已知:AB=AC,AD=AE.AB⊥AC,AD⊥AE,连接BE,CD,F为BE中点,问AF与CD的关系

AF=(1/2)CD,且AF⊥CD.证明:连接BD,DE.取DE的中点M,连接AM,FM,CE.∵AD=AE;,AD⊥AE.∴⊿ADE为等腰直角三角形,得:AM⊥DE;AM=(1/2)DE;∠ADE=∠AED=45°.同理:⊿ABC也为等腰直角三角形,AMAB=AC,∠BAC=90°=∠DAE.∴∠BAD=∠CAE;又AB=AC,AD=AE.∴⊿BAD≌⊿CAE(SAS),BD=CE;∠ADB=∠AEC.F,M分别为BE,DE的中点,则FM=(1/2)BD=(1/2)CE;FM∥BD,∠FME=∠BDE.∴∠AMF=∠AME-∠FME=90°-∠BDE=90°-(∠ADB+45°)=45°-∠ADB=45°-∠AEC;∠DEC=∠AED-∠AEC=45°-∠AEC.则:∠AMF=∠DEC;又AM:DE=1:2,FM:CE=1:2.∴⊿AMF∽⊿DEC,AF:CD=AM:DE=1:2,AF=(1/2)CD;且∠MAF=∠EDC,∠MAF+∠MAD+∠ADC=∠EDC+∠MAD+∠ADC=180°-∠AMD=90°.∴AF⊥CD.