Pythagorean theorem painting

As @DavidQuinn notes, the appearance of the black equilateral triangle forces the right triangle to be $30^\circ$-$60^\circ$-$90^\circ$. However, as @G.Sassatelli notes, the "orange+green+black" subdivisions aren't actually relevant to the dissection, since those elements are treated as a fixed grouping.

If we ignore the black equilateral triangle, and merge the "orange+green+black" into a single quadrilateral region, then the dissection ---which applies to arbitrary right triangles, so long as $a$ is the shorter leg--- can be understood by observing multiple appearances of the $a$-$b$-$c$ triangle throughout the figure. Labeling various edge-lengths should make this pretty clear:

As for proof of Cut-the-Knot's #72 ... I'll have to come back to that.