I know that an abelian triangulated category is semisimple,i.e.,any exact sequence splits.But Why does any distinguished triangle is isomorphic to a triangle of the form $X \stackrel{f}{\rightarrow} Y \stackrel{g}{\rightarrow} \operatorname{ker} f[1] \oplus \operatorname{coker} f \stackrel{h}{\rightarrow} X[1]$?(cf. Gelfand and Manin, Exercises to IV 1).
My try:Let $X \stackrel{f}{\rightarrow} Y \stackrel{g}{\rightarrow} Z \stackrel{h}{\rightarrow} X[1]$ be a distinguished triangle.Then we have $ \operatorname{coker} f \stackrel{g}{\rightarrow} Z \stackrel{h}{\rightarrow} \operatorname{ker} f[1]$.If it is exact at $\operatorname{coker} f,Z$,and $\operatorname{ker} f[1]$,then we have $Z\cong \operatorname{ker} f[1] \oplus \operatorname{coker} f$.But I can't show the exactness.
Split $X=Ker(f)\oplus X'$, and $Y\cong X'\oplus Coker(f)$. Now you can describe $f$ as a direct sum of 3 maps, zero in $Ker(f)$, identity on $X'$ and $0\to Coker(f)$. The result follows from the fact that identity fits into a triangle with 0 as cone, and the zero maps similarly (just traslate the triangle with identity)