= Solution
For the <Euclidean motion Lie algebra in two dimensions>, use the ordered basis $(J,E_1,E_2)$ and keep the printed rotation signs. The <Adjoint representation> matrices are
$$
\operatorname{ad}_J=\begin{pmatrix}0&0&0\\0&0&1\\0&-1&0\end{pmatrix},\qquad
\operatorname{ad}_{E_1}=\begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix},\qquad
\operatorname{ad}_{E_2}=\begin{pmatrix}0&0&0\\-1&0&0\\0&0&0\end{pmatrix}.
$$
Only the square of $\operatorname{ad}_J$ has a nonzero trace. Hence the <Killing form of the planar Euclidean motion Lie algebra> has matrix
$$
\boxed{[\kappa]_{(J,E_1,E_2)}=\operatorname{diag}(-2,0,0),\qquad\kappa(aJ+u_1E_1+u_2E_2,bJ+v_1E_1+v_2E_2)=-2ab.}
$$
Its <radical of the Killing form> is $\operatorname{span}\{E_1,E_2\}$, precisely the translation subspace. This is a nonzero abelian <ideal of a Lie algebra>, so the <Lie algebra> is neither simple nor semisimple. In fact its first <derived series of a Lie algebra> term is this translation ideal and its next is zero. Thus the degeneracy and the failure of semisimplicity agree with the structural proofs above; the adjoint action nevertheless remains nontrivial.
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