In the real-affine plan $\mathscr{A}_2$ consider ${A_1, A_2, A_3}$ independent points and $G_1$ its middle $(A_2A_3)$, $G_2$ its middle $(A_3A_1)$, $G_3$ its middle $(A_1A_2)$. Prove that the lines $\langle A_1G_1 \rangle$, $\langle A_2G_2\rangle$, $\langle A_3G_3\rangle$ are concurrent at a point $G$. Determine the barycentric coordinates of of $G$ in in relation to the affine mark $\{A_1, A_2, A_3\}$ and the simple ratios $(A_i, G_i; G)$, and $i\in {1, 2, 3}$. Where the simple ratio is given by: $(A_1,A_2;A_3)=\lambda\Longleftrightarrow \overrightarrow{A_1A_3}=\lambda\overrightarrow{A_3A_2}. $ I'm not sure how to approach this problem!
I would ask you to help me determine the solution of this particular case so that later I can make a generalization of it!
In the barycentric coordinate system $(A_1,A_2,A_3)$, we have
$$ G_1=\frac12(A_2+A_3),\quad G_2=\frac12(A_1+A_3),\quad G_3=\frac12(A_1+A_2). $$
The lines $\;A_1G_1\;$ and $\;A_2G_2\;$ are described respectively by
$$ \alpha A_1+(1-\alpha)G_1=\alpha A_1+\frac12(1-\alpha)(A_2+A_3) $$
and
$$ \beta A_2+(1-\beta)G_2=\beta A_2+\frac12(1-\beta)(A_1+A_3), $$
where $\alpha,\beta\in\mathbb R$.
These lines intersect in a point, say $G$, iff there exist $\alpha,\beta\in\mathbb R$ such that
$$ G=\alpha A_1+\frac12(1-\alpha)(A_2+A_3)=\beta A_2+\frac12(1-\beta)(A_1+A_3) $$
that is [see Note below] iff:
$$ \alpha=\frac 12(1-\beta)\qquad\quad \frac 12(1-\alpha)=\beta\qquad\quad \frac12 (1-\alpha)=\frac12 (1-\beta) $$
which gives
$$ \alpha=\beta=\frac13. $$
We conclude that there exist a unique intersection point $G$ given by
$$ G=\frac13 A_1+\frac12\big(1-\frac13\big)(A_2+A_3)=\frac13 A_1+\frac13 A_2+\frac13 A_3 $$
with barycentric coordinates $\displaystyle\big(\frac13,\frac13,\frac13\big)$.
As the expression of $G$ is symmetric in $A_1,A_2,A_3$, the remaining median $A_3G_3$ also passes through $G$, and we are done. In other words, if for example we want to calculate $A_1G_1\cap A_3G_3$, we just replace 2 with 3 in the previous calculation, and consequently we get the same result,$\quad$ q.e.d.
Regarding the calculation of simple ratios $(A_i,G_i;G)$ according to your definition (mine is a bit different, but only for its sign), we have, if $i=1$ for example:
$$ G-A_1=\frac13 (A_1-A_1)+\frac13 (A_2-A_1)+\frac13 (A_3-A_1)=\frac13 (A_2-A_1)+\frac13 (A_3-A_1) $$
and
\begin{align} G_1-G&=(G_1-A_1)-(G-A_1)=\\[1ex] &=\frac12(A_2-A_1)+\frac12 (A_3-A_1)-\frac13 (A_2-A_1)-\frac13 (A_3-A_1)=\\[1ex] &=\frac16 (A_2-A_1)+\frac16(A_3-A_1), \end{align}
hence
$$ G-A_1 = 2(G_1-G) $$
i.e.
$$ (A_1,G_1;G)=2. $$
Similarly for others ratios.
$\;$
$\bf \text{Note.}\quad$ We used the following observation: if $\;\sum_{i=1}^3\lambda_i=\sum_{i=1}^3\mu_i=1$, then
$$ \sum_{i=1}^3 \lambda_iA_i = \sum_{i=1}^3 \mu_iA_i\;\implies\; \lambda_i=\mu_i\quad[i=1,2,3]. $$
In fact, from the definition of barycentric coordinates, we have
$$ A_3+\sum_{i=1}^3 \lambda_i(A_i-A_3) = A_3+\sum_{i=1}^3 \mu_i(A_i-A_3) $$
hence
$$ (\lambda_1-\mu_1)(A_1-A_3)+(\lambda_2-\mu_2)(A_2-A_3) = 0. $$
Since the vectors $A_1-A_3,\,A_2-A_3$ are linearly independent, it follows $\lambda_1=\mu_1$ and $\lambda_2=\mu_2$, and hence $\lambda_3=1-\lambda_1-\lambda_2=1-\mu_1-\mu_2=\mu_3$.