The proof of the 2D case depends on the equidecomposability of two triangles of equal area, which can be split into two parts. The first part is a proof that two triangles with equal bases and equal altitudes are equivalent by dissection. This was proved in the Demonstration Two Equidecomposable Triangles.

It remains to prove that any two triangles of equal area are equidecomposable. So suppose that the two triangles and have the same area but unequal bases, and that . The aim is to construct a triangle , again of the same area , such that , so that the previous case can be applied.

Start with a triangle with the same area . (For example, if is at height from , choose arbitrarily and take at height .) Suppose . Draw a parallel to through . Let be the intersection of and the circle of radius with center . All three triangles , and have the same area. The triangles and have the same base and equal altitudes, so they are equidecomposable, say using a set of polygons . The triangles and have equal bases and , so they are equidecomposable, say using a set of polygons . Then and are equidecomposable using the intersections of and in .