在△ABC中,容易求得∠ABC=30°,∠ACB=15°根据正弦定理,得BC/sin135°=AC/sin30°即AC=BC*(sin30°/sin135°)=BC*(sin30°/sin45°)=BC*√2/2∵BC=DC∴AD=CD-AC=BC-AC=[（2-√2)/2]*BC=8-4√2
在△ABC中,容易求得∠ABC=30°,∠ACB=15°根据正弦定理,得BC/sin135°=AC/sin30°即AC=BC*(sin30°/sin135°)=BC*(sin30°/sin45°)=BC*√2/2∵BC=DC∴AD=CD-AC=BC-AC=[（2-√2)/2]*BC=8-4√2