The analysis is as follows:

Let the coordinates of point F be (-1 0), then point A is (-2,0), point C is (0,3), and point B is (0,3).

The slope of line AB is k1= ∠ 3/2 ∠ BAC = Arctan (∠ 3/2).

The slope of FC line is k2 =-√ 3 ∠ DFA = 60.

∠BDC=∠BAC+∠DFA

tan∠BDC=tan(∠BAC+∠DFA)=-3√3

So ∠BDC=-arctg3√3.