We know that the inscribed angle is half of the central angle. But, I believe that there are other points inside the circle (the point(s) B in the figure below other than the origin of the circle), such that: $\beta= \frac{1}{2}\alpha$.
I tried using geogebra to change the position of point B (without changing the points A, C, and D) in such a way the relation above holds but I cannot prove it. Any help is appreciated.
Denote $O$ to be the center, and draw a circle passing through $C,D,O$ (which is always possible given there are only 3 points). Then, if you "slide" $O$ along the arc, you will still get $\angle CBD=\alpha$.
(As in the diagram, $\angle CBD=\angle COD$):