Basic Theorems of Polygon

Theorem!
4.2 In the n-gon, we can build (n – 3) diagonals from a vertex.

Proof!
Look at figure below.

Fig. 4.6. Condition of th. 4.2

Let. P₁, P₂ , … ,P_n be the vertices of the polygon as shown in the figure.
Take any vertex as the certain vertex, let it be P_x.

The segments

build diagonals. So that there are 2 vertices those could not be connected from a chosen vertex, that is P_x–1 , and P_x+1 which are not build diagonals.
Obvious P_x couldn’t make diagonal to itself.
Because of P_x is any vertex, we conclude that the condition is suitable for all vertices.
So we can only build (n – 3) diagonal from a vertex in a n-gon.


Theorem!
4.3 The sum of diagonal in a n-gon is

Proof!
From theorem 3.2 we get there are (n – 3) diagonal for every vertex in a n-gon.
Obvious
from P₁ there are (n – 3) diagonals,
from P₂ there are (n – 3) diagonals,
from P₃ there are (n – 3) diagonals,
…
from P_n–1 there are (n – 3) diagonals, and
from P_n there are (n – 3) diagonals.

Take any P_x vertex in n-gon given.
Obvious there are (n – 3) diagonal out from the vertex and also there are (n – 3) diagonal in from another vertices.
It means every vertex in a n-gon is being counted twice.

So that the sum of diagonal in a n-gon should be
These prove the theorem.


Theorem!
4.4 The sum of exterior angle of a n-gon is 360^o.

Proof!
We leave the proof as exercise.