Alpha and Beta

Let the intersection point of $AD$ and $BC$ be point $I$ , from intersecting chords theorem in a circle we know that $EI \cdot DI = CI \cdot FI$ , and from the fact that $\triangle ICD$ and $\triangle IBA$ are similar it follows that $\dfrac{CI}{ DI} =\dfrac{ BI}{ AI}$ , hence $\dfrac{CI}{DI} = \dfrac{EI}{FI} = \dfrac{BI} {AI}$ ; this combined with the fact that the angle between $EI$ and $BI$ is the same angle between $FI$ and $AI$ , implies that $\triangle AFI$ is similar to $\triangle BEI$ . Thus $\angle AFI = \angle BEI$ , hence $\alpha = \beta$ .

Let $AD$ and $BC$ meet at $P$ . Since $AB$ is parallel to $CD$ , $\triangle ABP$ and $\triangle CDP$ are similar. Also $CDFE$ is a cyclic quadrilateral , $\triangle EFP$ is also similar to the other two triangles. This means that $\angle EFP = \angle EAB$ and opposite angles $\angle EAB + \angle BFE = 180^\circ$ . Therefore $ABFE$ is a cyclic quadrilateral and $\boxed{\alpha = \beta}$ .

maybe a little stupid, but nevertheless... construct EF. so angles DEF=DCF and EFC=FCD because they are inside angles of circle. FCD=ABF and CDE=EAB because of parallel AB and CD. DEF=ABF and AEF=180-DEF=180-ABF, so ABEF can be inscribed. If ABFE is inscribed - AEB=AFB inscribed angles to same horde.