similarity needed.

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3. Proof in Coordinate Geometry (Algebra)

Idea: Place the triangle on a Cartesian grid and compute distances using the distance formula, which itself embodies Pythagoras; however, we’ll exploit only the algebra of dot products.

1. Set $C=(0,0)$, $A=(a,0)$, $B=(0,b)$.

2. Vector form of the hypotenuse: $\overrightarrow{AB}=(-a,b)$.

3. Its squared length (dot product with itself):

   $c^{2}=(-a,b)\cdot(-a,b)=(-a)(-a)+b^{2}=a^{2}+b^{2}.$

Here the proof is a single line—the theorem is encoded inside the Euclidean norm. Conceptually, it says the square of the length equals the sum of squared coordinates, which is the algebraic counterpart of orthogonality.

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4. Why So Many Proofs Exist

1. Orthogonality is central in every approach—whether via similar‐triangle angle copying, right‐angle dissections, or perpendicular coordinate axes.

2. Squares of lengths behave additively under perpendicular composition; that principle manifests in geometry (areas), algebra (dot products), and even physics (energy components).

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5. How to Choose a Proof in Practice

• Teaching geometry basics? Use Proof 1 (similarity) because it reinforces angle and proportion concepts.
• Want a visual “aha”? Proof 2’s dissection is ideal; it needs minimal symbolic manipulation.
• Already in an analytic context (vectors, physics, linear algebra)? Proof 3 is fastest and sets up generalizations ($n$-dimensional Pythagoras, inner-product spaces).

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6. Generalizations You Can Explore Next

1. Law of Cosines ($a^{2}+b^{2}-2ab\cos\gamma=c^{2}$).
2. Inner-product spaces: $‖u+v‖^{2}=‖u‖^{2}+‖v‖^{2}$ iff $u\perp v$.
3. Differential geometry: Pythagoras corresponds to the metric tensor being diagonal in an orthonormal basis.

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Each proof above is conceptually self-contained—choose the one that best matches the tools you have at hand, and you’ll have a clear route to $a^{2}+b^{2}=c^{2}$.