Proof that the number of diagonals of a polygon is $\frac{n(n-3)}{2} $ For $n \geq 3$ proof that the number of diagonals of a polygon is $\frac{n(n-3)}{2} $ using induction. 
I don't know how to start this problem, can you give me a hint?

From any vertex we cannot draw a diagonal to its $2$ adjacent sides and the vertex itself. 
So for a polygon containing $n$ sides we cannot draw $3$ diagonals, it means we can draw $n-3$ diagonals. 
For $n$ vertices we can draw $n(n-3)$ diagonals.But each diagonal has two ends, so this would count each one twice.  So that we will divide $n(n-3)$ by $2$. So finally number of diagnols for a polygon containing $n$ sides is $\frac{n(n-3)}2$.
