= Solution
Work over $\mathbb R$ or $\mathbb C$, in finite dimension and characteristic zero. The <Killing form> of a <Lie algebra> is the trace form of its <Adjoint representation>:
$$
\boxed{\kappa(x,y)=\operatorname{tr}(\operatorname{ad}_x\operatorname{ad}_y),\qquad\operatorname{ad}_x(z)=[x,z].}
$$
It is bilinear and symmetric by the <cyclic property of the trace>. It is an <invariant bilinear form on a Lie algebra>, since the <Jacobi identity> gives $\operatorname{ad}_{[x,y]}=[\operatorname{ad}_x,\operatorname{ad}_y]$ and hence
$$
\begin{aligned}
\kappa([x,y],z)
&=\operatorname{tr}([\operatorname{ad}_x,\operatorname{ad}_y]\operatorname{ad}_z)\\
&=\operatorname{tr}(\operatorname{ad}_x[\operatorname{ad}_y,\operatorname{ad}_z])
=\kappa(x,[y,z]).
\end{aligned}
$$
Equivalently, $\kappa([z,x],y)+\kappa(x,[z,y])=0$. A Lie algebra automorphism preserves the form, because it conjugates the adjoint matrices. Its <radical of a bilinear form> $\mathfrak r_\kappa=\{x:\kappa(x,\mathfrak g)=0\}$ is an <ideal of a Lie algebra> by invariance. The <center of a Lie algebra> lies in this radical, so the form vanishes for an abelian algebra. On a direct sum of ideals the summands are orthogonal and the form restricts to their own Killing forms.
A short argument proves the requested implication without assuming the <Cartan criterion for semisimplicity>. Let $\mathfrak a$ be an abelian <ideal of a Lie algebra>, $x\in\mathfrak a$, and $y\in\mathfrak g$. The map $\operatorname{ad}_x$ takes $\mathfrak g$ into $\mathfrak a$ and vanishes on $\mathfrak a$, while $\operatorname{ad}_y$ preserves $\mathfrak a$. Therefore $\operatorname{ad}_x\operatorname{ad}_y$ has image in $\mathfrak a$ and is zero on $\mathfrak a$. In a basis extending a basis of $\mathfrak a$, both diagonal blocks are zero, so its trace vanishes. This proves that <abelian ideals lie in the radical of the Killing form>.
If the <solvable radical> $\mathfrak r$ were nonzero, its <derived series of a Lie algebra> would have a last nonzero term $\mathfrak a$. The <Jacobi identity> makes every derived term an ideal of $\mathfrak g$, and the last one is abelian. Thus $0\ne\mathfrak a\subseteq\mathfrak r_\kappa$, contradicting nondegeneracy. We obtain
$$
\boxed{\kappa\text{ nondegenerate}\ \Longrightarrow\ \mathfrak r=0\ \Longrightarrow\ \mathfrak g\text{ is a semisimple Lie algebra}.}
$$
The converse holds as well in characteristic zero; together these implications are the <Cartan criterion for semisimplicity>.
For a complex <simple Lie algebra>, that criterion gives nondegeneracy on $\mathfrak g$. Let $\mathfrak h$ be a <Cartan subalgebra>. Use the standard <root-space decomposition>
$$
\mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha,
\qquad[h,e_\alpha]=\alpha(h)e_\alpha.
$$
For $h,h'\in\mathfrak h$ and a <root vector> $e_\alpha$, choose $t\in\mathfrak h$ with $\alpha(t)\ne0$. Invariance gives
$$
\alpha(t)\kappa(h,e_\alpha)=\kappa(h,[t,e_\alpha])=\kappa([h,t],e_\alpha)=0.
$$
Thus $\mathfrak h$ is orthogonal to every nonzero <root space>. If $h\in\mathfrak h$ is also orthogonal to $\mathfrak h$, it is orthogonal to all of $\mathfrak g$ and hence is zero. This establishes <nondegeneracy of the Killing form on a Cartan subalgebra>:
$$
\boxed{\kappa|_{\mathfrak h\times\mathfrak h}\text{ is nondegenerate}.}
$$
The same invariance calculation shows that $\kappa(\mathfrak g_\alpha,\mathfrak g_\beta)=0$ unless $\alpha+\beta=0$. Opposite root spaces are therefore paired nondegenerately. Tracing the adjoint action on the root-space decomposition yields
$$
\kappa(h,h')=\sum_{\alpha\in\Phi}\alpha(h)\alpha(h'),
$$
since the root spaces of a complex semisimple algebra are one-dimensional and its adjoint action on $\mathfrak h$ is zero.
The relevant <Euclidean subspace of a Cartan subalgebra> is
$$
\mathfrak h_{\mathbb R}=\operatorname{span}_{\mathbb R}\{h_{\alpha_1},\ldots,h_{\alpha_\ell}\}
=\{h\in\mathfrak h:\alpha(h)\in\mathbb R\text{ for all }\alpha\in\Phi\},
$$
where the $h_{\alpha_i}$ are the simple <coroots>, normalized by $\alpha_i(h_{\alpha_i})=2$. The standard Euclidean property is that $\kappa$ is real and positive definite on this space. The displayed trace formula explains it: $\kappa(h,h)$ is a sum of real squares, and the roots span $\mathfrak h^*$, so all squares vanish only for $h=0$. Via the nondegenerate form, the roots can consequently be regarded as vectors in a real <Euclidean normed vector space>. The induced dual <inner product> makes <root reflections> orthogonal and allows root lengths and angles to be encoded by <Cartan integers> and the <Dynkin diagram>.
This positivity is on a specified real subspace; the complex Killing form is a bilinear form, not a positive <Hermitian inner product>. On a <compact real form> the Killing form is instead negative definite. For example the <Killing form of the special linear Lie algebra> gives $\kappa(X,Y)=4\operatorname{tr}(XY)$ for $\mathfrak{sl}_2$, so $\kappa(L_0,L_0)=8$ but $\kappa(iL_0,iL_0)=-8$.
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