Every closed convex cone in $ \mathbb{R}^2 $ is polyhedral

I recently came across the following problem:

Let $ K \subseteq \mathbb{R}^2 $ be a closed convex cone (meaning K is closed under non negative linear combinations) and I am asked to show it is polyhedral meaning it is the intersection of finitely many half spaces, or alternatively that it is finitely generated $ K = cone(A) $ where $ A $ is a finite set and cone(A) denotes all possible non-negative linear combinations of elements of A, or alternatively that K has a finite number of extreme vectors meaning vectors $x\in K$ such that if $ x=y+z $ where $ y,z\in K $ are non-negative multiples of x.

I have given three possible definitions of polyhedral cone but for some reason I cannot seem to see why a convex cone which is closed in $\mathbb{R}^2$ satisfies either one of the three possible definitions. I certainly need and appreciate all the help I can get on this.

Solutions Collecting From Web of "Every closed convex cone in $ \mathbb{R}^2 $ is polyhedral"

Let $K$ be a closed cone in the plane. If $K$ is not contained in a closed halfspace, then $K$ coincides with the plane. Suppose then that it is contained in a closed halfspace, whose boundary is a line $L$. Let now $L’$ be a line parallel to $L$ contained in the interior of that halfspace. The intersection $K\cap L’$ is a convex closed subset of $L’$, so it is a closed interval. Now consider cases: the interval may be the whole of $L’$, a closed halfline, or a bounded closed interval.