设顶点依次为 A,B,C,D.对角线的交点为O.根据三角形边长定理：AB^2=AO^2+OB^2-2AO*OB*COS<AOBBC^2=BO^2+OC^2-2BO*OC*COS<BOCCD^2=CO^2+OD^2-2CO*OD*COS<CODDA^2=DO^2+OA^2-2DO*OA*COS<DOA又应为...