AB is the chord of a circle with centre O and DOC is a line segment originating from a point D on the circle and intersecting AB produced at C such that BC = OD. If `angleBCD = 20^@`, then `angleAOD =` ?

Options

A. `20^@`
B. `30^@`
C. `40^@`
D. `60^@`

Answer

To solve the problem, we need to find the angle \( \angle AOD \) given the conditions of the circle and the angles involved. Let's break it down step by step.

### Step 1: Understand the Given Information
- We have a circle with center \( O \).
- \( AB \) is a chord of the circle.
- \( D \) is a point on the circle, and \( DOC \) is a line segment that intersects the extended line of \( AB \) at point \( C \).
- It is given that \( BC = OD \).
- The angle \( \angle BCD = 20^\circ \).

### Step 2: Analyze Triangle \( OBC \)
Since \( O \) is the center of the circle, the segments \( OA \), \( OB \), and \( OD \) are all radii of the circle and hence are equal:
- \( OA = OB = OD \).

Since \( BC = OD \), we can conclude that \( OB = OD = BC \). This means triangle \( OBC \) is isosceles with \( OB = OC \).

### Step 3: Find Angle \( \angle OBC \)
In triangle \( OBC \):
- Since \( OB = OC \), the angles opposite these sides are equal. Therefore, \( \angle OBC = \angle BOC \).
- Given \( \angle BCD = 20^\circ \), we can say \( \angle BCO = 20^\circ \) as well (since \( C \) is on the extension of \( AB \)).

Thus, we have:
- \( \angle OBC = 20^\circ \) and \( \angle BOC = 20^\circ \).

### Step 4: Find Angle \( \angle AOB \)
Now, we need to find \( \angle AOB \) in triangle \( AOB \):
- Since \( OA = OB \), triangle \( AOB \) is also isosceles.
- Let \( \angle OAB = \angle OBA = x \).

Using the triangle angle sum property:
\[ \angle AOB + \angle OAB + \angle OBA = 180^\circ \]
Substituting the known values:
\[ \angle AOB + x + x = 180^\circ \]
\[ \angle AOB + 2x = 180^\circ \]

### Step 5: Find \( x \)
From triangle \( OBC \):
\[ \angle OBC + \angle BOC + \angle BCO = 180^\circ \]
Substituting the known angles:
\[ 20^\circ + 20^\circ + \angle OBC = 180^\circ \]
Thus,
\[ \angle OBC = 180^\circ - 40^\circ = 140^\circ \]

Since \( \angle OBA = \angle OBC \) (as they are opposite angles in the isosceles triangle \( OAB \)):
\[ x = 140^\circ \]

### Step 6: Substitute Back to Find \( \angle AOB \)
Now substituting \( x \) back into the equation for \( \angle AOB \):
\[ \angle AOB + 2(20^\circ) = 180^\circ \]
\[ \angle AOB + 40^\circ = 180^\circ \]
\[ \angle AOB = 180^\circ - 40^\circ = 140^\circ \]

### Step 7: Find \( \angle AOD \)
Since \( DOC \) is a straight line, we can find \( \angle AOD \):
\[ \angle AOD + \angle AOB + \angle BOC = 180^\circ \]
Substituting the known values:
\[ \angle AOD + 100^\circ + 20^\circ = 180^\circ \]
\[ \angle AOD + 120^\circ = 180^\circ \]
Thus,
\[ \angle AOD = 180^\circ - 120^\circ = 60^\circ \]

### Final Answer
\[ \angle AOD = 60^\circ \]