4. As shown in the figure, construct a right-angled triangle outward with one side of the square as the hypotenuse, then construct squares outward on the two legs of the right-angled triangle, and repeat the above operation (where $\angle 1=\angle 2=\angle 3$), resulting in four smaller squares $A, B, C, D$, with their areas denoted as $S_{A}, S_{B}, S_{C}, S_{D}$, respectively. Which of the following conclusions is correct?
A. $S_{A}+S_{D}=S_{B}+S_{C}$
B. $S_{A} \cdot S_{D}=S_{B} \cdot S_{C}$
C. $S_{A}+S_{D} \geqslant 2 S_{B}$
D. $S_{A}+S_{D}<2 S_{C}$
Given the edge length of a cube is $1, \angle 1=\angle 2=\angle 3=\alpha$, then $S_{A}=\cos ^{4} \alpha, S_{B}=\cos ^{2} \alpha \sin ^{2} \alpha$, $S_{C}=\sin ^{2} \alpha \cos ^{2} \alpha, S_{D}=\sin ^{4} \alpha$, so the correct answers are $B, C$.