If D, E, F are respectively the mid-points of AB, AC and BC respectively in a ` Delta ABC, " then " vec (BE) + vec(AF) =`

Options

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

Answer

To solve the problem, we need to find the expression for \( \vec{BE} + \vec{AF} \) in triangle \( ABC \) where \( D, E, F \) are the midpoints of sides \( AB, AC, \) and \( BC \) respectively.

### Step-by-Step Solution:

1. **Identify the Points**:
Let \( A, B, C \) be the position vectors of points \( A, B, C \) respectively.
- The midpoint \( D \) of \( AB \) can be expressed as:
\[ \vec{D} = \frac{\vec{A} + \vec{B}}{2} \]
- The midpoint \( E \) of \( AC \) can be expressed as:
\[ \vec{E} = \frac{\vec{A} + \vec{C}}{2} \]
- The midpoint \( F \) of \( BC \) can be expressed as:
\[ \vec{F} = \frac{\vec{B} + \vec{C}}{2} \]

2. **Express \( \vec{AF} \)**:
The vector \( \vec{AF} \) can be expressed as:
\[ \vec{AF} = \vec{F} - \vec{A} = \left(\frac{\vec{B} + \vec{C}}{2}\right) - \vec{A} = \frac{\vec{B} + \vec{C} - 2\vec{A}}{2} \]

3. **Express \( \vec{BE} \)**:
The vector \( \vec{BE} \) can be expressed as:
\[ \vec{BE} = \vec{E} - \vec{B} = \left(\frac{\vec{A} + \vec{C}}{2}\right) - \vec{B} = \frac{\vec{A} + \vec{C} - 2\vec{B}}{2} \]

4. **Add \( \vec{BE} \) and \( \vec{AF} \)**:
Now, we can add \( \vec{BE} \) and \( \vec{AF} \):
\[ \vec{BE} + \vec{AF} = \left(\frac{\vec{A} + \vec{C} - 2\vec{B}}{2}\right) + \left(\frac{\vec{B} + \vec{C} - 2\vec{A}}{2}\right) \]
Combine the two fractions:
\[ = \frac{(\vec{A} + \vec{C} - 2\vec{B}) + (\vec{B} + \vec{C} - 2\vec{A})}{2} \]
Simplifying the numerator:
\[ = \frac{\vec{A} + \vec{C} - 2\vec{B} + \vec{B} + \vec{C} - 2\vec{A}}{2} \]
\[ = \frac{-\vec{A} + 2\vec{C} - \vec{B}}{2} \]

5. **Final Simplification**:
Rearranging gives:
\[ = \frac{2\vec{C} - \vec{A} - \vec{B}}{2} \]
\[ = \vec{C} - \frac{\vec{A} + \vec{B}}{2} \]
Recognizing that \( \frac{\vec{A} + \vec{B}}{2} \) is the position vector of point \( D \):
\[ = \vec{C} - \vec{D} \]

Thus, the final result is:
\[ \vec{BE} + \vec{AF} = \vec{C} - \vec{D} \]