Hi Guys, im not sure if this question belongs in here or in geometry, im really having problems understanding it. Any help would be appreicated.
Thanks
Consider a triangulated Triangle. Let the verticies of all the triangles be labled xy and z arbitarily. (Including the verticies on the outside perimiter). Show that the number of segments on the outside perimiter labled XY has the same parity as the triangulation labled XYZ.