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20 lines
770 B
20 lines
770 B
2 years ago
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**注意大招:见$45^{\circ}$想$90^{\circ}$!**
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### 方法1
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引辅助线$CE$,则$\triangle DEC \sim ABD$
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$\triangle ACE$是等腰直角三角形
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根据勾股定理，$AC=17,AE=EC=\frac{17}{\sqrt{2}}$
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设$CD=x$,则根据相似三角形，可得:
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$$\frac{x}{AD}=\frac{EC}{AB}$$
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$\therefore \frac{x}{\sqrt{15^2+(8+x)^2}}=\frac{\frac{17}{\sqrt{2}}}{15}$
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### 方法2
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<font color='red' size=4><b>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！</b></font>
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