If bisectors of `angleA and angleB ` of a quadrilateral ABCD intersect each other at P, of `angleB and angleC` at Q, of `angleC and angleD` of R and of `angleD and angleA` at S, then PQRS is a (A) Rectangle (B) Rhombus (C) Parallelogram (D) Quadrilateral whose opposite angles are supplementary

Options

A. rectangle
B. rhombus
C. parallelogram
D. quadrilateral whose opposite angles are supplementary

Explanation

To solve the problem, we need to analyze the properties of the angles formed by the angle bisectors of a quadrilateral ABCD. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Quadrilateral and the Bisectors**:
- We have a quadrilateral ABCD.
- The angle bisectors of angles A and B intersect at point P, B and C at Q, C and D at R, and D and A at S.

2. **Identifying Vertically Opposite Angles**:
- We know that angle QPS is equal to angle APB because they are vertically opposite angles.

3. **Applying the Angle Sum Property**:
- In triangle APB, we can apply the angle sum property:
\[ \text{Angle APB} + \frac{1}{2} \text{Angle A} + \frac{1}{2} \text{Angle B} = 180^\circ \]
- Rearranging gives us:
\[ \text{Angle APB} = 180^\circ - \left(\frac{1}{2} \text{Angle A} + \frac{1}{2} \text{Angle B}\right) \]
- We can label this as Equation (1).

4. **Finding Angle QPS**:
- Since angle QPS = angle APB, we can substitute from Equation (1):
\[ \text{Angle QPS} = 180^\circ - \left(\frac{1}{2} \text{Angle A} + \frac{1}{2} \text{Angle B}\right) \]
- We can label this as Equation (2).

5. **Repeating for Angle QRS**:
- Similarly, we can find angle QRS:
\[ \text{Angle QRS} = 180^\circ - \left(\frac{1}{2} \text{Angle C} + \frac{1}{2} \text{Angle D}\right) \]
- We can label this as Equation (3).

6. **Summing Angles QPS and QRS**:
- Now, we can sum angles QPS and QRS:
\[ \text{Angle QPS} + \text{Angle QRS} = \left(180^\circ - \left(\frac{1}{2} \text{Angle A} + \frac{1}{2} \text{Angle B}\right)\right) + \left(180^\circ - \left(\frac{1}{2} \text{Angle C} + \frac{1}{2} \text{Angle D}\right)\right) \]
- This simplifies to:
\[ \text{Angle QPS} + \text{Angle QRS} = 360^\circ - \left(\frac{1}{2} (\text{Angle A} + \text{Angle B} + \text{Angle C} + \text{Angle D})\right) \]

7. **Using the Angle Sum Property of Quadrilaterals**:
- We know that the sum of angles in a quadrilateral is 360 degrees:
\[ \text{Angle A} + \text{Angle B} + \text{Angle C} + \text{Angle D} = 360^\circ \]
- Therefore:
\[ \frac{1}{2} (\text{Angle A} + \text{Angle B} + \text{Angle C} + \text{Angle D}) = 180^\circ \]
- Substituting this back gives:
\[ \text{Angle QPS} + \text{Angle QRS} = 360^\circ - 180^\circ = 180^\circ \]

8. **Conclusion**:
- Since the sum of angles QPS and QRS is 180 degrees, we can conclude that the quadrilateral PQRS has opposite angles that are supplementary.
- Thus, the answer is:
**(D) Quadrilateral whose opposite angles are supplementary.**