 Solution

The Killing form is the symmetric bilinear formThe 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 $[\mathrm{ad}A,\mathrm{ad}C]=\mathrm{ad}[A,C]$ and cyclicity of the trace:Taking $A=H$, $C=X$, $D=Y$ provesNo 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.