Exact triangle in a derived category

An exact, or distinguished, triangle is one isomorphic to $E\stackrel{f}{\to }F\to \mathrm{Cone}(f)\to E[1]$. For a degreewise short exact sequence $0\to E\to F\to G\to 0$, the map $\mathrm{Cone}(f)\to G$ taking $(v,e)$ to the image of $v$ is a quasi-isomorphism: its kernel is the contractible cone of the identity on $E$. Consequently the short exact sequence defines such a triangle.