题目如图,在△ABC\triangle ABC△ABC中,∠BAC=90∘\angle BAC=90^{\circ}∠BAC=90∘,AE⊥BCAE\bot BCAE⊥BC于EEE,∠ABC\angle ABC∠ABC的平分线BDBDBD交AEAEAE于点FFF,交ACACAC于点DDD,AGAGAG平分∠EAC\angle EAC∠EAC,交BCBCBC于点GGG,连接GFGFGF,GDGDGD.下列结论一定正确的是( )A.∠EFB=∠ADB\angle EFB=\angle ADB∠EFB=∠ADBB.GF=GDGF=GDGF=GDC.DB=DCDB=DCDB=DCD.△AFD\triangle AFD△AFD是等边三角形知识点:直角三角形全等的判定章节:图形的性质 / 三角形