To prove the Polygon Interior-Angle Sum Theorem, we will use a polygon with \(n\) sides and demonstrate that the sum of the measures of the interior angles of this polygon is \((n - 2) \times 180^\circ\).
Steps of the Proof:
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Understanding the Polygon: Let's begin with a polygon that has \(n\) sides. This polygon can be a triangle, quadrilateral, pentagon, etc., up to an \(n\)-gon.
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Dividing the Polygon into Triangles: One effective method for finding the sum of the interior angles of a polygon is to divide it into triangles. We can do this by drawing diagonals from one vertex to all other non-adjacent vertices.
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Counting the Number of Triangles: In a polygon with \(n\) sides, if we select one vertex and connect it to the other vertices (excluding the adjacent vertices), we create \(n - 3\) triangles. This is because:
- There are \(n - 1\) vertices excluding the selected vertex.
- Out of these, \(2\) are adjacent to the selected vertex and cannot be connected to form a triangle.
- Hence, the number of diagonals that can be drawn from a single vertex is \(n - 3\).
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Total Number of Triangles Formed: The original polygon can now be seen as composed of these \(n - 2\) triangles (the original polygon plus the triangles formed by diagonals equals \(n - 2\) triangles in total).
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Sum of Interior Angles of the Triangles: Each triangle has a sum of interior angles equal to \(180^\circ\). Therefore, if there are \(n - 2\) triangles, the sum of the interior angles of all these triangles is: \[ (n - 2) \times 180^\circ \]
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Conclusion: Since the interior angles of the polygon are made up of the angles of these triangles, the interior angle sum of the polygon is also: \[ \text{Sum of interior angles of the polygon} = (n - 2) \times 180^\circ \] Thus, we have proven that for a polygon with \(n\) sides, the sum of the measures of the interior angles is \((n - 2) \times 180^\circ\).
\[ \text{Hence, } \text{the sum of the measures of the interior angles is } (n - 2) \times 180^\circ. \]