How was the angle sum of polygons derived?

[prasannapakkiam]

We all know that for the angle sum of the external angles of a non-concaved polygons is 360. How is/was this derived...

 

[Werg22]

The sum of the interior angles is 180n - 360. Let a set Q denote n angles whose sum is 180n - 360: [ {a_{1}, a_{2} , ... , a_{n} ]. For any of these angles, the external angle is equal to 180 - a_{k}. Since there are n angles, the sum of all external angles is

S = \sum_{k = 1}^{n} 180 - a_{k}

S is obviously equal to 180n - (180n - 360) = 360

 

[prasannapakkiam]

But... I see it as; the equation you started with was derived from the fact that the sum of exterior angles equalled 360... If not how was your starting equation derived?

 

[AlephZero]

Take any point inside the polygon. Join it to the vertices, to divide the polygon into n triangles.

The sum of the angles in a triangle is 180 (see any basic geometry textbook for a proof of that).

So the sum of the angles in all the triangles = 180n.

The sum of the angles round the interior point = 360.

So the sum of the interior angles if the polygon = 180n - 360.

 

[prasannapakkiam]

Brilliant; thanks for that.