M and N are the midpoints of the sides BC and | vedclass

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(D) Let the position vectors of $A, B, C, D$ be $\vec{0}, \vec{b}, \vec{c}, \vec{d}$ respectively. Since $ABCD$ is a parallelogram,we have $\vec{b} +

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$\frac{3}{2} \overline{AC}$

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(D) Let the position vectors of $A, B, C, D$ be $\vec{0}, \vec{b}, \vec{c}, \vec{d}$ respectively. Since $ABCD$ is a parallelogram,we have $\vec{b} + \vec{d} = \vec{c}$. $M$ is the midpoint of $BC$,so $\vec{M} = \frac{\vec{b} + \vec{c}}{2}$. $N$ is the midpoint of $CD$,so $\vec{N} = \frac{\vec{c} + \vec{d}}{2}$. Now,$\overline{AM} + \overline{AN} = \vec{M} + \vec{N} = \frac{\vec{b} + \vec{c}}{2} + \frac{\vec{c} + \vec{d}}{2}$. $= \frac{\vec{b} + \vec{d} + 2\vec{c}}{2}$. Substituting $\vec{b} + \vec{d} = \vec{c}$,we get: $= \frac{\vec{c} + 2\vec{c}}{2} = \frac{3\vec{c}}{2} = \frac{3}{2} \overline{AC}$.

$M$ and $N$ are the midpoints of the sides $BC$ and $CD$ of a parallelogram $ABCD$ respectively,then $\overline{AM} + \overline{AN} =$

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