在Rt△ABC中,∠ACB=90°,CD⊥AB,垂足为D,AF平分∠CAB交CD于E,交CB于F,且EG平行AB交CB于G求证：CF=GB

证明：过点G作GH∥AF交AB于H∵∠ACB＝90∴∠BAC+∠B＝90∵CD⊥AB∴∠BAC+∠ACD＝90∴∠ACD＝∠B∵AF平分∠BAC∴∠BAF＝∠CAF∵∠CEF＝∠ACD+∠CAF,∠CFE＝∠B+∠BAF∴∠CEF＝∠CFE∴CF＝CE∵GH∥AF∴∠BHG＝∠BAF∴...