Question

如图所示,在三角形ABC中,∠ACB=90°,高CD与∠CAB的平分线AE交与点F,试着说明∠CEF=∠CFE

Answer 1

过点E作EG⊥AB,垂足为G,CD平行于EG,∠CFE=∠AEG.因为角CAE被AE平分,∠CAE=∠GAE,∠EGA=∠ECA=90°,AE=AE,根据AAS证得△EGA全等于△ECA,∠CEA＝∠GEA,又因为∠CFE=∠AEG,证得∠CEF=∠CFE