`ABCD` is parallelogram. If `L and M` are the middle points of BC and CD, then `bar(AL)+bar(AM)` equals

Options

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

Answer

To solve the problem, we need to find the sum of the vectors \(\bar{AL} + \bar{AM}\) in the context of the parallelogram \(ABCD\) where \(L\) and \(M\) are the midpoints of sides \(BC\) and \(CD\) respectively.

### Step-by-step Solution:

1. **Define the Position Vectors**:
Let the position vector of point \(A\) be \(\vec{a}\), point \(B\) be \(\vec{b}\), point \(C\) be \(\vec{c}\), and point \(D\) be \(\vec{d}\). Since \(ABCD\) is a parallelogram, we have:
\[ \vec{c} = \vec{a} + \vec{b} \quad \text{(since } \vec{c} = \vec{a} + \vec{b} \text{)} \]
\[ \vec{d} = \vec{b} + \vec{a} \quad \text{(since } \vec{d} = \vec{b} + \vec{a} \text{)} \]

2. **Find the Midpoints \(L\) and \(M\)**:
The position vector of midpoint \(L\) of \(BC\) is given by:
\[ \vec{L} = \frac{\vec{b} + \vec{c}}{2} = \frac{\vec{b} + (\vec{a} + \vec{b})}{2} = \frac{2\vec{b} + \vec{a}}{2} = \vec{b} + \frac{\vec{a}}{2} \]
The position vector of midpoint \(M\) of \(CD\) is given by:
\[ \vec{M} = \frac{\vec{c} + \vec{d}}{2} = \frac{(\vec{a} + \vec{b}) + (\vec{b} + \vec{a})}{2} = \frac{2\vec{a} + 2\vec{b}}{2} = \vec{a} + \vec{b} \]

3. **Calculate Vectors \(\bar{AL}\) and \(\bar{AM}\)**:
The vector \(\bar{AL}\) is:
\[ \bar{AL} = \vec{L} - \vec{A} = \left(\vec{b} + \frac{\vec{a}}{2}\right) - \vec{a} = \vec{b} - \frac{\vec{a}}{2} \]
The vector \(\bar{AM}\) is:
\[ \bar{AM} = \vec{M} - \vec{A} = (\vec{a} + \vec{b}) - \vec{a} = \vec{b} \]

4. **Sum the Vectors**:
Now, we add \(\bar{AL}\) and \(\bar{AM}\):
\[ \bar{AL} + \bar{AM} = \left(\vec{b} - \frac{\vec{a}}{2}\right) + \vec{b} = 2\vec{b} - \frac{\vec{a}}{2} \]

5. **Express in terms of \(\vec{AC}\)**:
Since \(\vec{AC} = \vec{c} - \vec{a} = (\vec{a} + \vec{b}) - \vec{a} = \vec{b}\), we can express the final result in terms of \(\vec{AC}\):
\[ \bar{AL} + \bar{AM} = \frac{3}{2}\vec{AC} \]

### Final Result:
Thus, we conclude that:
\[ \bar{AL} + \bar{AM} = \frac{3}{2} \vec{AC} \]