The diagonals AC and BD of a parallelogram intersects at O. If `angleOAD=40^(@),angleOAB=20^(@)andangleCOD=65^(@)`, then evaluate the following,
`angleDBC`

Answer

To solve the problem, we need to find the angle \( \angle DBC \) in the given parallelogram where the diagonals intersect at point O. We are provided with the following angles:

- \( \angle OAD = 40^\circ \)
- \( \angle OAB = 20^\circ \)
- \( \angle COD = 65^\circ \)

### Step-by-Step Solution:

1. **Identify the Angles on a Straight Line:**
Since \( AC \) is a straight line, we can use the property that the sum of angles on a straight line is \( 180^\circ \). Therefore, we can write:
\[ \angle AOD + \angle COD = 180^\circ \]
Given \( \angle COD = 65^\circ \), we can find \( \angle AOD \):
\[ \angle AOD = 180^\circ - \angle COD = 180^\circ - 65^\circ = 115^\circ \]

2. **Use the Triangle Sum Property:**
In triangle \( AOD \), the sum of the angles is \( 180^\circ \). We know:
- \( \angle AOD = 115^\circ \)
- \( \angle OAD = 40^\circ \)
- Let \( \angle ODA = x \)

Therefore, we can write:
\[ \angle AOD + \angle OAD + \angle ODA = 180^\circ \]
Substituting the known values:
\[ 115^\circ + 40^\circ + x = 180^\circ \]
Solving for \( x \):
\[ 155^\circ + x = 180^\circ \implies x = 180^\circ - 155^\circ = 25^\circ \]
Hence, \( \angle ODA = 25^\circ \).

3. **Identify Alternate Angles:**
Since \( AD \) is parallel to \( BC \) in the parallelogram, we can use the property of alternate angles. The angle \( \angle ADB \) is equal to \( \angle DBC \):
\[ \angle ADB = \angle DBC \]
From triangle \( AOD \), we found that \( \angle ADB = \angle ODA = 25^\circ \).

4. **Conclusion:**
Thus, we have:
\[ \angle DBC = 25^\circ \]

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