Can the notion of Pythagorean cuts be extended to, say, regular *n*-gons? Consider, for example, the regular hexagons built on the sides of right triangle *ABC*, as shown in Figure 10, below.

**Figure 10: **Regular hexagons on the sides of a right triangle. Line segment *LJ,* shown in red, is one of six line segments forming a sort of generalized Pythagorean cut.

Can the hexagon on *AC* be divided into two parts each having area equal to that of one of the hexagons on *AB* and *BC*? Knowing how to divide triangles in this way gives us a way to divide the hexagon on *AC* into two sets of six pieces each, so that the areas of the pieces in each set add up to the area of the hexagon on *AB* or *BC*. If *BL* is perpendicular to *AC*, we know that *LJ* is a Pythagorean cut for equilateral triangle *AJC*, so that the area of triangle *ALJ* equals the area of triangle *AEB*. Marking off a segment equal to *AL* on each side of the hexagon with center *J* ensures that the sum of the areas of the triangles shaded in yellow in hexagon *J* will equal the area of the hexagon *E.* Consequently, since the area of triangle *LCJ* equals the area of triangle *BCG*, the sum of the areas of the triangles shaded in blue in hexagon *J* must equal the area of hexagon *G*. Thus,

area of hexagon *J* = area of hexagon *E* + area of hexagon *G*.

In this way, the Pythagorean cut can be replicated to ensure that the *n*-gon on the hypotenuse is subdivided into two collections, each consisting of *n* congruent triangles, such that the sum of the areas of the triangles in each collection equals the area of one of the two smaller *n*-gons. Please note that although the basic triangle for the hexagon is an equilateral triangle, the Pythagorean cut described above will hold for any isosceles triangle and thus for any regular *n-*gon.