在三角形abc中,D为BC边上一点,点P在AD上,过点P作PM平行AC叫AB于点M,作PN平行AB.求证AM/AB+AN/AC=AP/AD

问题描述:

在三角形abc中,D为BC边上一点,点P在AD上,过点P作PM平行AC叫AB于点M,作PN平行AB.求证AM/AB+AN/AC=AP/AD


证明:延长MP,交BC于H,延长NP,交BC于G;过点P作EF平行于BC.
∵PM∥AC,PN∥AB.

∴四边形AMPN为平行四边形,AN=PM,AM=PN.

∵⊿MEP∽⊿ABC.
∴PM/AC=EP/BC,则AN/AC=PM/AC=EP/BC;

同理:⊿NPF∽⊿ABC,NP/AB=PF/BC,则AM/AB=PF/BC.

∴AM/AB+AN/AC=EP/BC+PF/BC=EF/BC;

又EF∥BC,则EF/BC=AE/AB=AP/AD.
故:AM/AB+AN/AC=AP/AD.