XXV OBM

If we fix the length $BD$, then we obviously maximize area $BDA$ by taking the distance of $A$ from $BD$ as large as possible and hence by taking $BD$ perpendicular to $AO$ and on the opposite side of $O$ to $A$. We maximize $BCD$ by taking $C$ the midpoint of the arc $BD$. So suppose the radius of the circle is $R$, $OA=d$ and we take $BD=2x$. We find area $ABCD=x(R+d)$, which is maximized by maximizing $x$. Thus we take $BD$ to be the diameter perpendicular to $AO$ and $C$ the point of the circle on the ray $AO$.