(1)解:∵BD⊥CD,∠DCB=45°, ∴∠DBC=45°=∠DCB,∴BD=CD=2,在Rt△BDC中BC==2,∵CE⊥BE,点G为BC的中点,∴EG=BC=. 答:EG的长是. (2)证明:在线段CF上截取CH=BA,连接DH, ∵BD⊥CD,BE⊥CE,
∴∠EBF+∠EFB=90°,∠DFC+∠DCF=90° ∵∠EFB=∠DFC, ∴∠EBF=∠DCF, ∵DB=CD,BA=CH, ∴△ABD≌△HCD, ∴AD=DH,∠ADB=∠HDC, ∵AD∥BC, ∴∠ADB=∠DBC=45°, ∴∠HDC=45°,∴∠HDB=∠BDC﹣∠HDC=45°, ∴∠ADB=∠HDB, ∵AD=HD,DF=DF, ∴△ADF≌△HDF, ∴AF=HF, ∴CF=CH+HF=AB+AF, ∴CF=AB+AF. |