Prove the Polygon​ Interior-Angle Sum Theorem.

Given ​- A polygon with n sides.
Prove ​- The sum of the measures of the interior angles is ​(n - 2) 180 degrees.
5.​ Thus, the sum of the measures of the interior angles is__

1 answer

To prove the Polygon Interior-Angle Sum Theorem, which states that the sum of the measures of the interior angles of a polygon with \( n \) sides is \( (n - 2) \times 180^\circ \), we can use a method involving triangulation of the polygon.

Proof by Triangulation:

  1. Understand the Structure of Polygons:

    • A polygon with \( n \) sides is called an \( n \)-gon.
  2. Dividing the Polygon into Triangles:

    • We can divide a polygon with \( n \) sides into \( n - 2 \) triangles by drawing diagonals from one vertex to all other non-adjacent vertices.
    • For example, from one vertex of the polygon, you can connect to \( n - 3 \) other vertices (since one vertex is the starting point and two are adjacent and can't be connected directly).
  3. Calculating the Angles in the Triangles:

    • Each triangle has a sum of interior angles equal to \( 180^\circ \).
    • Since we have \( n - 2 \) triangles formed, the sum of the measures of the angles in these triangles is: \[ (n - 2) \times 180^\circ \]
  4. Relating Triangle Angles to Polygon Angles:

    • The angles of the triangles formed by triangulating the polygon are exactly the interior angles of the polygon.
    • Thus, the total sum of the interior angles of the polygon is equal to the total sum of the angles from the triangles.
  5. Conclusion:

    • Therefore, we conclude that the sum of the measures of the interior angles of a polygon with \( n \) sides is: \[ (n - 2) \times 180^\circ \]

Final Statement:

Thus, the sum of the measures of the interior angles of a polygon with \( n \) sides is \( (n - 2) \times 180^\circ \).