题目如图,△ABC\triangle ABC△ABC和△AGF\triangle AGF△AGF是等腰直角三角形,∠BAC=∠G=90∘\angle BAC=\angle G=90^{\circ}∠BAC=∠G=90∘,△AGF\triangle AGF△AGF的边AFAFAF,AGAGAG交边BCBCBC于点DDD,EEE.若BD=3BD=3BD=3,CE=4CE=4CE=4,则ADADAD的值是______.知识点:勾股定理章节:图形的性质 / 三角形