Inside the square $ABCD$ there is the square $A'B' C'D'$ so that the segments $AA', BB', CC'$ and $DD'$ do not intersect each other neither the sides of the smaller square (the sides of the larger and the smaller square do not need to be parallel). Prove that the sum of areas of the quadrangles $AA'B' B$ and $CC'D'D$ is equal to the sum of areas of the quadrangles $BB'C'C$ and $DD'A'A$.