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注意大招:见45^{\circ}想90^{\circ}!
方法1
引辅助线CE,则\triangle DEC \sim ABD
\triangle ACE是等腰直角三角形
根据勾股定理，AC=17,AE=EC=\frac{17}{\sqrt{2}}
设CD=x,则根据相似三角形，可得:
\frac{x}{AD}=\frac{EC}{AB}
\therefore \frac{x}{\sqrt{15^2+(8+x)^2}}=\frac{\frac{17}{\sqrt{2}}}{15}
方法2
个人理解：第二种方法更简单，第一种方法计算太麻烦，不能出现开方、二次！
