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?
 A: Let $d_n$ be the number of diagonals of a polygon.
$d_3=0$ (polygon is triangle)
Assume that the number of diagonals of a $n$-polygon is $\frac{n(n−3)}{2}.$
Сonsider $n+1-$poligon $A_1 A_2 \ldots A_{n+1}.$  Under the assumption the number of diagonals of  $n-$poligon $A_1 A_2 \ldots A_{n}$ is $\frac{n(n−3)}{2}.$ Besides there are $n-1$ diagonals $A_{n+1} A_2,\,\,$   $A_{n+1} A_3,\ldots , A_{n+1} A_{n-1},$ and $A_1A_n$ for  $n+1-$poligon $A_1 A_2 \ldots A_{n+1}.$ Therefore the number of diagonals of a $n+1$-polygon $A_1 A_2 \ldots A_{n+1}$ equal  to $d_{n+1}=\frac{n(n−3)}{2}+n-1=\frac{n^2−3n+2n-2}{2}=\frac{(n+1)((n+1)-3)}{2}.$
Sorry for my English.
A: Suppose you have an $n$-gon with vertices $V_1, \ldots, V_n$ which by induction hypothesis has $n (n - 3)/2$ diagonals. Now let us add another vertex $V_{n+1}$. How many new diagonals have appeared? Add that to the existing diagonals and verify that you get $(n + 1) (n + 1 - 3)/2$. (Do not forget that the side $V_n V_1$ becomes a diagonal when you add $V_{n+1}$ beween $V_n$ and $V_1$.)
A: Hint
Well start with the first step: prove, that $\frac{n(n-3)}{2}$ is the number of diagonals for the smallest possible polygon. In this case it is $n$-polygon, where $n=3$.
In the second step assume, that $\frac{k(k-3)}{2}$ is the number of diagonals for $k$-polygon and you must show that $\frac{(k+1)(k+1-3)}{2}$ is the number of diagonals for $(k+1)$-polygon. It might be helpful to draw a simple picture with arbitrary number of vertices and add one vertex and find out, what it does with the number of diagonals.
A: I don't know what you mean by induction. But it's simple: For each vertex (n), there is one diagonal linking for all vertex, except the same vertex and the two neighbours (n-3). Thus you have n(n-3), that you got to divide by two or else each diagonal will be counted going and coming back.
A: Let's assume that polygon has n sides, then number of diagonals(NOD) = $n(n-3)/2$.
Now for $n=3$, NOD=0, for $n=4$, NOD=2, . Now assume that for $n=k$ (or $k=n$ you can also assume), NOD=$n(n-3)/2$ is true. You got to prove that for $n=k+1$ (or $k=n$), this is also true. If your assumption is true, then for $n=k+1$, NOD=$(k+1)(k-2)/2$ (or $k=n$).
Let's look at increment (or decrement) of NOD as n increases (or decreases),
 $n=4$, NOD=2, 


*

*for $n=5$, extra 3 Diagonals are added to $n=4$ and NOD=$2+3=5$,

*for $n=6$, 4 extra Diagonals added to $n=5$ and NOD=$5+4=9$,

*for $n=7$, 5 extra Diagonals added to $n=6$ and NOD=$9+5=14$,

*for $n=8$, 6 extra Diagonals added to $n=7$ and NOD=$14+6=20$,

*for $n=9$, 7 extra Diagonals added to $n=8$ and NOD=$20+7=27$,


Do you see there is a relation between extra added diagonals and n? Therefore you can conclude that for $n+1$ side polygon, $(n+1)-2$ extra diagonals will added to n side polygons, So, $n(n-3)/2 + n-1= (n+1)(n-2)/2$. 
which implies that for $n=k+1$, NOD=$(k+1)((k+1)-3)/2$ , (or $k=n$)
A: 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 $\frac{n(n-3)}2$.
