= Solution
The <Killing form> of a finite-dimensional <Lie algebra> $\mathfrak g$ is the <symmetric bilinear form>
$$
\kappa(x,y)=\operatorname{tr}(\operatorname{ad}x\,\operatorname{ad}y).
$$
The cyclicity of the <trace> makes it invariant:
$$
\kappa([x,y],z)=\kappa(x,[y,z]).
$$
Consequently its <radical of a bilinear form> $R=\{x:\kappa(x,\mathfrak g)=0\}$ is an ideal, since $x\in R$ implies $\kappa([y,x],z)=\kappa(x,[z,y])=0$ for all $y,z\in\mathfrak g$.
If $\mathfrak g$ is simple, then $R$ is either $0$ or $\mathfrak g$. In the second case the Killing form vanishes identically, so the stated solvability criterion makes $\mathfrak g$ a <Solvable Lie algebra>. A nonabelian simple Lie algebra cannot be solvable: its first derived algebra is a nonzero ideal and hence equals $\mathfrak g$, after which the derived series never reaches zero. Thus $R=0$, and \b[the Killing form of a simple Lie algebra is nondegenerate].
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