The basic form is the Killing form divided by twice the dual Coxeter number

The adjoint module is irreducible because any invariant subspace of $\mathfrak{g}$ is an ideal and $\mathfrak{g}$ is simple. A highest-root vector generates this nonzero module, so [F5] with $\lambda =\theta$ says the Casimir [F3] acts on $\mathfrak{g}$ by a scalar ${\lambda }_{\mathrm{a}\mathrm{d}}=(\theta ,\theta +2\rho {)}_{\mathrm{K}\mathrm{i}\mathrm{l}\mathrm{l}}$. On the other hand, for any Killing-dual bases $(x_a),(x^a)$, its adjoint action is ${\sum }_a{\mathrm{ad}}_{x_a}{\mathrm{ad}}_{x^a}$; its trace is ${\sum }_a{B}_{\mathrm{K}\mathrm{i}\mathrm{l}\mathrm{l}}(x_a,x^a)=\dim \mathfrak{g}$ by [F1]. Since the action is scalar on the nonzero space $\mathfrak{g}$, this trace gives ${\lambda }_{\mathrm{a}\mathrm{d}}=1$.

Since $B_0=B_{\mathrm{K}\mathrm{i}\mathrm{l}\mathrm{l}}\mathrm{/}c$, the defining equations give $H_{\lambda ,0}=c{H}_{\lambda ,\mathrm{K}\mathrm{i}\mathrm{l}\mathrm{l}}$ for every $\lambda \in {\mathfrak{h}}^{\ast }$, and hence $(\lambda ,\mu {)}_0=c(\lambda ,\mu {)}_{\mathrm{K}\mathrm{i}\mathrm{l}\mathrm{l}}$. The choice of $c$ gives $(\theta ,\theta {)}_0=2$. By definition, ${\theta }^{\vee }=H_{\theta ,0}$, and $(\rho ,\theta {)}_0=B_0(H_{\rho ,0},H_{\theta ,0})=\rho (H_{\theta ,0})=\rho ({\theta }^{\vee })=h^{\vee }-1$. Therefore ${\lambda }_{\mathrm{a}\mathrm{d}}=(\theta ,\theta +2\rho {)}_{\mathrm{K}\mathrm{i}\mathrm{l}\mathrm{l}}=c^{-1}(2+2(h^{\vee }-1))=2h^{\vee }\mathrm{/}c$.

Equating the two expressions for ${\lambda }_{\mathrm{a}\mathrm{d}}$ from steps 1.1 and 2.1 gives $1=2h^{\vee }\mathrm{/}c$, hence $c=2h^{\vee }$. Substitution into $B_0=B_{\mathrm{K}\mathrm{i}\mathrm{l}\mathrm{l}}\mathrm{/}c$ proves the asserted form identity.