如图,已知正方形ABCD中,E是边AB上一点,AB=nBE(n>0),DF⊥CE于F,连接AF,AF⊥FG交CD于点G.

证明：（1）∵∠DCF+∠ECB=90°,∠DCF+∠CDF=90°∴∠ECB=∠CDF又∵∠B=∠CDA=90°∴△BCE相似于△DFC。当n=2时,AB=2BE=BC,可得DF/FC=BC/BE=2。（2）当n=3时AB=2BE=BC,可得DF/FC=BC/BE=3,又∵∠GFC+∠DFG=∠DFG+∠AFG=90°∴∠GFC=∠AFG,易得∠FCD=∠FDA∴△ADF相似于△CFG,∴CG/AD=FC/FD,因为DF/FC=BC/BE=3,AD=DC∴DC=3CG即DG=2CG 。过F点作BC的平行线。交DC,AB于M,H。S△AEF=（AE*FH)/2 ,S△DFG=(DG*FM)/2 , AE=（n-1）BE,DG=（n-1）CG,FH=FM=n-1∴S△AEF/S△DFG=(AE*FH)/(DG*FM)=n-1=2所以n=3百度地图