Apparently, a result of Lamé says the following. Suppose three lines $a$, $b$, $c$ intersect a non-singular cubic curve on points $a_1, a_2, a_3$, $b_1, b_2, b_3$ and $c_1, c_2,c_3$.
If $a_1, b_1,c_1$ are collinear and $a_2,b_2,c_2$ are collinear then so are $a_3, b_3,c_3$.
How can I prove this?
Is there a proof of this that doesn't go through the Cayley-Bacharach theorem (the 8 points -> 9 points result)?
I am aiming to use this fact to derive the associative law for the elliptic curves, so this proof needs to use only elementary things but not the associativity of elliptic curve addition.
If we denote the group operation for elliptic curves by $+$, then the nine points and three lines we're given correspond to three equations (remember that this operation is commutative): \begin{equation} a_1 + a_2 + a_3 = 0 \tag{1} \end{equation}
$$b_1 + b_2 + b_3 = 0\tag{2}$$
$$c_1 + c_2 + c_3 = 0\tag{3}$$
The antecedents in your theorem statement give us two further equations:
$$a_1 + b_1 + c_1 = 0\tag{4}$$
$$a_2 + b_2 + c_2 = 0\tag{5}$$
Add $(1)$, $(2)$, and $(3)$, and subtract $(4)$ and $(5)$. This gives the desired result:
$$a_3 + b_3 + c_3 = 0$$