Question

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

Answer 1

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