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**Generating a Pythagorean theorem proof** The user wants me to generate and explain a proof of the Pythagorean theorem. I need to create a clear proof, possibly using methods like rearrangement or similar triangles. It’s important to provide a step-by-step explanation to ensure the user understands. I’ll make sure to format any mathematical parts using KaTeX to enhance clarity. Let’s craft a well-structured response that answers their question thoroughly!

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.


Statement of the theorem

For any right triangle with legs of lengths aa and bb and hypotenuse of length cc,

a2+b2=c2.a^2 + b^2 = c^2.


Construction

  1. Draw a right triangle ABC\triangle ABC with right angle at CC.
    AC=bAC = b
    BC=aBC = a
    AB=cAB = c

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

  3. Arrange the four copies inside a large square of side length a+ba + 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.


Areas

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

  2. Area accounted for by the four triangles
    Each triangle has area 12ab\dfrac12 ab, so four of them have total area
    Area4 tri=4(12ab)=2ab.\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
    Areainner=(a+b)22ab.\text{Area}_{\text{inner}} = (a + b)^2 - 2ab.

  4. But the inner square’s side length is cc
    Because each side of the inner square is exactly the hypotenuse of one triangle,
    Areainner=c2.\text{Area}_{\text{inner}} = c^2.


Equating the two expressions for the inner square

Set the two formulas obtained for the inner square equal:

c2=(a+b)22ab.c^2 = (a + b)^2 - 2ab.

Expand the right–hand side:

(a+b)22ab=a2+2ab+b22ab=a2+b2.(a + b)^2 - 2ab = a^2 + 2ab + b^2 - 2ab = a^2 + b^2.

Hence,

c2=a2+b2.c^2 = a^2 + b^2.

This completes the proof.


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 a2+b2=c2a^2 + b^2 = c^2.


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.

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