**Quote:**

In the figure below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ACD= y degrees and AOD = x degrees such that x = ky, then the value of k is

A. 3

B. 2

C. 1

D. None of the above

OA and OB are radii, so OA=OB.

Since the prompt indicates that OB=BC, we get:

OA=OB=BC.

The following figure is yielded:

Let y=10.

Since the angles inside triangle BCO must sum to 180, and the angles opposite sides OB and BC must be equal, the following figure is yielded:

Since angles ABO and OBC must sum to 180, angle ABO=20.

Since the angles inside triangle ABO must sum to 180, and the angles opposite sides OA and OB must be equal, the following figure is yielded:

Since angles AOD, AOB and BOC must sum to 180, we get:

Since x=30, y=10, and \(k = \frac{x}{y}\), we get:

\(k = \frac{30}{10} = 3\)

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