One way to think about this proof is that there are two ways to break up an [math]\left(a+b\right)\times\left(a+b\right)[/math] square. One way to do it is to put an [math]a\times a[/math] square at the top left and a [math]\left(b\times b\right)[/math] square at the bottom right. That leaves two [math]a\times b[/math] rectangles at the bottom left and top right. Those can be divided along the diagonals into four right triangles with legs of [math]a[/math] and [math]b[/math] a hypotenuse of length [math]c[/math] (which length we are trying to discover).[br][br]Sliding the "twirl" slider, the four right triangles move along the edges of the large square, leaving all the hypotenuses of length [math]c[/math] forming a smaller square inside the large square. Since both arrangements contain those same four triangles, what remains inside the large square must have the same area, whether [math]a^2+b^2[/math] in the first square or [math]c^2[/math] in the second square. Thus [math]a^2+b^2=c^2[/math].[br]