20.59 Points $A, B, C$ are on a circle with radius $r$, and $A B=A C$ with $A B>r$, while the minor arc length of $B C$ is $r$. If angles are expressed in radians, then $\frac{A B}{B C}=$
(A) $\frac{1}{2} \csc \frac{1}{4}$.
(B) $2 \cos \frac{1}{2}$.
(C) $4 \sin \frac{1}{2}$.
(D) $\csc \frac{1}{2}$.
(E) $2 \sec \frac{1}{2}$.
[Solution] $\because \overparen{B C}=r, \therefore \angle B A C=\frac{1}{2}$. Connect $A O$ and extend it to intersect $B C$ at $D$, connect $O B, O C$, then
$$
\begin{array}{l}
B D=\frac{1}{2} B C, \text { and } \angle B A D=\frac{1}{2} \angle B A C=\frac{1}{4} . \\
\therefore \quad \frac{A B}{B C}=\frac{A B}{2 B D}=\frac{1}{2} \csc \angle B A D=\frac{1}{2} \csc \frac{1}{4} .
\end{array}
$$

Therefore, the answer is $(A)$.