Incircle of `DeltaABC` touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question
The length of side EF is

Options

A. `r sin.(A)/(2)`
B. `2r sin.(A)/(2)`
C. `r cos.(A)/(2)`
D. `2r cos.(A)/(2)`

Explanation

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

### Step 1: Understand the Geometry
The incircle of triangle \( ABC \) touches side \( BC \) at point \( D \), side \( AC \) at point \( E \), and side \( AB \) at point \( F \). The points \( D \), \( E \), and \( F \) are the points of tangency of the incircle with the respective sides.

### Step 2: Identify Relevant Angles
Since \( I \) is the incenter of the triangle, the angles \( \angle IFA \) and \( \angle IEA \) are both \( 90^\circ \) because the radius at the point of tangency is perpendicular to the tangent line at that point.

### Step 3: Use Quadrilateral Properties
In quadrilateral \( IAFE \), the sum of the angles is \( 360^\circ \). Since we have two angles equal to \( 90^\circ \), we can write:
\[ \angle FIE + \angle A = 360^\circ - 180^\circ = 180^\circ \]
Thus, we find:
\[ \angle FIE = 180^\circ - \angle A \]

### Step 4: Apply Sine Rule
Using the sine rule in triangle \( IEF \):
\[ \frac{EF}{\sin(\angle FIE)} = \frac{IF}{\sin(\angle IEF)} \]
From the previous step, we know \( \angle FIE = 180^\circ - \angle A \), so:
\[ \sin(\angle FIE) = \sin(A) \]
Also, since \( IF \) is the inradius \( R \) and \( \angle IEF = \frac{A}{2} \):
\[ \sin(\angle IEF) = \sin\left(\frac{A}{2}\right) \]

### Step 5: Substitute Values
Substituting these values into the sine rule gives us:
\[ \frac{EF}{\sin(A)} = \frac{R}{\sin\left(\frac{A}{2}\right)} \]
Rearranging this, we find:
\[ EF = R \cdot \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

### Step 6: Simplify Using Trigonometric Identities
Using the identity \( \sin(A) = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right) \), we can rewrite:
\[ EF = R \cdot \frac{2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
This simplifies to:
\[ EF = 2R \cos\left(\frac{A}{2}\right) \]

### Conclusion
Thus, the length of side \( EF \) is:
\[ EF = 2R \cos\left(\frac{A}{2}\right) \]
The correct option is **D: \( 2R \cos\left(\frac{A}{2}\right) \)**.