The candidate keys are A and CD, which you can determine by inspection in this case. (It's a simple problem; you can determine the closures in your head.) A->B, A->CD; that's all the attributes. CD->A, CD->B; again, that's all the attributes.
Since all the arrows are arrows out of candidate keys, it's in at least BCNF. (That's the informal definition of BCNF.)
All database textbooks include at least one algorithm for determining all the candidate keys, and for normalizing a set of relations to 5NF. If you don't have one, Maier's The Theory of Relational Databases is available free a the author's faculty web site.