$20 \cdot 17$ In a circle with center $O$, $AD$ is a diameter, $ABC$ is a chord, $BO=5$, and $\angle ABO \stackrel{\mathrm{m}}{=} \overparen{CD} \stackrel{\mathrm{m}}{=} 60^{\circ}$. Then the length of $BC$ is
(A) 3 .
(B) $3+\sqrt{3}$.
(C) $5-\frac{\sqrt{3}}{2}$.
(D) 5 .
(E) None of the above conclusions is correct.
[Solution] From the given $\overparen{C D} \stackrel{\mathrm{m}}{=} 60^{\circ}$, we know $\angle B A O=30^{\circ}$, thus $\triangle A B O$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
Given $B O=5$, then $A O=5 \sqrt{3}, A B=10$.
Connecting $C D$, by the given $A D$ is the diameter, then $\triangle A D C$ is also a $30^{\circ}-60^{\circ}-90^{\circ}$ right triangle.
$$
\begin{array}{l}
\because A D=2 A O=10 \sqrt{3}, \\
\therefore \quad A C=\frac{\sqrt{3}}{2} \cdot 10 \sqrt{3}=15, \\
\text { then } B C=A C-A B=15-10=5. \\
\end{array}
$$

Therefore, the answer is $(D)$.