Question

已知矩形ABCD,CF⊥BD于E,AE是∠DAB的平分线,交FC的延长线于E,求证BD=CE

Answer 1

设AC,BD交于O,AE交BD于H∵矩形ABCD ∴∠BAD=∠ADC=90°,OA=OB=OC=OD ∴∠DAE=∠EAB=45°,∠BAO=∠ABO ∴∠CAE=∠EAB-∠OAB=45°-∠