Small adjustment: Similar figures in the plane have areas proportional to the *squares* of the lengths of the corresponding sides. But: Is it safe to say that the theorem that a right triangle is split into two similar ones by the altitude to the hypotenuse does not rely on the Pythagorean theorem for its proof? —Dan
On Feb 14, 2016, at 8:17 PM, Andy Latto <andy.latto@pobox.com> wrote:
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.