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.
4. All the angles of the triangles formed at vertex P are not part of the polygon angle sum, so subtract them from the the angle sum of all the triangles.
Looking at the figure, notice that all these angles formed at vertex P have a sum of__ degrees.

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 \( (n - 2) \times 180 \) degrees, we will use a geometric approach involving triangulation.

Proof:

  1. Definition: Let's define a polygon \( P \) with \( n \) sides (or \( n \)-gon). For example, let’s take a polygon with vertices labeled \( P_1, P_2, \ldots, P_n \).

  2. Triangulation: To find the sum of the interior angles, we can triangulate the polygon. A triangulation is a division of the polygon into triangles by drawing non-intersecting diagonals.

  3. Number of Triangles: When an \( n \)-gon is triangulated, it can be divided into \( n - 2 \) triangles. This is because:

    • A polygon with 3 sides (a triangle) has 1 triangle.
    • A polygon with 4 sides (a quadrilateral) can be divided into 2 triangles.
    • A polygon with 5 sides (a pentagon) can be divided into 3 triangles.
    • Continuing this reasoning, we find that an \( n \)-sided polygon can be divided into \( n - 2 \) triangles.
  4. Sum of Angles in Triangles: The sum of the interior angles of each triangle is always \( 180 \) degrees. Therefore, for the \( n - 2 \) triangles formed: \[ \text{Sum of the angles in all triangles} = (n - 2) \times 180 \text{ degrees} \]

  5. Angles at Vertex P: In the process of dividing the polygon into triangles, we do not include angles formed at the vertices of these triangles with \( n \)-sides in our final sum, but rather just those angles of the triangle itself. Each vertex contributes to the triangle outside of the polygon, and the angles formed at the intersection (vertex \( P \), in this case) are extraneous to the calculation of the polygon's interior angles.

Conclusion:

Since every triangle formed at the vertices of the polygon contributes to the total angle of the polygon's internal angles, and these angles have been counted, we conclude that:

\[ \text{Sum of the interior angles of the polygon} = (n - 2) \times 180 \text{ degrees} \]

Thus, we have proven the Polygon Interior-Angle Sum Theorem.

To Fill In the Blank:

In the statement, "Looking at the figure, notice that all these angles formed at vertex P have a sum of __ degrees," the answer would be \( 360 \) degrees because the angles at vertex \( P \) (which connects to the other triangles) collectively complete the circle around that point (vertex).