If D, E and F be the middle points of the sides BC,CA and AB of the `DeltaABC`, then `AD+BE+CF` is

Options

A. A. a zero vector
B. B. a unit vector
C. C. 0
D. D. none of these

Explanation

To solve the problem, we need to find the value of \( AD + BE + CF \) where D, E, and F are the midpoints of the sides BC, CA, and AB of triangle ABC respectively.

### Step-by-Step Solution:

1. **Identify the Points and Midpoints**:
- Let the position vectors of points A, B, and C be represented as \( \vec{A}, \vec{B}, \vec{C} \).
- The midpoints are defined as follows:
- \( D \) is the midpoint of \( BC \):
\[ \vec{D} = \frac{\vec{B} + \vec{C}}{2} \]
- \( E \) is the midpoint of \( CA \):
\[ \vec{E} = \frac{\vec{C} + \vec{A}}{2} \]
- \( F \) is the midpoint of \( AB \):
\[ \vec{F} = \frac{\vec{A} + \vec{B}}{2} \]

2. **Calculate Vectors AD, BE, and CF**:
- The vector \( \vec{AD} \) is given by:
\[ \vec{AD} = \vec{D} - \vec{A} = \left(\frac{\vec{B} + \vec{C}}{2}\right) - \vec{A} = \frac{\vec{B} + \vec{C} - 2\vec{A}}{2} \]
- The vector \( \vec{BE} \) is given by:
\[ \vec{BE} = \vec{E} - \vec{B} = \left(\frac{\vec{C} + \vec{A}}{2}\right) - \vec{B} = \frac{\vec{C} + \vec{A} - 2\vec{B}}{2} \]
- The vector \( \vec{CF} \) is given by:
\[ \vec{CF} = \vec{F} - \vec{C} = \left(\frac{\vec{A} + \vec{B}}{2}\right) - \vec{C} = \frac{\vec{A} + \vec{B} - 2\vec{C}}{2} \]

3. **Sum the Vectors**:
- Now, we sum \( \vec{AD} + \vec{BE} + \vec{CF} \):
\[ \vec{AD} + \vec{BE} + \vec{CF} = \left(\frac{\vec{B} + \vec{C} - 2\vec{A}}{2}\right) + \left(\frac{\vec{C} + \vec{A} - 2\vec{B}}{2}\right) + \left(\frac{\vec{A} + \vec{B} - 2\vec{C}}{2}\right) \]
- Combine the fractions:
\[ = \frac{(\vec{B} + \vec{C} - 2\vec{A}) + (\vec{C} + \vec{A} - 2\vec{B}) + (\vec{A} + \vec{B} - 2\vec{C})}{2} \]
- Simplifying the numerator:
\[ = \vec{B} + \vec{C} - 2\vec{A} + \vec{C} + \vec{A} - 2\vec{B} + \vec{A} + \vec{B} - 2\vec{C} \]
- Combine like terms:
\[ = (1 - 2 + 1)\vec{A} + (1 - 2 + 1)\vec{B} + (1 - 2 + 1)\vec{C} = 0\vec{A} + 0\vec{B} + 0\vec{C} = \vec{0} \]

4. **Conclusion**:
- Therefore, we find that:
\[ AD + BE + CF = \vec{0} \]

### Final Answer:
The value of \( AD + BE + CF \) is the zero vector \( \vec{0} \).