Let the centres of the squares be $C_1$ and $C_2$. Clearly if $C_1 = C_2$ then we are done. Consider fixing the square with center $C_1$ and consider the translation that maps $C_2$ to $C_1$. It is not too hard to see by similar triangles that this translation does not change the area of a shaded quadrilateral [details omitted but similar triangles pretty much], and so we are done.