I am an open university second-year math undergrad student. I am taking a first course on multivariable calculus - I am solving problems from Vector analysis by Louis Brand to gain a physics-based intuition to what's going on.
I would like someone to verify my simple proof - or possible suggest alternate line of thought.
Show that $\vec{AB}+\vec{CD}=2\vec{MN}$, where $M$ and $N$ are the mid-points of $AC$ and $BD$.
Solution.
$\begin{aligned} \vec{MN} &= \vec{AN}-\vec{AM}\\ &= (\vec{AB} + \vec{BN}) - \frac{1}{2}\vec{AC}\\ &= \vec{AB} + \frac{1}{2}\vec{BD} - \frac{1}{2} \vec{AC}\\ &= \vec{AB} + \frac{1}{2}(\vec{BC} + \vec{CD}) - \frac{1}{2}(\vec{AB} + \vec{BC}) \\ &= \frac{1}{2}(\vec{AB} + \vec{CD})\\ 2\vec{MN} &= \vec{AB} + \vec{CD} \end{aligned}$
The posted proof is correct.
For an alternative, express everything in terms of position vectors relative to an arbitrary origin $O$. With the notation $\,x=\vec{OX}\,$, and using that $\,m = (a+c)/2\,$ and $\,n = (b+d)/2\,$:
$$ \begin{align} \vec{AB}+\vec{CD} &= (b-a)+(d-c) = (b+d)-(a+c) = 2n -2m =2(n-m)=2\vec{MN} \end{align} $$