nice square problem

Triangle ABE & AHG are similar

$\triangle {ABE}$ and $\triangle {ADF}$ are similar. So $\angle {DAF}=\angle {BAE}=\frac{1}{2}\times (90\degree-60\degree)=15\degree$ .

Let the length of each side of the square be $a$ . Then

$\tan 15\degree=\dfrac{\sqrt 3-1}{\sqrt 3+1}=\dfrac{a}{1-a}\implies a=\dfrac{3-\sqrt 3}{6}$ .

So $a=3,b=3,c=6$ and $a+b+c=3+3+6=\boxed {12}$ .