A cyclic quadrilateral is a quadrilateral whose vertices all lie on a circle. So they are necessarily equidistant from the point $O$, aren't they?
And you have two arcs: $CD=120$, and $BC=60$. So, $BD=180$.
• Of course. Look, in your example you have vertices $B$ and $D$ on the diameter. Then you have $C$ on one third of the arc beginning at $B$ ($BC=60$, $CD=120$). And now you can put $A$ at any place on the other (with respect to $C$) half of circumference. Nothing about arcs will depend on it but the diagonals will. – sas Feb 17 '14 at 7:25