Let $C$ be a irreducible cubic in the projective plane and let $F,F^\prime$ be two algebraic curves of degree $m$ satisfying $(C,F)=\Sigma_{i=1}^{3m}p_i$ and $(C,F^\prime)=\Sigma_{i=1}^{3m-1}p_i+q$, where all $p_i$ are smooth points of $C$ (not necessarily different from each other). Prove that $p_{3m}=q$.
I have been trying to solve the following problem and I know that it has to be solved with the AF+BG Theorem from Max Noether, but I don't know how to do it. Some hints?
Only a sketch, but I hope it is helpful:
Let $L$ be the line through $p_{3m}$ and $q$. It intersects $C$ in a third point, call it $r$.
Let $L'$ be the tangent line to $C$ at $q$. It follows from the $AF+BG$ theorem that $L'$ passes through $r$.
I claim that $L=L'$ and hence $p_{3m}=q$ (since $L'$ is tangent to $C$ at $q$). If $q \neq r$, then the two lines $L,L'$ have two distinct points in common, hence must be equal. If on the other hand $q=r$, then $L$ is also tangent to $C$ at $q$, and hence $L=L'$, because $q$ is necessarily a simple point on $C$.
Edit: Let me expand on the use of the $AF+BG$ theorem in the above argument. With notation as above, we have $$ (L,C)=p_{3m}+q+r $$ and therefore $$ (LF',C)=(L,C)+(F',C)=p_{3m}+q+r+\sum_{i=1}^{3m-1}p_i+q=\sum_{i=1}^{3m}p_i+2q+r=(F,C)+2q+r $$ hence $LF'$ intersects $C$ in a bigger cycle than $F$ does. Now by a Corollary of the $AF+BG$ theorem (cf. Fulton, Algebraic Curves, Corollary to Proposition 5.5), there exists a line $L'$ such that $$ (L',C)=2q+r $$ i.e. $L'$ is the tangent line to $C$ at $q$, since $q$ is a simple point of $C$ (this in turn follows from properties of the intersection pairing).
Is it now helpful?