Succinct Proof: All Pentagons Are Star Shaped 
Question: What is a succinct proof that all pentagons are star shaped?

In case the term star shaped (or star convex) is unfamiliar or forgotten:
Definition Reminder: A subset $X$ of $\mathbb{R}^n$ is star shaped if there exists an $x \in X$ such that the line segment from $x$ to any point in $X$ is contained in $X$.  
This topic has arisen in the past in my class discussions around interior angle sums; specifically, for star shaped polygons in $\mathbb{R}^2$ we can find their sum of interior angles as follows:
By assumption, there is an interior point $x$ that can be connected to each of the $n$ vertices. Drawing in these line segments, we construct $n$ triangles; summing across all of their interior angles gives a total of $180n^\circ$, but this over-counts the angle sum for the polygon by the $360^\circ$ around $x$. Therefore, the sum of interior angles is $(180n - 360)^\circ = 180(n-2)^\circ$.
This formula for the sum of interior angles holds more generally (often shown by triangulating polygons) but the proof strategy above already fails for some (obviously concave) hexagons. 
For example:

The above depicted polygon is not star shaped.
Moreover, it is a fact that any polygon with five or fewer sides is star shaped. And so I re-paste:

Question: What is a succinct proof that all pentagons are star shaped?


Edit: Since it has come up as a counterexample (of sorts) for each of the first two responses, here is an example of a concave pentagon that may be worth examining in thinking through a proof.


This is several years late, but I think I have a nicer proof. I thought I would post it.
Let's believe we can triangulate the 5-gon into three triangles. Note that these three triangles all must have a vertex $v$ in common: To see this, remove one of the triangles $T$, s.t. we have a 4-gon left. Then whichever way we triangulate the 4-gon every edge will contain a vertex from both of its triangles. In particular, the joint edge of $T$ and the 4-gon will have such a vertex $v$.
Then $v$ sees all three triangles and hence the entire 5-gon. 
