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These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof.

This same configuration could be observed in a proof by tessellation.) Now we start with four copies of the same triangle.

I include it on a separate page with Jim's kind permission.

The proof below is a somewhat shortened version of the original Euclidean proof as it appears in Sir Thomas Heath's translation. This is because, and ∠BAF = ∠BAC ∠CAF = ∠CAB ∠BAE = ∠CAE.

If, in a triangle, angles α, β, γ lie opposite the sides of length a, b, c, then (EWD) sign(α β - γ) = sign(a² b² - c²), where sign(t) is the The most famous of right-angled triangles, the one with dimensions 3:4:5, has been sighted in Gothic Art and can be obtained by paper folding.

Obviously the resulting shape is a square with the side c and area c². (A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. The proof itself starts with noting the presence of four equal right triangles surrounding a strangely looking shape as in the current proof #2.Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB. At this point we therefore have two triangles and a strange looking shape.The configuration at hand admits numerous variations. As a last step, we rotate the triangles 90°, each around its top vertex.There is a small collection of rather elementray facts whose proof may be based on the Pythagorean Theorem.There is a more recent page with a list of properties of the Euclidian diagram for I.47.

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Curiously, nowhere in the book does Loomis mention Euclid's VI.31 even when offering it and the variants as algebraic proofs 1 and 93 or as geometric proof 230. I'll give an example of their approach in proof #56.

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