$20 \cdot 86$ Quadrilateral $A B C D$ is inscribed in a circle with the side $A D$ of length 4 as the diameter. If the lengths of $A B$ and $B C$ are each 1, then the length of $C D$ is 
(A) $\frac{7}{2}$.
(B) $\frac{5 \sqrt{2}}{2}$.
(C) $\sqrt{11}$.
(D) $\sqrt{13}$.
(E) $2 \sqrt{3}$.
[Solution] Draw radius $O B, O B$ perpendicular bisects chord $A C$ at $G$, connect $B D$. Side $C D$ is parallel to $G O$ and is twice the length of segment $G O$.
Also, from Rt $\triangle B G A \backsim \operatorname{Rt} \triangle A B D$, we have $\frac{B G}{A B}=\frac{A B}{A D}$.
That is, $\frac{B G}{1}=\frac{1}{A D}=\frac{1}{4}$.
$\therefore G O=B O-B G=2-\frac{1}{4}=\frac{7}{4}$,

and $\square$ $C D=2 G O=\frac{7}{2}$.
Therefore, the answer is $(A)$.