If the bisector of the angles A and B of a quadrilateral ABCD meet at `O`, then `angle AOB` is equal to:

Options

A. `angle C + angle D`
B. `(1)/(2) ( angle C + angle D)`
C. `(1)/(2) angle C + (1)/(3) angle D`
D. `(1)/(3) angle C + (1)/(2) angle D`

Answer

To find the angle \( AOB \) in quadrilateral \( ABCD \) where the angle bisectors of angles \( A \) and \( B \) meet at point \( O \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Angles**:
- Let \( \angle A = A \) and \( \angle B = B \).
- The angle bisector of \( A \) divides it into two equal angles, so \( \angle OAB = \frac{A}{2} \).
- The angle bisector of \( B \) divides it into two equal angles, so \( \angle OBA = \frac{B}{2} \).

2. **Apply the Triangle Angle Sum Property**:
- In triangle \( AOB \), the sum of the angles is \( 180^\circ \).
- Therefore, we can write the equation:
\[ \angle OAB + \angle OBA + \angle AOB = 180^\circ \]

3. **Substitute the Known Angles**:
- Substitute \( \angle OAB \) and \( \angle OBA \) into the equation:
\[ \frac{A}{2} + \frac{B}{2} + \angle AOB = 180^\circ \]

4. **Combine the Terms**:
- Combine the terms on the left side:
\[ \frac{A + B}{2} + \angle AOB = 180^\circ \]

5. **Isolate \( \angle AOB \)**:
- Rearranging the equation gives:
\[ \angle AOB = 180^\circ - \frac{A + B}{2} \]

6. **Use the Quadrilateral Angle Sum Property**:
- In quadrilateral \( ABCD \), the sum of all angles is \( 360^\circ \):
\[ A + B + C + D = 360^\circ \]
- Rearranging gives:
\[ A + B = 360^\circ - (C + D) \]

7. **Substitute Back**:
- Substitute \( A + B \) back into the equation for \( \angle AOB \):
\[ \angle AOB = 180^\circ - \frac{360^\circ - (C + D)}{2} \]

8. **Simplify**:
- Simplifying gives:
\[ \angle AOB = 180^\circ - 180^\circ + \frac{C + D}{2} \]
- Thus, we have:
\[ \angle AOB = \frac{C + D}{2} \]

### Final Answer:
\[ \angle AOB = 90^\circ - \frac{A + B}{2} \]