# 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**