设AB=AC=1,则BC=√2,BD平分∠ABC,由角平分线性质,AD/DC=AB/BC=1/√2,
由比例性质,AD/(AD+DC)=1/(1+√2)=√2-1,
∴AD=√2-1,
由勾股定理,BD=√[(√2-1)^2+1]=√(4-2√2),
BD平分∠ABC,CE⊥BD,
∴△BAD∽△BEC,
∴BD/BC=AD/EC,
∴CE=AD*BC/BD=(√2-1)√2/√(4-2√2)=(2-√2)√(4-2√2)/(4-2√2)=√(4-2√2)/2=BD/2.
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