= Isotropic orthogonality consequence of the Hodge index theorem
If $D$ is nonzero numerically with $D^2=0$ and $D\cdot A>0$ for ample $A$, then $D\cdot E=0$ implies $E^2\leq0$, with equality only when $[E]$ is proportional to $[D]$. To see this, use the <Hodge index theorem for algebraic surfaces> to split off the positive $A$ direction and diagonalize the remaining negative definite form. Orthogonality to a nonzero isotropic vector then leaves a negative semidefinite form whose radical is precisely that vector's span.
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