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865 B
865 B
- 看到两条边的乘积,需要考虑相似三角形
BD \cdot CD=2\sqrt{3}
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
\because \frac{AD}{DC}=\frac{BD}{DE}
再加上\angle BDA=\angle EDC
\therefore \triangle ABD \sim \triangle EDC
但是这样,好像离结果也没啥进展,还需要继续思考:
有一个结论:
\angle AEC=\angle ABC=\angle 1
母子相似三角形!
\because \angle 2=\angle 2
\because \angle CBA=\angle 1
\therefore \angle AEC=\angle 1
\triangle ACD \sim \triangle AEC
\therefore \frac{AC}{AE}=\frac{AD}{AC}
将数值代入。解得:
AC=\sqrt{3}+1