Induction proof that for every convex n-corner there are n(n-3)/2 diagonals

Question

I have to proof that that for every convex n-corner there are $n(n-3)/2$ diagonals.

1.First step is to find n for which the sentence is correct. If $n0 = 3 => n(n-3)/2 = 0$. It is true because triangle has no diagonals.

2.Let's assume that $n(n-3)/2$ = p where $p$ is an amount of diagonals. Thesis is that for $((n+1)(n-2))/2 = z$ where $z$ is an amount of diagonals.

Proof: $L=((n+1)(n-2))/2 = (k^2 - 2k+k-2)/2 = (k^2-3k)/2 + (2k-2)/2$

$(k^2-3k)/2 = p$ but what should I do with this $(2k-2)/2$ expression?

Answer 1

We want to show increasing the number of vertices from $k$ to $k+1$ adds $k-1$ diagonals. The new vertex is adjacent to $2$ of the original $k$ vertices, so is joined to $k-2$ vertices by diagonals. However, adding a new vertex also causes $2$ previously adjacent vertices to no longer be adjacent. A new diagonal therefore exists between them too, so the total number of new diagonals is $k-1$ as required.

An alternative approach is to note that the required result is equivalent to the total number of diagonals and sides being $n\left(n-1\right)/2$. If you want to prove that by induction, the inductive step needs to verify increasing the number of vertices from $k$ to $k+1$ adds $$k\left(k+1\right)/2-k\left(k-1\right)/2=k$$lines, which it obviously does (the lines joining the new vertex to the old ones).

Answer 2

Here's an alternative approach without using induction:

There are a total of $n$ vertices and we can connect a each vertice to the other $(n-1)$ vertices. As there are $n$ vertices, the number should be $n(n-1)$. But, now we've counted each line twice (once for each of the vertices it joins). So, the actual number of lines would be $\frac{n(n-1)}{2}$.

Now, in an $n$-corner, there are $n$ sides (I'll leave the proof, if you need it though, ask me in the comments and I'll add it). So, the number of diagonals would be $\frac{n(n-1)}{2}-n= \frac{n(n-3)}{2}$.