checking the normal form of the current table
You are correct, the relation is in 2NF, 3NF, BCNF.

The reason is that the relation has two keys, A and CD.

So the relation is in BCNF (which is a property stronger than 3NF and 2NF) since each determinant of the minimal conver of R1 is a key. Here is one minimal cover:

A → C
A → D
C D → B
C D → A