First please let me introduce some terminology in order to avoid misunderstandings (forgive me if it is a bit tedious).
Consider a collection of finite number of planar, consecutive line segments, with no internal intersections and forming a simple , closed curve. As such a curve, it separates the plane in one bounded and one unbounded domain. We consider a $n$-polygon to be such a curve including the bounded domain. The perimeter of $P$ is the above mentioned line.
A triangulation (without adding extra vertices) of a polygon $P$ is its partition into a set of triangles that have the same vertices with $P$ and these triangles have pairwise non-intersecting interiors and also their union is $P$.
Now comes the question: Is it possible to triangulate any polygon?
After searching a bit, I found that this is just a corrollary of the famous 2 -Ear Theorem by Max Dehn which gives an afformative answer.
But now I would like to demonstrate my idea- (if proof) and you to tell me if it is ok or not.
Proof (by Induction).
The case for $3$- polygons (triangles) is trivial.
Consider now that all $3,4,...n-1$-polygons can be triangulated and $P$ is an $n$ -Polygon. Then $P$ should have at least one convex angle $ABC$, otherwise $P$ is unbounded.Inside angle $ABC$ can be some other vertices too. Select those that have the least Euclidian distance from vertice $B$. And from those vertices select one, lets say $B'$ that is on the outmost left (or right). See picture below:
Now the segment $BB'$ is a diagonal of $P$ and the perimeter of $P$ is divided into $2$ lines $BC...B'$ and $BB'...A$, with lengths $l,m<n-1$ respectively. Clearly $n=m+l$. Then the polygons $BC...B'$, $BB'...A$ are $l+1, m+1<n$ and by the assumption of the induction they can be triangulated. So can be $P$.
Is it oK ? Thanks.
EDIT $1$
After some really usefull comments of @Jaap Scherphuis, I edit this question as follows:
In the proof instead of selecting the nearest and outmost right (or left) vertice, take a semi line from vertice $B$ and segment $BC$. Rotate this semi line untill it reaches a vertex (it maybe more than one). From all such vertices select this one which is closer to vertice $B$ (Let' s say $b$). The rest of the proof remains the same.
Given a polygon, the first step is to determine the extrema vertices. Can be done along the $X$ or $Y$ axes. Then we can choose one to begin the process. Lets begin with vertex $B$. This vertex opens a region from left to right, limited with by segments linking it to the nearest companions $C, A$. Note that the sequence $A\to B\to C$ is oriented CCW (counterclockwise). The next step is to verify if the triangle $CBA$ contains any of the remaining vertices. If not, the vertex $B$ is eliminated and a first triangulation is obtained: otherwise, the process continues counterclockwise to the next vertex $C$. Now considering the sequence $B\to C\to ?$ we can observe that it is CW oriented so we jump until the three corresponding vertices are CCW oriented, repeating the eliminate/non eliminate process, until the number of vertices is less or equal to $3$. Included three triangulations obtained with this method.
Follows the action sequence when applying the procedure in the first example. In dotted red, the remaining vertices after a new triangle classification. In this case we choose as the first vertex, the rightmost vertex.