如图,在△ABC中,AB=AC,∠BAC=90°,∠C=45°,D为AC的中点,AE⊥BD于F交于BC于E,AM⊥BC于

(1)∠DAF与∠ABD相等∵∠BAC=90° AE⊥BD∴∠ABD+∠ADB=90° ∠DAF+∠ADF=90°∵∠ADB=∠ADF∴∠DAF=∠ABD(2）∵∠BAC=90° ∠C=45°∴△ABC是等腰RT△ ∵AM⊥BC于M交BD于点N∴AM=BM ∠ABM=∠BAM=45°∴∠BAM=∠C=45°∵由上题可知∠DAF=∠ABDAB=AC∠BAM=∠C∴△BNA≌△AEC（ASA）（3）∠CDE≠∠ADN∵△BNA≌△AEC∴AD=CE∵D为AC的中点 ∠BAC=90°∴AD=DC AD=1/2AB∴CE=DC ∠ADN=∠ADB=2∠ABD=60°∴△CDE是等腰△∵∠C=45°∴∠DCE=∠DEC=1/2(180°-∠C)=1/2(180°-45°)=67。5°∴∠CDE≠∠ADN