本题好象还需AB=AC,连接AD,∵AB=AC,∠A=90°,D为BC中点,∴AD⊥BC,∠DAE=45°=∠C,AD=1/2BC=CD,∴∠ADF+∠CDF=90°,∵DE⊥DF,∴∠ADE+∠ADF=90°,∴∠ADE=∠CDF,∴ΔADE≌ΔCDF,∴AE=CF=1,∴AB=AC=4,∴AF=3,∴EF=√(A...