Question

如图,已知在△ABC中,∠BAC=90°,高AD与∠ABC的平分线BE相交于点E,EF∥AC交BC于点F.求证AE=EF

Answer 1

证明：延长FE交AB与点G

EF//AC

∠BAC=90°,即AC⊥AB

∴EF⊥AB

即EG⊥AB

AD⊥BC

BE为∠ABC的角平分线

∴EG＝ED

在Rt△ABC中,∠BAC＝90°,AD⊥BC

∴∠ABC=∠DAC,∠C=∠BAD

EF//AC

∴∠DEF=∠DAC,∠DFE=∠C

∴∠DEF=∠ABC ,∠DFE＝∠BAD

∵EG⊥AB

∴∠ABC=∠AEG

在△DEF和△GEA中

∠DEF=∠AEG,∠DFE=∠GAE,DE=EG

∴△DEF全等于△GEA

∴EF=EA