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22 lines
865 B
22 lines
865 B
2 years ago
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- 看到两条边的乘积，需要考虑相似三角形
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- $BD \cdot CD=2\sqrt{3}$
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如果这是在两个相似三角形中，那么可能需要延长$AD$,然后截取$DE$,使得$AD\cdot DE=BD \cdot DC$
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$\because \frac{AD}{DC}=\frac{BD}{DE}$
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再加上$\angle BDA=\angle EDC$
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$\therefore \triangle ABD \sim \triangle EDC$
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但是这样，好像离结果也没啥进展，还需要继续思考：
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有一个结论：
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$\angle AEC=\angle ABC=\angle 1$
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<font color='red' size=4><b>母子相似三角形！</b></font>
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$\because \angle 2=\angle 2$
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$\because \angle CBA=\angle 1$
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$\therefore \angle AEC=\angle 1$
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$\triangle ACD \sim \triangle AEC$
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$\therefore \frac{AC}{AE}=\frac{AD}{AC}$
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将数值代入。解得：
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$AC=\sqrt{3}+1$
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