Question

用梅涅劳斯定理M、N分别在△ABC的AB、AC上,且MN‖BC,在MN上取一点P,直线AP、BP、CP交BC、AC、AB

Answer 1

由梅涅劳斯定理得：(AD/CD)×（CB/FB）×（FP/AP)=1,(AE/BE)×(BC/FC)×(FP/AP)=1

∴AD/CD=(AP×FB)/(CB×FP),AE/BF=(FC×AP)/(BC×FP)

∴（AD/CD)+(AE/BE)=(BC×AP)/(BC×FP)=AP/FP

∵MN∥BC

∴AM/MB=AP/PF=(AD/CD)+(AE/BF)

∴(AD/DC)+(AE/EB)与点P没关