在△ABC中,角BAC=90度,AD⊥BC于D,CE平分角ACB于G,交AB于E,EF⊥BC于F,证四边形AEFG是菱形

证明：∵∠C+DAC=90°,∠BAD+∠DAC=90°∴∠C=∠BAD∵BE平分∠ABC∴∠ABE=∠CBE∵∠AGE=∠BAD+∠ABE,∠AEG=∠C+∠CBE∴∠AGE=∠AEG∴AG=AE∵BE平分∠ABC,EF⊥BC,EA⊥AB∴EA=EF=AG∵AD⊥BC,EF⊥BC∴AD//EF∵AG=EF∴AG...