That is $\angle DAB + \angle DCB =\angle ADC + \angle ABC=\pi$
Is this condition sufficient for $ABCD$ to be cyclic?
If we draw a circumcircle to the $\Delta ABC$ and pick a point $X$ on the arc $AC$ such that $AXCB$ is a quadrilateral whose sides do not intersect, then by necessary condition, we immediately have $\beta = \pi-\alpha$. We also know that $\gamma =\pi -\alpha$.
Is it necessary for $D$ to lie on the common circle with $X$ in order for $\beta = \gamma$?
So, assume for a contradiction $D$ does not lie on the circumcircle, it's either outside or inside.
How does one make progress? I read everywhere that these conditions are equivalent, but haven't found a proof to the converse