用同一法证设三角形ABC三边AB、BC、CA上（或延长线上）分别有三点D、E、F若AD/DB*BE/EC*CF/AF=1下面证明D、E、F共线延长DE交AC于F'由Menelaus定理有，AD/DB*BE/EC*CF‘/AF’=1于是CF/AF=CF'/AF'于是F与F'重合证毕
