R(ABCD)
AB-->C C-AD
we can see candidate key here is{AB,BC}
For BCNF the FD Set:
X-->Y
where x should be super key
but we can see that C-->AD failed to satisfy bcnf.
Decompose into
R1(ABC) R2(CD) R3(AC)
AB-->C. C-->D. C-->A
Key{AB} Key{C} Key{C}
Minimum 3 table required ,2 Foreign key
required.
BCNF not satisfied ,dp decomposition satisfied.
If you want to satisfy BCNF
You will definitely loose dp decomposition.