Poly ∠ sum — Polygon angle sum = (n − 2)·180° · Principia

Dependencies: Triangle interior angles sum to 180° (Triangle interior angles sum to 180°).

Statement

Let $P$ be a convex $n$ -gon ( $n\ge 3$ ). The sum of its $n$ interior angles is

\sum_{i=1}^{n} \angle V_i \;=\; (n-2)\cdot 180^{\circ}.

Convexity is the key: from any one vertex one can "see" all the other vertices, and the resulting $n-3$ diagonals partition $P$ into $n-2$ non-overlapping triangles.

$A hexagon split from V_1 by 3 diagonals into 4 triangles: interior-angle sum = 4\cdot 180^\circ = 720^\circ$

Polygon angle sum = (n − 2)·180°

Statement

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