
Kindly consider this rough figure and assume that $\angle A=\theta$ with BF=BC=DF and AB=BD.
Since AB=BD, then $\angle ADB=\theta$ ( isosceles triangle)
and $\angle C=\pi-\theta$ (Parallelogram's opposite vertices have supplementary angles).
Now since BF=BC,
$\angle C=\theta=\angle BFC$ (isosceles triangles again) making $\angle FBC=2\theta-\pi$
Now, $\angle DFB=\theta$ (Linear angle with $\angle CFB$)
also BF=DF making $\angle FDB=\frac{\pi-\theta}{2}=\angle FBD$
Now, $\angle ABD=\angle FBD$
$$\frac{\pi-\theta}{2}=\pi-2\theta$$
$$\theta=\frac{\pi}{3}$$
Thus $\angle ABC=\frac{\pi-\theta}{2}=\frac{\pi}{3}=60^o$