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Current time: July 23, 2019, 3:41 pm

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Proof by Rearrangement of the Pythagorean Theorem
#11
RE: Proof by Rearrangement of the Pythagorean Theorem
Expanding on what Polymath said...

From earlier theorems (if we're talking Euclidean geometry, that is, the Pythagorean theorem doesn't hold in non-euclidean geometry), we know the sum of the angles of a triangle to be 180 degrees. I'll name those three angles A, B, and C, which are opposite of the sides of lengths a, b, and c respectively Since we design the triangles in the proof by rearrangement to be congruent right triangles, we know that C is 90 degrees, and so A and B (WHATEVER they are, it will vary with the values of a and b) have to add up to 90 degrees. Then we rearrange the triangles so that each leg of length a is collinear and flush against the next leg b. Because they are collinear, the three angles there must total 180 degrees. Since we know that A and B together total 90 degrees, that means the unknown interior angle must be 90 degrees. Either appeal to similarity or just repeat the process four times, and ALL of the unknown interior angles are 90 degrees. Since all four sides of the interior figure are the same length, c, and all the angles are 90 degrees, we do in fact have a square.
Being an antipistevist is like being an antipastovist, only with epistemic responsibility instead of bruschetta.

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