Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram. Problem
Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram.

My Solution

B (Conclusion): The midpoints of the sides of a space quadrilateral form a parallelogram.
A (Hypothesis): Let $A$, $B$, $C$, $D$ be four points such that they form a space quadrilateral. 
B1: $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ where $\dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B}$ and $\dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ are congruent sides. The same can be said for the other two sides.
A1: $\mathbf{A} + \mathbf{B} = \mathbf{C} + \mathbf{D}$ by the definition of quadrilaterals.
$\implies \dfrac{1}{2} \left( \mathbf{A} + \mathbf{B} \right) = \dfrac{1}{2} \left( \mathbf{C} + \mathbf{D} \right)$
$\implies \dfrac{1}{2} \mathbf{A} + \dfrac{1}{2} \mathbf{B} = \dfrac{1}{2} \mathbf{C} + \dfrac{1}{2} \mathbf{D}$ 
$Q.E.D.$

I would greatly appreciate it if people could please review my proof for correctness.
 A: Hint: If your four points are $a, b, c, d$, then the midpoints, in order around the quad, are 
$$
p = \frac{1}{2}(a+b), q = \frac{1}{2}(b+c), r = \frac{1}{2}(c+d), s = \frac{1}{2}(d+a).
$$
For $pqrs$ to be a parallelogram, you need the edge from $p$ to $q$ to have the same direction vector as the edge from $s$ to $r$; you need a similar thing to hold for the edges from $q$ to $r$ and $p$ to $s$. 
What's the direction vector of the edge from $p$ to $q$? Can you express it in terms of $a, b, c, d$? 
A: Hint:
since $\vec A+\vec B=\vec C+ \vec D$ we have:
$$
\frac{1}{2}(\vec A+\vec B)=\frac{1}{2}(\vec C+ \vec D)\iff \frac{1}{2}\vec A +\frac{1}{2}\vec B=\frac{1}{2}\vec C+\frac{1}{2}\vec D
$$
A: In general, the midpoints of any convex quadrilateral form a parallelogram, and you can prove that quite easily by drawing diagonals of the initial quadrilateral, but I'm not exactly sure what a space parallelogram is either, nor do I know how to prove this using vectors or check your proof as I have close to none understanding of them.
