在Rt三角形ABC中,∠BAC=90°,AB=AC=2,点D在BC上运动,过D作∠ADE=45°,DE交AC于E.若BD=1,求AE的长

∵∠BAC=90°,AB=AC=2∴BC=2√2 ∠B=∠C∵∠ADE=45° ∠ADB=∠C+∠ADC=∠ADC+45°∴∠DEC=∠DAC+∠ADE=∠DAC+45°=∠ADB∴△ABD∽△DCE∴AB/DC=BD/CE∵BD=1∴CD=BC-BD=2√2-1∴CE=1*(2√2-1)/2=√2-1/2∴AE=AC-CE=2-...