= Solution
The <Killing form> is the <symmetric bilinear form>
$$
\boxed{B(A,C)=\operatorname{tr}_{\mathfrak g}(\operatorname{ad}A\,\operatorname{ad}C).}
$$
The facts needed here are its invariance, $B([A,C],D)=B(A,[C,D])$, and the <root space> identity $[H,X]=\alpha(H)X$. Invariance follows from $[\operatorname{ad}A,\operatorname{ad}C]=\operatorname{ad}[A,C]$ and cyclicity of the <trace>:
$$
\operatorname{tr}([\operatorname{ad}A,\operatorname{ad}C]\operatorname{ad}D)
=\operatorname{tr}(\operatorname{ad}A[\operatorname{ad}C,\operatorname{ad}D]).
$$
Taking $A=H$, $C=X$, $D=Y$ proves
$$
\boxed{B(H,[X,Y])=B([H,X],Y)=\alpha(H)B(X,Y).}
$$
No normalization of $X,Y$ is needed for this identity. Semisimplicity gives nondegeneracy of the <Killing form>, but that fact is not needed for this particular equality.
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