Generic translates of a divisor intersect curves

Question

Let $G$ be a connected algebraic group and suppose $G$ acts transitively on a proper variety $X$ (say over the complex numbers). If you pick a curve $C$ and an effective divisor $D$, I have seen a claim that there is a nonempty open $U \subseteq G$ such that $gD \cdot C \geq 0$ for $g \in U$. Why does such a $U$ exist?

Answer 1

Consider the natural morphism $f:G\times C\to X$ and let $E=f^{-1}D$. Then $E$ is a closed subset of $G\times C$. Then consider the proper map, $E\to G$. Transitivity says that $E$ is codimension one and thus the morphism $E\to G$ has fibers either finite or one dimensional. Not all fibers are one dimensional by $\dim E\leq \dim G$. Then the set of points where $f:E\to G$ has fiber dimension one is a proper closed subset. Take $U$ to be the complement.