Vertices and angles

Points $A,B,C,D$ and $F$ all lie on a circle with center $E$ . As $\angle FAB$ and $\angle FEB$ are subtended by the same chord $FB$ we then know from our knowledge of inscribed angles that

$\angle FAB = \dfrac{1}{2} \angle FEB = \dfrac{1}{2} \times 60^{\circ} = \boxed{30^{\circ}}$ .

Comment: Let $M$ be the point of intersection of $AB$ and $EF$ . As $\angle EBM = 45^{\circ}$ we have that $\angle AMF = \angle EMB = 75^{\circ}$ , and so as $\angle FAB = 30^{\circ}$ we have that $\angle AMF = 75^{\circ}$ as well. Thus $\Delta AMF$ is isosceles, which I thought was an outcome worth noting.

As we should know, any square can be inscribed inside a circle, in this case, of center in $E$ . Here $\overline{EF} \cong \overline{ED}$ are radii of the circle and, therefore $\angle BAF$ is inscribed in the circle, enclosing the same arc as $\angle BEF$ , then: $\angle BEF = 2\cdot \angle BAF \Rightarrow \angle BAF = 30^{\circ}$

the image explains all ;

Extend BF and EA, so that these two lines meet at a point G.FE=BE=AE, so we have isosceles triangle AFE. Angle EAF= 75, so angle FAB=30.