Proof that exterior angles of a triangle sum to 360

[Mr Davis 97]

So I am working on this simple proof, but am confused about the term "external angle." The problem says that if $a$, $b$, and $c$ are external angles to a triangle, then $a + b + c = 360$. However, is seems that the vertex of each triangle has two possible external angles, since there are two line segments that converge to that point (the vertex). So this would mean that there are 6 possible external angles for the triangle. So when it says to sum 3 of them, I'm not sure which ones, or how.

 

[fresh_42]

Here we have $a+b+c= 180°$ as it is always the case with planar triangles.
Therefore $3\cdot 360° = 1080°$ and $1080° -180° = 900° = a' + b' +c'$.

The only chance I see to get $360°$ by somehow external angles would be this:

Now we have $a'+b'+c' = \frac{1}{2} (3 \cdot 360° - (2a+2b+2c)) = 540° - (a+b+c) = 540° - 180° = 360°$
or $a'+b'+c' = 3\cdot 180° - (a+b+c) = 2 \cdot 180° = 360°$