Let $ABC$ be a triangle and the points $M\in(BC), N\in (CA), P\in(AB)$, different from the vertices of the triangle. If the lines $AM, BN, CP$ are concurrent in a point $S$, then the relation takes place: \begin{equation} \frac{AP}{PB}+\frac{AN}{NC}=\frac{AS}{SM}. \end{equation} I tried to formulate this theorem for the affine case (barycentric case). And I just managed to write that: In a given plan, the points $A_1, A_2, A_3$ are considered to be independent and the points $B_1, B_2, B_2$ different from $A_1, A_2, A_3$ so that: \begin{align} (A_1,A_2;B_1)&=\lambda_1\\ (A_2,A_3,B_2)&=\lambda_2\\ (A_3,A_1;B_3)&=\lambda_3\\. \end{align} If the lines $AA_1, BB_1, CC_1$ are concurrent in a point $S$, then Van Aubel's formula takes place.
To prove this theorem I thought that if it has one thing in common then I can say that: $\exists \alpha_1,\alpha_2,\alpha_3\in \mathbb{K}$ such that:\begin{equation}\alpha_1A_1+(1-\alpha_1)M_1=\alpha_2A_2+(1-\alpha_2)M_2=\alpha_3A_3+(1-\alpha_3)M_3. \end{equation} And according to the hypothesis, it turns out that: \begin{align} (A_1,A_2;B_1)&=\lambda_1\Rightarrow M_1=\frac{1}{1+\lambda_1}A_2+\frac{\lambda_1}{1+\lambda_1}A_3\\ (A_2,A_3,B_2)&=\lambda_2\Rightarrow M_2=\frac{1}{1+\lambda_2}A_3+\frac{\lambda_2}{1+\lambda_2}A_1\\ (A_3,A_1;B_3)&=\lambda_3\Rightarrow M_3=\frac{1}{1+\lambda_3}A_1+\frac{\lambda_3}{1+\lambda_3}A_2. \end{align} Substituting all these $M_i, i\in\{1,2,3\}$ in the previous relation we will obtain the values for $\lambda_i, i=\overline{1,3}$. But I do not know if I could reach a conclusion of the theorem with this reasoning. Specifically, I do not even realize what the conclusion looks like in this case. I don't know exactly how the problem should be formulated and what the conclusion would be. For example, in the case of Ceva's theorem, the exact same reasoning is made, but the conclusion is $\lambda_1\cdot\lambda_2\cdot\lambda_3=1$.