设BE=X,
∵ΔABC是等腰直角三角形,∠ACB=90°,∴∠B=45°,
又DE⊥AB,∴ΔBDE也是等腰直角三角形,DE=BE=X,BD=√2X,∴AC=BC=2√2X,
AB=√2AC=4X,∴AE=3X,
过E作EF⊥BC于F,则EF=BF=1/2BD=√2X/2,CF=2√2X-BF=3√2X/2,
∴CE=√(EF^2+CF^2)=√5X,
接着可用正弦定理求得sin∠ACE,下面用初中方法:
过A作AH⊥CE于H,设CH=Y,则EH=√5X-Y,
∵AC^2-CH^2=AH^2=AE^2-EH^2,
∴8X^2-Y^2=9X^2-(√5X-Y)^2,
Y=2√5X/5,
∴AH=√(AC^2-Y^2)=√(8X^2-4/5X^2)=√(36X^2/5)=6√5X/5
∴sin∠ACE=AH/AC=3√10/10