如图,过D作DF⊥AF于F,∵点B的坐标为(1,3),∴AO=1,AB=3, 根据折叠可知:CD=OA,而∠D=∠AOE=90°,∠DEC=∠AEO,∴△CDE≌△AOE,∴OE=DE,OA=CD=1, 设OE=x,那么CE=3-x,DE=x,∴在Rt△DCE中,CE2=DE2+CD2,∴(3-x)2=x2+12,∴x= , 又DF⊥AF,∴DF∥EO,∴△AEO∽△ADF,而AD=AB=3,∴AE=CE=3- = , ∴ ,即 ,∴DF= ,AF= ,∴OF= -1= , ∴D的坐标为(- , .故选A.
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