∴DC=CE=2CF=4，
∵ quadrilateral ABCD is a parallelogram,
∴AB=CD=4，
∵AE⊥BC，
∴∠AEB=90，
In Rt△ABE, we can know from Pythagorean theorem that BE=42? 32=7;
(2) It is proved that G is GM⊥AE in M,
∵AE⊥BE,GM⊥AE，
∴GM∥BC∥AD，
∫In△DCF and△ △ECG,
∠ 1=∠2∠C=∠CCD=CE，
∴△DCF≌△ECG(AAS)，
∴CG=CF，
CE = CD，CE=2CF，
∴CD=2CG
That is, g is the midpoint of the CD,
∫AD∨GM∨BC，
∴M is the midpoint of AE,
∴AM=EM，
∵GM⊥AE，
∴AG=EG，
∴∠AGM=∠EGM，
∴∠AGE=2∠MGE，
∫GM∨BC，
∴∠EGM=∠CEG，
∴∠CEG= 12∠AGE.