Proving $\vec{MN}=\frac12(\vec{AD}+\vec{BC})$ $M$ and $N$ are the midpoints of the sides $AB$ and $CD$ of a quadrilateral $ABCD$. How to prove that $\vec{MN}=\frac12(\vec{AD}+\vec{BC})$?
 A: We have that


*

*$\vec{OM}=\frac12(\vec{OA}+\vec{OB})$

*$\vec{ON}=\frac12(\vec{OC}+\vec{OD})$
and therefore consider
$$\vec{MN}=\vec{ON}-\vec{OM}=\frac12(\vec{OC}+\vec{OD})-\frac12(\vec{OA}+\vec{OB})=$$
$$=\frac12(\vec{OD}-\vec{OA})+\frac12(\vec{OC}-\vec{OB})=\frac12(\vec{AD}+\vec{BC})$$
A: Hint:
\begin{align}
\overrightarrow{MN}&=\overrightarrow{MA}+\overrightarrow{AD}+\overrightarrow{DN} \\
&=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN},
\end{align}
so
$$\overrightarrow{MN}=\frac12\Bigl(\overrightarrow{MA}+\overrightarrow{AD}+\overrightarrow{DN}+\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}\Bigr).$$
A: Since $M$ is a midpoint of $AB$, we have $\vec{AM} = \vec{MB} = \frac12 \vec{AB}$. Also, since $N$ is a midpoint of $CD$, we have $\vec{DN} = \vec{NC} = \frac12 \vec{DC}$.
Notice that
$$\vec{MN} = \vec{MA} + \vec{AD} + \vec{DN} = \vec{MB} + \vec{BC} + \vec{CN}$$
Therefore
\begin{align}
\vec{MN} &= \frac12\big(\vec{MA} + \vec{AD} + \vec{DN} + \vec{MB} + \vec{BC} + \vec{CN}\big)\\
&= \frac12\left(-\frac12 \vec{AB} + \vec{AD} + \frac12\vec{DC} + \frac12\vec{AB} + \vec{BC} - \frac12\vec{DC}\right)\\
&= \frac12\big( \vec{AD} + \vec{BC}\big)
\end{align}
