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 = ∠HAE .
在△ECG和△AHE中,∠ECG = ∠AHE ,EC = AH ,∠CEG = ∠HAE ,
所以,△ECG ≌ △AHE ,
可得：EG = AE .

EG=AE