如图
MM'⊥ADAD交BC于N
FC/MM' =NF/NM' =NC/NM = (NM+MC)/NM = 1+MC/NM =1+MB/NM = 1+(BN+NM)/NM = 2+ BN/NM = 2+EN/NM' =1+(EN+NM')/NM' =1+EM'/NM'
所以:NF/NM' = 1+M'F =1+EM'/NM',既M'F=EM',△EMM'和△FMM'为相等三角形,ME=MF.
如图
MM'⊥ADAD交BC于N
FC/MM' =NF/NM' =NC/NM = (NM+MC)/NM = 1+MC/NM =1+MB/NM = 1+(BN+NM)/NM = 2+ BN/NM = 2+EN/NM' =1+(EN+NM')/NM' =1+EM'/NM'
所以:NF/NM' = 1+M'F =1+EM'/NM',既M'F=EM',△EMM'和△FMM'为相等三角形,ME=MF.