OSN 2013 Problem 7

Let $ABCD$ be a parallelogram. Construct squares $ABC_1D_1, BCD_2A_2, CDA_3B_3, DAB_4C_4$ on the outer side of the parallelogram. Construct a square having $B_4D_1$ as one of its sides and it is on the outer side of $AB_4D_1$ and call its center $O_A$ . Similarly do it for $C_1A_2, D_2B_3, A_3C_4$ to obtain $O_B, O_C, O_D$ . Prove that $AO_A = BO_B = CO_C = DO_D$ .