Doubt on 2NF decomposition

I have read that 1NF, 2NF and 3NF decompositions are lossless and dependency-preserving. Consider this example on a relation **R(A,B,C,D)** with functional dependencies set as **FD ={ AB->CD, A->C, BC->D}** Here when we do 2NF decomposition we get **R1(A,C)** with **FD ={A->C}** and **R2(A,B,D)** with **FD ={AB->D}** The functional dependency **BC->D** is lost when we join but we know that 2NF is dependency preserving so why is it that we are unable to preserve the original FD?