Let $P, A, B, C, A', B' , C' \in \mathbb{R}^2$ be points such that $PAB'C, PBC'A, PCA'B$ form parallelograms. Prove that there is a unique point $Q$ so that $QA'BC' , QB'CA' , QC'AB'$ form parallelograms.
I drew a picture to convince myself and I have point Q turns our connected points into a cube. However I am not sure how to go about proving it. Since $PAB'C, PBC'A, PCA'B$ are parallelograms we know that $$ P = A + C -B' = B + A - C' = C + B - A'$$ I'm not sure if this is the way to go about it or instead go about it with knowing what lines are parallel.
Hint: the necessary and sufficient condition for $\,QA'BC'\,$ to be a parallelogram is:
$$ Q+B = A'+C' = (B+C-P)+(A+B-P) \;\;\iff\;\; Q = A+B+C-2P $$
What remains to be shown is that this is the same point $\,Q\,$ that forms the other parallelograms.