How to prove that the exterior angle sum of an n-sided polygon is 360 degrees?

Just wondering about the smartest and best way to prove such a question. I know of many ways, and also I don't want to use anything related to $180(n - 2)$.

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By dividing the n-gon into n-2 triangles. –  ᴊ ᴀ s ᴏ ɴ Sep 3 '12 at 8:17

I don't know exactly what are you looking for, but perhaps try this:

You can divide the n-sided polygon into n triangles. Every triangle has the angle sum of 180°. We are not interested in the 360° from the circle around the middle point, so we have: $n \cdot 180°-360°$

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you should read my title - How to prove that the exterior angle sum of an n-sided polygon is 360 degrees? –  think123 Sep 3 '12 at 8:20

Intuitively it isn't too hard to see since as you trace round the shape you make one complete full turn so must go through a total of 360 degrees.

See this picture, link. Imagine moving the angles to a point along the lines, they make a full turn.

For a formal proof the only way I can think of at the moment is to use the fact that the interior angle sum is $180(n-2)$. Given the exterior angles $\alpha_1, \alpha_2, ..., \alpha_n$, the interior angles are $180-\alpha_1, 180-\alpha_2, ..., 180-\alpha_n$.

Summing these gives the interior angle sum to be $180n - \sum_{i=1}^n \alpha_n$. But this has to equal $180(n-2)$. Rearranging gives $\sum_{i=1}^n \alpha_i = 360$.

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This may be completely wrong, it seems far too easy. But I can't see any error, so here goes. This proof tries to bring the theorem closer to the "walking around the polygon" intuition.

Let $(u_1...u_n)$ be an n-tuple of vectors in $\mathbb{R}^2$, with $u_1 = u_n$. The angle between two successive vectors in this sequence is $\angle u_{i+1} - \angle u_i$, where $\angle v$ is the angle that $v$ makes with the positive x-axis. The sum of these angles is:

$$\sum_{1}^{n-1}{(\angle u_{i+1} - \angle u_i)}$$ $$=\sum_{1}^{n-1}{\angle u_{i+1} - \sum_{1}^{n-1}\angle u_i}$$ $$=\sum_{2}^{n}{\angle u_{i} - \sum_{1}^{n-1}\angle u_i}$$ $$=\sum_{2}^{n-1}{\angle u_{i} + \angle u_n - \sum_{2}^{n-1}\angle u_i} - \angle u_1$$ $$=0$$

The sides of a polygon and the exterior angles between them are simply a special case of this proposition.

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