Original problem

We have $BA=BC$ , $BA=AD$ . Then $\triangle ABC$ , $\triangle ABD$ are Equilateral triangles. So $\angle ABD$ = $\angle ADB$ and $\angle BAC$ = $\angle BCA$ And we have $\triangle BAF$ , $\triangle AFC$ , $\triangle BFD$ are Right-angle triangles in $\angle F$ Then we have $\angle FAC$ + $\angle ACB$ =90, $\angle BAF$ + $\angle ABC$ =90, $\angle FBD$ + $\angle BDA$ =90... adding them together to have $\angle FAC+\angle ACB+ \angle BAF+\angle ABC+ \angle FBD+\angle BDA=90+90+90$ And we know that $\angle ABD=\angle ABC +\angle FBD$ and $\angle BAC=\angle FAC+\angle BAF$ So $2\angle BAC+2\angle BDA=270$ Because $\angle ABD=\angle ADB, \angle BAC=\angle BCA$ And finally we get $\angle BAC+\angle BDA = \boxed{135}$

In the special case of equilateral triangle ABC, the sum would be $60 + \dfrac{1}{2}(180-30)=135$

Proving this in the general case is another matter.