Did I ever post my favorite proof of the Pythagorean Theorem to math-fun? It deserves to be better known. Start by observing that there's nothing special about squares; similar figures have areas proportional to the length of a corresponding side. So the Pythagorean theorem, about the areas of 3 squares with sides the lengths of the sides of a right triangle, is equivalent to theorem where instead of squares, we put equilateral triangles, or regular pentagons, or semicircles, on each side of a right triangle; if the sum of the areas of the two smaller figures is equal to the area of the figure on the hypotenuse for any figure, it's true for all sets of three similar figures. So the proof, once you accept the lemma above, is simple enough that I don't even have to bother to draw the diagram, I'll just describe it.
From the right angle of the triangle, draw an altitude, perpendicular to the hypotenuse. This divides the triangle into two smaller ones, and all 3 are similar.
QED. On Sun, Feb 14, 2016 at 3:32 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-02-13 15:11, Gareth McCaughan wrote:
On 13/02/2016 18:38, Bill Gosper wrote:
gosper.org/Perigal.gif --rwg
Cute picture, but is proving that it actually proves Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg
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