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Why is the sum of all external angles in a convex polygon $360^\circ$?

From my understanding, for each vertex in a convex polygon, there exist exactly $2$ exterior angles corresponding to it, which are both equal, vertically opposite, and add up to $180^\circ$ with the interior angle. If we take as true that sum of interior angles in a triangle is $(n-2)180^\circ$ degrees, then $$\sum_i 2\cdot (180^\circ-\alpha_i) = n\cdot 360^\circ - (n-2)\cdot 360^\circ = 720^\circ.$$ Am I missing something here?

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  • $\begingroup$ You insist on counting each external angle twice.... $\endgroup$ Jan 2 '19 at 16:29
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    $\begingroup$ The convention is to count just one of the pair of exterior angles at each vertex. $\endgroup$ Jan 2 '19 at 16:29
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    $\begingroup$ The taxicab proof is nice - going round the polygon you turn through the exterior angles (taken one per vertex) in turn, and end up pointing in the same direction having completed one full urn. $\endgroup$ Jan 2 '19 at 16:34
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    $\begingroup$ Conventions just are what they are and changing them can be hard. This one is a good one because the nicest proof that the exterior angles sum to a circle is watching what happens as you take a line containing one of the edges and swing it around the polygon, counting the turnings. Works for nonconvex polygons too. (That's @MarkBennet 's taxicab proof.) $\endgroup$ Jan 2 '19 at 16:46
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    $\begingroup$ @EthanBolker I see. In my country, there's no such convention, and exterior angles aren't defined for concave interior angles. This was really confusing for me when trying to refresh geometry in English. I'm going to leave this question though, for future people that might be confused with it like I was $\endgroup$
    – Jakobian
    Jan 2 '19 at 16:50
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As the comment says, there are two equal exterior angles at each vertex, one on the left of the vertex and one on the right. When we say that "the sum of the exterior angles is 360°", we mean that the sum of the left-side angles is 360° and that the sum of the right-side angles is 360°, not that the sum of the two sets together is 360°.

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I recommend you look at Geometry by Jurgensen under the angles of a polygon section.

Draw a picture, it will help. Start out with pentagon to get some intuition. Now separate the inside of the polygon into non-overlapping triangles, and observe how you found the total sum of the interior angles of the polygon, and then of the sum of the exterior angles of it.

Basically, given a convex polygon, we can form a linear pair -- a pair of supplementary angles , at each vertex.

Taking all $n$ vertices of the polygon into account, we have $n \cdot 180 $ degrees.

The polygon can be partitioned/separated into exactly $n-2$ triangles.

So, the sum of all the interior angles of the polygon is $$(n-2) \cdot 180 \text{ degrees} $$

Hence, the sum of all the external angles is the difference

$$ n \cdot 180 - (n-2) \cdot 180 = 2\cdot 180 = 360 \text{ degrees} $$

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