<s>[INST] 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.
 [/INST] Hint:
since $\vec A+\vec B=\vec C+ \vec D$ we have:
$$
\frac{1}{2}(\vec A+\vec B)=\frac{1}{2}(\vec C+ \vec D)\iff \frac{1}{2}\vec A +\frac{1}{2}\vec B=\frac{1}{2}\vec C+\frac{1}{2}\vec D
$$
</s>