The Chinese proof is by dissection. Inside square ABCD (figure 5.2) inscribe a smaller, tilted squaren KLMN, as shownin (a). This leaves four congruent right triangles (shaded in the figure). By reassembling these triangles as in (b), we see that the remaining (unahaded) area is the sum of the areas of squares 1 and 2, that is, the squares built on the sides of each of the right triangles.