问题描述：
如图,在三角形ABC中,M是BC中点,过点A做射线AD,并作BE垂直于AD,CF垂直于AD,连接ME、MF,求证：ME=MF
问题解答：
我来补答
如图
MM'⊥AD AD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.
MM'⊥AD AD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.
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