∵FA=BA GA=CA ∠FAC=∠FAB+∠BAC=∠CAG+∠BAC=∠BAG
∴△FAC≌△BAG
∴CE=BG
2. 设FC与BG的交点为H,AC与BG的交点为M
∵∠AGB=∠ACF ∠BMC=∠AMG
∵∠AGB+∠AMG=90°
∴∠BMC+∠ACF=90°
∴∠BHC=90
∴CE⊥BG
3.∵DA=BA EA=CA ∠BAE=∠BAC+∠CAE=∠BAC+∠BAD=∠BAG
∴△DAC≌△BAE
∴CD=BE
4.∠DCA=∠BEA∠∠
∠DCE+∠CEB=∠ACE+∠CEA=120°
CD和BE所成的锐角为60°