Question

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__

Answer 1

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 \( (n - 2) \times 180^\circ \).