Hence, a polygon with n = 5, produces a triangulation of 2 diagonals and 3 triangles. When placing one guard anywhere (not adhering to Fisk's algorithm), it will at least cover the triangle in which the guard is placed. However, in order to have the boundary of P covered, we will need another guard since n has 5 sides, just one guard placed arbitrarily will cover up at most 2 sides of the polygon within the triangle it sits in, if that triangle is an ear. Sometimes, you may need 3 arbitrarily placed guards (one in each triangle of the triangulation), as in the polygon with two adjacent reflex vertices.
In the case above, two guards may not cover the whole interior, but then there would also be a part of the boundary not covered. So at least, we would need 2 arbitrarily placed guards in order to cover the entire boundary of the polygon.
So if one guard sits on one ear of the triangle, and the other sits somewhere between its ear and the fifth side of the polygon (so as to have visibility), then the whole interior is covered and constructing a placement of guards such that every point on the boundary is covered and some point interior is not seen by either guard is not possible.