If D, E and F are respectively, the mid-points of AB, AC and BC in `DeltaABC`, then BE + AF is equal to

Options

A. DC
B. `(1)/(2)BF`
C. `2BF`
D. `(3)/(2)BF`

Explanation

To solve the problem, we need to find the sum of the lengths \( BE \) and \( AF \) in triangle \( ABC \) where \( D, E, F \) are the midpoints of sides \( AB, AC, \) and \( BC \) respectively.

### Step-by-Step Solution:

1. **Identify the Midpoints**:
Let \( D, E, F \) be the midpoints of sides \( AB, AC, \) and \( BC \) respectively. By the midpoint theorem, we know that:
\[ AD = DB, \quad AE = EC, \quad BF = FC \]

2. **Use Vector Representation**:
We can represent the points in vector form:
- Let \( A \) be represented by vector \( \vec{A} \),
- Let \( B \) be represented by vector \( \vec{B} \),
- Let \( C \) be represented by vector \( \vec{C} \).

The midpoints can then be expressed as:
\[ \vec{D} = \frac{\vec{A} + \vec{B}}{2}, \quad \vec{E} = \frac{\vec{A} + \vec{C}}{2}, \quad \vec{F} = \frac{\vec{B} + \vec{C}}{2} \]

3. **Express \( BE \) and \( AF \)**:
Using the vector representation, we can find the vectors \( \vec{BE} \) and \( \vec{AF} \):
\[ \vec{BE} = \vec{E} - \vec{B} = \left(\frac{\vec{A} + \vec{C}}{2}\right) - \vec{B} = \frac{\vec{A} + \vec{C} - 2\vec{B}}{2} \]
\[ \vec{AF} = \vec{F} - \vec{A} = \left(\frac{\vec{B} + \vec{C}}{2}\right) - \vec{A} = \frac{\vec{B} + \vec{C} - 2\vec{A}}{2} \]

4. **Sum \( BE \) and \( AF \)**:
Now, we can add the two vectors:
\[ \vec{BE} + \vec{AF} = \frac{\vec{A} + \vec{C} - 2\vec{B}}{2} + \frac{\vec{B} + \vec{C} - 2\vec{A}}{2} \]
Combining the terms:
\[ = \frac{(\vec{A} - 2\vec{A}) + (\vec{B} - 2\vec{B}) + 2\vec{C}}{2} = \frac{-\vec{A} - \vec{B} + 2\vec{C}}{2} \]

5. **Relate to \( CD \)**:
From the triangle properties and the midpoint theorem, we can conclude that:
\[ BE + AF = CD \]
where \( CD \) is the segment connecting midpoint \( D \) to point \( C \).

### Final Result:
Thus, we find that:
\[ BE + AF = CD \]