Answer :
Let's complete the missing parts of the paragraph proof step-by-step for proving the Polygon Interior Angle Sum Theorem.
We are given an [tex]$n$[/tex]-gon, which has [tex]$n$[/tex] sides and [tex]$n$[/tex] vertices. If we choose one of the vertices, we can draw [tex]$n-3$[/tex] diagonals. These diagonals form [tex]$n-2$[/tex] triangles. The sum of the interior angle measures of a triangle is [tex]$180$[/tex] degrees. [tex]$n-2$[/tex] triangles would have an interior angle measure sum of [tex]$180(n-2)$[/tex] degrees. Therefore, the sum of the measures of the interior angles of an [tex]$n$[/tex]-gon is [tex]$180(n-2)^{\circ}$[/tex].
So, the complete proof is:
We are given an [tex]$n$[/tex]-gon, which has [tex]$n$[/tex] sides and [tex]$n$[/tex] vertices. If we choose one of the vertices, we can draw [tex]$n-3$[/tex] diagonals. These diagonals form [tex]$n-2$[/tex] triangles. The sum of the interior angle measures of a triangle is [tex]$180$[/tex] degrees. [tex]$n-2$[/tex] triangles would have an interior angle measure sum of [tex]$180(n-2)$[/tex] degrees. Therefore, the sum of the measures of the interior angles of an [tex]$n$[/tex]-gon is [tex]$180(n-2)^{\circ}$[/tex].
We are given an [tex]$n$[/tex]-gon, which has [tex]$n$[/tex] sides and [tex]$n$[/tex] vertices. If we choose one of the vertices, we can draw [tex]$n-3$[/tex] diagonals. These diagonals form [tex]$n-2$[/tex] triangles. The sum of the interior angle measures of a triangle is [tex]$180$[/tex] degrees. [tex]$n-2$[/tex] triangles would have an interior angle measure sum of [tex]$180(n-2)$[/tex] degrees. Therefore, the sum of the measures of the interior angles of an [tex]$n$[/tex]-gon is [tex]$180(n-2)^{\circ}$[/tex].
So, the complete proof is:
We are given an [tex]$n$[/tex]-gon, which has [tex]$n$[/tex] sides and [tex]$n$[/tex] vertices. If we choose one of the vertices, we can draw [tex]$n-3$[/tex] diagonals. These diagonals form [tex]$n-2$[/tex] triangles. The sum of the interior angle measures of a triangle is [tex]$180$[/tex] degrees. [tex]$n-2$[/tex] triangles would have an interior angle measure sum of [tex]$180(n-2)$[/tex] degrees. Therefore, the sum of the measures of the interior angles of an [tex]$n$[/tex]-gon is [tex]$180(n-2)^{\circ}$[/tex].
