A circle, with centre O, circumscribs a pentagon ABCDE. IF `AB- BC= CD and angle BCD = 126^(@),` find:
`angle AOC`

Answer

To solve the problem step by step, we will find the angle AOC in the given pentagon ABCDE circumscribed by a circle with center O.

### Step-by-Step Solution:

1. **Identify Given Information:**
- The pentagon ABCDE is circumscribed by a circle with center O.
- We know that \( AB - BC = CD \).
- We are given that \( \angle BCD = 126^\circ \).

2. **Establish Relationships in Triangles:**
- In triangles BOC and COD, we note that:
- \( OC \) is common to both triangles.
- \( OB = OD \) (radii of the circle).
- Since \( AB - BC = CD \), we can infer that \( BC = CD \).

3. **Use the Given Angle:**
- Since \( \angle BCD = 126^\circ \), we can find angles \( OCB \) and \( OCD \):
- By the properties of the circle, \( \angle OCB = \angle OCD = \frac{1}{2} \angle BCD \).
- Thus, \( \angle OCB = \angle OCD = \frac{126^\circ}{2} = 63^\circ \).

4. **Establish Angles in Triangle BOC:**
- In triangle BOC:
- \( OB = OC \) (both are radii).
- Therefore, \( \angle OCB = \angle OBC \) (angles opposite to equal sides).
- Hence, \( \angle OBC = 63^\circ \).

5. **Apply the Angle Sum Property:**
- The sum of angles in triangle BOC gives us:
\[ \angle OBC + \angle OCB + \angle BOC = 180^\circ \]
- Substituting the known values:
\[ 63^\circ + 63^\circ + \angle BOC = 180^\circ \]
- This simplifies to:
\[ 126^\circ + \angle BOC = 180^\circ \]
- Therefore,
\[ \angle BOC = 180^\circ - 126^\circ = 54^\circ \]

6. **Relate Angles AOB and COD:**
- Since \( AB = BC = CD \), we have:
\[ \angle AOB = \angle BOC = \angle COD = 54^\circ \]

7. **Find Angle AOC:**
- Now, we can find \( \angle AOC \):
\[ \angle AOC = \angle AOB + \angle BOC = 54^\circ + 54^\circ = 108^\circ \]

### Final Answer:
\[ \angle AOC = 108^\circ \]