= Solution
For a finite-dimensional <Lie algebra>, write $A_X=\operatorname{ad}X$ for its <Adjoint representation>. The <Killing form> is the symmetric <bilinear form>
$$
\kappa(X,Y)=\operatorname{tr}(A_XA_Y).
$$
The <Jacobi identity> gives $A_{[X,Y]}=[A_X,A_Y]$. Using the <cyclic property of the trace>,
$$
\kappa([X,Y],Z)=\operatorname{tr}([A_X,A_Y]A_Z)
=\operatorname{tr}(A_X[A_Y,A_Z])=\kappa(X,[Y,Z]).
$$
Thus it is an <invariant bilinear form on a Lie algebra>. If $[T_a,T_b]=f_{ab}{}^cT_c$, the adjoint matrix has entries $(A_{T_a})^c{}_b=f_{ab}{}^c$. Its trace product yields the <structure constants of a Lie algebra> formula
$$
\boxed{\kappa_{ab}=\sum_{c,d}f_{ac}{}^d f_{bd}{}^c.}
$$
A <semisimple Lie algebra> has zero <solvable radical>, equivalently no nonzero solvable ideal. In characteristic zero, the <Cartan criterion for semisimplicity> characterizes it by nondegeneracy of its <Killing form>. A real <compact Lie algebra> is the Lie algebra of a <compact Lie group>, equivalently one admitting a positive-definite invariant inner product. For a real semisimple algebra, the <compactness criterion from the Killing form> says compactness is equivalent to negative-definiteness of $\kappa$. An abelian compact-type algebra has zero Killing form, so this last characterization specifically concerns the semisimple case.
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