If $A′,B′,C′$ are the midpoints of $BC, CA, AB$ respectively, then show that
$$4A′ ∧ B′ ∧ C′ = A ∧ B ∧ C.$$
So to begin with I have $4A' = 2BC, B' = 1/2(CA), C' = 1/2(AB)$ and therefore $4A′ ∧ B′ ∧ C′ = 2(B-C) ∧ 1/2(A-C) ∧ 1/2(B-A)$, then I try to manipulate this by expanding and other various ways but can't get it right, is there something obvious that I am missing?
Do you really believe the midpoint of $BC$ is $\frac{B-C}{2}$? That would mean the midpoint of $(1,0)$ and $(2,0)$ is $(-0.5, 0)$ and that the midpoint of the points in the other order is $(0.5,0)$.
So the first big problem must be that you need to use the right expression for the midpoint: $\frac{B+C}{2}$. (Notice how the order of the endpoints doesn't matter.)
After that, it's a simple matter to compute that
$$ (B+C)\wedge(C+A)\wedge (A+B)=2(A\wedge B\wedge C) $$
and then juggling the scalar factors that are missing:
$$ 4\left(\frac{B+C}{2}\wedge\frac{C+A}{2}\wedge \frac{A+B}{2}\right)=A\wedge B\wedge C $$