已知在△ABC中，AD是∠ABC的平分线，DE⊥AB，DF⊥AC。求证：AD⊥EF

千问 · 2012-1-14 03:24:15

Rt△ADE与Rt△ADF中有∠DAE=∠DAF∠DEA=∠DFAAD=AD由AAS可证得：Rt△ADE≌Rt△ADFAE=AF△AEF是等腰三角形，AD平分∠ABC三线合一，可得：AD⊥EF

千问 · 2012-1-14 03:24:15

①证：∵AD是∠BAC的平分线∴∠EAD=∠FAD∵DE⊥AB，DF⊥AC∴∠BFD=∠CFD=90°∴∠AED与∠AFD=90°在△AED与△AFD中∠EAD=∠FADAD=AD∠AED=∠AFD∴△AED≌△AFD（AAS）∴AE=AF在△AEO与△AFO中∠EAO=∠FAOAO=AOAE=AF∴△AEO≌△AFO（SAS）∴∠AOE=∠AOF=90°∴AD⊥EF②证：∵AD是∠BAC的平分线∴∠BAD=∠CAD又∵DE⊥AB，DF⊥AC∴∠AED=∠AFD=90°在△AED与△AFD中∠BAD=∠CAD∠AED=∠AFDAD=AD∴△AED≌△AFD（AAS）∴∠EDO=∠FDO，ED=FD在△EDO=△FOD中DO=OD∠EDO=∠FDOED=FD∴△EDO≌△FOD（SAS）∴∠1=∠2∵∠1∠2=180°∴∠1=∠2=90°∴∠3=∠4=90°即AD⊥EF赞同

千问 · 2012-1-14 03:24:15

证：∵AD是∠BAC的平分线∴∠BAD=∠CAD又∵DE⊥AB，DF⊥AC∴∠AED=∠AFD=90°在△AED与△AFD中∠BAD=∠CAD∠AED=∠AFDAD=A∴△AED≌△AFD∴∠EDO=∠FDO，ED=FD在△EDO=△FOD中DO=OD∠EDO=∠FDOED=FD∴△EDO≌△FOD∴∠1=∠2∵∠1∠2=180°∴∠1=∠2=90°∴∠3=∠4=90°即AD⊥EF赞同