`Delta ABC` is an isosceles triangle and AB = AC. If all the sides AB. BC and CA of a `Delta ABC` touch a circle at D, E and F respectively. Then BE is equal to :

Options

A. `(1)/(2)` BC
B. `(1)/(3)` BC
C. `(1)/(4)` BC
D. `(2)/(3)` BC

Answer

To solve the problem, we will use the properties of tangents drawn from an external point to a circle.

### Step-by-Step Solution:

1. **Identify the Triangle and Circle**:
- We have an isosceles triangle \( \Delta ABC \) where \( AB = AC \).
- A circle is inscribed within the triangle, touching the sides at points \( D \), \( E \), and \( F \).

2. **Understand Tangent Properties**:
- When a tangent is drawn from an external point to a circle, the lengths of the tangents from that point to the points of tangency are equal.
- Therefore, if we consider point \( B \) which touches the circle at point \( E \), we have:
\[ BE = BD \]
- Similarly, from point \( C \) touching the circle at point \( F \):
\[ CF = CE \]

3. **Express Side Lengths in Terms of Tangents**:
- For side \( AB \):
\[ AB = AD + BD \]
- For side \( AC \):
\[ AC = AF + CF \]
- Since \( AB = AC \), we can set up the equation:
\[ AD + BD = AF + CF \]

4. **Substituting Tangent Lengths**:
- Let \( AD = x \), \( BD = BE = y \), \( AF = z \), and \( CF = CE = w \).
- Then, we can rewrite the sides as:
\[ AB = x + y \]
\[ AC = z + w \]
- Since \( AB = AC \), we have:
\[ x + y = z + w \]

5. **Using the Isosceles Property**:
- Since \( AB = AC \) and \( BE = CE \) (because \( B \) and \( C \) are equidistant from the circle), we can say:
\[ BE + CE = BC \]
- Thus, we can express \( BC \) as:
\[ BC = BE + CE = y + w \]

6. **Finding the Relationship**:
- Since \( BE = BD \) and \( CE = CF \), we can express \( BC \) in terms of \( BE \):
\[ BC = 2BE \]
- Therefore, we can write:
\[ BE = \frac{1}{2} BC \]

7. **Conclusion**:
- Hence, the length \( BE \) is equal to:
\[ BE = \frac{1}{2} BC \]

### Final Answer:
Thus, the value of \( BE \) is \( \frac{1}{2} BC \).