延长FE和BA交于G∵EF⊥BC，∠BAC=∠CAG=90°∴∠EAG=∠EFC∠AEG=∠FEC∴△AEG∽△FCE∴EG/EC=AE/EF即EG×EF=AE×EC∵AD⊥BC，GF⊥BC∴AD∥GF∴DM/EF=BM/BEAM/EG=BM/BE∴AM/EG=DM/EF∵M是AD的中点即AM=DM∴EG=EF∴EF²=AE×EC
