19(2)∵等腰Rt△ABC
∴∠ACB=45°
∠AEB=∠CAE+∠ACB=30°+45°=75°
又∵△ABE≌△CBD
∴∠BDC=∠AEB=75°
20(2)∵△EAG≌△EAF
∴AG=AF,EG=EF
又∵EB=EC
∴Rt△EGB≌Rt△EFC
∴GB=FC
∴AG+AB=AC-AF
又∵AG=AF
∴AF+AB=AC-AF
∴2AF=AC-AB=5-3=2
∴AF=1