Question

1 如图,在正方形ABCD中,BE//AC,CA=CE,EC的延长线与BA的延长线相交于点F,求证:AE=AF

Answer 1

1、取AC的中点O,过C点作CH⊥BE于H

则CH＝OB＝1/2AC＝1/2CE

∴∠CEH＝30°

∵CA＝CE

∴∠CAE＝∠CEA

∵BE‖AC

∴∠CAE＝∠AEB

∴∠AEB＝∠CEA＝1/2∠CEH＝15°

∴∠ACF＝∠CAE＋∠CEA＝30°

∴∠FCD＝∠ACD－∠ACF＝15°

∵AF‖CD

∴∠F＝∠FCD＝15°＝∠CEA

∴AE＝AF

2、延长BH至M,使HM＝BH,连结ME

易证△MEH≌△BHF

∴ME＝BF＝BA

∵BE＝CB

∴△BME≌△CAB

∴MB＝AC

∴AC＝HM＋BH＝2BH