In circle $ O$ chord $ AB$ is produced so that $ BC$ equals a radius of the circle.  $ CO$ is drawn and extended to $ D$.  $ AO$ is drawn.  Which of the following expresses the relationship between $ x$ and $ y$?

[asy]size(200);defaultpen(linewidth(0.7)+fontsize(10));
pair O=origin, D=dir(195), A=dir(150), B=dir(30), C=B+1*dir(0);
draw(O--A--C--D);
dot(A^^B^^C^^D^^O);
pair point=O;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$O$", O, dir(285));
label("$x$", O+0.1*dir(172.5), dir(172.5));
label("$y$", C+0.4*dir(187.5), dir(187.5));
draw(Circle(O,1));
[/asy]


$ \textbf{(A)}\ x\equal{}3y \\
\textbf{(B)}\ x\equal{}2y \\
\textbf{(C)}\ x\equal{}60^\circ \\
\textbf{(D)}\ \text{there is no special relationship between }x\text{ and }y \\
\textbf{(E)}\ x\equal{}2y \text{ or }x\equal{}3y\text{, depending upon the length of }AB$
1. Let \( E \) be the intersection of the circle and \( CD \).
2. Consider the triangle \( \triangle AOD \). Since \( |OA| = |OD| \) (both are radii of the circle), \( \triangle AOD \) is isosceles.
3. In an isosceles triangle, the base angles are equal. Therefore, \( \angle DAO = \angle ODA \).
4. Let \( \angle DAO = \angle ODA = \alpha \). Since the sum of the angles in a triangle is \( 180^\circ \), we have:
   \[
   \alpha + \alpha + x = 180^\circ \implies 2\alpha + x = 180^\circ \implies \alpha = 90^\circ - \frac{x}{2}
   \]
5. By Thales' theorem, \( \angle DAE = 90^\circ \) because \( DE \) is a diameter of the circle.
6. Therefore, \( \angle OAE = \angle DAE - \angle DAO = 90^\circ - (90^\circ - \frac{x}{2}) = \frac{x}{2} \).
7. Now consider the triangle \( \triangle CBO \). Since \( |BO| = |BC| \) (both are radii of the circle), \( \triangle CBO \) is isosceles.
8. In an isosceles triangle, the base angles are equal. Therefore, \( \angle BOE = \angle OCB = y \).
9. The angle \( \angle BAE \) is half of \( \angle BOE \) because \( BE \) is a common chord. Thus, \( \angle BAE = \frac{y}{2} \).
10. The angle \( \angle AOE \) is supplementary to \( \angle AOD \) because they form a straight line. Therefore, \( \angle AOE = 180^\circ - x \).
11. Now we have all the angles of \( \triangle AOC \):
    \[
    \angle OAE + \angle BAE + \angle OCB + \angle AOE = \frac{x}{2} + \frac{y}{2} + y + (180^\circ - x) = 180^\circ
    \]
12. Simplifying the equation:
    \[
    \frac{x}{2} + \frac{y}{2} + y + 180^\circ - x = 180^\circ
    \]
    \[
    \frac{x}{2} + \frac{3y}{2} - x = 0
    \]
    \[
    -\frac{x}{2} + \frac{3y}{2} = 0
    \]
    \[
    \frac{3y}{2} = \frac{x}{2}
    \]
    \[
    x = 3y
    \]

The final answer is \( \boxed{ x = 3y } \)