$\mathrm{M}$ and $\mathrm{N}$ are the mid points of the sides $\mathrm{BC}$ and $\mathrm{CD}$ of a…
$\mathrm{M}$ and $\mathrm{N}$ are the mid points of the sides $\mathrm{BC}$ and $\mathrm{CD}$ of a parallelogram $\mathrm{ABCD}$ respectively then $\overline{A M}+\overline{A N}=$
$\frac{1}{3} \overline{A C}$
$\frac{2}{3} \overline{A C}$
$\frac{3}{4} \overline{A C}$
$\frac{3}{2} \overline{A C}$
Solution
Let position vectors of A, B, C, D are $\overrightarrow{0}, \vec{b}, \vec{c}, \vec{d}$ respectively.
$\begin{aligned} & \overrightarrow{A C}=\vec{c} \\ & \overrightarrow{B C}=\vec{c}-\vec{b}\end{aligned}$
$\mathrm{M}$ is mid point of $\mathrm{BC}$
$\begin{aligned} & \overrightarrow{B M}=\frac{1}{2} \overrightarrow{B C}=\frac{1}{2}(\vec{c}-\vec{b}) \\ & \overrightarrow{A M}=\overrightarrow{A B}+\overrightarrow{B M}\end{aligned}$
$=\vec{b}+\frac{1}{2}(\vec{c}-\vec{b})=\frac{1}{2}(\vec{b}+\vec{c})$
$\begin{aligned} & \overrightarrow{A D}=\vec{d} \\ & \overrightarrow{D C}=\vec{c}-\vec{d}\end{aligned}$
$\mathrm{N}$ is mid point of $\mathrm{DC}$
$\overrightarrow{D N}=\frac{1}{2}(\vec{c}-\vec{d})$
$\begin{aligned} & \overrightarrow{A N}=\overrightarrow{A D}+\overrightarrow{D N}=\vec{d}+\frac{1}{2}(\vec{c}-\vec{d})=\frac{1}{2}(\vec{c}+\vec{d}) \\ & \overrightarrow{A M}+\overrightarrow{A N}=\frac{1}{2}(\vec{b}+\vec{c})+\frac{1}{2}(\vec{c}+\vec{d})\end{aligned}$
$=\vec{c}+\frac{1}{2}(\vec{b}+\vec{d})$
$\mathrm{ABCD}$ is parallelogram $\Rightarrow \vec{b}+\vec{d}=\vec{c}$
$\overrightarrow{A M}+\overrightarrow{A N}=\vec{c}+\frac{1}{2} \vec{c}=\frac{3}{2} \vec{c}=\frac{3}{2} \overrightarrow{A C}$