Exchanging objects in distinguished triangles

Question

Let $\mathcal{D}$ be a triangulated category and assume we are given two distinguished triangles $$ \begin{align} A \rightarrow B \rightarrow C \rightarrow +1\\ D \rightarrow C \rightarrow E \rightarrow + 1 \end{align} $$ does there exist a d.t. $$ D \oplus A \rightarrow B \rightarrow E \rightarrow +1$$ where $B \rightarrow E = B \rightarrow C \rightarrow E$? The question is motivated by the following: if $\mathcal{D}$ were the homotopy category of complexes of an abelian category, and we had two exact sequences $$ \begin{align} 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0\\ 0 \rightarrow D \rightarrow C \rightarrow E \rightarrow 0 \end{align} $$ then we could regard $C$ as $C \simeq \left[ A \rightarrow B \right]$, with $B$ in degree zero, and therefore the second exact sequence can be considered as $$ D \rightarrow \left[ A \rightarrow B \right] \rightarrow E \rightarrow +1 $$ i.e. $$ E = Cone(D \rightarrow \left[A \rightarrow B \right]) = \left[ D \oplus A \rightarrow B \right]$$ which gives what we want. In the case of a general triangulated category, using the following d.t. $$ \begin{align} B \rightarrow C \rightarrow A[1] \rightarrow +1\\ C \rightarrow E \rightarrow D[1] \rightarrow +1\\ B \rightarrow E \rightarrow Z \rightarrow +1 \end{align} $$ and the octahedron axiom, one finds the d.t. $$ A[1] \rightarrow Z \rightarrow D[1] \rightarrow +1 $$ but how can I show that $Z[-1] \simeq A \oplus D$?