= Solution
The event separation is timelike precisely when
$$
\boxed{(D-B)^2=c^2(t_d-t_b)^2-|\mathbf x_d-\mathbf x_b|^2>0.}
$$
A simultaneous pair in an inertial frame would instead have $(D-B)^2=-|\Delta\mathbf x'|^2\le0$. Since a <Lorentz transformation> preserves this <interval>, no such simultaneous frame exists. A timelike pair can be brought to the same spatial position: the required <velocity> is $\mathbf u=\Delta\mathbf x/\Delta t$, whose magnitude is less than $c$.
The separate condition $D\cdot B=0$ says that the two event-position <four-vectors>, measured from the chosen spacetime origin, are orthogonal in the <Minkowski inner product>. It is invariant under <Lorentz transformations> about that origin but not under arbitrary <translations>. It is not a general criterion for their causal separation.
For a concrete example, put $B=(cT,0,0,0)$ and $D=(0,R,0,0)$. Their inner product vanishes, but their separation has squared interval $c^2T^2-R^2$, which can be positive, zero or negative. If $B$ is timelike, a rest frame for $B$ makes $D^0=0$; this means that $D$ and the chosen origin are simultaneous in that frame, not that $D$ and $B$ are simultaneous.
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