Question 31Triangle $ABC$ 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: $CD=BD=O'D$$AO=CO=OD$$CD=CO=BD$$CD=OD=BD$$O'B=O'C=OD$
