As before, simplifications yield a In a private correspondence, Dr.
The construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c.But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC. Playing with the applet that demonstrates the Euclid's proof (#7), I have discovered another one which, although ugly, serves the purpose nonetheless.Thus starting with the triangle 1 we add three more in the way suggested in proof #7: similar and similarly described triangles 2, 3, and 4.This same configuration could be observed in a proof by tesselation.) The fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a b) and a hole with the side c. Garfield in 1876 [Pappas], is a variation on the previous one. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a b)/2·(a b).We can compute the area of the big square in two ways. Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 ab/2 c·c/2.