Decomposing a relation into 3NF/BCNF
No, your decomposition is noteven in 2nd normal form.


{A -> B,C}
{E -> F}
{D -> I}
{A,D -> G}
{G -> H}


A, D and E cannot be found on the right side of a dependency, so they are members of every key. {A, D, E}  generate all attributes, so it is a superkey and therefore the only minimal key
But your decomposition is wrong AA is not a key of R1A, because G does not depend on A. So the key of R1A is {A,G}. So R1A is not in 2nd Normal Form because B depends on {A} which is a proper subset of the candidate key.
So R should be decomposed to {A,B,C}, {E,F}, {D,I}, {A,D,G}, {G,H}. These relations are in 2nd, 3rd and even in Boyce-Codd normal form.