I am reading a proof about the above that goes like this:
"Let $(A, B, C, D)$ be a tetrahedron and let $X ∈ (A, B, C, D)$. Let $m_X$ be the median line going from $X$ to the centroid of the opposite face. We will show that the two medians intersect in the centroid $S$
The centroid of a triangle $(X, Y, Z)$ is $S_{XYZ} = (X+Y+Z) / 3$ so we can write the medians as:
$m_A = {A + λ[ (B+C+D)/3 - A]}$
$m_D = {D + μ[ (A+B+C)/3 - D]}$
where both $μ$ and $λ$ are between $0$ and $1$
Intersecting both sets, we get $λ = μ = 3/4$"
I really do not understand the "intersecting both sets" part. What is going on and how did they obtain $3/4$?
We can suppose that midpoint of $BC$ is origin of the position vector. Then $B+C=0$ and then we have (if medians intersect)
$$ G = (1-\lambda)A+{\lambda \over 3}D = (1-\mu)D+{\mu \over 3}A$$
Since $A$ and $D$ are linearly independent we have
$$1-\lambda= {\mu \over 3}\;\;\;{\rm and}\;\;\;{\lambda \over 3} = 1-\mu$$
From here we get $\mu = \lambda =3/4$. So we get $$G = {1\over 4}(A+D)$$