First, denote the intersection of $\overline{AC}$ with $\overline{OD}$ as $X$. Then, note that proving the statement is equivalent to showing that $\angle{AXO}=90$, since $AX$ is the angle bisector. Now, let $\angle{CAO}=\theta$, and since $AO=CO$, $\angle{OCA}=\theta$. Also, because $\angle{AOC}=180-2\theta$, reflex angle $\angle{AOC}=180+2\theta \rightarrow \angle{ABC}=90+\theta$. Now since $BDCO$ is cyclic, then $\angle{DOC}=90-\theta$, and since $\angle{ACO}=\theta$, we see that $\angle{CXO}=90=\angle{AXO}$, as desired