[asy] draw(circle((0,0),10),black+linewidth(1)); draw(circle((-1.25,2.5),4.5),black+linewidth(1)); dot((0,0)); dot((-1.25,2.5)); draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5)); draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5)); draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5)); draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5)); MP("O'", (0,0), W); MP("O", (-2,2), W); MP("A", (-10,-2), W); MP("B", (10,-2), E); MP("C", (-2,sqrt(96)), N); MP("D", (sqrt(96)-2.5,7), NE); [/asy]
Triangle $ABC$ is inscribed in a circle with center $O'$. A circle with center $O$ is inscribed in triangle $ABC$. $AO$ is drawn, and extended to intersect the larger circle in $D$. Then we must have:
$\text{(A) } CD=BD=O'D \quad \text{(B) } AO=CO=OD \quad \text{(C) } CD=CO=BD \\ \text{(D) } CD=OD=BD \quad \text{(E) } O'B=O'C=OD$
We will prove that $\triangle DOB$ and $\triangle COD$ is isosceles, meaning that $CD=OD=BD$ and hence $\fbox{D}$. 
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 $\triangle DOB$ 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 $\triangle COD$ is isosceles, and hence $CD=OD=BD$.
Thus our answer is $\fbox{D.}$