Diagonals.The fourth week of our course on the British known geometry will be devoted to polygon triangulations. Complex geometric shapes are often approximated by polygons or other polygonal measures. This is especially true for computer graphics and computer aided design. But those polygons may still be too complex to process. To facilitate and speed-up their handling, they are typically triangulated or split into triangles. This implies a need in development of efficient algorithms for triangulation polygons. But prior to doing that, we need to formalize the definition of triangulation of polygon. As a first step, we need to define a diagonal of polygon. Let P be a polygon with n vertices. A diagonal of p is an open segments with the endpoints at two distinct vertices of P, which internalize inside the polygon. In the figure you see on the left, there is a polygon P with seven vertices, and p1 p5 represents a diagonal. On the contrary, the segment with the endpoints at the vertices p3 and p6 is not a diagonal of p because it partially lies outside the polygon. None of the segment before p6, which entirely lies outside the polygon is diagonal of P. Finally, take a look at this element p3, p7. It is also not a diagonal because it passes through vertex p5. This is not allowed according to our definition. Now, let us prove that any polygon with at least four vertices is a diagonal. Let P be polygon with n vertices where N is at least four. We will indicate a way to find a diagonal for this polygon. Let us proceed as follows. First of all, we will find that the left-most vertex of P. Let me denote this vertex. Next, let us take a look at its neighbor vertices in the boundary of P which we should denote by U and W. If the segment U, W comprises a diagonal of P, then we have found a diagonal and we are done. Now, suppose this was not the case. If soon, some vertices of P must lie inside the triangle V, U, W. Let us consider all those vertices and choose from them the one which is the farthest from the segment U, W that is denoted by the prime. It is easy to see that no vertex from P can lie inside the triangle cut off from the polygon P by a line passing through the prime which is parallel to the segment U, W. Finally, let us consider the segment VP prime. If local allies inside the polygon P in the vicinity of both v and v prime and it can intersect no edges of the polygon, we conclude that the segment v prime represents a diagonal of the polygon P. This means that we're done. Note that our proof of the existence of the diagonal is constructive. In other words, it provides us with an algorithm for finding a diagonal in a simple polygon.