A,B,C and D are 4 points on the circumference of the circle and O is the centre of circle `angle OBD - angle CBD = angle BDC - angle ODB` find `angle A `

Answer

To solve the problem, we need to analyze the given information about the angles formed by points A, B, C, and D on the circumference of the circle with center O.

### Step-by-Step Solution:

1. **Identify the Angles**:
We are given the equation:
\[ \angle OBD - \angle CBD = \angle BDC - \angle ODB \]
Let's denote:
- \(\angle OBD = x\)
- \(\angle CBD = y\)
- \(\angle BDC = z\)
- \(\angle ODB = w\)

We can rewrite the equation as:
\[ x - y = z - w \]

2. **Use Properties of Isosceles Triangles**:
Since O is the center of the circle, triangles OBD and OBC are isosceles (OB = OD = radius of the circle). Therefore:
\[ \angle OBD = \angle ODB = x \]
This means:
\[ w = x \]

3. **Substituting Back**:
Substitute \(w\) in the equation:
\[ x - y = z - x \]
Rearranging gives:
\[ 2x = y + z \]
Thus, we can express \(y\) in terms of \(x\) and \(z\):
\[ y = 2x - z \]

4. **Using the Angles in Triangle BCD**:
In triangle BCD, the sum of the angles is 180 degrees:
\[ \angle BDC + \angle CBD + \angle CDB = 180^\circ \]
Substituting the known angles:
\[ z + y + \angle CDB = 180^\circ \]
Using \(y = 2x - z\):
\[ z + (2x - z) + \angle CDB = 180^\circ \]
Simplifying gives:
\[ 2x + \angle CDB = 180^\circ \]
Therefore:
\[ \angle CDB = 180^\circ - 2x \]

5. **Finding Angle A**:
Since A, B, C, and D are points on the circumference, the angles subtended by the same arc are equal. Therefore:
\[ \angle A = \angle CDB \]
Thus:
\[ \angle A = 180^\circ - 2x \]

6. **Conclusion**:
To find a specific value for \(\angle A\), we need additional information about \(x\). However, we have established the relationship:
\[ \angle A = 180^\circ - 2x \]