It is proved that the crossing of point H GH⊥AB, the extension of point I HG intersection CD, and the crossing of point K GK⊥AD.

The quadrilateral GIDK is a square, the quadrilateral AKGH is a rectangle,

∴AK=HG,KD=DI=GI=AH，

AD = CD，

∴IC=HG，

∫AD∨GH∨EF, g is the midpoint of DF,

∴HA=HE，

∴HE=GI，

∵ In Rt△HGE and Rt△ICG,

He =GI

∠GHE=∠CIG

HG=IC

∴Rt△HGE≌Rt△ICG(SAS)，

∴EG=CG,∠HGE=∠GCI,∠HEG=∠CGI，

∴∠HGE+∠CGI=90，

∴∠EGC=90，

∴eg⊥cg；