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

$A\stackrel{0}{\to}B\to A\oplus B\to$ 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.
 A: The distinguished triangle is $B\to A\oplus B\to A \overset{0}{\to}$ is the exact triangle (this differs from what you have above by a shift).  I suspect that what you should should use is the map axiom (TR3) in Wikipedia's labeling to get a map from the cone  of the zero map to A (by applying this axiom for the identity on A and 0 on B) and to B (by doing the reverse) and the use the universal property of product (you might need to throw in the octahedral axiom for that).
A: So if I understand correctly the question you wanted to ask was:
Is it true that a triangle $$X \stackrel{u}{\to} Y \stackrel{v}{\to} Z \stackrel{w}{\to} \Sigma X$$ is split if and only if one of $u$, $v$, or $w$ is zero. The answer to this is yes.
It is clear (I think I can add details if someone wants) that if the triangle is split then one map must be zero (basically since we have an epi composing to zero). Conversely suppose that $w$ is zero, which is sufficient since we can always just rotate. Now we know by the axioms that $$Z \stackrel{-1}{\to} Z \to 0 \to \Sigma Z$$ and hence $$0 \to Z \stackrel{1}{\to} Z \to 0$$ are triangles (as an exercise check that any sequence of this form given by an isomorphism is necessarily a distinguished triangle), and $$X \stackrel{1}{\to} X \to 0 \to \Sigma X$$ is also a triangle. It is easy to check then that so is the direct sum $$ X \to X\oplus Z \to Z \stackrel{0}{\to} \Sigma X$$. The identity maps on $X$ and $Z$ then induce a map via [TR3] from this to the original triangle (since we have $w=0$ this trivially satisfies the necessary commutativity to apply [TR3]) and since two of the maps are isomorphisms so is the third. Hence any triangle as above with $w=0$ is isomorphic to one obtained by summing triangles on identity maps.
Proof of the claim that the direct sum of triangles is a triangle:
Let $X \to Y \to Z \to \Sigma X$ and $X' \to Y' \to Z' \to \Sigma X'$ be two distinguished triangles. We can complete the map $X\oplus X' \to Y \oplus Y'$ to a triangle $$X\oplus X' \to Y \oplus Y' \to Q \to \Sigma(X\oplus X')$$. We then get two diagrams whose rows are triangles by projecting to the factors $X, Y$ and $X', Y'$ respectively which we can complete to maps of triangles by the mapping axiom [TR3]. The maps $Q\to Z$ and $Q\to Z'$ such obtained induce a map $Q\to Z\oplus Z'$ by the universal property which gives a map from the triangle $X\oplus X' \to Y\oplus Y' \to Q$ to the pretriangle (a pretriangle is one where the maps compose to zero and it plays nicely with homological functors, that is they take it to a long exact sequence, this is clear from the fact that homological functors are additive and it is a sum of distinguished triangles) $X\oplus X' \to Y\oplus Y' \to Z\oplus Z'$. Two of the maps are isomorphims, namely the identities on the terms $X\oplus X'$ and $Y\oplus Y'$ so that the third must be also - this follows from the fact that $Hom(A,-)$ is a homological functor for any $A$, Yoneda, and the 5 lemma. So the triangle in question is isomorphic to a distinguished triangle and hence itself distinguished.
This works in the generality of arbitrary coproducts/products provided the coproducts/products in question exist and we consider a pretriangle to be one which homological/cohomological functors preserving the coproducts/products take to long exact sequences.
I'd recommend reading through the first chapter of Neeman's book Triangulated Categories - this is certainly covered in there as well as a bunch of other facts you might find useful in reading that paper. The reference for this result is Corollary 1.2.7 (it seemed lazy not to check since it is on my shelf) and the proof there is pretty much identical to the one here except that the facts I glossed over are proved earlier.
