 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 $E^2\le 0$, 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.