Now, the cross section of these pieces, excluding the curvature, can represent your bisectors. Choosing a point x on any of those lines, and letting it be on the triangle makes this a simple problem. However, the triangle XYZ denotes a triangle formed my drawing lines from X to Y, Y to Z and Z to X. Therefore, X is treated as a vertex of the triangle. The bisector is the line that lies on the point exactly in the middle of two vertices. Therefore, to make this situation valid, the point X should lie in the center of the bisectors/risk-army/triangle effectively making the triangle non-existent because it has an absolute area of zero.