$20.56 \triangle ABC$ is inscribed in circle $O'$, circle $O$ is inscribed in $\triangle ABC$, connect and extend $AO$ to intersect the larger circle at $D$, then it must be true that
(A) $CD = BD = O'D$.
(B) $AO = CO = OD$.
(C) $CD = CO = BD$.
(D) $CD = OD = BD$.
(E) $O'B = O'C = OD$.
(17th American High School Mathematics Examination, 1966)
[Solution] Connect $C D, D B, O C$.
Given that $A C, A B$ are tangents to $\odot O$, then $A O$ bisects $\angle C A B$. Let $\angle C A D=$ $\angle B A D=\alpha$.
Similarly, $\angle A C O=\angle B C O=\beta$.
Therefore, $B D=C D$,
thus $\angle D C B=\angle D B C=\alpha$.
And $B D=C D$.
Hence, $\angle O C D=\alpha+\beta$,
and $\angle C O D$ is the exterior angle of $\triangle O A C$, so $\angle C O D=$ $\alpha+\beta$.
$\therefore \triangle C D O$ is an isosceles triangle. Thus, $C D=O D$. Therefore, the answer is $(D)$.