Polygon exterior sum = 360°

Depends on: Polygon angle sum = (n − 2)·180° = $(n-2)\cdot 180^\circ$ (Polygon angle sum = (n − 2)·180°), Linear pair sums to 180° (Linear pair sums to 180°).

Statement

Let $P$ be a convex $n$-gon ($n \ge 3$). At each vertex $V_i$, extend the edge entering that vertex along its original direction; together with the edge leaving the vertex, this forms an exterior angle $\beta_i$. Then

The counterintuitive part: the right-hand side is independent of $n$ — the sum of the three exterior angles of a triangle, the four of a quadrilateral, and the twelve of a 12-gon are all equal to $360^\circ$.