Incircle of `DeltaABC` touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question
`angleDEF` is equal to

Options

A. `(pi -B)/(2)`
B. `pi - 2B`
C. `A -C`
D. none of these

Explanation

To find the angle \( \angle DEF \) in triangle \( \Delta ABC \) where the incircle touches the sides at points \( D, E, \) and \( F \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Incenter and Points of Contact**:
- Let \( I \) be the incenter of triangle \( ABC \).
- The incircle touches side \( BC \) at point \( D \), side \( AC \) at point \( E \), and side \( AB \) at point \( F \).

2. **Understand the Angles Involved**:
- The angles \( \angle IFA \), \( \angle IEA \), \( \angle IDC \), and \( \angle IEC \) are all \( 90^\circ \) because the radius at the point of contact is perpendicular to the tangent (the sides of the triangle).

3. **Consider Quadrilaterals**:
- The quadrilateral \( IFAE \) and \( IECD \) are formed. Since \( IFAE \) is a cyclic quadrilateral, we can use the property of cyclic quadrilaterals.

4. **Use the Angle Sum Property**:
- In quadrilateral \( IFAE \):
\[ \angle IFA + \angle IEA + \angle A = 360^\circ \]
Since \( \angle IFA = 90^\circ \) and \( \angle IEA = 90^\circ \):
\[ 90^\circ + 90^\circ + \angle A = 360^\circ \implies \angle FIE = 180^\circ - \angle A \]

5. **Isosceles Triangle Properties**:
- Since \( IF = IE = R \) (the inradius), triangle \( IFE \) is isosceles. Thus, \( \angle IEF = \angle IFE = X \).
- From the angle sum property in triangle \( IFE \):
\[ 2X + \angle FIE = 180^\circ \]
Substituting \( \angle FIE = 180^\circ - \angle A \):
\[ 2X + (180^\circ - \angle A) = 180^\circ \implies 2X = \angle A \implies X = \frac{\angle A}{2} \]

6. **Repeat for the Other Angles**:
- By similar reasoning in triangle \( IED \):
\[ \angle IED = \angle IDC = Y \implies 2Y + (180^\circ - \angle C) = 180^\circ \implies 2Y = \angle C \implies Y = \frac{\angle C}{2} \]

7. **Calculate \( \angle DEF \)**:
- Now, \( \angle DEF = \angle IEF + \angle IED = X + Y \):
\[ \angle DEF = \frac{\angle A}{2} + \frac{\angle C}{2} = \frac{\angle A + \angle C}{2} \]
- Since \( \angle A + \angle C + \angle B = 180^\circ \):
\[ \angle A + \angle C = 180^\circ - \angle B \implies \angle DEF = \frac{180^\circ - \angle B}{2} = 90^\circ - \frac{\angle B}{2} \]

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