Prerequisites:
Angle sum property of a triangle (proof)
Proof:
We prove this result by mathematical induction:
For $n = 3$, polygon is a triangle,
hence the sum total of interior angles $\:= 180^o\\
\begin{align}\!\!\!\! &= (3-2)\times 180^o\\
&= (n-2)180^o\end{align}$
Thus, the result holds for $n=3$.
Now, let us assume that the result holds for $n=k$.
Therefore, for a $k$-sided polygon, sum total of interior angles $S_k = (k-2)180^o$.
For $n = k+1$:
Let us consider a $k+1$-sided polygon. Let us draw a line joining the $(k-1)^{th}$ and $(k+1)^{th}$ vertices, as shown in the figure. Then the resulting polygon $A_1A_2\cdots A_{k-1}A_{k+1}$ is a $k$-sided polygon.
Hence, the sum of its interior angles $= S_k = (k-2)180^o$ (from $n=k$ case).
Also by the angle sum property of a triangle, sum total of interior angles of the triangle $\triangle A_{k-1}A_kA_{k+1} = 180^o$.
Hence, sum total of all the angles of the $k+1$-sided polygon '$S_{k+1}$'$=$Sum of angles of polygon $A_1A_2\cdots A_{k-1}A_{k+1} +$ sum of angles of the triangle $\triangle A_{k-1}A_kA_{k+1}$, i.e.,
$\begin{align}\qquad\quad S_{k+1} &= S_k + 180^o\\
&= (k-2)180^o + 180^o\\
&= ((k+1) - 2)180^o\end{align}$
Thus, the result holds for $n=k+1$.
Hence, by the principle of mathematical induction, the result holds for all $n\geq 3$.
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