Sum of Internal Angles of Polygon
Theorem
The sum $S$ of all internal angles of a polygon with $n$ sides is given by the formula $S = \paren {n - 2} 180 \degrees$.
Corollary
The size $A$ of each internal angle of a regular $n$-gon is given by:
- $A = \dfrac {\paren {n - 2} 180 \degrees} n$
Proof 1
The Polygon Triangulation Theorem shows that there exists a triangulation of the polygon that consists of $n - 2$ triangles.
The sides of these triangles are sides and chords of the polygon, where the chords lie completely in the interior of $P$.
Hence the vertices of the triangles are vertices of the polygon.
Sum of Angles of Triangle equals Two Right Angles shows that the sum of the internal angles of a triangle is $180 \degrees$.
As the triangulation covers the polygon, the sum of the internal angles of the vertices of the triangles in the triangulation is equal to $S$.
So:
- $S = \paren {n - 2} 180 \degrees$
$\blacksquare$
Proof 2
This proof assumes that the polygon is convex.
Name a vertex as $A_1$, go clockwise and name the vertices as $A_2, A_3, \ldots, A_n$.
By joining $A_1$ to every vertex except $A_2$ and $A_n$, one can form $\paren {n - 2}$ triangles in a fan triangulation of the convex polygon.
From Sum of Angles of Triangle equals Two Right Angles, the sum of the internal angles of a triangle is $180 \degrees$.
Therefore, the sum of internal angles of a polygon with $n$ sides is $\paren {n - 2} 180 \degrees$.
$\blacksquare$
Sources
- 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.9$: Corollary
- 1987: Joseph O'Rourke: Art Gallery Theorems and Algorithms: $\S 1.3.1$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): polygon
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): polygon