Using the axioms for a triangulated category, is it possible to prove the following:

If 0:A-->B, then $A\stackrel{0}{\to}B\to A\oplus B\to$ is a distinguished triangle.

From the first axiom, the map 0:A-->B extends to its cone, but there is no guarantee I see that the direct sum fits into a triangle. If it does, however, clearly they are (should be?) isomorphic.

I have tried using the universal properties of the direct sum, in that finite coproducts and finite products coincide in additive categories, so that I have two diagrams, and extended each of these diagrams into triangles in every way I can imagine, but I think I'm just getting lost in the plethora of sequences. 0:A-->B extends to a triangle A-->B-->X--> and so I can get things like $X\to A\oplus B\to \Sigma^{-1}X\cong X$, but by moving away from triangles, I have lost notions of exactness (so that this sequence of maps merely commutes..)

I ask this because the proof of Lemma 3.3(2) of this paper seems to use this without reference.