∠ACB=90°,AB=2,设AC=BC=X,则:AC^2+BC^2=AB^2,X^2+X^2=4,X=√2.
则BC=AC=√2;三角形ACD为等边三角形,故AC=AD=CD=√2.
作DF垂直BC的延长线于F.
∠BCD=∠BCA+∠ACD=150°,则∠DCF=30°,DF=CD/2=√2/2,CD=√(CD^2-DF^2)=√6/2.
BD=√(BF^2+DF^2)=√[(BC+CF)^2+DF^2]=√(√3+1)^2=√3+1.
cos∠CBE=BF/BD=(√2+√6/2)/(√3+1)=(√6+√2)/4;
CE/BC=DF/BF,即:CE/√2=(√2/2)/(√2+√6/2),CE=2√2-√6;
故AE=AC-CE=√6-√2.