# Why is the sum of all external angles in a convex polygon $360^\circ$ and not $720^\circ$?

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?

• You insist on counting each external angle twice.... – Angina Seng Jan 2 '19 at 16:29
• The convention is to count just one of the pair of exterior angles at each vertex. – Ethan Bolker Jan 2 '19 at 16:29
• 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. – Mark Bennet Jan 2 '19 at 16:34
• 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.) – Ethan Bolker Jan 2 '19 at 16:46
• @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 – Jakobian Jan 2 '19 at 16:50