E是正方形ABCD的BC边上的一点,CG平分∠DCF,连结AE,并在CG上取一点G,使AE垂直EG,求证：EG=AE

在AB上取一点H,使得：AH = CE .则有：BH = AB-AH = BC-CE = BE ,可得：∠BHE = 45° ,∠AHE = 180°-∠BHE = 135° .∠ECG = ∠DCE+∠DCG = 90°+45° = 135° = ∠AHE ,∠CEG = 180°-∠AEG-∠AEB = 90°-∠AEB = ...