smc

We will prove that $△DOB$ and $△COD$ is isosceles, meaning that $CD=OD=BD$ and hence $\boxed{D}$. Let $\angle A=2\alpha$ and $\angle B=2\beta$. Since the incentre of a triangle is the intersection of its angle bisectors, $\angle OAB=\alpha$ and $\angle ABO=\beta$. Hence $\angle DOB=\alpha +\beta$. Since quadrilateral $ABCD$ is cyclic, $\angle CAD=\alpha =\angle CBD$. So $\angle OBD=\angle OBC+\angle CBD=\alpha +\beta =\angle DOB$. This means that $△DOB$ is isosceles, and hence $BD=OD$. Now let $\angle C=2\gamma$ which means $\angle ACO=COD=\gamma$. Since $ABCD$ is cyclic, $\angle DAB=\alpha =\angle DCB.$ Also, $\angle DAC\alpha$ so $\angle DOC=\alpha +\gamma$. Thus, $\angle OCD=\angle OCB+\angle BCD=\gamma +\alpha =\angle DOC$ which means $△COD$ is isosceles, and hence $CD=OD=BD$. Thus our answer is $\boxed{CD=OD=BD}$