Take any simple polygon. Extend all sides in both directions. Note the angles where these sides meet, if at all. What is the sum of these angles?
For example, consider the figure formed from a regular hexagon, the Star of David:
Given that the angles of a regular hexagon are each $120$ degrees, it’s easy to calculate that the angles in question are $60$ degrees each, since the base angles of the triangle are supplementary to the angles of the hexagon, and the angles of a triangle must add to $180$.
Therefore, since there are six of these, they add up to $360$ degrees in total.
Exchanging the $6$ for $n$ (for general case), this can be written as
$$n\left(180-2\left(180-\frac{180(n-2)}{n}\right)\right)$$
Distributing and simplifying:
$$180n-2n\left(180-\frac{180(n-2)}{n}\right)$$
$$180n-360n+360n-720$$
$$180n-720$$
$$180(n-4)$$
Another way of wording this is that this gives a measure of how far any two adjacent angles in a polygon are from forming parallel lines; that is, how far their angles are from $90$ degrees, with the result being positive if they’re slanted inward, and negative if they’re slanted outward. You can calculate this by taking each angle in the pair, subtracting $90$ from them, and adding the results together.
Mathematically, it’s the same as above, but it’s conceptually very different. The benefit of thinking of it this way is that this justifies a square giving an output of $0$ and an equilateral triangle giving an output of $-180$. And…well, when I discuss crazier cases later, thinking of the angles this way will make more sense, especially when they don’t exist.
This proof hinges on the original polygon being regular, however. The formula is derived from multiplying the angle by the number of triangles, and that’s only true if there are that many triangles.
Consider the following pentagon, with its lines extended:
This pentagon has three right angles and two $135$-degree angles. By the same method as above, the four triangles formed can be shown to have points of $45$ degrees each. The bottom of the pentagon doesn’t form a triangle at all, but, more specifically, it forms two parallel lines; as this doesn’t stray from $90$ degrees at all, this gets us a $0$ for the bottom, and $180$ overall.
This is exactly what you’d get from the above formula: $180(5-4)$.
Consider the figure formed from a trapezoid:
Two pairs of angles yield parallel lines and therefore output $0$ each. One pair actually yields a triangle, with a third angle measuring $90$; that is to say, each of those angles on the trapezoid leans inward from a perfectly horizontal line by $45$ degrees. The fourth pair doesn’t form a triangle, but more importantly, the lines lean outward by $45$ degrees. These two effects cancel out, and overall, the figure nets a $0$ – no different than a square.
It’s not just irregular polygons – concave ones follow this pattern as well. It would be too hard to draw how the “triangles” work here, but consider a quadrilateral with angles $90, 30, 30, 210$. By the same logic as above – by subtracting $90$ from each angle in each adjacent pair – you still end up with a result of $0$!
I can justify $180(n-4)$ holding true by regular polygons. How do I prove that this is the case for all polygons?
Best Answer
For an n-gon, considering a line rotating around it rotates 360. This implies the standard result that the sum of the exterior angles in the direction of the rotation is 360 so the sum of the interior angles is n*180-360 = 180(n-2).
Therefore the sum of the exterior angles on both sides of each vertex is 720.
From your first diagram, each angle of the extended polygon is 180 minus the sum of the adjacent exterior angles, so their sum is 180*n-720 =180(n-4).