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 \(\vec{BE} + \vec{AF}\) given that \(D\), \(E\), and \(F\) are the midpoints of the sides of triangle \(ABC\).

### Step-by-Step Solution:

1. **Identify the Midpoints:**
- Let \(D\) be the midpoint of \(AB\).
- Let \(E\) be the midpoint of \(AC\).
- Let \(F\) be the midpoint of \(BC\).

Using the midpoint formula:
\[ \vec{D} = \frac{\vec{A} + \vec{B}}{2}, \quad \vec{E} = \frac{\vec{A} + \vec{C}}{2}, \quad \vec{F} = \frac{\vec{B} + \vec{C}}{2} \]

2. **Calculate \(\vec{BE}\):**
\[ \vec{BE} = \vec{E} - \vec{B} \]
Substituting for \(\vec{E}\):
\[ \vec{BE} = \left(\frac{\vec{A} + \vec{C}}{2}\right) - \vec{B} = \frac{\vec{A} + \vec{C} - 2\vec{B}}{2} \]

3. **Calculate \(\vec{AF}\):**
\[ \vec{AF} = \vec{F} - \vec{A} \]
Substituting for \(\vec{F}\):
\[ \vec{AF} = \left(\frac{\vec{B} + \vec{C}}{2}\right) - \vec{A} = \frac{\vec{B} + \vec{C} - 2\vec{A}}{2} \]

4. **Add \(\vec{BE}\) and \(\vec{AF}\):**
\[ \vec{BE} + \vec{AF} = \frac{\vec{A} + \vec{C} - 2\vec{B}}{2} + \frac{\vec{B} + \vec{C} - 2\vec{A}}{2} \]
Combine the fractions:
\[ \vec{BE} + \vec{AF} = \frac{(\vec{A} + \vec{C} - 2\vec{B}) + (\vec{B} + \vec{C} - 2\vec{A})}{2} \]
Simplifying the numerator:
\[ = \frac{\vec{A} - 2\vec{A} + \vec{B} - 2\vec{B} + 2\vec{C}}{2} = \frac{-\vec{A} - \vec{B} + 2\vec{C}}{2} \]

5. **Final Result:**
\[ \vec{BE} + \vec{AF} = \frac{2\vec{C} - \vec{A} - \vec{B}}{2} \]

This can be interpreted as:
\[ \vec{BE} + \vec{AF} = \vec{C} - \frac{\vec{A} + \vec{B}}{2} \]
Which indicates that \(\vec{BE} + \vec{AF}\) is the vector from the midpoint \(D\) to point \(C\).