正方形ABCD的边BC,CD上取E,F两点使∠EAF=45°,求证EF=BE+DF

延长CB至M 使BM=DF 连结AM三角形ABM全等于三角形DFA(∠ABM=90°=∠ADF,AB=AD,BM=DF)(S.A.S.) 则AM=AF,且∠DAF=∠BAM又∠EAF=45°则∠DAF+∠EAB=45°则∠BAM+∠EAB=45°即∠EAM=45所以∠EAF=∠EAM 而AE=AE AM=AF三角...