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

证明:延长AD到M,使DM=DA,连接BM.
又BD=CD,∠BDM=∠CDA,则⊿BDM≌⊿CDA,BM=AC=AF;∠BMD=∠CAD.
∴BM∥AC,∠ABM+∠BAC=180°;
又∠BAE=∠CAF=90°,则∠EAF+∠BAC=180°.
∴∠ABM=∠EAF(同角的补角相等).
又AB=AE.故⊿ABM≌⊿EAF(SAS),AM=EF.
即2AD=EF,AD=(1/2)EF.

Answer 2

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