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Below is a classic, purely geometric proof of the Pythagorean Theorem (often called the “re-arrangement” or “Bhāskara” proof), followed by an explanation of each step.

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Statement of the theorem

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

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

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Construction

1. Draw a right triangle $\triangle ABC$ with right angle at $C$.
   • $AC = b$
   • $BC = a$
   • $AB = c$

2. Make four identical copies of $\triangle ABC$.

3. Arrange the four copies inside a large square of side length $a + b$ in the following way:
   • Place one triangle in each corner so that the right angles touch the corners of the large square.
   • Orient each triangle the same way (legs along the square’s edges, hypotenuse facing inward).

   After the placement, the four hypotenuses form a smaller, tilted square in the center.

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Areas

1. Area of the big square
   Each side is $a + b$, so
   $\text{Area}_{\text{big}} = (a + b)^2.$

2. Area accounted for by the four triangles
   Each triangle has area $\dfrac12 ab$, so four of them have total area
   $\text{Area}_{\text{4 tri}} = 4 \left(\dfrac12 ab\right) = 2ab.$

3. Area of the inner (tilted) square
   What remains after removing the triangles from the big square is the inner square. Its area is therefore
   $\text{Area}_{\text{inner}} = (a + b)^2 - 2ab.$

4. But the inner square’s side length is $c$
   Because each side of the inner square is exactly the hypotenuse of one triangle,
   $\text{Area}_{\text{inner}} = c^2.$

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Equating the two expressions for the inner square

Set the two formulas obtained for the inner square equal:

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

Expand the right–hand side:

$(a + b)^2 - 2ab = a^2 + 2ab + b^2 - 2ab = a^2 + b^2.$

Hence,

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

This completes the proof.

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Why the argument works

• Each “piece” (big square, four triangles, inner square) is counted exactly once; no overlaps or gaps exist because the triangles’ right angles exactly fill the corners.
• The hypotenuse of every triangle is straight, so the four together form a true square centrally.
• Equating the two descriptions of the same inner region’s area forces the algebraic relationship $a^2 + b^2 = c^2$.

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Historical note

A version of this rearrangement proof appears in Chinese texts (the Zhoubi Suanjing) and in Bhāskara II’s 12-century work in India. Bhāskara famously wrote a single word next to the diagram: “Behold!” — inviting the reader to see the result almost at a glance.