O is the centre on the circle. PA and PB are tangent segments. Then the quadrilateral AOBP is :

Options

A. Rectangle
B. Square
C. Cyclic
D. None of these

Answer

To solve the problem, we need to analyze the given information about the circle and the tangents. Let's break it down step by step:

### Step-by-Step Solution:

1. **Identify the Elements**:
- Let O be the center of the circle.
- PA and PB are tangent segments from point P to points A and B on the circle.

2. **Understanding Tangents**:
- By the property of tangents, we know that the tangent to a circle is perpendicular to the radius at the point of tangency.
- Therefore, \( OA \perp PA \) and \( OB \perp PB \). This means that \( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).

3. **Analyzing Quadrilateral AOBP**:
- The quadrilateral AOBP consists of points A, O, B, and P.
- We have established that \( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).

4. **Sum of Angles in Quadrilateral**:
- The sum of the interior angles of any quadrilateral is \( 360^\circ \).
- Therefore, we can express this as:
\[ \angle AOB + \angle OBP + \angle BPA + \angle PAO = 360^\circ \]

5. **Substituting Known Angles**:
- We know:
- \( \angle OBP = 90^\circ \)
- \( \angle PAO = 90^\circ \)
- Substituting these values into the equation:
\[ \angle AOB + 90^\circ + \angle BPA + 90^\circ = 360^\circ \]
- Simplifying this gives:
\[ \angle AOB + \angle BPA + 180^\circ = 360^\circ \]
- Thus, we have:
\[ \angle AOB + \angle BPA = 180^\circ \]

6. **Conclusion on Opposite Angles**:
- Since the sum of the opposite angles \( \angle AOB \) and \( \angle BPA \) is \( 180^\circ \), we can conclude that quadrilateral AOBP is a cyclic quadrilateral.
- In a cyclic quadrilateral, the sum of the opposite angles is always \( 180^\circ \).

### Final Answer:
Therefore, the quadrilateral AOBP is a **cyclic quadrilateral**.

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