解:过G作GA"⊥DB,垂足为A"则△DAG≌DA"G, AG=A"G,DA" = DA=BC= 2设AG= x,则GA"=x,DB=A"B=DB-DA"=2-2,BG = AB-AG = 4-x. 在 Rt△BGA"中,x2+解得 x=故AG=
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