已知AD是△ABC的中线AE⊥AB,AF⊥AC且AE=AB,AF=AC连接EF求证AD=1/2EF

延长AD至M使AD=DM
连接BM
△ACD≌△BDM
AC=BM=AF
∠C=∠MBD
∠ABM =∠ABC +∠MBD=∠ABC +∠C =180°-∠BAC
∠FAE=90+90-∠BAC=180°-∠BAC
∠ABM =∠FAE
AF=AC=BM,AB=AE
△AEF≌△ABM
AM=EF
AD=1/2EF