Gergonne's theorem Theorem: Given C and D, plane curves of degree $n$ that intersect in exactly $n^2$ points, if $nm$ of these points are in a curve E of degree $m\le n$, then the remaining $n(n-m)$ are in a curve H of degree $n-m$. 
Initially we prove that C,D,E are reduced curves.
Then we consider the pencil generated by C and D, and call $C_t$, with $t\in \mathbb{P^1}$ the curves of the pencil. Let $E=E_1+\dots + E_r$ be the decomposition of E in irreducible components and $q_1\in E_1-(C\cap D)$. Then there exists a curve $C_{t_1}$ such that $q_1\in C_{t_1}$.
I would like to understand why for Bézout's theorem $E_1 \subset C_{t_1}$,  $E_2 \subset C_{t_2}$... and why $C_{t_1}=C_{t_2}$.
Thanks a lot for any help.
 A: Interesting, I only knew this statement for irreducible $E$, but it seems to be correct in general. I think that the argument sketched is slightly imprecise, let me explain why. 
Bézout's Theorem asserts that either $E$ and $C_{t_1}$ share a component or that $E$ and $C_{t_1}$ intersect in $n \cdot m$ points, counted with multiplicity (note that $\deg(C_{t_1}) = n$, as it is an element of the pencil generated by $C$ and $D$).
Since the $n\cdot m$ distinct intersection points of $C$ and $D$ also lie on $E$, and moreover $q_1 \in E \cap C_{t_1}$ is another intersection point, we have that $E$ and $C_{t_1}$ intersect in $\geq n \cdot m + 1$ points. Hence, by Bézout's Theorem, they must share a component.
Since the $E_j\subset E$ are irreducible, we have that $E_{j_0} \subset C_{t_1}$ for some $j_0 \in \{1,...,r\}$. Now, we would like to conclude that $j_0 = 1$ is possible. If
$$q_1 \in E_1 \setminus \sum_{j=2}^r E_j,$$
then we are fine - indeed, in this case, Bézout's Theorem proves that the (unique!) component of $E$ containing $q_1$ is contained in $C_{t_1}$. But if (say) $q_1 \in E_1 \cap E_2$, then we cannot conclude in this way (or I do not see how). So I think one has to choose $q_1$ to not be an intersection point of two of the $E_j$ (or just to be a smooth point of $E$ - this suffices and is of course possible) and not to be an intersection point of $C$ and $D$. Hence, we proved that $E_1 \subset C_{t_1}$, ..., $E_r \subset C_{t_r}$.
To conclude, we want to show that $C_{t_1} =\ ...\ = C_{t_r}$. I show that $t_1 = t_2$, the rest of course being proved in the same way. 
Observe that, by Bézout's Theorem, none of the $n \cdot m$ intersection points of $C \cap D$ can be singular points on $E$. In particular, they are not equal to intersection points of two of the $E_j$. Since $\deg(C_{t_1}) = \deg(C_{t_2}) = n$ and the $n^2$ intersection points of $C$ and $D$, as well as intersection points of $E_1$ and $E_2$ lie on both $C_{t_1}$ and $C_{t_2}$, Bézout's Theorem implies that $C_{t_1}$ and $C_{t_2}$ must have a common component.
Since $C$ and $D$ do not have a common component, it suffices to prove the following
Lemma: Suppose that $\{F_0 = 0\}$ and $\{F_1 = 0\}$ are two plane curves. If there exist two different points $(t_0:t_1), (s_0:s_1) \in \mathbb P^1$ such that $t_0 F_0 + t_1 F_1$ and $s_0 F_0 + s_1 F_1$ have a common factor $F$, then $F_0$ and $F_1$ have a common factor.
Proof of the Lemma: Without loss of generality, $t_0 =1 $. If $s_0 = 0$, then we are done, so we can assume that $s_0 = 1$. Then $t_1 \neq s_1$ and $F$ is a factor of
$$(F_0 + t_1 F_1)-(F_0 + s_1 F_1) = (t_1-s_1)F_1,$$
hence $F$ divides $F_1$. But then $F$ also divides $F_0 = F_0 + t_1F_1 - t_1F_1$. $\;\Box$
The Lemma hence implies that $t_1 = t_2$, as we wanted to show. 
