Let ABCD be a quadrilateral such that there exists a point E inside the quadrilateral satisfying AE = BE = CE = DE. Suppose `angleDAB,angleABC, angleBCD` is an arithmetic progression. Then the median of the set `(angleDAB,angleABC,angleBCD)` is :-

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `pi/2`

Answer

To solve the problem, we need to find the median of the angles \( \angle DAB, \angle ABC, \angle BCD \) given that they are in an arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding the Angles in AP**:
Since \( \angle DAB, \angle ABC, \angle BCD \) are in AP, we can express them as:
- Let \( \angle DAB = \theta - \alpha \)
- Let \( \angle ABC = \theta \)
- Let \( \angle BCD = \theta + \alpha \)

2. **Finding the Median**:
The median of three numbers in AP is simply the middle term, which is \( \angle ABC \). Thus, the median is:
\[ \text{Median} = \theta \]

3. **Using the Property of Cyclic Quadrilaterals**:
Since point \( E \) is equidistant from all vertices \( A, B, C, D \), quadrilateral \( ABCD \) is cyclic. For a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \) or \( \pi \) radians.

4. **Finding \( \angle ADC \)**:
The sum of angles in quadrilateral \( ABCD \) is \( 360^\circ \) or \( 2\pi \) radians. Therefore:
\[ \angle DAB + \angle ABC + \angle BCD + \angle ADC = 2\pi \]
Substituting the values:
\[ (\theta - \alpha) + \theta + (\theta + \alpha) + \angle ADC = 2\pi \]
Simplifying this gives:
\[ 3\theta + \angle ADC = 2\pi \]
Thus,
\[ \angle ADC = 2\pi - 3\theta \]

5. **Using the Property of Cyclic Quadrilaterals Again**:
For cyclic quadrilaterals, we have:
\[ \angle ADC + \angle ABC = \pi \]
Substituting \( \angle ADC \):
\[ (2\pi - 3\theta) + \theta = \pi \]
Simplifying this gives:
\[ 2\pi - 2\theta = \pi \]
Rearranging gives:
\[ 2\theta = \pi \implies \theta = \frac{\pi}{2} \]

6. **Conclusion**:
The median of the angles \( \angle DAB, \angle ABC, \angle BCD \) is:
\[ \theta = \frac{\pi}{2} \]

### Final Answer:
The median of the set \( (\angle DAB, \angle ABC, \angle BCD) \) is \( \frac{\pi}{2} \).