ABCD is a cyclic quadrilateral in which AB is a diameter, BC = CD and `angleA=50^(@)` then `angleDBC` is :

Options

A. `25^(@)`
B. `20^(@)`
C. `45^(@)`
D. `30^(@)`

Answer

To solve the problem, we need to find the angle \( \angle DBC \) in the cyclic quadrilateral \( ABCD \) where \( AB \) is a diameter, \( BC = CD \), and \( \angle A = 50^\circ \).

### Step-by-step Solution:

1. **Identify the Given Information:**
- \( ABCD \) is a cyclic quadrilateral.
- \( AB \) is a diameter of the circle.
- \( BC = CD \).
- \( \angle A = 50^\circ \).

2. **Use the Property of Cyclic Quadrilaterals:**
- In a cyclic quadrilateral, the opposite angles are supplementary. Therefore:
\[ \angle A + \angle C = 180^\circ \]
- Substituting the value of \( \angle A \):
\[ 50^\circ + \angle C = 180^\circ \]
- Solving for \( \angle C \):
\[ \angle C = 180^\circ - 50^\circ = 130^\circ \]

3. **Analyze Triangle \( BCD \):**
- Since \( BC = CD \), triangle \( BCD \) is an isosceles triangle.
- Let \( \angle DBC = \theta \) and \( \angle CDB = \theta \) (the angles opposite to the equal sides).

4. **Set Up the Equation for the Angles in Triangle \( BCD \):**
- The sum of angles in triangle \( BCD \) is \( 180^\circ \):
\[ \angle DBC + \angle CDB + \angle BCD = 180^\circ \]
- Substituting the known angles:
\[ \theta + \theta + 130^\circ = 180^\circ \]
- Simplifying the equation:
\[ 2\theta + 130^\circ = 180^\circ \]
- Solving for \( \theta \):
\[ 2\theta = 180^\circ - 130^\circ \]
\[ 2\theta = 50^\circ \]
\[ \theta = 25^\circ \]

5. **Conclusion:**
- Therefore, \( \angle DBC = \theta = 25^\circ \).

### Final Answer:
\[ \angle DBC = 25^\circ \]