∠ABC=90°,BD=CD,∠DAB+∠CDE=90°,求BFEF\angle ABC=90\degree, BD=CD, \angle DAB + \angle CDE=90\degree,\text{求}\dfrac{BF}{EF}∠ABC=90°,BD=CD,∠DAB+∠CDE=90°,求EFBF
