∵∠BAC=90度,AB=AC=2倍根号2,
∴BC=√(AB²+AC²)=√(8+8)=4
∠B=∠C=45°
∴在△ABD中,余弦定理：
AD²=AB²+BD²-2AB×BD×cos45°=8+1-2×2√2×1×√2/2=5
AD=√5
∴正弦定理
BD/sin∠BAD=AD/sin45°
sin∠BAD=BD×sin45°/AD=(1×√2/2)/√5=√10/10
∴cos∠BAD=3√10/10(利用sin²∠BAD+cos²∠BAD=1)
∴sin∠BAE
=sin(∠BAD+45°)
=sin∠BADcos45°+cos∠BADsin45°
=√10/10×√2/2+3√10/10×√2/2
=2√5/5
∴cos∠BAE=√5/5
∵∠ADE=∠B+∠BAD=45°+∠BAD=∠BAD+∠DAE=∠BAE
∴sin∠ADE=2√5/5,cos∠ADE=√5/5
∴sin∠AED
=sin(180°-45°-∠ADE)
=sin135°cos∠ADE-cos135°sin∠ADE
=√2/2×√5/5+√2/2×2√5/5
=√10/10+2√10/10
=3√10/10
∴在△ADE中,正弦定理
DE/sin45°=AD/sin∠AED
DE=AD×sin45°/sin∠AED
=(√5×√2/2)/(3√10/10)
=5/3