题目【问题背景】如图111,在Rt△ABCRt\triangle ABCRt△ABC和Rt△ADERt\triangle ADERt△ADE中,AB=ACAB=ACAB=AC,AD=AEAD=AEAD=AE,由已知可以得到:①△______\triangle \_\_\_\_\_\_△______≌△______\triangle \_\_\_\_\_\_△______;②△______\triangle \_\_\_\_\_\_△______∽△______.\triangle \_\_\_\_\_\_.△______.【尝试应用】如图222,在△ABC\triangle ABC△ABC和△ADE\triangle ADE△ADE中,∠ACB=∠AED=90∘\angle ACB=\angle AED=90^{\circ}∠ACB=∠AED=90∘,∠ABC=∠ADE=30∘\angle ABC=\angle ADE=30^{\circ}∠ABC=∠ADE=30∘,求证:△ACE\triangle ACE△ACE∽△ABD.\triangle ABD.△ABD.【问题解决】如图333,在△ABC\triangle ABC△ABC和△ADE\triangle ADE△ADE中,∠BAC=∠DAE=90∘\angle BAC=\angle DAE=90^{\circ}∠BAC=∠DAE=90∘,∠ABC=∠ADE=30∘\angle ABC=\angle ADE=30^{\circ}∠ABC=∠ADE=30∘,ACACAC与DEDEDE相交于点FFF,点DDD在BCBCBC上,ADBD=3\frac{{AD}}{{BD}}=\sqrt{3}BDAD=3,求DFCF\frac{{DF}}{{CF}}CFDF的值.知识点:勾股定理章节:图形的性质 / 三角形